Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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Details For UPSC Maths Optional Solved Papers (2018-2022)

UPSC Maths Optional Question Papers

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Section:- A
Question:-01 (a) Show that the multiplicative group G = { 1 , − 1 , i , − i } G = { 1 , − 1 , i , − i } G={1,-1,i,-i}G=\{1,-1, i,-i\}G={1,−1,i,−i}, where i = ( − 1 ) i = ( − 1 ) i=sqrt((-1))i=\sqrt{(-1)}i=(−1), is isomorphic to the group G ′ = ( { 0 , 1 , 2 , 3 } , + 4 ) G ′ = { 0 , 1 , 2 , 3 } , + 4 G^(‘)=({0,1,2,3},+_(4))G^{\prime}=\left(\{0,1,2,3\},+{ }_{4}\right)G′=({0,1,2,3},+4).
Question:-01 (b) If f ( z ) = u + i v f ( z ) = u + i v f(z)=u+ivf(z)=u+i vf(z)=u+iv is an analytic function of z z zzz, and u − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − y u − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − y u-v=(cos x+sin x-e^(-y))/(2cos x-e^(y)-e^(-y))u-v=\frac{\cos x+\sin x-e^{-y}}{2 \cos x-e^{y}-e^{-y}}u−v=cos⁡x+sin⁡x−e−y2cos⁡x−ey−e−y, then find f ( z ) f ( z ) f(z)f(z)f(z) subject to the condition f ( π 2 ) = 0 f π 2 = 0 f((pi)/(2))=0f\left(\frac{\pi}{2}\right)=0f(π2)=0.
Question:-01 (c) Test the convergence of ∫ 0 ∞ cos ⁡ x 1 + x 2 d x ∫ 0 ∞   cos ⁡ x 1 + x 2 d x int_(0)^(oo)(cos x)/(1+x^(2))dx\int_{0}^{\infty} \frac{\cos x}{1+x^{2}} d x∫0∞cos⁡x1+x2dx.
Question:-01 (d) Expand f ( z ) = 1 ( z − 1 ) 2 ( z − 3 ) f ( z ) = 1 ( z − 1 ) 2 ( z − 3 ) f(z)=(1)/((z-1)^(2)(z-3))f(z)=\frac{1}{(z-1)^{2}(z-3)}f(z)=1(z−1)2(z−3) in a Laurent series valid for the regions
(i) 0 < | z − 1 | < 2 0 < | z − 1 | < 2 0 < |z-1| < 20<|z-1|<20<|z−1|<2 and (ii) 0 < | z − 3 | < 2 0 < | z − 3 | < 2 0 < |z-3| < 20<|z-3|<20<|z−3|<2.
Question:-01 (e) Use two-phase method to solve the following linear programming problem :
Minimize Z = x 1 + x 2 subject to 2 x 1 + x 2 ≥ 4 x 1 + 7 x 2 ≥ 7 x 1 , x 2 ≥ 0  Minimize  Z = x 1 + x 2  subject to  2 x 1 + x 2 ≥ 4 x 1 + 7 x 2 ≥ 7 x 1 , x 2 ≥ 0 {:[” Minimize “Z=x_(1)+x_(2)],[” subject to “],[2x_(1)+x_(2) >= 4],[x_(1)+7x_(2) >= 7],[x_(1)”,”x_(2) >= 0]:}\begin{array}{r} \text { Minimize } Z=x_{1}+x_{2} \\ \text { subject to } \\ 2 x_{1}+x_{2} \geq 4 \\ x_{1}+7 x_{2} \geq 7 \\ x_{1}, x_{2} \geq 0 \end{array} Minimize Z=x1+x2 subject to 2x1+x2≥4x1+7x2≥7x1,x2≥0
Question:-02 (a) Let f ( x ) = x 2 f ( x ) = x 2 f(x)=x^(2)f(x)=x^{2}f(x)=x2 on [ 0 , k ] , k > 0 [ 0 , k ] , k > 0 [0,k],k > 0[0, k], k>0[0,k],k>0. Show that f f fff is Riemann integrable on the closed interval [ 0 , k ] [ 0 , k ] [0,k][0, k][0,k] and ∫ 0 k f d x = k 3 3 ∫ 0 k   f d x = k 3 3 int_(0)^(k)fdx=(k^(3))/(3)\int_{0}^{k} f d x=\frac{k^{3}}{3}∫0kfdx=k33.
Question:-02 (b) Prove that every homomorphic image of a group G G GGG is isomorphic to some quotient group of G G GGG.
Question:-02 (c) Apply the calculus of residues to evaluate ∫ − ∞ ∞ cos ⁡ x d x ( x 2 + a 2 ) ( x 2 + b 2 ) , a > b > 0 ∫ − ∞ ∞   cos ⁡ x d x x 2 + a 2 x 2 + b 2 , a > b > 0 int_(-oo)^(oo)(cos xdx)/((x^(2)+a^(2))(x^(2)+b^(2))),a > b > 0\int_{-\infty}^{\infty} \frac{\cos x d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}, a>b>0∫−∞∞cos⁡xdx(x2+a2)(x2+b2),a>b>0.
Question:-03 (a) Evaluate ∫ C z + 4 z 2 + 2 z + 5 d z ∫ C   z + 4 z 2 + 2 z + 5 d z int_(C)(z+4)/(z^(2)+2z+5)dz\int_{C} \frac{z+4}{z^{2}+2 z+5} d z∫Cz+4z2+2z+5dz, where C C CCC is | z + 1 − i | = 2 | z + 1 − i | = 2 |z+1-i|=2|z+1-i|=2|z+1−i|=2
Question:-03 (b) Find the maximum and minimum values of x 2 a 4 + y 2 b 4 + z 2 c 4 x 2 a 4 + y 2 b 4 + z 2 c 4 (x^(2))/(a^(4))+(y^(2))/(b^(4))+(z^(2))/(c^(4))\frac{x^{2}}{a^{4}}+\frac{y^{2}}{b^{4}}+\frac{z^{2}}{c^{4}}x2a4+y2b4+z2c4, when l x + m y + n z = 0 l x + m y + n z = 0 lx+my+nz=0l x+m y+n z=0lx+my+nz=0 and x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1x2a2+y2b2+z2c2=1. Interpret the result geometrically.
Question:-03 (c) Solve the following linear programming problem by the simplex method. Write its dual. Also, write the optimal solution of the dual from the optimal table of the given problem :
Maximize Z = x 1 + x 2 + x 3 subject to 2 x 1 + x 2 + x 3 ≤ 2 4 x 1 + 2 x 2 + x 3 ≤ 2 x 1 , x 2 , x 3 ≥ 0  Maximize  Z = x 1 + x 2 + x 3  subject to  2 x 1 + x 2 + x 3 ≤ 2 4 x 1 + 2 x 2 + x 3 ≤ 2 x 1 , x 2 , x 3 ≥ 0 {:[” Maximize “Z=x_(1)+x_(2)+x_(3)],[” subject to “],[2x_(1)+x_(2)+x_(3) <= 2],[4x_(1)+2x_(2)+x_(3) <= 2],[x_(1)”,”x_(2)”,”x_(3) >= 0]:}\begin{array}{r} \text { Maximize } Z=x_{1}+x_{2}+x_{3} \\ \text { subject to } \\ 2 x_{1}+x_{2}+x_{3} \leq 2 \\ 4 x_{1}+2 x_{2}+x_{3} \leq 2 \\ x_{1}, x_{2}, x_{3} \geq 0 \end{array} Maximize Z=x1+x2+x3 subject to 2x1+x2+x3≤24x1+2x2+x3≤2x1,x2,x3≥0
Question:-04 (a) Let R R RRR be a field of real numbers and S S SSS, the field of all those polynomials f ( x ) ∈ R [ x ] f ( x ) ∈ R [ x ] f(x)in R[x]f(x) \in R[x]f(x)∈R[x] such that f ( 0 ) = 0 = f ( 1 ) f ( 0 ) = 0 = f ( 1 ) f(0)=0=f(1)f(0)=0=f(1)f(0)=0=f(1). Prove that S S SSS is an ideal of R [ x ] R [ x ] R[x]R[x]R[x]. Is the residue class ring R [ x ] / S R [ x ] / S R[x]//SR[x] / SR[x]/S an integral domain? Give justification for your answer.
Question:-04 (b) Test for convergence or divergence of the series
x + 2 2 x 2 2 ! + 3 3 x 3 3 ! + 4 4 x 4 4 ! + 5 5 x 5 5 ! + ⋯ ( x > 0 ) x + 2 2 x 2 2 ! + 3 3 x 3 3 ! + 4 4 x 4 4 ! + 5 5 x 5 5 ! + ⋯ ( x > 0 ) x+(2^(2)x^(2))/(2!)+(3^(3)x^(3))/(3!)+(4^(4)x^(4))/(4!)+(5^(5)x^(5))/(5!)+cdotsquad(x > 0)x+\frac{2^{2} x^{2}}{2 !}+\frac{3^{3} x^{3}}{3 !}+\frac{4^{4} x^{4}}{4 !}+\frac{5^{5} x^{5}}{5 !}+\cdots \quad(x>0)x+22x22!+33x33!+44x44!+55x55!+⋯(x>0)
Question:-04 (c) Find the initial basic feasible solution of the following transportation problem by Vogel’s approximation method and use it to find the optimal solution and the transportation cost of the problem :
A A AAA B B BBB C C CCC D D DDD
S 1 S 1 S_(1)S_1S12116251311
S 2 S 2 S_(2)S_2S21718142313
S 3 S 3 S_(3)S_3S33227184119
610121543
A B C D S_(1) 21 16 25 13 11 S_(2) 17 18 14 23 13 S_(3) 32 27 18 41 19 6 10 12 15 43| | $A$ | $B$ | $C$ | $D$ | | | :— | :—: | :—: | :—: | :—: | :—: | | $S_1$ | 21 | 16 | 25 | 13 | 11 | | $S_2$ | 17 | 18 | 14 | 23 | 13 | | $S_3$ | 32 | 27 | 18 | 41 | 19 | | | 6 | 10 | 12 | 15 | 43 |
Section:- B
Question:-05 (a) It is given that the equation of any cone with vertex at ( a , b , c ) ( a , b , c ) (a,b,c)(a, b, c)(a,b,c) is f ( x − a z − c , y − b z − c ) = 0 f x − a z − c , y − b z − c = 0 f((x-a)/(z-c),(y-b)/(z-c))=0f\left(\frac{x-a}{z-c}, \frac{y-b}{z-c}\right)=0f(x−az−c,y−bz−c)=0. Find the differential equation of the cone.
Question:-05(b) Solve, by Gauss elimination method, the system of equations
2 x + 2 y + 4 z = 18 x + 3 y + 2 z = 13 . 3 x + y + 3 z = 14 2 x + 2 y + 4 z = 18 x + 3 y + 2 z = 13 . 3 x + y + 3 z = 14 {:[2x+2y+4z=18],[x+3y+2z=13.],[3x+y+3z=14]:}\begin{array}{r} 2 x+2 y+4 z=18 \\ x+3 y+2 z=13 . \\ 3 x+y+3 z=14 \end{array}2x+2y+4z=18x+3y+2z=13.3x+y+3z=14
Question:-05 (c) (i) Convert the number ( 1093.21875 ) 10 ( 1093.21875 ) 10 (1093.21875)_(10)(1093.21875)_{10}(1093.21875)10 into octal and the number ( 1693 ⋅ 0628 ) 10 ( 1693 ⋅ 0628 ) 10 (1693*0628)_(10)(1693 \cdot 0628)_{10}(1693⋅0628)10 into hexadecimal systems.
(ii) Express the Boolean function F ( x , y , z ) = x y + x ′ z F ( x , y , z ) = x y + x ′ z F(x,y,z)=xy+x^(‘)zF(x, y, z)=x y+x^{\prime} zF(x,y,z)=xy+x′z in a product of maxterms form.
Question:-05 (d) A particle at a distance r r rrr from the centre of force moves under the influence of the central force F = − k r 2 F = − k r 2 F=-(k)/(r^(2))F=-\frac{k}{r^{2}}F=−kr2, where k k kkk is a constant. Obtain the Lagrangian and derive the equations of motion.
Question:-05 (e) The velocity components of an incompressible fluid in spherical polar coordinates ( r , θ , ψ ) ( r , θ , ψ ) (r,theta,psi)(r, \theta, \psi)(r,θ,ψ) are ( 2 M r − 3 cos ⁡ θ , M r − 2 sin ⁡ θ , 0 ) 2 M r − 3 cos ⁡ θ , M r − 2 sin ⁡ θ , 0 (2Mr^(-3)cos theta,Mr^(-2)sin theta,0)\left(2 M r^{-3} \cos \theta, M r^{-2} \sin \theta, 0\right)(2Mr−3cos⁡θ,Mr−2sin⁡θ,0), where M M MMM is a constant. Show that the velocity is of the potential kind. Find the velocity potential and the equations of the streamlines.
Question:-06 (a) Solve the heat equation ∂ u ∂ t = ∂ 2 u ∂ x 2 , 0 < x < l , t > 0 ∂ u ∂ t = ∂ 2 u ∂ x 2 , 0 < x < l , t > 0 (del u)/(del t)=(del^(2)u)/(delx^(2)),0 < x < l,t > 0\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}, 0<x<l, t>0∂u∂t=∂2u∂x2,0<x<l,t>0 subject to the conditions
u ( 0 , t ) = u ( l , t ) = 0 u ( x , 0 ) = x ( l − x ) , 0 ≤ x ≤ l u ( 0 , t ) = u ( l , t ) = 0 u ( x , 0 ) = x ( l − x ) , 0 ≤ x ≤ l {:[u(0″,”t)=u(l”,”t)=0],[u(x”,”0)=x(l-x)”,”quad0 <= x <= l]:}\begin{aligned} &u(0, t)=u(l, t)=0 \\ &u(x, 0)=x(l-x), \quad 0 \leq x \leq l \end{aligned}u(0,t)=u(l,t)=0u(x,0)=x(l−x),0≤x≤l
Question:-06 (b) Find a combinatorial circuit corresponding to the Boolean function
f ( x , y , z ) = [ x ⋅ ( y ¯ + z ) ] + y f ( x , y , z ) = [ x ⋅ ( y ¯ + z ) ] + y f(x,y,z)=[x*( bar(y)+z)]+yf(x, y, z)=[x \cdot(\bar{y}+z)]+yf(x,y,z)=[x⋅(y¯+z)]+y
and write the input/output table for the circuit.
Question:-06 (c) Find the moment of inertia of a right circular solid cone about one of its slant sides (generator) in terms of its mass M M MMM, height h h hhh and the radius of base as a a aaa.
Question:-07 (a) Find the general solution of the partial differential equation
( D 2 + D D ′ − 6 D ′ 2 ) z = x 2 sin ⁡ ( x + y ) D 2 + D D ′ − 6 D ′ 2 z = x 2 sin ⁡ ( x + y ) (D^(2)+DD^(‘)-6D^(‘2))z=x^(2)sin(x+y)\left(D^{2}+D D^{\prime}-6 D^{\prime 2}\right) z=x^{2} \sin (x+y)(D2+DD′−6D′2)z=x2sin⁡(x+y)
where D ≡ ∂ ∂ x D ≡ ∂ ∂ x D-=(del)/(del x)D \equiv \frac{\partial}{\partial x}D≡∂∂x and D ′ ≡ ∂ ∂ y D ′ ≡ ∂ ∂ y D^(‘)-=(del)/(del y)D^{\prime} \equiv \frac{\partial}{\partial y}D′≡∂∂y.
Question:-07 (b) The velocity of a train which starts from rest is given by the following table, the time being reckoned in minutes from the start and the velocity in k m / k m / km//\mathrm{km} /km/ hour :
t t ttt (minutes)2468101214161820
v ( k m / v ( k m / v(km//v(\mathrm{~km} /v( km/ hour ) ) )))16 28 ⋅ 8 28 ⋅ 8 28*828 \cdot 828⋅840 46 ⋅ 4 46 ⋅ 4 46*446 \cdot 446⋅4 51 ⋅ 2 51 ⋅ 2 51*251 \cdot 251⋅232 17 ⋅ 6 17 ⋅ 6 17*617 \cdot 617⋅68 3 ⋅ 2 3 ⋅ 2 3*23 \cdot 23⋅20
t (minutes) 2 4 6 8 10 12 14 16 18 20 v(km// hour ) 16 28*8 40 46*4 51*2 32 17*6 8 3*2 0| $t$ (minutes) | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | | :— | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $v(\mathrm{~km} /$ hour $)$ | 16 | $28 \cdot 8$ | 40 | $46 \cdot 4$ | $51 \cdot 2$ | 32 | $17 \cdot 6$ | 8 | $3 \cdot 2$ | 0 |
Using Simpson’s 1 3 r d 1 3 r d (1)/(3)rd\frac{1}{3} \mathrm{rd}13rd rule, estimate approximately in k m k m km\mathrm{km}km the total distance run in 20 minutes.
Question:-07 (c) Two point vortices each of strength k k kkk are situated at ( ± a , 0 ) ( ± a , 0 ) (+-a,0)(\pm a, 0)(±a,0) and a point vortex of strength − k 2 − k 2 -(k)/(2)-\frac{k}{2}−k2 is situated at the origin. Show that the fluid motion is stationary and also find the equations of streamlines. If the streamlines, which pass through the stagnation points, meet the x x xxx-axis at ( ± b , 0 ) ( ± b , 0 ) (+-b,0)(\pm b, 0)(±b,0), then show that 3 3 ( b 2 − a 2 ) 2 = 16 a 3 b 3 3 b 2 − a 2 2 = 16 a 3 b 3sqrt3(b^(2)-a^(2))^(2)=16a^(3)b3 \sqrt{3}\left(b^{2}-a^{2}\right)^{2}=16 a^{3} b33(b2−a2)2=16a3b
Question:-08 (a) Reduce the following partial differential equation to a canonical form and hence solve it :
y u x x + ( x + y ) u x y + x u y y = 0 y u x x + ( x + y ) u x y + x u y y = 0 yu_(xx)+(x+y)u_(xy)+xu_(yy)=0y u_{x x}+(x+y) u_{x y}+x u_{y y}=0yuxx+(x+y)uxy+xuyy=0
Question:-08 (b) Using Runge-Kutta method of fourth order, solve the differential equation d y d x = x + y 2 d y d x = x + y 2 (dy)/(dx)=x+y^(2)\frac{d y}{d x}=x+y^{2}dydx=x+y2 with y ( 0 ) = 1 y ( 0 ) = 1 y(0)=1y(0)=1y(0)=1, at x = 0 ⋅ 2 x = 0 ⋅ 2 x=0*2x=0 \cdot 2x=0⋅2. Use four decimal places for calculation and step length 0 ⋅ 1 0 ⋅ 1 0*10 \cdot 10⋅1.
Question:-08 (c) Verify that w = i k log ⁡ { ( z − i a ) / ( z + i a ) } w = i k log ⁡ { ( z − i a ) / ( z + i a ) } w=ik log{(z-ia)//(z+ia)}w=i k \log \{(z-i a) /(z+i a)\}w=iklog⁡{(z−ia)/(z+ia)} is the complex potential of a steady flow of fluid about a circular cylinder, where the plane y = 0 y = 0 y=0y=0y=0 is a rigid boundary. Find also the force exerted by the fluid on unit length of the cylinder.
upsc-m2021-2-4843690d-2e01-49d3-ba3c-bc046aac2819
खण्ड-A / SECTION-A
1(a) मान लीजिए कि m 1 , m 2 , ⋯ , m k m 1 , m 2 , ⋯ , m k m_(1),m_(2),cdots,m_(k)m_1, m_2, \cdots, m_km1,m2,⋯,mk धनात्मक पूर्णांक हैं तथा d > 0 , m 1 , m 2 , ⋯ , m k d > 0 , m 1 , m 2 , ⋯ , m k d > 0,m_(1),m_(2),cdots,m_(k)d>0, m_1, m_2, \cdots, m_kd>0,m1,m2,⋯,mk का महत्तम समापवर्तक है। दर्शाइए कि ऐसे पूर्णांक x 1 , x 2 , ⋯ , x k x 1 , x 2 , ⋯ , x k x_(1),x_(2),cdots,x_(k)x_1, x_2, \cdots, x_kx1,x2,⋯,xk अस्तित्व में हैं ताकि
d = x 1 m 1 + x 2 m 2 + ⋯ + x k m k d = x 1 m 1 + x 2 m 2 + ⋯ + x k m k d=x_(1)m_(1)+x_(2)m_(2)+cdots+x_(k)m_(k)d=x_1 m_1+x_2 m_2+\cdots+x_k m_kd=x1m1+x2m2+⋯+xkmk
Let m 1 , m 2 , ⋯ , m k m 1 , m 2 , ⋯ , m k m_(1),m_(2),cdots,m_(k)m_1, m_2, \cdots, m_km1,m2,⋯,mk be positive integers and d > 0 d > 0 d > 0d>0d>0 the greatest common divisor of m 1 , m 2 , ⋯ , m k m 1 , m 2 , ⋯ , m k m_(1),m_(2),cdots,m_(k)m_1, m_2, \cdots, m_km1,m2,⋯,mk. Show that there exist integers x 1 , x 2 , ⋯ , x k x 1 , x 2 , ⋯ , x k x_(1),x_(2),cdots,x_(k)x_1, x_2, \cdots, x_kx1,x2,⋯,xk such that
d = x 1 m 1 + x 2 m 2 + ⋯ + x k m k d = x 1 m 1 + x 2 m 2 + ⋯ + x k m k d=x_(1)m_(1)+x_(2)m_(2)+cdots+x_(k)m_(k)d=x_1 m_1+x_2 m_2+\cdots+x_k m_kd=x1m1+x2m2+⋯+xkmk
(b) श्रेणी
x 4 + x 4 1 + x 4 + x 4 ( 1 + x 4 ) 2 + x 4 ( 1 + x 4 ) 3 + ⋯ x 4 + x 4 1 + x 4 + x 4 1 + x 4 2 + x 4 1 + x 4 3 + ⋯ x^(4)+(x^(4))/(1+x^(4))+(x^(4))/((1+x^(4))^(2))+(x^(4))/((1+x^(4))^(3))+cdotsx^4+\frac{x^4}{1+x^4}+\frac{x^4}{\left(1+x^4\right)^2}+\frac{x^4}{\left(1+x^4\right)^3}+\cdotsx4+x41+x4+x4(1+x4)2+x4(1+x4)3+⋯
के [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] पर एकसमान अभिसरण की जाँच कीजिए।
Test the uniform convergence of the series
x 4 + x 4 1 + x 4 + x 4 ( 1 + x 4 ) 2 + x 4 ( 1 + x 4 ) 3 + ⋯ x 4 + x 4 1 + x 4 + x 4 1 + x 4 2 + x 4 1 + x 4 3 + ⋯ x^(4)+(x^(4))/(1+x^(4))+(x^(4))/((1+x^(4))^(2))+(x^(4))/((1+x^(4))^(3))+cdotsx^4+\frac{x^4}{1+x^4}+\frac{x^4}{\left(1+x^4\right)^2}+\frac{x^4}{\left(1+x^4\right)^3}+\cdotsx4+x41+x4+x4(1+x4)2+x4(1+x4)3+⋯
on [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1].
(c) यदि एक फलन f f fff, अन्तराल [ a , b ] [ a , b ] [a,b][a, b][a,b] में एकदिष्ट है, तब सिद्ध कीजिए कि f , [ a , b ] f , [ a , b ] f,[a,b]f,[a, b]f,[a,b] में रीमान समाकलनीय है।
If a function f f fff is monotonic in the interval [ a , b ] [ a , b ] [a,b][a, b][a,b], then prove that f f fff is Riemann integrable in [ a , b ] [ a , b ] [a,b][a, b][a,b].
(d) मान लीजिए कि c : [ 0 , 1 ] → C , c ( t ) = e 4 π i t , 0 ≤ t ≤ 1 c : [ 0 , 1 ] → C , c ( t ) = e 4 π i t , 0 ≤ t ≤ 1 c:[0,1]rarrC,c(t)=e^(4pi it),0 <= t <= 1c:[0,1] \rightarrow \mathbb{C}, c(t)=e^{4 \pi i t}, 0 \leq t \leq 1c:[0,1]→C,c(t)=e4πit,0≤t≤1 के द्वारा परिभाषित एक वक्र है। कन्दूर समाकल ∫ c d z 2 z 2 − 5 z + 2 ∫ c   d z 2 z 2 − 5 z + 2 int _(c)(dz)/(2z^(2)-5z+2)\int_c \frac{d z}{2 z^2-5 z+2}∫cdz2z2−5z+2 का मान निकालिए।
Let c : [ 0 , 1 ] → C c : [ 0 , 1 ] → C c:[0,1]rarrCc:[0,1] \rightarrow \mathbb{C}c:[0,1]→C be the curve, where c ( t ) = e 4 π i t , 0 ≤ t ≤ 1 c ( t ) = e 4 π i t , 0 ≤ t ≤ 1 c(t)=e^(4pi it),0 <= t <= 1c(t)=e^{4 \pi i t}, 0 \leq t \leq 1c(t)=e4πit,0≤t≤1. Evaluate the contour integral ∫ c d z 2 z 2 − 5 z + 2 ∫ c   d z 2 z 2 − 5 z + 2 int _(c)(dz)/(2z^(2)-5z+2)\int_c \frac{d z}{2 z^2-5 z+2}∫cdz2z2−5z+2.
(e) एक कम्पनी के एक विभाग के पाँच कर्मचारियों को पाँच कार्य सम्पन्न करने हैं। जितना समय (घंटों में) एक व्यक्ति एक कार्य को सम्पन्न करने के लिए लेता है, वह प्रभाविता आव्यूह में दिया गया है। इन पाँच कर्मचारियों को इन सभी कार्यों को इस तरह निर्धारित कीजिए जिससे कि समस्त कार्य सम्पन्न करने का समय न्यूनतम हो :
A department of a company has five employees with five jobs to be performed. The time (in hours) that each man takes to perform each job is given in the effectiveness matrix. Assign all the jobs to these five employees to minimize the total processing time :
2(a) f ( x ) = x 3 − 9 x 2 + 26 x − 24 f ( x ) = x 3 − 9 x 2 + 26 x − 24 f(x)=x^(3)-9x^(2)+26 x-24f(x)=x^3-9 x^2+26 x-24f(x)=x3−9x2+26x−24 का, 0 ≤ x ≤ 1 0 ≤ x ≤ 1 0 <= x <= 10 \leq x \leq 10≤x≤1 के लिए, अधिकतम तथा न्यूनतम मान निकालिए।
Find the maximum and minimum values of f ( x ) = x 3 − 9 x 2 + 26 x − 24 f ( x ) = x 3 − 9 x 2 + 26 x − 24 f(x)=x^(3)-9x^(2)+26 x-24f(x)=x^3-9 x^2+26 x-24f(x)=x3−9x2+26x−24 for 0 ≤ x ≤ 1 0 ≤ x ≤ 1 0 <= x <= 10 \leq x \leq 10≤x≤1
(b) मान लीजिए कि F F FFF एक क्षेत्र है तथा f ( x ) ∈ F [ x ] f ( x ) ∈ F [ x ] f(x)in F[x]f(x) \in F[x]f(x)∈F[x], क्षेत्र F F FFF के ऊपर घात > 0 > 0 > 0>0>0 का एक बहुपद है। दर्शाइए कि एक क्षेत्र F ′ F ′ F^(‘)F^{\prime}F′ तथा एक अंतःस्थापन q : F → F ′ q : F → F ′ q:F rarrF^(‘)q: F \rightarrow F^{\prime}q:F→F′ इस प्रकार से अस्तित्व में हैं कि बहुपद f q ∈ F ′ [ x ] f q ∈ F ′ [ x ] f^(q)inF^(‘)[x]f^q \in F^{\prime}[x]fq∈F′[x] का एक मूल F ′ F ′ F^(‘)F^{\prime}F′ में है, जहाँ f q , f f q , f f^(q),ff^q, ffq,f के प्रत्येक गुणांक a a aaa को q ( a ) q ( a ) q(a)q(a)q(a) द्वारा प्रतिस्थापित करने से प्राप्त होता है।
Let F F FFF be a field and f ( x ) ∈ F [ x ] f ( x ) ∈ F [ x ] f(x)in F[x]f(x) \in F[x]f(x)∈F[x] a polynomial of degree > 0 > 0 > 0>0>0 over F F FFF. Show that there is a field F ′ F ′ F^(‘)F^{\prime}F′ and an imbedding q : F → F ′ q : F → F ′ q:F rarrF^(‘)q: F \rightarrow F^{\prime}q:F→F′ s.t. the polynomial f q ∈ F ′ [ x ] f q ∈ F ′ [ x ] f^(q)inF^(‘)[x]f^q \in F^{\prime}[x]fq∈F′[x] has a root in F ′ F ′ F^(‘)F^{\prime}F′, where f q f q f^(q)f^qfq is obtained by replacing each coefficient a a aaa of f f fff by q ( a ) q ( a ) q(a)q(a)q(a).
(c) क्षेत्र | z + 1 | > 3 | z + 1 | > 3 |z+1| > 3|z+1|>3|z+1|>3 में f ( z ) = z 2 − z + 1 z ( z 2 − 3 z + 2 ) f ( z ) = z 2 − z + 1 z z 2 − 3 z + 2 f(z)=(z^(2)-z+1)/(z(z^(2)-3z+2))f(z)=\frac{z^2-z+1}{z\left(z^2-3 z+2\right)}f(z)=z2−z+1z(z2−3z+2) का लौराँ श्रेणी प्रसार, ( z + 1 ) ( z + 1 ) (z+1)(z+1)(z+1) की घातों में ज्ञात कीजिए।
Find the Laurent series expansion of f ( z ) = z 2 − z + 1 z ( z 2 − 3 z + 2 ) f ( z ) = z 2 − z + 1 z z 2 − 3 z + 2 f(z)=(z^(2)-z+1)/(z(z^(2)-3z+2))f(z)=\frac{z^2-z+1}{z\left(z^2-3 z+2\right)}f(z)=z2−z+1z(z2−3z+2) in the powers of ( z + 1 ) ( z + 1 ) (z+1)(z+1)(z+1) in the region | z + 1 | > 3 | z + 1 | > 3 |z+1| > 3|z+1|>3|z+1|>3
3(a) मान लीजिए कि f f fff एक सर्वत्र वैश्लेषिक फलन है जिसके केन्द्र z = 0 z = 0 z=0z=0z=0 पर टेलर श्रेणी प्रसार में अपरिमित रूप से अनेक पद हैं। दर्शाइए कि f ( 1 z ) f 1 z f((1)/(z))f\left(\frac{1}{z}\right)f(1z) की z = 0 z = 0 z=0z=0z=0 एक अनिवार्य विचित्रता है।
Let f f fff be an entire function whose Taylor series expansion with centre z = 0 z = 0 z=0z=0z=0 has infinitely many terms. Show that z = 0 z = 0 z=0z=0z=0 is an essential singularity of ( f ( 1 z ) ) f 1 z ) f((1)/(z)))f\left(\frac{1}{z}\right))f(1z)) .
(b) शर्तों a x 2 + b y 2 + c z 2 = 1 a x 2 + b y 2 + c z 2 = 1 ax^(2)+by^(2)+cz^(2)=1a x^2+b y^2+c z^2=1ax2+by2+cz2=1 तथा l x + m y + n z = 0 l x + m y + n z = 0 lx+my+nz=0l x+m y+n z=0lx+my+nz=0 से प्रतिबन्धित x 2 + y 2 + z 2 x 2 + y 2 + z 2 x^(2)+y^(2)+z^(2)x^2+y^2+z^2x2+y2+z2 के स्तब्ध (अचर) मान निकालिए। परिणाम की ज्यामितीय व्याख्या कीजिए।
Find the stationary values of x 2 + y 2 + z 2 x 2 + y 2 + z 2 x^(2)+y^(2)+z^(2)x^2+y^2+z^2x2+y2+z2 subject to the conditions a x 2 + b y 2 + c z 2 = 1 a x 2 + b y 2 + c z 2 = 1 ax^(2)+by^(2)+cz^(2)=1a x^2+b y^2+c z^2=1ax2+by2+cz2=1 and l x + m y + n z = 0 l x + m y + n z = 0 lx+my+nz=0l x+m y+n z=0lx+my+nz=0. Interpret the result
(c) निम्न रैखिक प्रोग्रामन समस्या को द्वैती रैखिक प्रोग्रामन सम न्यूनतमीकरण कीजिए Z = x 1 − 3 x 2 − 2 x 3 Z = x 1 − 3 x 2 − 2 x 3 Z=x_(1)-3x_(2)-2x_(3)Z=x_1-3 x_2-2 x_3Z=x1−3x2−2x3
बशर्ते कि
3 x 1 − x 2 + 2 x 3 ≤ 7 2 x 1 − 4 x 2 ≥ 12 − 4 x 1 + 3 x 2 + 8 x 3 = 10 3 x 1 − x 2 + 2 x 3 ≤ 7 2 x 1 − 4 x 2 ≥ 12 − 4 x 1 + 3 x 2 + 8 x 3 = 10 {:[3x_(1)-x_(2)+2x_(3) <= 7],[2x_(1)-4x_(2) >= 12],[-4x_(1)+3x_(2)+8x_(3)=10]:}\begin{aligned} 3 x_1-x_2+2 x_3 & \leq 7 \\ 2 x_1-4 x_2 & \geq 12 \\ -4 x_1+3 x_2+8 x_3 &=10 \end{aligned}3x1−x2+2x3≤72x1−4x2≥12−4x1+3x2+8x3=10
जहाँ x 1 , x 2 ≥ 0 x 1 , x 2 ≥ 0 x_(1),x_(2) >= 0x_1, x_2 \geq 0x1,x2≥0 तथा x 3 x 3 x_(3)x_3x3 का चिह्न अप्रतिबंधित है।
Convert the following LPP into dual LPP :
Minimize Z = x 1 − 3 x 2 − 2 x 3 Z = x 1 − 3 x 2 − 2 x 3 quad Z=x_(1)-3x_(2)-2x_(3)\quad Z=x_1-3 x_2-2 x_3Z=x1−3x2−2x3
subject to
3 x 1 − x 2 + 2 x 3 ≤ 7 2 x 1 − 4 x 2 ≥ 12 − 4 x 1 + 3 x 2 + 8 x 3 = 10 3 x 1 − x 2 + 2 x 3 ≤ 7 2 x 1 − 4 x 2 ≥ 12 − 4 x 1 + 3 x 2 + 8 x 3 = 10 {:[3x_(1)-x_(2)+2x_(3) <= 7],[2x_(1)-4x_(2) >= 12],[-4x_(1)+3x_(2)+8x_(3)=10]:}\begin{aligned} 3 x_1-x_2+2 x_3 & \leq 7 \\ 2 x_1-4 x_2 & \geq 12 \\ -4 x_1+3 x_2+8 x_3 &=10 \end{aligned}3x1−x2+2x3≤72x1−4x2≥12−4x1+3x2+8x3=10
where x 1 , x 2 ≥ 0 x 1 , x 2 ≥ 0 x_(1),x_(2) >= 0x_1, x_2 \geq 0x1,x2≥0 and x 3 x 3 x_(3)x_3x3 is unrestricted in sign.
4(a) दर्शाइए कि परिमेय संख्याओं के योज्य समूह Q Q Q\mathbb{Q}Q के अपरिमित रूप से अनेक उपसमूह हैं।
Show that there are infinitely many subgroups of the additive group Q Q Q\mathbb{Q}Q of rational numbers.
(b) कन्टूर समाकलन का उपयोग कर समाकल ∫ − ∞ ∞ sin ⁡ x d x x ( x 2 + a 2 ) , a > 0 ∫ − ∞ ∞   sin ⁡ x d x x x 2 + a 2 , a > 0 int_(-oo)^(oo)(sin xdx)/(x(x^(2)+a^(2))),a > 0\int_{-\infty}^{\infty} \frac{\sin x d x}{x\left(x^2+a^2\right)}, a>0∫−∞∞sin⁡xdxx(x2+a2),a>0 का मान ज्ञात कीजिए।
Using contour integration, evaluate the integral ∫ − ∞ ∞ sin ⁡ x d x x ( x 2 + a 2 ) , a > 0 . 20 ∫ − ∞ ∞   sin ⁡ x d x x x 2 + a 2 , a > 0 . 20 int_(-oo)^(oo)(sin xdx)/(x(x^(2)+a^(2))),a > 0.quad20\int_{-\infty}^{\infty} \frac{\sin x d x}{x\left(x^2+a^2\right)}, a>0 . \quad 20∫−∞∞sin⁡xdxx(x2+a2),a>0.20
(c) बड़ा M M MMM (बिग M M MMM ) विधि का उपयोग करके निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
अधिकतमीकरण कीजिए Z = 4 x 1 + 5 x 2 + 2 x 3 Z = 4 x 1 + 5 x 2 + 2 x 3 Z=4x_(1)+5x_(2)+2x_(3)Z=4 x_1+5 x_2+2 x_3Z=4x1+5x2+2x3
बशर्ते कि
2 x 1 + x 2 + x 3 ≥ 10 x 1 + 3 x 2 + x 3 ≤ 12 x 1 + x 2 + x 3 = 6 x 1 , x 2 , x 3 ≥ 0 2 x 1 + x 2 + x 3 ≥ 10 x 1 + 3 x 2 + x 3 ≤ 12 x 1 + x 2 + x 3 = 6 x 1 , x 2 , x 3 ≥ 0 {:[2x_(1)+x_(2)+x_(3) >= 10],[x_(1)+3x_(2)+x_(3) <= 12],[x_(1)+x_(2)+x_(3)=6],[x_(1)”,”x_(2)”,”x_(3) >= 0]:}\begin{aligned} 2 x_1+x_2+x_3 & \geq 10 \\ x_1+3 x_2+x_3 & \leq 12 \\ x_1+x_2+x_3 &=6 \\ x_1, x_2, x_3 & \geq 0 \end{aligned}2x1+x2+x3≥10x1+3x2+x3≤12x1+x2+x3=6x1,x2,x3≥0
Solve the following linear programming problem using Big M method :
Maximize Z = 4 x 1 + 5 x 2 + 2 x 3 Z = 4 x 1 + 5 x 2 + 2 x 3 quad Z=4x_(1)+5x_(2)+2x_(3)\quad Z=4 x_1+5 x_2+2 x_3Z=4x1+5x2+2x3
subject to
2 x 1 + x 2 + x 3 ≥ 10 x 1 + 3 x 2 + x 3 ≤ 12 x 1 + x 2 + x 3 = 6 x 1 , x 2 , x 3 ≥ 0 2 x 1 + x 2 + x 3 ≥ 10 x 1 + 3 x 2 + x 3 ≤ 12 x 1 + x 2 + x 3 = 6 x 1 , x 2 , x 3 ≥ 0 {:[2x_(1)+x_(2)+x_(3) >= 10],[x_(1)+3x_(2)+x_(3) <= 12],[x_(1)+x_(2)+x_(3)=6],[x_(1)”,”x_(2)”,”x_(3) >= 0]:}\begin{aligned} 2 x_1+x_2+x_3 & \geq 10 \\ x_1+3 x_2+x_3 & \leq 12 \\ x_1+x_2+x_3 &=6 \\ x_1, x_2, x_3 & \geq 0 \end{aligned}2x1+x2+x3≥10x1+3x2+x3≤12x1+x2+x3=6x1,x2,x3≥0
खण्ड-B / SECTION-B
5(a) समीकरण f ( x + y + z , x 2 + y 2 + z 2 ) = 0 f x + y + z , x 2 + y 2 + z 2 = 0 f(x+y+z,x^(2)+y^(2)+z^(2))=0f\left(x+y+z, x^2+y^2+z^2\right)=0f(x+y+z,x2+y2+z2)=0 से स्वैच्छिक फलन f f fff का विलोपन कर आंशिक अवकल समीकरण को प्राप्त कीजिए।
Obtain the partial differential equation by eliminating arbitrary function f f fff from the equation f ( x + y + z , x 2 + y 2 + z 2 ) = 0 f x + y + z , x 2 + y 2 + z 2 = 0 f(x+y+z,x^(2)+y^(2)+z^(2))=0f\left(x+y+z, x^2+y^2+z^2\right)=0f(x+y+z,x2+y2+z2)=0.
(b) प्रारंभिक मानों 0 , π 2 0 , π 2 0,(pi)/(2)0, \frac{\pi}{2}0,π2 का उपयोग करके एक संख्यात्मक तकनीक के द्वारा समीकरण 3 x = 1 + cos ⁡ x 3 x = 1 + cos ⁡ x 3x=1+cos x3 x=1+\cos x3x=1+cos⁡x का एक धनात्मक मूल ज्ञात कीजिए, तथा न्यूटन-राफ्सन विधि के द्वारा परिणाम को 8 सार्थक अंकों तक और शुद्ध मान के निकट लाइए।
Find a positive root of the equation 3 x = 1 + cos ⁡ x 3 x = 1 + cos ⁡ x 3x=1+cos x3 x=1+\cos x3x=1+cos⁡x by a numerical technique using initial values 0 , π 2 0 , π 2 0,(pi)/(2)0, \frac{\pi}{2}0,π2; and further improve the result using Newton-Raphson method correct to 8 significant figures.
(c) (i) ( 3798 ⋅ 3875 ) 10 ( 3798 ⋅ 3875 ) 10 (3798*3875)_(10)(3798 \cdot 3875)_{10}(3798⋅3875)10 को अष्टधारी तथा षोडशाधारी तुल्यमानों में बदलिए।
(ii) (\rceil P → R ) ∧ ( Q ⇄ P ) P → R ) ∧ ( Q ⇄ P ) P rarr R)^^(Q⇄P)P \rightarrow R) \wedge(Q \rightleftarrows P)P→R)∧(Q⇄P) का मुख्य संयोजक सामान्य रूप (प्रिंसिपल कंजंक्टिव नॉर्मल फॉर्म) प्राप्त कीजिए।
(i) Convert ( 3798 ⋅ 3875 ) 10 ( 3798 ⋅ 3875 ) 10 (3798*3875)_(10)(3798 \cdot 3875)_{10}(3798⋅3875)10 into octal and hexadecimal equivalents.
(ii) Obtain the principal conjunctive normal form of
(\rceil P → R ) ∧ ( Q ⇄ P ) P → R ) ∧ ( Q ⇄ P ) P rarr R)^^(Q⇄P)P \rightarrow R) \wedge(Q \rightleftarrows P)P→R)∧(Q⇄P).
(d) ऊर्ध्वर्धर x y x y xyx yxy-तल में स्थित एक वृत्त के अनुदिश एक कण गति के लिए व्यवरुद्ध है। डी’एलम्बर्ट के नियम की सहायता से दर्शाइए कि इसकी गति का समीकरण x ¨ y − y ¨ x − g x = 0 x ¨ y − y ¨ x − g x = 0 x^(¨)y-y^(¨)x-gx=0\ddot{x} y-\ddot{y} x-g x=0x¨y−y¨x−gx=0 है, जहाँ g g ggg गुरुत्वीय त्वरण है।
A particle is constrained to move along a circle lying in the vertical x y x y xyx yxy-plane. With the help of the D’Alembert’s principle, show that its equation of motion is x ¨ y − y ¨ x − g x = 0 x ¨ y − y ¨ x − g x = 0 x^(¨)y-y^(¨)x-gx=0\ddot{x} y-\ddot{y} x-g x=0x¨y−y¨x−gx=0, where g g ggg is the acceleration due to gravity.
(e) उद्गमों (स्रोतों) व अभिगमों (सिंकों) के किस विन्यास से वेग विभव w = log e ⁡ ( z − a 2 z ) w = log e ⁡ z − a 2 z w=log _(e)(z-(a^(2))/(z))w=\log _e\left(z-\frac{a^2}{z}\right)w=loge⁡(z−a2z) हो सकता है? संगत धारा-रेखाओं का ख़ाका खींचिए और सिद्ध कीजिए कि उनमें से दो, वृत्त r = a r = a r=ar=ar=a तथा y y yyy-अक्ष में प्रविभाजित होती हैं।
What arrangements of sources and sinks can have the velocity potential w = log e ⁡ ( z − a 2 z ) w = log e ⁡ z − a 2 z w=log _(e)(z-(a^(2))/(z))w=\log _e\left(z-\frac{a^2}{z}\right)w=loge⁡(z−a2z) ? Draw the corresponding sketch of the streamlines and prove that two of them subdivide into the circle r = a r = a r=ar=ar=a and the axis of y y yyy.
6(a) तरंग समीकरण
a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 , 0 < x < L , t > 0 a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 , 0 < x < L , t > 0 a^(2)(del^(2)u)/(delx^(2))=(del^(2)u)/(delt^(2)),quad0 < x < L,quad t > 0a^2 \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}, \quad 0<x<L, \quad t>0a2∂2u∂x2=∂2u∂t2,0<x<L,t>0
का शर्तों
u ( 0 , t ) = 0 , u ( L , t ) = 0 u ( x , 0 ) = 1 4 x ( L − x ) , ∂ u ∂ t | t = 0 = 0 u ( 0 , t ) = 0 , u ( L , t ) = 0 u ( x , 0 ) = 1 4 x ( L − x ) , ∂ u ∂ t t = 0 = 0 {:[u(0″,”t)=0″,”quad u(L”,”t)=0],[u(x”,”0)=(1)/(4)x(L-x)”,”(del u)/(del t)|_(t=0)=0]:}\begin{gathered} u(0, t)=0, \quad u(L, t)=0 \\ u(x, 0)=\frac{1}{4} x(L-x),\left.\frac{\partial u}{\partial t}\right|_{t=0}=0 \end{gathered}u(0,t)=0,u(L,t)=0u(x,0)=14x(L−x),∂u∂t|t=0=0
से प्रतिबन्धित हल ज्ञात कीजिए।
Solve the wave equation
a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 , 0 < x < L , t > 0 a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 , 0 < x < L , t > 0 a^(2)(del^(2)u)/(delx^(2))=(del^(2)u)/(delt^(2)),quad0 < x < L,quad t > 0a^2 \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}, \quad 0<x<L, \quad t>0a2∂2u∂x2=∂2u∂t2,0<x<L,t>0
subject to the conditions
u ( 0 , t ) = 0 , u ( L , t ) = 0 u ( x , 0 ) = 1 4 x ( L − x ) , ∂ u ∂ t | t = 0 = 0 u ( 0 , t ) = 0 , u ( L , t ) = 0 u ( x , 0 ) = 1 4 x ( L − x ) , ∂ u ∂ t t = 0 = 0 {:[u(0″,”t)=0″,”u(L”,”t)=0],[u(x”,”0)=(1)/(4)x(L-x)”,”(del u)/(del t)|_(t=0)=0]:}\begin{gathered} u(0, t)=0, u(L, t)=0 \\ u(x, 0)=\frac{1}{4} x(L-x),\left.\frac{\partial u}{\partial t}\right|_{t=0}=0 \end{gathered}u(0,t)=0,u(L,t)=0u(x,0)=14x(L−x),∂u∂t|t=0=0
(b) नीचे दी गई सारणी पर आधारित बूलीय फलन F ( x , y , z ) F ( x , y , z ) F(x,y,z)F(x, y, z)F(x,y,z) को निकालिए और तब F ( x , y , z ) F ( x , y , z ) F(x,y,z)F(x, y, z)F(x,y,z) को सरल कीजिए तथा उसके अनुरूप GATE परिपथ खींचिए :
x x xxx y y yyy z z zzz F ( x , y , z ) F ( x , y , z ) F(x,y,z)F(x, y, z)F(x,y,z)
1111
1101
1011
1000
0111
0100
0010
0000
x y z F(x,y,z) 1 1 1 1 1 1 0 1 1 0 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 0 0 0| $x$ | $y$ | $z$ | $F(x, y, z)$ | | :—: | :—: | :—: | :—: | | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 1 | | 1 | 0 | 1 | 1 | | 1 | 0 | 0 | 0 | | 0 | 1 | 1 | 1 | | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 0 | | 0 | 0 | 0 | 0 |
Obtain the Boolean function F ( x , y , z ) F ( x , y , z ) F(x,y,z)F(x, y, z)F(x,y,z) based on the table given below. Then simplify F ( x , y , z ) F ( x , y , z ) F(x,y,z)F(x, y, z)F(x,y,z) and draw the corresponding GATE network :
x x xxx y y yyy z z zzz F ( x , y , z ) F ( x , y , z ) F(x,y,z)F(x, y, z)F(x,y,z)
1111
1101
1011
1000
0111
0100
0010
0000
x y z F(x,y,z) 1 1 1 1 1 1 0 1 1 0 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 0 0 0| $x$ | $y$ | $z$ | $F(x, y, z)$ | | :—: | :—: | :—: | :—: | | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 1 | | 1 | 0 | 1 | 1 | | 1 | 0 | 0 | 0 | | 0 | 1 | 1 | 1 | | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 0 | | 0 | 0 | 0 | 0 |
(c) एक छोटी चिकनी घिरनी के ऊपर से गुजरने वाली एक अवितान्य डोरी के सिरों से बंधे असमान संहति वाले दो कणों के निकाय की गति के लिए लग्रांजी समीकरण ज्ञात कीजिए।
Obtain the Lagrangian equation for the motion of a system of two particles of unequal masses connected by an inextensible string passing over a small smooth pulley.
7(a) आंशिक अवकल समीकरण
( D 2 − D ′ 2 − 3 D + 3 D ′ ) z = x y + e x + 2 y D 2 − D ′ 2 − 3 D + 3 D ′ z = x y + e x + 2 y (D^(2)-D^(‘2)-3D+3D^(‘))z=xy+e^(x+2y)\left(D^2-D^{\prime 2}-3 D+3 D^{\prime}\right) z=x y+e^{x+2 y}(D2−D′2−3D+3D′)z=xy+ex+2y
का व्यापक हल ज्ञात कीजिए, जहाँ D ≡ ∂ ∂ x D ≡ ∂ ∂ x D-=(del)/(del x)D \equiv \frac{\partial}{\partial x}D≡∂∂x तथा D ′ ≡ ∂ ∂ y D ′ ≡ ∂ ∂ y D^(‘)-=(del)/(del y)D^{\prime} \equiv \frac{\partial}{\partial y}D′≡∂∂y हैं।
Find the general solution of the partial differential equation
( D 2 − D ′ 2 − 3 D + 3 D ′ ) z = x y + e x + 2 y D 2 − D ′ 2 − 3 D + 3 D ′ z = x y + e x + 2 y (D^(2)-D^(‘2)-3D+3D^(‘))z=xy+e^(x+2y)\left(D^2-D^{\prime 2}-3 D+3 D^{\prime}\right) z=x y+e^{x+2 y}(D2−D′2−3D+3D′)z=xy+ex+2y
where D ≡ ∂ ∂ x D ≡ ∂ ∂ x D-=(del)/(del x)D \equiv \frac{\partial}{\partial x}D≡∂∂x and D ′ ≡ ∂ ∂ y D ′ ≡ ∂ ∂ y D^(‘)-=(del)/(del y)D^{\prime} \equiv \frac{\partial}{\partial y}D′≡∂∂y.
(b) समीकरणों के निकाय
3 x 1 + 9 x 2 − 2 x 3 = 11 4 x 1 + 2 x 2 + 13 x 3 = 24 4 x 1 − 2 x 2 + x 3 = − 8 3 x 1 + 9 x 2 − 2 x 3 = 11 4 x 1 + 2 x 2 + 13 x 3 = 24 4 x 1 − 2 x 2 + x 3 = − 8 {:[3x_(1)+9x_(2)-2x_(3)=11],[4x_(1)+2x_(2)+13x_(3)=24],[4x_(1)-2x_(2)+x_(3)=-8]:}\begin{aligned} 3 x_1+9 x_2-2 x_3 &=11 \\ 4 x_1+2 x_2+13 x_3 &=24 \\ 4 x_1-2 x_2+x_3 &=-8 \end{aligned}3x1+9x2−2x3=114x1+2x2+13x3=244x1−2x2+x3=−8
का गाउस-सीडल विधि द्वारा 4 सार्थक अंकों तक सही हल ज्ञात कीजिए, यह सत्यापन करने के बाद कि क्या यह विधि आपके द्वारा निकाय के रूपांतरित रूप में अनुप्रयोज्य है।
Solve the system of equations
3 x 1 + 9 x 2 − 2 x 3 = 11 4 x 1 + 2 x 2 + 13 x 3 = 24 4 x 1 − 2 x 2 + x 3 = − 8 3 x 1 + 9 x 2 − 2 x 3 = 11 4 x 1 + 2 x 2 + 13 x 3 = 24 4 x 1 − 2 x 2 + x 3 = − 8 {:[3x_(1)+9x_(2)-2x_(3)=11],[4x_(1)+2x_(2)+13x_(3)=24],[4x_(1)-2x_(2)+x_(3)=-8]:}\begin{aligned} 3 x_1+9 x_2-2 x_3 &=11 \\ 4 x_1+2 x_2+13 x_3 &=24 \\ 4 x_1-2 x_2+x_3 &=-8 \end{aligned}3x1+9x2−2x3=114x1+2x2+13x3=244x1−2x2+x3=−8
correct up to 4 significant figures by using Gauss-Seidel method after verifying whether the method is applicable in your transformed form of the system. 15
(c) दर्शाइए कि q → = λ ( − y i ^ + x j ^ ) x 2 + y 2 , ( λ = q → = λ ( − y i ^ + x j ^ ) x 2 + y 2 , ( λ = vec(q)=(lambda(-y( hat(i))+x( hat(j))))/(x^(2)+y^(2)),(lambda=\vec{q}=\frac{\lambda(-y \hat{i}+x \hat{j})}{x^2+y^2},(\lambda=q→=λ(−yi^+xj^)x2+y2,(λ= स्थिरांक) एक संभाव्य असंपीड्य तरल गति है। धारा-रेखाएँ निकालिए। क्या गति का प्रकार विभव है? यदि हाँ, तो वेग विभव निकालिए।
Show that q → = λ ( − y i ^ + x j ^ ) x 2 + y 2 , ( λ = q → = λ ( − y i ^ + x j ^ ) x 2 + y 2 , ( λ = vec(q)=(lambda(-y( hat(i))+x( hat(j))))/(x^(2)+y^(2)),(lambda=\vec{q}=\frac{\lambda(-y \hat{i}+x \hat{j})}{x^2+y^2},(\lambda=q→=λ(−yi^+xj^)x2+y2,(λ= constant ) ) ))) is a possible incompressible fluid motion. Determine the streamlines. Is the kind of the motion potential? If yes, then find the velocity potential.
8 (a) चार्पिट विधि का उपयोग करके आंशिक अवकल समीकरण p = ( z + q y ) 2 p = ( z + q y ) 2 p=(z+qy)^(2)p=(z+q y)^2p=(z+qy)2 का पूर्ण समाकल ज्ञात कीजिए।
Find a complete integral of the partial differential equation p = ( z + q y ) 2 p = ( z + q y ) 2 p=(z+qy)^(2)p=(z+q y)^2p=(z+qy)2 by using Charpit’s method.
(b) न्यूटन के पश्चांतर अंतर्वेशन सूत्र की व्युत्पत्ति कीजिए तथा त्रुटि-विश्लेषण भी कीजिए।
Derive Newton’s backward difference interpolation formula and also do error analysis.
(c) दर्शाइए कि सम्मिश्र विभव tan − 1 ⁡ z tan − 1 ⁡ z tan^(-1)z\tan ^{-1} ztan−1⁡z के लिए धारा-रेखाएँ तथा समविभव वक्र, वृत्त हैं। किसी भी बिन्दु पर वेग निकालिए तथा z = ± i z = ± i z=+-iz=\pm iz=±i पर विचित्रता जाँचिए।
Show that for the complex potential tan − 1 ⁡ z tan − 1 ⁡ z tan^(-1)z\tan ^{-1} ztan−1⁡z, the streamlines and equipotential curves are circles. Find the velocity at any point and check the singularities at z = ± i z = ± i z=+-iz=\pm iz=±i.
upsc-m2020-2-8516f8b5-1b27-4d2a-a0f1-a49d9b15de17
खण्ड ‘ A ′ A ′ A^(‘)A^{\prime}A′ SECTION ‘ A ′ A ′ A^(‘)A^{\prime}A′
1.(a) मान लीजिए कि S 3 S 3 S_(3)S_3S3 व Z 3 Z 3 Z_(3)Z_3Z3 क्रमशः 3 प्रतीकों का क्रमचय समूह एवं मॉड्यूल 3 अवशिष्ट वर्गों के समूह हैं। दर्शाइए कि S 3 S 3 S_(3)S_3S3 का Z 3 Z 3 Z_(3)Z_3Z3 में तुच्छ समाकारिता के अतिरिक्त कोई भी समाकारिता नहीं है ।
Let S 3 S 3 S_(3)S_3S3 and Z 3 Z 3 Z_(3)Z_3Z3 be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of S 3 S 3 S_(3)S_3S3 in Z 3 Z 3 Z_(3)Z_3Z3 except the trivial homomorphism.
1.(b) मान लीजिए R R RRR मुख्य गुणजावली प्रान्त है । दर्शाइए कि R R RRR के विभाग-वलय की प्रत्येक गुणजावली, मुख्य गुणजावली है तथा R / P , R R / P , R R//P,RR / P, RR/P,R के अभाज्यगुणजावली P P PPP के लिए मुख्य गुणजावली प्रान्त है ।
Let R R RRR be a principal ideal domain. Show that every ideal of a quotient ring of R R RRR is principal ideal and R / P R / P R//PR / PR/P is a principal ideal domain for a prime ideal P P PPP of R R RRR.
1.(c) सिद्ध कीजिए कि शर्त
| a n + 1 − a n | ⩽ α | a n − a n − 1 | a n + 1 − a n ⩽ α a n − a n − 1 |a_(n+1)-a_(n)| <= alpha|a_(n)-a_(n-1)|\left|a_{n+1}-a_n\right| \leqslant \alpha\left|a_n-a_{n-1}\right||an+1−an|⩽α|an−an−1|, जहाँ पर 0 < α < 1 0 < α < 1 0 < alpha < 10<\alpha<10<α<1 को सभी प्राकृतिक संख्याओं n ⩾ 2 n ⩾ 2 n >= 2n \geqslant 2n⩾2 के लिए सन्तुष्ट करने वाला अनुक्रम ( a n ′ ) a n ′ (a_(n)^(‘))\left(a_n^{\prime}\right)(an′), कॉशी-अनुक्रम होता है ।
Prove that the sequence ( a n ) a n (a_(n))\left(a_n\right)(an) satisfying the condition | a n + 1 − a n | ⩽ α | a n − a n − 1 | , 0 < α < 1 a n + 1 − a n ⩽ α a n − a n − 1 , 0 < α < 1 |a_(n+1)-a_(n)| <= alpha|a_(n)-a_(n-1)|,0 < alpha < 1\left|a_{n+1}-a_n\right| \leqslant \alpha\left|a_n-a_{n-1}\right|, 0<\alpha<1|an+1−an|⩽α|an−an−1|,0<α<1 for all natural numbers n ⩾ 2 n ⩾ 2 n >= 2n \geqslant 2n⩾2, is a Cauchy sequence.
1.(d) समाकल ∫ C ( z 2 + 3 z ) d z ∫ C   z 2 + 3 z d z int _(C)(z^(2)+3z)dz\int_C\left(z^2+3 z\right) d z∫C(z2+3z)dz का, ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) से ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) तक वक्र C C CCC के वामावर्त अनुगत जहाँ पर C C CCC वृत्त | z | = 2 | z | = 2 |z|=2|z|=2|z|=2 है, मान निकालिए ।
Evaluate the integral ∫ C ( z 2 + 3 z ) d z ∫ C   z 2 + 3 z d z int _(C)(z^(2)+3z)dz\int_C\left(z^2+3 z\right) d z∫C(z2+3z)dz counterclockwise from ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) to ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) along the curve C C CCC, where C C CCC is the circle | z | = 2 | z | = 2 |z|=2|z|=2|z|=2.
1.(e) यू.पी.एस.सी. के रखरखाव विभाग ने भवन में पर्दों की आवश्यकता-पूर्ति हेतु पर्दा-कपडे के पर्याप्त संख्या में टुकड़े खरीदे हैं। प्रत्येक टुकड़े की लम्बाई 17 फुट है। पर्दों की लम्बाई के अनुसार आवश्यकता निम्नलिखित है :
Curtain length (in feet) Number required 5 700 9 400 7 300  Curtain length (in feet)   Number required  5 700 9 400 7 300 {:[” Curtain length (in feet) “,” Number required “],[5,700],[9,400],[7,300]:}\begin{array}{cc}\text { Curtain length (in feet) } & \text { Number required } \\ 5 & 700 \\ 9 & 400 \\ 7 & 300\end{array} Curtain length (in feet)  Number required 570094007300
टुकडों एवं सभी पर्दों की चौड़ाइयाँ समान हैं। विभिन्न रूप से काटे गये टुकड़ों की संख्या का निर्णय इस प्रकार करने हेतु कि कुल कटान-हानि न्यूनतम हो, एक रैखिक प्रोग्रामन समस्या का प्रामाणिक इस प्रकार करने हेतु कि कुल कटान-हानि न्यूनतम हो, एक रैखिक प्रोग्रामन समस्या का प्रामाणिक रूप में निर्धारण कीजिए । इसका एक आधारी सुसंगत हल भी दीजिए ।
UPSC maintenance section has purchased sufficient number of curtain cloth pieces to meet the curtain requirement of its building. The length of each piece is 17 feet. The requirement according to curtain length is as follows:
Curtain length (in feet) Number required 5 700 9 400 7 300  Curtain length (in feet)   Number required  5 700 9 400 7 300 {:[” Curtain length (in feet) “,” Number required “],[5,700],[9,400],[7,300]:}\begin{array}{cc}\text { Curtain length (in feet) } & \text { Number required } \\ 5 & 700 \\ 9 & 400 \\ 7 & 300\end{array} Curtain length (in feet)  Number required 570094007300
The width of all curtains is same as that of available pieces. Form a linear programming problem in standard form that decides the number of pieces cut in different ways so that the total trim loss is minimum. Also give a basic feasible solution to it.
2.(a) मान लीजिए G , n G , n G,nG, nG,n समूहांक का परिमित चक्रीय समूह है। तब सिद्ध कीजिए कि G G GGG के ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) जनक हैं (जहाँ पर ϕ ϕ phi\phiϕ ऑयलर ϕ ϕ phi\phiϕ-फलन है) ।
Let G G GGG be a finite cyclic group of order n n nnn. Then prove that G G GGG has ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) generators (where ϕ ϕ phi\phiϕ is Euler’s ϕ ϕ phi\phiϕ-function).
2.(b) सिद्ध कीजिए कि फलन f ( x ) = sin ⁡ x 2 f ( x ) = sin ⁡ x 2 f(x)=sin x^(2)f(x)=\sin x^2f(x)=sin⁡x2 अंतराल [ 0 , ∞ [ 0 , ∞ [0,oo[0, \infty[0,∞ [ पर एकसमान संतत नहीं है।
Prove that the function f ( x ) = sin ⁡ x 2 f ( x ) = sin ⁡ x 2 f(x)=sin x^(2)f(x)=\sin x^2f(x)=sin⁡x2 is not uniformly continuous on the interval [ 0 , ∞ [ [ 0 , ∞ [ [0,oo[[0, \infty[[0,∞[.
2.(c) कन्टूर समाकलन का उपयोग कर, समाकल ∫ 0 2 π 1 3 + 2 sin ⁡ θ d θ ∫ 0 2 π   1 3 + 2 sin ⁡ θ d θ int_(0)^(2pi)(1)/(3+2sin theta)d theta\int_0^{2 \pi} \frac{1}{3+2 \sin \theta} d \theta∫02π13+2sin⁡θdθ का मान ज्ञात कीजिए ।
Using contour integration, evaluate the integral ∫ 0 2 π 1 3 + 2 sin ⁡ θ d θ ∫ 0 2 π   1 3 + 2 sin ⁡ θ d θ int_(0)^(2pi)(1)/(3+2sin theta)d theta\int_0^{2 \pi} \frac{1}{3+2 \sin \theta} d \theta∫02π13+2sin⁡θdθ.
3(a) मान लीजिए R , p ( > 0 ) R , p ( > 0 ) R,p( > 0)R, p(>0)R,p(>0) अभिलक्षण का एक परिमित क्षेत्र है । दर्शाइए कि f ( a ) = a p , ∀ a ∈ R f ( a ) = a p , ∀ a ∈ R f(a)=a^(p),AA a in Rf(a)=a^p, \forall a \in Rf(a)=ap,∀a∈R द्वारा परिभाषित प्रतिचित्रण f : R → R f : R → R f:R rarr Rf: R \rightarrow Rf:R→R एकैक समाकारी है ।
Let R R RRR be a finite field of characteristic p ( > 0 ) p ( > 0 ) p( > 0)p(>0)p(>0). Show that the mapping f : R → R f : R → R f:R rarr Rf: R \rightarrow Rf:R→R defined by f ( a ) = a p , ∀ a ∈ R f ( a ) = a p , ∀ a ∈ R f(a)=a^(p),AA a in Rf(a)=a^p, \forall a \in Rf(a)=ap,∀a∈R is an isomorphism.
3.(b) एकधा विधि के द्वारा निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
न्यूनतमीकरण कीजिए z = − 6 x 1 − 2 x 2 − 5 x 3 z = − 6 x 1 − 2 x 2 − 5 x 3 z=-6x_(1)-2x_(2)-5x_(3)z=-6 x_1-2 x_2-5 x_3z=−6x1−2x2−5x3
बशर्ते कि
2 x 1 − 3 x 2 + x 3 ⩽ 14 − 4 x 1 + 4 x 2 + 10 x 3 ⩽ 46 2 x 1 + 2 x 2 − 4 x 3 ⩽ 37 x 1 ⩾ 2 , x 2 ⩾ 1 , x 3 ⩾ 3 2 x 1 − 3 x 2 + x 3 ⩽ 14 − 4 x 1 + 4 x 2 + 10 x 3 ⩽ 46 2 x 1 + 2 x 2 − 4 x 3 ⩽ 37 x 1 ⩾ 2 , x 2 ⩾ 1 , x 3 ⩾ 3 {:[2x_(1)-3x_(2)+x_(3) <= 14],[-4x_(1)+4x_(2)+10x_(3) <= 46],[2x_(1)+2x_(2)-4x_(3) <= 37],[x_(1) >= 2″,”x_(2) >= 1″,”x_(3) >= 3]:}\begin{gathered} 2 x_1-3 x_2+x_3 \leqslant 14 \\ -4 x_1+4 x_2+10 x_3 \leqslant 46 \\ 2 x_1+2 x_2-4 x_3 \leqslant 37 \\ x_1 \geqslant 2, x_2 \geqslant 1, x_3 \geqslant 3 \end{gathered}2x1−3x2+x3⩽14−4x1+4x2+10x3⩽462x1+2x2−4x3⩽37x1⩾2,x2⩾1,x3⩾3
Solve the linear programming problem using simplex method:
Minimize z = − 6 x 1 − 2 x 2 − 5 x 3 z = − 6 x 1 − 2 x 2 − 5 x 3 z=-6x_(1)-2x_(2)-5x_(3)z=-6 x_1-2 x_2-5 x_3z=−6x1−2x2−5x3
subject to 2 x 1 − 3 x 2 + x 3 ⩽ 14 2 x 1 − 3 x 2 + x 3 ⩽ 14 quad2x_(1)-3x_(2)+x_(3) <= 14\quad 2 x_1-3 x_2+x_3 \leqslant 142x1−3x2+x3⩽14
− 4 x 1 + 4 x 2 + 10 x 3 ⩽ 46 2 x 1 + 2 x 2 − 4 x 3 ⩽ 37 x 1 ⩾ 2 , x 2 ⩾ 1 , x 3 ⩾ 3 − 4 x 1 + 4 x 2 + 10 x 3 ⩽ 46 2 x 1 + 2 x 2 − 4 x 3 ⩽ 37 x 1 ⩾ 2 , x 2 ⩾ 1 , x 3 ⩾ 3 {:[-4x_(1)+4x_(2)+10x_(3) <= 46],[2x_(1)+2x_(2)-4x_(3) <= 37],[x_(1) >= 2″,”x_(2) >= 1″,”x_(3) >= 3]:}\begin{gathered} -4 x_1+4 x_2+10 x_3 \leqslant 46 \\ 2 x_1+2 x_2-4 x_3 \leqslant 37 \\ x_1 \geqslant 2, x_2 \geqslant 1, x_3 \geqslant 3 \end{gathered}−4x1+4x2+10x3⩽462x1+2x2−4x3⩽37x1⩾2,x2⩾1,x3⩾3
3.(c) यदि u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u=tan^(-1)(x^(3)+y^(3))/(x-y),x!=yu=\tan ^{-1} \frac{x^3+y^3}{x-y}, x \neq yu=tan−1⁡x3+y3x−y,x≠y
तब दर्शाइए कि x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = ( 1 − 4 sin 2 ⁡ u ) sin ⁡ 2 u x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = 1 − 4 sin 2 ⁡ u sin ⁡ 2 u x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=(1-4sin^(2)u)sin 2ux^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}=\left(1-4 \sin ^2 u\right) \sin 2 ux2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=(1−4sin2⁡u)sin⁡2u
If u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u=tan^(-1)(x^(3)+y^(3))/(x-y),x!=yu=\tan ^{-1} \frac{x^3+y^3}{x-y}, x \neq yu=tan−1⁡x3+y3x−y,x≠y
then show that x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = ( 1 − 4 sin 2 ⁡ u ) sin ⁡ 2 u x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = 1 − 4 sin 2 ⁡ u sin ⁡ 2 u x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=(1-4sin^(2)u)sin 2ux^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}=\left(1-4 \sin ^2 u\right) \sin 2 ux2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=(1−4sin2⁡u)sin⁡2u
4.(a) यदि v ( r , θ ) = ( r − 1 r ) sin ⁡ θ , r ≠ 0 v ( r , θ ) = r − 1 r sin ⁡ θ , r ≠ 0 v(r,theta)=(r-(1)/(r))sin theta,r!=0v(r, \theta)=\left(r-\frac{1}{r}\right) \sin \theta, r \neq 0v(r,θ)=(r−1r)sin⁡θ,r≠0,
तब विश्लेषिक फलन f ( z ) = u ( r , θ ) + i v ( r , θ ) f ( z ) = u ( r , θ ) + i v ( r , θ ) f(z)=u(r,theta)+iv(r,theta)f(z)=u(r, \theta)+i v(r, \theta)f(z)=u(r,θ)+iv(r,θ) ज्ञात कीजिए ।
If v ( r , θ ) = ( r − 1 r ) sin ⁡ θ , r ≠ 0 v ( r , θ ) = r − 1 r sin ⁡ θ , r ≠ 0 v(r,theta)=(r-(1)/(r))sin theta,r!=0v(r, \theta)=\left(r-\frac{1}{r}\right) \sin \theta, r \neq 0v(r,θ)=(r−1r)sin⁡θ,r≠0,
then find an analytic function f ( z ) = u ( r , θ ) + i v ( r , θ ) f ( z ) = u ( r , θ ) + i v ( r , θ ) f(z)=u(r,theta)+iv(r,theta)f(z)=u(r, \theta)+i v(r, \theta)f(z)=u(r,θ)+iv(r,θ)
4.(b) दर्शाइए कि ∫ 0 π / 2 sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) ∫ 0 π / 2   sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) int_(0)^(pi//2)(sin^(2)x)/(sin x+cos x)dx=(1)/(sqrt2)log _(e)(1+sqrt2)\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x=\frac{1}{\sqrt{2}} \log _e(1+\sqrt{2})∫0π/2sin2⁡xsin⁡x+cos⁡xdx=12loge⁡(1+2)
Show that ∫ 0 π / 2 sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) ∫ 0 π / 2   sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) int_(0)^(pi//2)(sin^(2)x)/(sin x+cos x)dx=(1)/(sqrt2)log _(e)(1+sqrt2)\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x=\frac{1}{\sqrt{2}} \log _e(1+\sqrt{2})∫0π/2sin2⁡xsin⁡x+cos⁡xdx=12loge⁡(1+2)
4(c) वोगेल की सम्निकटन विधि से निम्नलिखित परिवहन समस्या का आरंभिक आधारिक सुसंगत हल ज्ञात कीजिए । इस हल का उपयोग कर समस्या का इष्टतम हल एवं परिवहन लागत ज्ञात कीजिए ।
D 1 D 2 D 3 D 4 Supply S 1 10 0 20 11 15 S 2 12 8 9 20 25 S 3 0 14 16 18 10 Demand 5 20 15 10      D 1      D 2      D 3      D 4       Supply  S 1      10      0      20      11      15 S 2      12      8      9      20      25 S 3      0      14      16      18      10  Demand       5      20      15      10      {:[,D_(1),D_(2),D_(3),D_(4),” Supply “],[S_(1),10,0,20,11,15],[S_(2),12,8,9,20,25],[S_(3),0,14,16,18,10],[” Demand “,5,20,15,10,]:}\begin{array}{|l|l|l|l|l|l|} \hline & D_1 & D_2 & D_3 & D_4 & \text { Supply } \\ \hline S_1 & 10 & 0 & 20 & 11 & 15 \\ \hline S_2 & 12 & 8 & 9 & 20 & 25 \\ \hline S_3 & 0 & 14 & 16 & 18 & 10 \\ \hline \text { Demand } & 5 & 20 & 15 & 10 & \\ \hline \end{array}D1D2D3D4 Supply S1100201115S212892025S3014161810 Demand 5201510
Find the initial basic feasible solution of the following transportation problem by Vogel’s approximation method and use it to find the optimal solution and the transportation cost of the problem.
D 1 D 2 D 3 D 4 Supply S 1 10 0 20 11 15 S 2 12 8 9 20 25 S 3 0 14 16 18 10 Demand 5 20 15 10      D 1      D 2      D 3      D 4       Supply  S 1      10      0      20      11      15 S 2      12      8      9      20      25 S 3      0      14      16      18      10  Demand       5      20      15      10      {:[,D_(1),D_(2),D_(3),D_(4),” Supply “],[S_(1),10,0,20,11,15],[S_(2),12,8,9,20,25],[S_(3),0,14,16,18,10],[” Demand “,5,20,15,10,]:}\begin{array}{|l|l|l|l|l|l|} \hline & D_1 & D_2 & D_3 & D_4 & \text { Supply } \\ \hline S_1 & 10 & 0 & 20 & 11 & 15 \\ \hline S_2 & 12 & 8 & 9 & 20 & 25 \\ \hline S_3 & 0 & 14 & 16 & 18 & 10 \\ \hline \text { Demand } & 5 & 20 & 15 & 10 & \\ \hline \end{array}D1D2D3D4 Supply S1100201115S212892025S3014161810 Demand 5201510
खण्ड ‘B’ SECTION ‘B’
5.(a) z = y f ( x ) + x g ( y ) z = y f ( x ) + x g ( y ) z=yf(x)+xg(y)z=y f(x)+x g(y)z=yf(x)+xg(y) से स्वैच्छिक फलनों f ( x ) f ( x ) f(x)f(x)f(x) व g ( y ) g ( y ) g(y)g(y)g(y) का विलोपन कर आंशिक अवकल समीकरण बनाइए तथा इसकी प्रकृति (दीर्घवृत्तीय, अतिपरवलीय या परवलीय) x > 0 , y > 0 x > 0 , y > 0 x > 0,y > 0x>0, y>0x>0,y>0 क्षेत्र में इंगित कीजिए ।
Form a partial differential equation by eliminating the arbitrary functions f ( x ) f ( x ) f(x)f(x)f(x) and g ( y ) g ( y ) g(y)g(y)g(y) from z = y f ( x ) + x g ( y ) z = y f ( x ) + x g ( y ) z=yf(x)+xg(y)z=y f(x)+x g(y)z=yf(x)+xg(y) and specify its nature (elliptic, hyperbolic or parabolic) in the region x > 0 , y > 0 x > 0 , y > 0 x > 0,y > 0x>0, y>0x>0,y>0.
5.(b) दर्शाइए कि समीकरण : f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f(x)=cos (pi(x+1))/(8)+0*148 x-0*9062=0f(x)=\cos \frac{\pi(x+1)}{8}+0 \cdot 148 x-0 \cdot 9062=0f(x)=cos⁡π(x+1)8+0⋅148x−0⋅9062=0
का एक मूल अन्तराल ( − 1 , 0 ) ( − 1 , 0 ) (-1,0)(-1,0)(−1,0) में तथा एक मूल ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1) में है । ऋणात्मक मूल की न्यूटन-रॉफसन विधि से दशमलव के चार स्थान तक सही गणना कीजिए।
Show that the equation: f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f(x)=cos (pi(x+1))/(8)+0*148 x-0*9062=0f(x)=\cos \frac{\pi(x+1)}{8}+0 \cdot 148 x-0 \cdot 9062=0f(x)=cos⁡π(x+1)8+0⋅148x−0⋅9062=0
has one root in the interval ( − 1 , 0 ) ( − 1 , 0 ) (-1,0)(-1,0)(−1,0) and one in ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1). Calculate the negative root correct to four decimal places using Newton-Raphson method.
5(c) मान लीजिए g ( w , x , y , z ) = ( w + x + y ) ( x + y ¯ + z ) ( w + y ¯ ) g ( w , x , y , z ) = ( w + x + y ) ( x + y ¯ + z ) ( w + y ¯ ) g(w,x,y,z)=(w+x+y)(x+ bar(y)+z)(w+ bar(y))g(w, x, y, z)=(w+x+y)(x+\bar{y}+z)(w+\bar{y})g(w,x,y,z)=(w+x+y)(x+y¯+z)(w+y¯) एक बूलीय-फलन है । g ( w , x , y , z ) g ( w , x , y , z ) g(w,x,y,z)g(w, x, y, z)g(w,x,y,z) का योगात्मक प्रसामान्य स्वरूप (कन्जंकटिव नार्मल फॉर्म) प्राप्त कीजिए । g ( w , x , y , z ) g ( w , x , y , z ) g(w,x,y,z)g(w, x, y, z)g(w,x,y,z) को उच्च-पदों (मैक्स टर्म्स) के गुणन के रूप में भी व्यक्त कीजिए ।
Let g ( w , x , y , z ) = ( w + x + y ) ( x + y ¯ + z ) ( w + y ¯ ) g ( w , x , y , z ) = ( w + x + y ) ( x + y ¯ + z ) ( w + y ¯ ) g(w,x,y,z)=(w+x+y)(x+ bar(y)+z)(w+ bar(y))g(w, x, y, z)=(w+x+y)(x+\bar{y}+z)(w+\bar{y})g(w,x,y,z)=(w+x+y)(x+y¯+z)(w+y¯) be a Boolean function. Obtain the conjunctive normal form for g ( w , x , y , z ) g ( w , x , y , z ) g(w,x,y,z)g(w, x, y, z)g(w,x,y,z). Also express g ( w , x , y , z ) g ( w , x , y , z ) g(w,x,y,z)g(w, x, y, z)g(w,x,y,z) as a product of maxterms.
5.(d) आंशिक अवकल समीकरण :
( D 3 − 2 D 2 D ′ − D D ′ 2 + 2 D ′ 3 ) z = e 2 x + y ′ + sin ⁡ ( x − 2 y ) D 3 − 2 D 2 D ′ − D D ′ 2 + 2 D ′ 3 z = e 2 x + y ′ + sin ⁡ ( x − 2 y ) (D^(3)-2D^(2)D^(‘)-DD^(‘2)+2D^(‘3))z=e^(2x+y^(‘))+sin(x-2y)\left(D^3-2 D^2 D^{\prime}-D D^{\prime 2}+2 D^{\prime 3}\right) z=e^{2 x+y^{\prime}}+\sin (x-2 y)(D3−2D2D′−DD′2+2D′3)z=e2x+y′+sin⁡(x−2y)
D ≡ ∂ ∂ x , D ′ ≡ ∂ ∂ y D ≡ ∂ ∂ x , D ′ ≡ ∂ ∂ y D-=(del)/(del x),quadD^(‘)-=(del)/(del y)D \equiv \frac{\partial}{\partial x}, \quad D^{\prime} \equiv \frac{\partial}{\partial y}D≡∂∂x,D′≡∂∂y
को हल कीजिए ।
Solve the partial differential equation :
( D 3 − 2 D 2 D ′ − D D ′ 2 + 2 D ′ 3 ) z = e 2 x + y + sin ⁡ ( x − 2 y ) D 3 − 2 D 2 D ′ − D D ′ 2 + 2 D ′ 3 z = e 2 x + y + sin ⁡ ( x − 2 y ) (D^(3)-2D^(2)D^(‘)-DD^(‘2)+2D^(‘3))z=e^(2x+y)+sin(x-2y)\left(D^3-2 D^2 D^{\prime}-D D^{\prime 2}+2 D^{\prime 3}\right) z=e^{2 x+y}+\sin (x-2 y)(D3−2D2D′−DD′2+2D′3)z=e2x+y+sin⁡(x−2y)
D ≡ ∂ ∂ x , D ′ ≡ ∂ ∂ y D ≡ ∂ ∂ x , D ′ ≡ ∂ ∂ y D-=(del)/(del x),quadD^(‘)-=(del)/(del y)D \equiv \frac{\partial}{\partial x}, \quad D^{\prime} \equiv \frac{\partial}{\partial y}D≡∂∂x,D′≡∂∂y
5.(e) सिद्ध कीजिए कि त्रिभुजाकार पटल A B C A B C ABCA B CABC का इसके तल में A A AAA से होकर जाने वाली किसी भी अक्ष के सापेक्ष जड़त्व-आघूर्ण
M 6 ( β 2 + β γ + γ 2 ) M 6 β 2 + β γ + γ 2 (M)/(6)(beta^(2)+beta gamma+gamma^(2))\frac{M}{6}\left(\beta^2+\beta \gamma+\gamma^2\right)M6(β2+βγ+γ2)
है, जहाँ पर M M MMM पटल की संहति तथा β β beta\betaβ व γ γ gamma\gammaγ क्रमशः B B BBB व C C CCC से अक्ष पर डाले गये लम्बों की लम्बाइयां हैं।
Prove that the moment of inertia of a triangular lamina A B C A B C ABCA B CABC about any axis through A A AAA in its plane is
M 6 ( β 2 + β γ + γ 2 ) M 6 β 2 + β γ + γ 2 (M)/(6)(beta^(2)+beta gamma+gamma^(2))\frac{M}{6}\left(\beta^2+\beta \gamma+\gamma^2\right)M6(β2+βγ+γ2)
where M M MMM is the mass of the lamina and β , γ β , γ beta,gamma\beta, \gammaβ,γ are respectively the length of perpendiculars from B B BBB and C C CCC on the axis.
6.(a) आंशिक अवकल समीकरण :
( x − y ) y 2 ∂ z ∂ x + ( y − x ) x 2 ∂ z ∂ y = ( x 2 + y 2 ) z ( x − y ) y 2 ∂ z ∂ x + ( y − x ) x 2 ∂ z ∂ y = x 2 + y 2 z (x-y)y^(2)(del z)/(del x)+(y-x)x^(2)(del z)/(del y)=(x^(2)+y^(2))z(x-y) y^2 \frac{\partial z}{\partial x}+(y-x) x^2 \frac{\partial z}{\partial y}=\left(x^2+y^2\right) z(x−y)y2∂z∂x+(y−x)x2∂z∂y=(x2+y2)z
के वक्र : x z = a 3 , y = 0 x z = a 3 , y = 0 xz=a^(3),y=0x z=a^3, y=0xz=a3,y=0 को अपने ऊपर समाहित करने वाले समाकल पृष्ठ को ज्ञात कीजिए ।
Find the integral surface of the partial differential equation :
( x − y ) y 2 ∂ z ∂ x + ( y − x ) x 2 ∂ z ∂ y = ( x 2 + y 2 ) z ( x − y ) y 2 ∂ z ∂ x + ( y − x ) x 2 ∂ z ∂ y = x 2 + y 2 z (x-y)y^(2)(del z)/(del x)+(y-x)x^(2)(del z)/(del y)=(x^(2)+y^(2))z(x-y) y^2 \frac{\partial z}{\partial x}+(y-x) x^2 \frac{\partial z}{\partial y}=\left(x^2+y^2\right) z(x−y)y2∂z∂x+(y−x)x2∂z∂y=(x2+y2)z
that contains the curve : x z = a 3 , y = 0 x z = a 3 , y = 0 xz=a^(3),y=0x z=a^3, y=0xz=a3,y=0 on it.
6.(b) समीकरण निकाय : 4 x + y + 2 z = 4 4 x + y + 2 z = 4 4x+y+2z=44 x+y+2 z=44x+y+2z=4 3 x + 5 y + z = 7 3 x + 5 y + z = 7 3x+5y+z=73 x+5 y+z=73x+5y+z=7 x + y + 3 z = 3 x + y + 3 z = 3 x+y+3z=3x+y+3 z=3x+y+3z=3
के हल के लिए गाउस-सीडल पुनरावर्ती क्रिया-विधि निर्धारित कीजिए तथा आरंभिक सदिश X ( 0 ) = 0 X ( 0 ) = 0 X^((0))=0X^{(0)}=0X(0)=0 से प्रारंभ करके तीन बार पुनरावर्त कीजिए । यथातथ (बिल्कुल ठीक) हल भी निकालिए और पुनरावर्त हलों से तुलना कीजिए।
For the solution of the system of equations : 4 x + y + 2 z = 4 4 x + y + 2 z = 4 4x+y+2z=44 x+y+2 z=44x+y+2z=4
3 x + 5 y + z = 7 x + y + 3 z = 3 3 x + 5 y + z = 7 x + y + 3 z = 3 {:[3x+5y+z=7],[x+y+3z=3]:}\begin{aligned} &3 x+5 y+z=7 \\ &x+y+3 z=3 \end{aligned}3x+5y+z=7x+y+3z=3
set up the Gauss-Seidel iterative scheme and iterate three times starting with the initial vector X ( 0 ) = 0 X ( 0 ) = 0 X^((0))=0X^{(0)}=0X(0)=0. Also find the exact solutions and compare with the iterated solutions.
6(c) एक कण जिसकी संहति m m mmm है, x 2 + y 2 = R 2 x 2 + y 2 = R 2 x^(2)+y^(2)=R^(2)x^2+y^2=R^2x2+y2=R2, जहाँ पर R R RRR अचर है, द्वारा परिभाषित बेलन पर गति के लिए व्यवरोधित है। कण मूल बिन्दु की ओर लगे बल जो कण की मूल बिन्दु से दूरी r r rrr के अनुपाती है, से प्रतिबन्धित है । बल F → = − k r → F → = − k r → vec(F)=-k vec(r)\vec{F}=-k \vec{r}F→=−kr→, जहाँ पर k k kkk अचर है, से दिया गया है ।
By writing down the Hamiltonian, find the equations of motion of a particle of mass m m mmm constrained to move on the surface of a cylinder defined by x 2 + y 2 = R 2 x 2 + y 2 = R 2 x^(2)+y^(2)=R^(2)x^2+y^2=R^2x2+y2=R2, R R RRR is a constant. The particle is subject to a force directed towards the origin and proportional to the distance r r rrr of the particle from the origin given by F → = − k r → , k F → = − k r → , k vec(F)=-k vec(r),k\vec{F}=-k \vec{r}, kF→=−kr→,k is a constant.
7.(a) आंशिक अवकल समीकरण :
z = 1 2 ( p 2 + q 2 ) + ( p − x ) ( q − y ) ; p ≡ ∂ z ∂ x , q ≡ ∂ z ∂ y z = 1 2 p 2 + q 2 + ( p − x ) ( q − y ) ; p ≡ ∂ z ∂ x , q ≡ ∂ z ∂ y z=(1)/(2)(p^(2)+q^(2))+(p-x)(q-y);quad p-=(del z)/(del x),q-=(del z)/(del y)z=\frac{1}{2}\left(p^2+q^2\right)+(p-x)(q-y) ; \quad p \equiv \frac{\partial z}{\partial x}, q \equiv \frac{\partial z}{\partial y}z=12(p2+q2)+(p−x)(q−y);p≡∂z∂x,q≡∂z∂y
का हल ज्ञात कीजिये जो कि x x xxx-अक्ष से गुजरता हो ।
Find the solution of the partial differential equation:
z = 1 2 ( p 2 + q 2 ) + ( p − x ) ( q − y ) ; p ≡ ∂ z ∂ x , q ≡ ∂ z ∂ y z = 1 2 p 2 + q 2 + ( p − x ) ( q − y ) ; p ≡ ∂ z ∂ x , q ≡ ∂ z ∂ y z=(1)/(2)(p^(2)+q^(2))+(p-x)(q-y);quad p-=(del z)/(del x),q-=(del z)/(del y)z=\frac{1}{2}\left(p^2+q^2\right)+(p-x)(q-y) ; \quad p \equiv \frac{\partial z}{\partial x}, q \equiv \frac{\partial z}{\partial y}z=12(p2+q2)+(p−x)(q−y);p≡∂z∂x,q≡∂z∂y
which passes through the x x xxx-axis.
7.(b) क्षेत्रकलन के लिए ∫ 0 1 f ( x ) d x x ( 1 − x ) = α 1 f ( 0 ) + α 2 f ( 1 2 ) + α 3 f ( 1 ) ∫ 0 1   f ( x ) d x x ( 1 − x ) = α 1 f ( 0 ) + α 2 f 1 2 + α 3 f ( 1 ) int_(0)^(1)f(x)(dx)/(sqrt(x(1-x)))=alpha_(1)f(0)+alpha_(2)f((1)/(2))+alpha_(3)f(1)\int_0^1 f(x) \frac{d x}{\sqrt{x(1-x)}}=\alpha_1 f(0)+\alpha_2 f\left(\frac{1}{2}\right)+\alpha_3 f(1)∫01f(x)dxx(1−x)=α1f(0)+α2f(12)+α3f(1) द्वारा उस सूत्र को ज्ञात कीजिए जो अधिकतम सम्भव घात के बहुपद के लिए यथातथ (बिल्कुल ठीक) हो । सूत्र का उपयोग ∫ 0 1 d x x − x 3 ∫ 0 1   d x x − x 3 int_(0)^(1)(dx)/(sqrt(x-x^(3)))\int_0^1 \frac{d x}{\sqrt{x-x^3}}∫01dxx−x3 का (दशमलव के तीन स्थानों तक सही) मूल्यांकन के लिए कीजिए ।
Find a quadrature formula ∫ 0 1 f ( x ) d x x ( 1 − x ) = α 1 f ( 0 ) + α 2 f ( 1 2 ) + α 3 f ( 1 ) ∫ 0 1   f ( x ) d x x ( 1 − x ) = α 1 f ( 0 ) + α 2 f 1 2 + α 3 f ( 1 ) int_(0)^(1)f(x)(dx)/(sqrt(x(1-x)))=alpha_(1)f(0)+alpha_(2)f((1)/(2))+alpha_(3)f(1)\int_0^1 f(x) \frac{d x}{\sqrt{x(1-x)}}=\alpha_1 f(0)+\alpha_2 f\left(\frac{1}{2}\right)+\alpha_3 f(1)∫01f(x)dxx(1−x)=α1f(0)+α2f(12)+α3f(1)
which is exact for polynomials of highest possible degree. Then use the formula to evaluate ∫ 0 1 d x x − x 3 ∫ 0 1   d x x − x 3 int_(0)^(1)(dx)/(sqrt(x-x^(3)))\int_0^1 \frac{d x}{\sqrt{x-x^3}}∫01dxx−x3 (correct up to three decimal places).
7(c) एक द्विविमीय द्रव्य-प्रवाह का वेग विभव ϕ ( x , y ) = x y + x 2 − y ˙ 2 ϕ ( x , y ) = x y + x 2 − y ˙ 2 phi(x,y)=xy+x^(2)-y^(˙)^(2)\phi(x, y)=x y+x^2-\dot{y}^2ϕ(x,y)=xy+x2−y˙2 द्वारा दिया गया है । इस प्रवाह का धारा-फलन ज्ञात कीजिए ।
A velocity potential in a two-dimensional fluid flow is given by ϕ ( x , y ) = x y + x 2 − y 2 ϕ ( x , y ) = x y + x 2 − y 2 phi(x,y)=xy+x^(2)-y^(2)\phi(x, y)=x y+x^2-y^2ϕ(x,y)=xy+x2−y2. Find the stream function for this flow.
8.(a) लम्बाई l l lll की कसकर खींची गई लचीली-पतली डोरी का एक सिरा मूल बिन्दु पर तथा दूसरा x = l x = l x=lx=lx=l पर बंधा है। आरंभिक अवस्था में इसे x = l 3 x = l 3 x=(l)/(3)x=\frac{l}{3}x=l3 बिन्दु से ऐसे खींचकर छोड़ा जाता है ताकि यह x − y x − y x-yx-yx−y तल में h h hhh ऊँचाई के त्रिभुज का आकार लेता है। किसी भी दूरी x x xxx तथा समय t t ttt, डोरी को विरामावस्था से छोड़ने के बाद, पर विस्थापन y y yyy को ज्ञात कीजिए ।
डोरी में horizontal tension mass per unit length = c 2  horizontal tension   mass per unit length  = c 2 (” horizontal tension “)/(” mass per unit length “)=c^(2)\frac{\text { horizontal tension }}{\text { mass per unit length }}=c^2 horizontal tension  mass per unit length =c2 लीजिए ।
One end of a tightly stretched flexible thin string of length l l lll is fixed at the origin and the other at x = l x = l x=lx=lx=l. It is plucked at x = l 3 x = l 3 x=(l)/(3)x=\frac{l}{3}x=l3 so that it assumes initially the shape of a triangle of height h h hhh in the x − y x − y x-yx-yx−y plane. Find the displacement y y yyy at any distance x x xxx and at any time t t ttt after the string is released from rest. Take, horizontal tension mass per unit length = c 2  horizontal tension   mass per unit length  = c 2 (” horizontal tension “)/(” mass per unit length “)=c^(2)\frac{\text { horizontal tension }}{\text { mass per unit length }}=c^2 horizontal tension  mass per unit length =c2.
8.(b) बिन्दुओं x 0 , x 0 + ε x 0 , x 0 + ε x_(0),x_(0)+epsix_0, x_0+\varepsilonx0,x0+ε तथा x 1 x 1 x_(1)x_1x1 के सापेक्ष तीन-बिन्दु लेगरान्ज-अन्तर्वेशन बहुपद को लिखिए । तदुपरान्त limit ε → 0 ε → 0 epsi rarr0\varepsilon \rightarrow 0ε→0 करने पर निम्नलिखित सम्बन्ध को स्थापित कीजिए :
f ( x ) = ( x 1 − x ) ( x + x 1 − 2 x 0 ) ( x 1 − x 0 ) 2 f ( x 0 ) + ( x − x 0 ) ( x 1 − x ) ( x 1 − x 0 ) f ′ ( x 0 ) + ( x − x 0 ) 2 ( x 1 − x 0 ) f ( x 1 ) + E ( x ) f ( x ) = x 1 − x x + x 1 − 2 x 0 x 1 − x 0 2 f x 0 + x − x 0 x 1 − x x 1 − x 0 f ′ x 0 + x − x 0 2 x 1 − x 0 f x 1 + E ( x ) f(x)=((x_(1)-x)(x+x_(1)-2x_(0)))/((x_(1)-x_(0))^(2))f(x_(0))+((x-x_(0))(x_(1)-x))/((x_(1)-x_(0)))f^(‘)(x_(0))+((x-x_(0))^(2))/((x_(1)-x_(0)))f(x_(1))+E(x)f(x)=\frac{\left(x_1-x\right)\left(x+x_1-2 x_0\right)}{\left(x_1-x_0\right)^2} f\left(x_0\right)+\frac{\left(x-x_0\right)\left(x_1-x\right)}{\left(x_1-x_0\right)} f^{\prime}\left(x_0\right)+\frac{\left(x-x_0\right)^2}{\left(x_1-x_0\right)} f\left(x_1\right)+E(x)f(x)=(x1−x)(x+x1−2x0)(x1−x0)2f(x0)+(x−x0)(x1−x)(x1−x0)f′(x0)+(x−x0)2(x1−x0)f(x1)+E(x)
जहाँ पर E ( x ) = 1 6 ( x − x 0 ) 2 ( x − x 1 ) f ′ ′ ′ ( ξ ) E ( x ) = 1 6 x − x 0 2 x − x 1 f ′ ′ ′ ( ξ ) E(x)=(1)/(6)(x-x_(0))^(2)(x-x_(1))f^(”’)(xi)E(x)=\frac{1}{6}\left(x-x_0\right)^2\left(x-x_1\right) f^{\prime \prime \prime}(\xi)E(x)=16(x−x0)2(x−x1)f′′′(ξ) त्रुटि-फलन है और
न्यूनतम ( x 0 , x 0 + ε , x 1 ) < ξ < x 0 , x 0 + ε , x 1 < ξ < (x_(0),x_(0)+epsi,x_(1)) < xi <\left(x_0, x_0+\varepsilon, x_1\right)<\xi<(x0,x0+ε,x1)<ξ< उच्चतम ( x 0 , x 0 + ε , x 1 ) x 0 , x 0 + ε , x 1 (x_(0),x_(0)+epsi,x_(1))\left(x_0, x_0+\varepsilon, x_1\right)(x0,x0+ε,x1)
Write the three point Lagrangian interpolating polynomial relative to the points x 0 , x 0 + ε x 0 , x 0 + ε x_(0),x_(0)+epsix_0, x_0+\varepsilonx0,x0+ε and x 1 x 1 x_(1)x_1x1. Then by taking the limit ε → 0 ε → 0 epsi rarr0\varepsilon \rightarrow 0ε→0, establish the relation
f ( x ) = ( x 1 − x ) ( x + x 1 − 2 x 0 ) ( x 1 − x 0 ) 2 f ( x 0 ) + ( x − x 0 ) ( x 1 − x ) ( x 1 − x 0 ) f ′ ( x 0 ) + ( x − x 0 ) 2 ( x 1 − x 0 ) f ( x 1 ) + E ( x ) f ( x ) = x 1 − x x + x 1 − 2 x 0 x 1 − x 0 2 f x 0 + x − x 0 x 1 − x x 1 − x 0 f ′ x 0 + x − x 0 2 x 1 − x 0 f x 1 + E ( x ) f(x)=((x_(1)-x)(x+x_(1)-2x_(0)))/((x_(1)-x_(0))^(2))f(x_(0))+((x-x_(0))(x_(1)-x))/((x_(1)-x_(0)))f^(‘)(x_(0))+((x-x_(0))^(2))/((x_(1)-x_(0)))f(x_(1))+E(x)f(x)=\frac{\left(x_1-x\right)\left(x+x_1-2 x_0\right)}{\left(x_1-x_0\right)^2} f\left(x_0\right)+\frac{\left(x-x_0\right)\left(x_1-x\right)}{\left(x_1-x_0\right)} f^{\prime}\left(x_0\right)+\frac{\left(x-x_0\right)^2}{\left(x_1-x_0\right)} f\left(x_1\right)+E(x)f(x)=(x1−x)(x+x1−2x0)(x1−x0)2f(x0)+(x−x0)(x1−x)(x1−x0)f′(x0)+(x−x0)2(x1−x0)f(x1)+E(x)
where E ( x ) = 1 6 ( x − x 0 ) 2 ( x − x 1 ) f ′ ′ ′ ( ξ ) E ( x ) = 1 6 x − x 0 2 x − x 1 f ′ ′ ′ ( ξ ) E(x)=(1)/(6)(x-x_(0))^(2)(x-x_(1))f^(”’)(xi)E(x)=\frac{1}{6}\left(x-x_0\right)^2\left(x-x_1\right) f^{\prime \prime \prime}(\xi)E(x)=16(x−x0)2(x−x1)f′′′(ξ)
is the error function and min. ( x 0 , x 0 + ε , x 1 ) < ξ < max . ( x 0 , x 0 + ε , x 1 ) x 0 , x 0 + ε , x 1 < ξ < max . x 0 , x 0 + ε , x 1 (x_(0),x_(0)+epsi,x_(1)) < xi < max.(x_(0),x_(0)+epsi,x_(1))\left(x_0, x_0+\varepsilon, x_1\right)<\xi<\max .\left(x_0, x_0+\varepsilon, x_1\right)(x0,x0+ε,x1)<ξ<max.(x0,x0+ε,x1)
  1. (c) m 2 m 2 (m)/(2)\frac{m}{2}m2 शक्ति वाले दो स्रोत, बिन्दुओं ( ± a , 0 ) ( ± a , 0 ) (+-a,0)(\pm a, 0)(±a,0) पर स्थित हैं। दर्शाइए कि वृत x 2 + y 2 = a 2 x 2 + y 2 = a 2 x^(2)+y^(2)=a^(2)x^2+y^2=a^2x2+y2=a2 के किसी भी बिन्दु पर वेग y y yyy-अक्ष के समान्तर तथा y y yyy के व्युत्क्रमानुपाती है।
Two sources of strength m 2 m 2 (m)/(2)\frac{m}{2}m2 are placed at the points ( ± a , 0 ) ( ± a , 0 ) (+-a,0)(\pm a, 0)(±a,0). Show that at any point on the circle x 2 + y 2 = a 2 x 2 + y 2 = a 2 x^(2)+y^(2)=a^(2)x^2+y^2=a^2x2+y2=a2, the velocity is parallel to the y y yyy-axis and is inversely proportional to y y yyy.
upsc-m2019-2-d4349598-fdd2-4519-8497-a330d7fac251
खण्ड ‘A’ SECTION ‘ A A AAA ‘
1.(a) मान लीजिए कि G G GGG एक परिमित समूह है और H H HHH तथा K , G K , G K,GK, GK,G के उप-समूह हैं, ऐसा कि K ⊂ H K ⊂ H K sub HK \subset HK⊂H । दर्शाइए ( G : K ) = ( G : H ) ( H : K ) ( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K)=(G: H)(H: K)(G:K)=(G:H)(H:K) ।
Let G G GGG be a finite group, H H HHH and K K KKK subgroups of G G GGG such that K ⊂ H K ⊂ H K sub HK \subset HK⊂H. Show that ( G : K ) = ( G : H ) ( H : K ) ( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K)=(G: H)(H: K)(G:K)=(G:H)(H:K).
1.(b) दर्शाईए कि फलन
f ( x , y ) = { x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 ) f ( x , y ) = x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 ) f(x,y)={[(x^(2)-y^(2))/(x-y)”,”,(x”,”y)!=(1″,”-1)”,”(1″,”1)],[0″,”,(x”,”y)=(1″,”1)”,”(1″,”-1)]:}f(x, y)=\left\{\begin{array}{cl} \frac{x^2-y^2}{x-y}, & (x, y) \neq(1,-1),(1,1) \\ 0, & (x, y)=(1,1),(1,-1) \end{array}\right.f(x,y)={x2−y2x−y,(x,y)≠(1,−1),(1,1)0,(x,y)=(1,1),(1,−1)
संतत और बिन्दु ( 1 , − 1 ) ( 1 , − 1 ) (1,-1)(1,-1)(1,−1) पर अवकलनीय है ।
Show that the function
f ( x , y ) = { x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 ) f ( x , y ) = x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 ) f(x,y)={[(x^(2)-y^(2))/(x-y)”,”,(x”,”y)!=(1″,”-1)”,”(1″,”1)],[0″,”,(x”,”y)=(1″,”1)”,”(1″,”-1)]:}f(x, y)=\left\{\begin{array}{cc} \frac{x^2-y^2}{x-y}, & (x, y) \neq(1,-1),(1,1) \\ 0, & (x, y)=(1,1),(1,-1) \end{array}\right.f(x,y)={x2−y2x−y,(x,y)≠(1,−1),(1,1)0,(x,y)=(1,1),(1,−1)
is continuous and differentiable at ( 1 , − 1 ) ( 1 , − 1 ) (1,-1)(1,-1)(1,−1).
1.(c) मूल्यांकन कीजिए :
∫ 0 ∞ tan − 1 ⁡ ( a x ) x ( 1 + x 2 ) d x , a > 0 , a ≠ 1 . ∫ 0 ∞   tan − 1 ⁡ ( a x ) x 1 + x 2 d x , a > 0 , a ≠ 1 . int_(0)^(oo)(tan^(-1)(ax))/(x(1+x^(2)))dx,a > 0,a!=1.\int_0^{\infty} \frac{\tan ^{-1}(a x)}{x\left(1+x^2\right)} d x, a>0, a \neq 1 .∫0∞tan−1⁡(ax)x(1+x2)dx,a>0,a≠1.
Evaluate
∫ 0 ∞ tan − 1 ⁡ ( a x ) x ( 1 + x 2 ) d x , a > 0 , a ≠ 1 . ∫ 0 ∞   tan − 1 ⁡ ( a x ) x 1 + x 2 d x , a > 0 , a ≠ 1 . int_(0)^(oo)(tan^(-1)(ax))/(x(1+x^(2)))dx,a > 0,a!=1.\int_0^{\infty} \frac{\tan ^{-1}(a x)}{x\left(1+x^2\right)} d x, a>0, a \neq 1 .∫0∞tan−1⁡(ax)x(1+x2)dx,a>0,a≠1.
1(d) मान लीजिये C C C\mathbb{C}C में D D DDD प्रक्षेत्र पर f ( z ) f ( z ) f(z)f(z)f(z) एक विश्लेषिक फलन है और समीकरण Im ⁡ f ( z ) = ( Re ⁡ f ( z ) ) 2 , Z ∈ D Im ⁡ f ( z ) = ( Re ⁡ f ( z ) ) 2 , Z ∈ D Im f(z)=(Re f(z))^(2),Z in D\operatorname{Im} f(z)=(\operatorname{Re} f(z))^2, Z \in DIm⁡f(z)=(Re⁡f(z))2,Z∈D को संतुष्ट करता है । दर्शाइए कि D D DDD में f ( z ) f ( z ) f(z)f(z)f(z) अचर है ।
Suppose f ( z ) f ( z ) f(z)f(z)f(z) is analytic function on a domain D D DDD in ₫ ₫ ₫₫₫ and satisfies the equation Im f ( z ) = ( Re ⁡ f ( z ) ) 2 , Z ∈ D f ( z ) = ( Re ⁡ f ( z ) ) 2 , Z ∈ D f(z)=(Re f(z))^(2),Z in Df(z)=(\operatorname{Re} f(z))^2, Z \in Df(z)=(Re⁡f(z))2,Z∈D. Show that f ( z ) f ( z ) f(z)f(z)f(z) is constant in D D DDD.
1.(e) ग्राफी विधि के इस्तेमाल के द्वारा रैखिक प्रोग्रामन समस्या को हल कीजिए । अधिकतमीकरण कीजिए Z = 3 x 1 + 2 x 2 Z = 3 x 1 + 2 x 2 Z=3x_(1)+2x_(2)Z=3 x_1+2 x_2Z=3x1+2x2 बशर्ते कि
x 1 − x 2 ⩾ 1 x 1 − x 2 ⩾ 1 x_(1)-x_(2) >= 1x_1-x_2 \geqslant 1x1−x2⩾1,
x 1 + x 3 ⩾ 3 x 1 + x 3 ⩾ 3 x_(1)+x_(3) >= 3x_1+x_3 \geqslant 3x1+x3⩾3
और x 1 , x 2 , x 3 ≥ 0 x 1 , x 2 , x 3 ≥ 0 x_(1),x_(2),x_(3) >= 0x_1, x_2, x_3 \geq 0x1,x2,x3≥0
Use graphical method to solve the linear programming problem.
Maximize Z = 3 x 1 + 2 x 2 Z = 3 x 1 + 2 x 2 Z=3x_(1)+2x_(2)Z=3 x_1+2 x_2Z=3x1+2x2
subject to
x 1 − x 2 ⩾ 1 , x 1 + x 3 ⩾ 3 and x 1 , x 2 , x 3 ⩾ 0      x 1 − x 2 ⩾ 1 ,      x 1 + x 3 ⩾ 3  and       x 1 , x 2 , x 3 ⩾ 0 {:[,x_(1)-x_(2) >= 1″,”],[,x_(1)+x_(3) >= 3],[” and “,x_(1)”,”x_(2)”,”x_(3) >= 0]:}\begin{array}{ll} & x_1-x_2 \geqslant 1, \\ & x_1+x_3 \geqslant 3 \\ \text { and } & x_1, x_2, x_3 \geqslant 0 \end{array}x1−x2⩾1,x1+x3⩾3 and x1,x2,x3⩾0
2(a) यदि G G GGG और H H HHH परिमित समूह हैं जिनकी कोटियां सापेक्षतः अभाज्य हैं, तो सिद्ध करें कि G G GGG से H H HHH तक केवल एक ही समाकारिता होमोमोर्फिज्म है जो कि तुच्छ है।
If G G GGG and H H HHH are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from G G GGG to H H HHH, the trivial one.
2.(b) समूह Z 12 Z 12 Z_(12)Z_{12}Z12 के सभी विभाग समूह लिखिए ।
Write down all quotient groups of the group Z 12 Z 12 Z_(12)Z_{12}Z12.
2.(c) अवकलों का उपयोग करते हुए, f ( 4 ⋅ 1 , 4 ⋅ 9 ) f ( 4 ⋅ 1 , 4 ⋅ 9 ) f(4*1,4*9)f(4 \cdot 1,4 \cdot 9)f(4⋅1,4⋅9) का सन्निकट मान ज्ञात करें, जहाँ
f ( x , y ) = ( x 3 + x 2 y ) 1 2 है। f ( x , y ) = x 3 + x 2 y 1 2  है।  f(x,y)=(x^(3)+x^(2)y)^((1)/(2))” है। “f(x, y)=\left(x^3+x^2 y\right)^{\frac{1}{2}} \text { है। }f(x,y)=(x3+x2y)12 है। 
Using differentials, find an approximate value of f ( 4 ⋅ 1 , 4.9 ) f ( 4 ⋅ 1 , 4.9 ) f(4*1,4.9)f(4 \cdot 1,4.9)f(4⋅1,4.9) where f ( x , y ) = ( x 3 + x 2 y ) 1 2 f ( x , y ) = x 3 + x 2 y 1 2 f(x,y)=(x^(3)+x^(2)y)^((1)/(2))f(x, y)=\left(x^3+x^2 y\right)^{\frac{1}{2}}f(x,y)=(x3+x2y)12.
2.(d) दर्शाइए कि वियुक्त विचित्र बिन्दु z 0 z 0 z_(0)z_0z0, फलन f ( z ) f ( z ) f(z)f(z)f(z) का m m mmm कोटि का पोल होगा यदि और केवल यदि f ( z ) f ( z ) f(z)f(z)f(z) को f ( z ) = ϕ ( z ) ( z − z 0 ) m f ( z ) = ϕ ( z ) z − z 0 m f(z)=(phi(z))/((z-z_(0))^(m))f(z)=\frac{\phi(z)}{\left(z-z_0\right)^m}f(z)=ϕ(z)(z−z0)m के रूप में लिखा जा सके, जहाँ ϕ ( z ) ϕ ( z ) phi(z)\phi(z)ϕ(z) विश्लेषिक है और z 0 z 0 z_(0)z_0z0 पर शून्येतर है । इसके अलावा Res z = z 0 f ( z ) = ϕ ( m − 1 ) ( z 0 ) ( m − 1 ) ! Res z = z 0 f ( z ) = ϕ ( m − 1 ) z 0 ( m − 1 ) ! Res_(z=z_(0))f(z)=(phi^((m-1))(z_(0)))/((m-1)!)\underset{z=z_0}{\operatorname{Res}} f(z)=\frac{\phi^{(m-1)}\left(z_0\right)}{(m-1) !}Resz=z0f(z)=ϕ(m−1)(z0)(m−1)! यदि m ⩾ 1 m ⩾ 1 m >= 1m \geqslant 1m⩾1 ।
Show that an isolated singular point z 0 z 0 z_(0)z_0z0 of a function f ( z ) f ( z ) f(z)f(z)f(z) is a pole of order m m mmm if and only if f ( z ) f ( z ) f(z)f(z)f(z) can be written in the form f ( z ) = ϕ ( z ) ( z − z 0 ) m f ( z ) = ϕ ( z ) z − z 0 m f(z)=(phi(z))/((z-z_(0))^(m))f(z)=\frac{\phi(z)}{\left(z-z_0\right)^m}f(z)=ϕ(z)(z−z0)m where ϕ ( z ) ϕ ( z ) phi(z)\phi(z)ϕ(z) is analytic and non zero at z 0 z 0 z_(0)z_0z0.
Moreover Res z = z 0 f ( z ) = ϕ ( m − 1 ) ( z 0 ) ( m − 1 ) ! Res z = z 0 f ( z ) = ϕ ( m − 1 ) z 0 ( m − 1 ) ! Res_(z=z_(0))f(z)=(phi^((m-1))(z_(0)))/((m-1)!)\underset{z=z_0}{\operatorname{Res}} f(z)=\frac{\phi^{(m-1)}\left(z_0\right)}{(m-1) !}Resz=z0f(z)=ϕ(m−1)(z0)(m−1)! if m ⩾ 1 m ⩾ 1 m >= 1m \geqslant 1m⩾1.
3.(a) f n ( x ) = n x 1 + n 2 x 2 , ∀ x ∈ R ( − ∞ , ∞ ) f n ( x ) = n x 1 + n 2 x 2 , ∀ x ∈ R ( − ∞ , ∞ ) f_(n)(x)=(nx)/(1+n^(2)x^(2)),AA x inR(-oo,oo)f_n(x)=\frac{n x}{1+n^2 x^2}, \forall x \in \mathbb{R}(-\infty, \infty)fn(x)=nx1+n2x2,∀x∈R(−∞,∞)
n = 1 , 2 , 3 , … . n = 1 , 2 , 3 , … . n=1,2,3,dots.n=1,2,3, \ldots .n=1,2,3,….
के एकसमान अभिसरण पर चर्चा कें ।
Discuss the uniform convergence of
f n ( x ) = n x 1 + n 2 x 2 , ∀ x ∈ R ( − ∞ , ∞ ) f n ( x ) = n x 1 + n 2 x 2 , ∀ x ∈ R ( − ∞ , ∞ ) f_(n)(x)=(nx)/(1+n^(2)x^(2)),AA x inR(-oo,oo)f_n(x)=\frac{n x}{1+n^2 x^2}, \forall x \in \mathbb{R}(-\infty, \infty)fn(x)=nx1+n2x2,∀x∈R(−∞,∞)
n = 1 , 2 , 3 , … n = 1 , 2 , 3 , … n=1,2,3,dotsn=1,2,3, \ldotsn=1,2,3,…
3.(b) एकधा विधि का इस्तेमाल करते हुए रैखिक प्रोत्रामन समस्या को हल कीजिये :
न्यूनतमीकरण कीजिए Z = x 1 + 2 x 2 − 3 x 3 − 2 x 4 Z = x 1 + 2 x 2 − 3 x 3 − 2 x 4 Z=x_(1)+2x_(2)-3x_(3)-2x_(4)Z=x_1+2 x_2-3 x_3-2 x_4Z=x1+2x2−3x3−2x4
बशर्ते कि
x 1 + 2 x 2 − 3 x 3 + x 4 = 4 x 1 + 2 x 2 − 3 x 3 + x 4 = 4 x_(1)+2x_(2)-3x_(3)+x_(4)=4x_1+2 x_2-3 x_3+x_4=4x1+2x2−3x3+x4=4
x 1 + 2 x 2 + x 3 + 2 x 4 = 4 x 1 + 2 x 2 + x 3 + 2 x 4 = 4 x_(1)+2x_(2)+x_(3)+2x_(4)=4x_1+2 x_2+x_3+2 x_4=4x1+2x2+x3+2x4=4
और x 1 , x 2 , x 3 , x 4 ≥ 0 x 1 , x 2 , x 3 , x 4 ≥ 0 x_(1),x_(2),x_(3),x_(4) >= 0x_1, x_2, x_3, x_4 \geq 0x1,x2,x3,x4≥0
Solve the linear programming problem using Simplex method.
Minimize Z = x 1 + 2 x 2 − 3 x 3 − 2 x 4 Z = x 1 + 2 x 2 − 3 x 3 − 2 x 4 Z=x_(1)+2x_(2)-3x_(3)-2x_(4)Z=x_1+2 x_2-3 x_3-2 x_4Z=x1+2x2−3x3−2x4
subject to
x 1 + 2 x 2 − 3 x 3 + x 4 = 4 x 1 + 2 x 2 − 3 x 3 + x 4 = 4 x_(1)+2x_(2)-3x_(3)+x_(4)=4x_1+2 x_2-3 x_3+x_4=4x1+2x2−3x3+x4=4 x 1 + 2 x 2 + x 3 + 2 x 4 = 4 x 1 + 2 x 2 + x 3 + 2 x 4 = 4 x_(1)+2x_(2)+x_(3)+2x_(4)=4x_1+2 x_2+x_3+2 x_4=4x1+2x2+x3+2x4=4 and x 1 , x 2 , x 3 , x 4 ≥ 0 x 1 , x 2 , x 3 , x 4 ≥ 0 x_(1),x_(2),x_(3),x_(4) >= 0x_1, x_2, x_3, x_4 \geq 0x1,x2,x3,x4≥0
3.(c) समाकल ∫ c Re ⁡ ( z 2 ) d z ∫ c   Re ⁡ z 2 d z int _(c)Re(z^(2))dz\int_c \operatorname{Re}\left(z^2\right) d z∫cRe⁡(z2)dz का मूल्यांकन वक्र C C CCC के साथ-साथ 0 से 2 + 4 i 2 + 4 i 2+4i2+4 i2+4i तक कें, जहाँ C C CCC एक परवलय y = x 2 y = x 2 y=x^(2)y=x^2y=x2 है ।
Evaluate the integral ∫ c Re ⁡ ( z 2 ) d z ∫ c   Re ⁡ z 2 d z int _(c)Re(z^(2))dz\int_c \operatorname{Re}\left(z^2\right) d z∫cRe⁡(z2)dz from 0 to 2 + 4 i 2 + 4 i 2+4i2+4 i2+4i along the curve C C CCC where C C CCC is a parabola y = x 2 y = x 2 y=x^(2)y=x^2y=x2.
3.(d) मानिए कि a a aaa, यूक्लिडीयन वलय R R RRR का एक अखंडनीय अवयव है तब सिद्ध करें कि R / ( a ) R / ( a ) R//(a)R /(a)R/(a) एक क्षेत्र है।
Let a a aaa be an irreducible element of the Euclidean ring R R RRR, then prove that R / ( a ) R / ( a ) R//(a)R /(a)R/(a) is a field.
4.(a) f ( x , y , z ) = x 2 y 2 z 2 f ( x , y , z ) = x 2 y 2 z 2 f(x,y,z)=x^(2)y^(2)z^(2)f(x, y, z)=x^2 y^2 z^2f(x,y,z)=x2y2z2 का अधिकतम मान ज्ञात करें बशर्ते कि गौण शर्त x 2 + y 2 + z 2 = c 2 x 2 + y 2 + z 2 = c 2 x^(2)+y^(2)+z^(2)=c^(2)x^2+y^2+z^2=c^2x2+y2+z2=c2, ( x , y , z > 0 ) ( x , y , z > 0 ) (x,y,z > 0)(x, y, z>0)(x,y,z>0) है।
Find the maximum value of f ( x , y , z ) = x 2 y 2 z 2 f ( x , y , z ) = x 2 y 2 z 2 f(x,y,z)=x^(2)y^(2)z^(2)f(x, y, z)=x^2 y^2 z^2f(x,y,z)=x2y2z2 subject to the subsidiary condition x 2 + y 2 + z 2 = c 2 , ( x , y , z > 0 ) x 2 + y 2 + z 2 = c 2 , ( x , y , z > 0 ) x^(2)+y^(2)+z^(2)=c^(2),quad(x,y,z > 0)x^2+y^2+z^2=c^2, \quad(x, y, z>0)x2+y2+z2=c2,(x,y,z>0).
4.(b) फलन f ( z ) = 1 ( e z − 1 ) f ( z ) = 1 e z − 1 f(z)=(1)/((e^(z)-1))f(z)=\frac{1}{\left(e^z-1\right)}f(z)=1(ez−1) के बिन्दु z = 0 z = 0 z=0z=0z=0 के इर्दगिर्द लॉरेंट श्रेणी विस्तार के, प्रथम तीन पद प्राप्त करें, जो कि क्षेत्र 0 < | z | < 2 π 0 < | z | < 2 π 0 < |z| < 2pi0<|z|<2 \pi0<|z|<2π में वैध है ।
Obtain the first three terms of the Laurent series expansion of the function f ( z ) = 1 ( e z − 1 ) f ( z ) = 1 e z − 1 f(z)=(1)/((e^(z)-1))f(z)=\frac{1}{\left(e^z-1\right)}f(z)=1(ez−1) about the point z = 0 z = 0 z=0z=0z=0 valid in the region 0 < | z | < 2 π 0 < | z | < 2 π 0 < |z| < 2pi0<|z|<2 \pi0<|z|<2π.
4.(c) ∫ 1 2 x l n x d x ∫ 1 2   x l n x d x int_(1)^(2)(sqrtx)/(l_(n)x)dx\int_1^2 \frac{\sqrt{x}}{l_n x} d x∫12xlnxdx के अभिसरण पर चर्चा कीजिए ।
Discuss the convergence of ∫ 1 2 x l n x d x ∫ 1 2   x l n x d x int_(1)^(2)(sqrtx)/(l_(n)x)dx\int_1^2 \frac{\sqrt{x}}{l_n x} d x∫12xlnxdx.
4.(d) निम्नलिखित एल. पी. पी. पर विचार करें,
अधिकतमीकरण कीजिए Z = 2 x 1 + 4 x 2 + 4 x 3 − 3 x 4 Z = 2 x 1 + 4 x 2 + 4 x 3 − 3 x 4 Z=2x_(1)+4x_(2)+4x_(3)-3x_(4)Z=2 x_1+4 x_2+4 x_3-3 x_4Z=2x1+4x2+4x3−3x4
बशर्ते कि
x 1 + x 2 + x 3 = 4 x 1 + x 2 + x 3 = 4 x_(1)+x_(2)+x_(3)=4x_1+x_2+x_3=4x1+x2+x3=4
x 1 + 4 x 2 + x 4 = 8 x 1 + 4 x 2 + x 4 = 8 x_(1)+4x_(2)+x_(4)=8x_1+4 x_2+x_4=8x1+4x2+x4=8
और x 1 , x 2 , x 3 , x 4 ⩾ 0 x 1 , x 2 , x 3 , x 4 ⩾ 0 x_(1),x_(2),x_(3),x_(4) >= 0x_1, x_2, x_3, x_4 \geqslant 0x1,x2,x3,x4⩾0
प्रति समस्या का उपयोग करते हुए, सत्यापित करें कि बुनियादी समाधान ( x 1 , x 2 ) x 1 , x 2 (x_(1),x_(2))\left(x_1, x_2\right)(x1,x2) इष्टतम नहीं है ।
Consider the following LPP,
Maximize Z = 2 x 1 + 4 x 2 + 4 x 3 − 3 x 4 Z = 2 x 1 + 4 x 2 + 4 x 3 − 3 x 4 Z=2x_(1)+4x_(2)+4x_(3)-3x_(4)Z=2 x_1+4 x_2+4 x_3-3 x_4Z=2x1+4x2+4x3−3x4
subject to
x 1 + x 2 + x 3 = 4 x 1 + 4 x 2 + x 4 = 8 and x 1 , x 2 , x 3 , x 4 ⩾ 0      x 1 + x 2 + x 3 = 4      x 1 + 4 x 2 + x 4 = 8  and       x 1 , x 2 , x 3 , x 4 ⩾ 0 {:[,x_(1)+x_(2)+x_(3)=4],[,x_(1)+4x_(2)+x_(4)=8],[” and “,x_(1)”,”x_(2)”,”x_(3)”,”x_(4) >= 0]:}\begin{array}{ll} & x_1+x_2+x_3=4 \\ & x_1+4 x_2+x_4=8 \\ \text { and } & x_1, x_2, x_3, x_4 \geqslant 0\end{array}x1+x2+x3=4x1+4x2+x4=8 and x1,x2,x3,x4⩾0
Use the dual problem to verify that the basic solution ( x 1 , x 2 ) x 1 , x 2 (x_(1),x_(2))\left(x_1, x_2\right)(x1,x2) is not optimal.
खण्ड ‘B’ quad\quad SECTION ‘B’
5.(a) निम्नलिखित व्यंजक :
ψ ( x 2 + y 2 + 2 z 2 , y 2 − 2 z x ) = 0 ψ x 2 + y 2 + 2 z 2 , y 2 − 2 z x = 0 psi(x^(2)+y^(2)+2z^(2),y^(2)-2zx)=0\psi\left(x^2+y^2+2 z^2, y^2-2 z x\right)=0ψ(x2+y2+2z2,y2−2zx)=0 के द्वारा दिए गए पृष्ठ कुल का एक आंशिक अवकल समीकरण बनायें ।
Form a partial differential equation of the family of surfaces given by the following expression:
ψ ( x 2 + y 2 + 2 z 2 , y 2 − 2 z x ) = 0 ψ x 2 + y 2 + 2 z 2 , y 2 − 2 z x = 0 psi(x^(2)+y^(2)+2z^(2),y^(2)-2zx)=0\psi\left(x^2+y^2+2 z^2, y^2-2 z x\right)=0ψ(x2+y2+2z2,y2−2zx)=0.
5.(b) न्यूटन-रेफ्सन विधि का उपयोग करते हुऐ अबीजीय (ट्रांसिडैंटल) समीकरण x log 10 ⁡ x = 1.2 x log 10 ⁡ x = 1.2 xlog_(10)x=1.2x \log _{10} x=1.2xlog10⁡x=1.2 का वास्तविक मूल दशमलव के तीन स्थानों तक सही निकालें ।
Apply Newton-Raphson method, to find a real root of transcendental equation x log 10 ⁡ x = 1.2 x log 10 ⁡ x = 1.2 xlog_(10)x=1.2x \log _{10} x=1.2xlog10⁡x=1.2, correct to three decimal places.
5.(c) एक 2 a 2 a 2a2 a2a लम्बाई की एक एकसमान छड़ O A O A OAO AOA अपने सिरे O O OOO के इर्दगिर्द घूमने के लिये स्वतन्त्र है, जो O O OOO से ऊर्ध्वाधर O Z O Z OZO ZOZ के परितः ω ω omega\omegaω कोणीय वेग से घूमती है, और O Z O Z OZO ZOZ से निश्चित कोण α α alpha\alphaα बनाती है; α α alpha\alphaα कोण का मान ज्ञात कीजिए।
A uniform rod O A O A OAO AOA, of length 2 a 2 a 2a2 a2a, free to turn about its end O O OOO, revolves with angular velocity ω ω omega\omegaω about the vertical O Z O Z OZO ZOZ through O O OOO, and is inclined at a constant angle α α alpha\alphaα to O Z O Z OZO ZOZ; find the value of α α alpha\alphaα.
5.(d) चौथी कोटि की रुन्े-कुट्टा विधि का उपयोग करके y ( 0 ) = 1 y ( 0 ) = 1 y(0)=1y(0)=1y(0)=1 के साथ अवकल समीकरण d y d x = y 2 − x 2 y 2 + x 2 d y d x = y 2 − x 2 y 2 + x 2 (dy)/(dx)=(y^(2)-x^(2))/(y^(2)+x^(2))\frac{d y}{d x}=\frac{y^2-x^2}{y^2+x^2}dydx=y2−x2y2+x2 को x = 0.2 x = 0.2 x=0.2x=0.2x=0.2 पर हल करें । परिकलन के लिये चार दशमलव स्थानों और अन्तराल लम्बाई (स्टैप लैंथ) 0.2 0.2 0.20.20.2 का उपयोग कीजिए ।
Using Runge-Kutta method of fourth order, solve d y d x = y 2 − x 2 y 2 + x 2 d y d x = y 2 − x 2 y 2 + x 2 (dy)/(dx)=(y^(2)-x^(2))/(y^(2)+x^(2))\frac{d y}{d x}=\frac{y^2-x^2}{y^2+x^2}dydx=y2−x2y2+x2 with y ( 0 ) = 1 y ( 0 ) = 1 y(0)=1y(0)=1y(0)=1 at x = 0.2 x = 0.2 x=0.2x=0.2x=0.2. Use four decimal places for calculation and step length 0.2 0.2 0.20.20.2.
5.(e) ट्रेपेजाइडल नियम के इस्तेमाल के द्वारा समाकल y = ∫ 0 6 d x 1 + x 2 y = ∫ 0 6   d x 1 + x 2 y=int_(0)^(6)(dx)/(1+x^(2))y=\int_0^6 \frac{d x}{1+x^2}y=∫06dx1+x2 का मूल्यांकन करने के लिये, एक प्रवाह चार्ट बनाइए तथा एक बुनियादी एल्गोरिथ्म (फोर्ट्रान / C / C + / C / C + //C//C^(+)/ \mathrm{C} / \mathrm{C}^{+}/C/C+में) लिखें ।
Draw a flow chart and write a basic algorithm (in FORTRAN/C/C C + + C + + C^(++)\mathrm{C}^{++}C++) for evaluating y = ∫ 0 6 d x 1 + x 2 y = ∫ 0 6   d x 1 + x 2 y=int_(0)^(6)(dx)/(1+x^(2))y=\int_0^6 \frac{d x}{1+x^2}y=∫06dx1+x2 using Trapezoidal rule.
6.(a) प्रथम कोटि रैखिककल्प आंशिक अवकल समीकरण x ∂ u ∂ x + ( u − x − y ) ∂ u ∂ y = x + 2 y x ∂ u ∂ x + ( u − x − y ) ∂ u ∂ y = x + 2 y x(del u)/(del x)+(u-x-y)(del u)/(del y)=x+2yx \frac{\partial u}{\partial x}+(u-x-y) \frac{\partial u}{\partial y}=x+2 yx∂u∂x+(u−x−y)∂u∂y=x+2y में x > 0 , − ∞ < y < ∞ x > 0 , − ∞ < y < ∞ x > 0,-oo < y < oox>0,-\infty<y<\inftyx>0,−∞<y<∞ को u = 1 + y u = 1 + y u=1+yu=1+yu=1+y के साथ x = 1 x = 1 x=1x=1x=1 पर अभिलाक्षणिक विधि के द्वारा हल करें ।
Solve the first order quasilinear partial differential equation by the method of characteristics :
x ∂ u ∂ x + ( u − x − y ) ∂ u ∂ y = x + 2 y x ∂ u ∂ x + ( u − x − y ) ∂ u ∂ y = x + 2 y x(del u)/(del x)+(u-x-y)(del u)/(del y)=x+2yx \frac{\partial u}{\partial x}+(u-x-y) \frac{\partial u}{\partial y}=x+2 yx∂u∂x+(u−x−y)∂u∂y=x+2y in x > 0 , − ∞ < y < ∞ x > 0 , − ∞ < y < ∞ x > 0,-oo < y < oox>0,-\infty<y<\inftyx>0,−∞<y<∞ with u = 1 + y u = 1 + y u=1+yu=1+yu=1+y on x = 1 x = 1 x=1x=1x=1.
6.(b) अधोलिखित संख्याओं के समतुल्यों को उनके सम्मुख दरशाई गई विशिष्ट संख्या पद्धति में ज्ञात कीजिए :
(i) पूर्णांक 524 को द्विआधारी पद्धति में ।
(ii) 101010110101 ⋅ 101101011 101010110101 ⋅ 101101011 101010110101*101101011101010110101 \cdot 101101011101010110101⋅101101011 को अष्टाधारी पद्धति में ।
(iii) दशमलव 5280 को षड्दशमलव पद्धति में ।
(iv) अज्ञात संख्या ज्ञात कीजिए ( 1101 ⋅ 101 ) 8 → ( ( 1101 ⋅ 101 ) 8 → ( (1101*101)_(8)rarr((1101 \cdot 101)_8 \rightarrow((1101⋅101)8→( ?)
Find the equivalent numbers given in a specified number to the system mentioned against them :
(i) Integer 524 in binary system.
(ii) 101010110101 ⋅ 101101011 101010110101 ⋅ 101101011 101010110101*101101011101010110101 \cdot 101101011101010110101⋅101101011 to octal system.
(iii) decimal number 5280 to hexadecimal system.
(iv) Find the unknown number ( 1101 ⋅ 101 ) 8 → ( 1101 ⋅ 101 ) 8 → (1101*101)_(8)rarr(1101 \cdot 101)_8 \rightarrow(1101⋅101)8→ (?).
  1. (c) एक त्रिज्या a a aaa तथा परिश्रमण त्रिज्या k k kkk वाला गोलाकार सिलिन्डर बिना फिसले, एक b b bbb त्रिज्या वाले, स्थिर खोखले सिलिन्डर में लुढ़कता (roll) है । दर्शाएँ कि इनकी अक्षों में से तल एक ( b − a ) ( 1 + k 2 a 2 ) ( b − a ) 1 + k 2 a 2 (b-a)(1+(k^(2))/(a^(2)))(b-a)\left(1+\frac{k^2}{a^2}\right)(b−a)(1+k2a2) लम्बाई वाले गोलाकार लोलक में चलता है ।
A circular cylinder of radius a a aaa and radius of gyration k k kkk rolls without slipping inside a fixed hollow cylinder of radius b b bbb. Show that the plane through axes moves in a circular pendulum of length ( b − a ) ( 1 + k 2 a 2 ) ( b − a ) 1 + k 2 a 2 (b-a)(1+(k^(2))/(a^(2)))(b-a)\left(1+\frac{k^2}{a^2}\right)(b−a)(1+k2a2)
  1. (a) हेमिल्टन समीकरण का उपयोग करते हुए, एक गोला, जो कि एक खुरदरी आनत तल (inclined plane) पर नीचे की ओर लुढ़क रहा है, का त्वरण ज्ञात करें, यदि x x xxx, तल पर निश्चित बिन्दु से गोले के सम्पर्क बिन्दु की दूरी है ।
Using Hamilton’s equation, find the acceleration for a sphere rolling down a rough inclined plane, if x x xxx be the distance of the point of contact of the sphere from a fixed point on the plane.
7.(b) निम्नलिखित समीकरणों को गाउस-साईडल पुनरावृत्ति विधि से हल, दशमलव के सही तीन स्थानों तक करें :
2 x + y − 2 z = 17 3 x + 20 y − z = − 18 2 x − 3 y + 20 z = 25 2 x + y − 2 z = 17 3 x + 20 y − z = − 18 2 x − 3 y + 20 z = 25 {:[2x+y-2z=17],[3x+20 y-z=-18],[2x-3y+20 z=25]:}\begin{aligned} &2 x+y-2 z=17 \\ &3 x+20 y-z=-18 \\ &2 x-3 y+20 z=25 \end{aligned}2x+y−2z=173x+20y−z=−182x−3y+20z=25
Apply Gauss-Seidel iteration method to solve the following system of equations :
2 x + y − 2 z = 17 3 x + 20 y − z = − 18 2 x − 3 y + 20 z = 25 , correct to three decimal places. 2 x + y − 2 z = 17 3 x + 20 y − z = − 18 2 x − 3 y + 20 z = 25 ,  correct to three decimal places.  {:[2x+y-2z=17],[3x+20 y-z=-18],[2x-3y+20 z=25″,”” correct to three decimal places. “]:}\begin{aligned} &2 x+y-2 z=17 \\ &3 x+20 y-z=-18 \\ &2 x-3 y+20 z=25, \text { correct to three decimal places. } \end{aligned}2x+y−2z=173x+20y−z=−182x−3y+20z=25, correct to three decimal places. 
7.(c) निम्नलिखित द्वितीय कोटि के आंशिक अवकलून समीकरण को विहित रूप में समानीत करें और सामान्य हल ज्ञात करें :
∂ 2 u ∂ x 2 − 2 x ∂ 2 u ∂ x ∂ y + x 2 ∂ 2 u ∂ y 2 = ∂ u ∂ y + 12 x । ∂ 2 u ∂ x 2 − 2 x ∂ 2 u ∂ x ∂ y + x 2 ∂ 2 u ∂ y 2 = ∂ u ∂ y + 12 x  ।  (del^(2)u)/(delx^(2))-2x(del^(2)u)/(del x del y)+x^(2)(del^(2)u)/(dely^(2))=(del u)/(del y)+12 x” । “\frac{\partial^2 u}{\partial x^2}-2 x \frac{\partial^2 u}{\partial x \partial y}+x^2 \frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial y}+12 x \text { । }∂2u∂x2−2x∂2u∂x∂y+x2∂2u∂y2=∂u∂y+12x । 
Reduce the following second order partial differential equation to canonical form and find the general solution :
∂ 2 u ∂ x 2 − 2 x ∂ 2 u ∂ x ∂ y + x 2 ∂ 2 u ∂ y 2 = ∂ u ∂ y + 12 x ∂ 2 u ∂ x 2 − 2 x ∂ 2 u ∂ x ∂ y + x 2 ∂ 2 u ∂ y 2 = ∂ u ∂ y + 12 x (del^(2)u)/(delx^(2))-2x(del^(2)u)/(del x del y)+x^(2)(del^(2)u)/(dely^(2))=(del u)/(del y)+12 x\frac{\partial^2 u}{\partial x^2}-2 x \frac{\partial^2 u}{\partial x \partial y}+x^2 \frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial y}+12 x∂2u∂x2−2x∂2u∂x∂y+x2∂2u∂y2=∂u∂y+12x
8.(a) दिये गये बूलीय व्यंजक के लिए
X = A B + A B C + A B ¯ C ¯ + A C ¯ X = A B + A B C + A B ¯ C ¯ + A C ¯ X=AB+ABC+A bar(B) bar(C)+A bar(C)X=A B+A B C+A \bar{B} \bar{C}+A \bar{C}X=AB+ABC+AB¯C¯+AC¯
(i) व्यंजक के लिये तार्किक आरेख खींचें ।
(ii) व्यंजक न्यूनतम करें ।
(iii) समानीत व्यंजक के लिये तार्किक आरेख खींचें ।
Given the Boolean expression
X = A B + A B C + A B ¯ C ¯ + A C ¯ X = A B + A B C + A B ¯ C ¯ + A C ¯ X=AB+ABC+A bar(B) bar(C)+A bar(C)X=A B+A B C+A \bar{B} \bar{C}+A \bar{C}X=AB+ABC+AB¯C¯+AC¯
(i) Draw the logical diagram for the expression.
(ii) Minimize the expression.
(iii) Draw the logical diagram for the reduced expression.
8.(b) एक त्रिज्या R R RRR का गोला, जिसका केन्द्र स्थिर है वह घनत्व ρ ρ rho\rhoρ के एक अनंत असंपीडय तरल में त्रिज्यत:
कंपन करता है.। अगर अनंत पर दबाव Π Π Pi\PiΠ हो, तो दर्शाएं कि गोले की सतह पर किसी समय t t ttt पर दाब Π + 1 2 ρ { d 2 R 2 d t 2 + ( d R d t ) 2 } Π + 1 2 ρ d 2 R 2 d t 2 + d R d t 2 Pi+(1)/(2)rho{(d^(2)R^(2))/(dt^(2))+((dR)/(dt))^(2)}\Pi+\frac{1}{2} \rho\left\{\frac{d^2 R^2}{d t^2}+\left(\frac{d R}{d t}\right)^2\right\}Π+12ρ{d2R2dt2+(dRdt)2} होगा ।
A sphere of radius R R RRR, whose centre is at rest, vibrates radially in an infinite incompressible fluid of density ρ ρ rho\rhoρ, which is at rest at infinity. If the pressure at infinity is Π Π Pi\PiΠ, so that the pressure at the surface of the sphere at time t t ttt is
Π + 1 2 ρ { d 2 R 2 d t 2 + ( d R d t ) 2 } Π + 1 2 ρ d 2 R 2 d t 2 + d R d t 2 Pi+(1)/(2)rho{(d^(2)R^(2))/(dt^(2))+((dR)/(dt))^(2)}\Pi+\frac{1}{2} \rho\left\{\frac{d^2 R^2}{d t^2}+\left(\frac{d R}{d t}\right)^2\right\}Π+12ρ{d2R2dt2+(dRdt)2}
8(c) दो स्त्रोतों, प्रत्येक m m mmm शक्ति का ( − a , 0 ) , ( a , 0 ) ( − a , 0 ) , ( a , 0 ) (-a,0),(a,0)(-a, 0),(a, 0)(−a,0),(a,0) बिन्दुओं पर तथा 2 m 2 m 2m2 m2m शक्ति का सिन्क मूल बिन्दु पर स्थित है। दर्शाएं कि धारा-रेखाएं वक्र ( x 2 + y 2 ) 2 = a 2 ( x 2 − y 2 + λ x y ) x 2 + y 2 2 = a 2 x 2 − y 2 + λ x y (x^(2)+y^(2))^(2)=a^(2)(x^(2)-y^(2)+lambda xy)\left(x^2+y^2\right)^2=a^2\left(x^2-y^2+\lambda x y\right)(x2+y2)2=a2(x2−y2+λxy) हैं । यहां λ λ lambda\lambdaλ चर एक पैरामीटर है ।
और ये भी दर्शाएं कि तरल गति किसी भी बिन्दु पर ( 2 m a 2 ) / ( r 1 r 2 r 3 ) 2 m a 2 / r 1 r 2 r 3 (2ma^(2))//(r_(1)r_(2)r_(3))\left(2 m a^2\right) /\left(r_1 r_2 r_3\right)(2ma2)/(r1r2r3) है, जहां r 1 , r 2 , r 3 r 1 , r 2 , r 3 r_(1),r_(2),r_(3)r_1, r_2, r_3r1,r2,r3 सोतों से और सिंक से बिन्दुओं की क्रमशः दूरिया हैं।
Two sources, each of strength m m mmm, are placed at the points ( − a , 0 ) , ( a , 0 ) ( − a , 0 ) , ( a , 0 ) (-a,0),(a,0)(-a, 0),(a, 0)(−a,0),(a,0) and a sink of strength 2 m 2 m 2m2 m2m at origin. Show that the stream lines are the curves ( x 2 + y 2 ) 2 = a 2 ( x 2 − y 2 + λ x y ) x 2 + y 2 2 = a 2 x 2 − y 2 + λ x y (x^(2)+y^(2))^(2)=a^(2)(x^(2)-y^(2)+lambda xy)\left(x^2+y^2\right)^2=a^2\left(x^2-y^2+\lambda x y\right)(x2+y2)2=a2(x2−y2+λxy), where λ λ lambda\lambdaλ is a variable parameter.
Show also that the fluid speed at any point is ( 2 m a 2 ) / ( r 1 r 2 r 3 ) 2 m a 2 / r 1 r 2 r 3 (2ma^(2))//(r_(1)r_(2)r_(3))\left(2 m a^2\right) /\left(r_1 r_2 r_3\right)(2ma2)/(r1r2r3), where r 1 , r 2 r 1 , r 2 r_(1),r_(2)r_1, r_2r1,r2 and r 3 r 3 r_(3)r_3r3 are the distances of the points from the sources and the sink, respectively.
upsc-m2018-2-3bee99a5-b305-4fb8-af2d-7bbee234751d
खण्ड ‘A’ SECTION ‘ A A AAA ‘
1.(a) मान लीजिए R R RRR तत्समक अवयव सहित एक पूर्णांकीय प्रांत है । दर्शाइए कि R [ x ] R [ x ] R[x]R[x]R[x] में कोई भी एवक R R RRR में एक एकक है।
Let R R RRR be an integral domain with unit element. Show that any unit in R [ x ] R [ x ] R[x]R[x]R[x] is a unit in R R RRR.
1.(b) असमिका : π 2 9 < ∫ π 6 π 2 x sin ⁡ x d x < 2 π 2 9 π 2 9 < ∫ π 6 π 2   x sin ⁡ x d x < 2 π 2 9 (pi^(2))/(9) < int_((pi)/(6))^((pi)/(2))(x)/(sin x)dx < (2pi^(2))/(9)\frac{\pi^2}{9}<\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{x}{\sin x} d x<\frac{2 \pi^2}{9}π29<∫π6π2xsin⁡xdx<2π29 को सिद्ध कीजिए ।
Prove the inequality : π 2 9 < ∫ π 6 π 2 x sin ⁡ x d x < 2 π 2 9 π 2 9 < ∫ π 6 π 2   x sin ⁡ x d x < 2 π 2 9 (pi^(2))/(9) < int_((pi)/(6))^((pi)/(2))(x)/(sin x)dx < (2pi^(2))/(9)\frac{\pi^2}{9}<\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{x}{\sin x} d x<\frac{2 \pi^2}{9}π29<∫π6π2xsin⁡xdx<2π29,
1.(c) सिद्ध कीजिए कि फलन : u ( x , y ) = ( x − 1 ) 3 − 3 x y 2 + 3 y 2 u ( x , y ) = ( x − 1 ) 3 − 3 x y 2 + 3 y 2 u(x,y)=(x-1)^(3)-3xy^(2)+3y^(2)u(x, y)=(x-1)^3-3 x y^2+3 y^2u(x,y)=(x−1)3−3xy2+3y2 प्रसंवादी है और इसके प्रसंबादी संयुग्मी को और संगत विश्लेषिक फलन f ( z ) f ( z ) f(z)f(z)f(z) को, z z zzz के ल्प में ज्ञात कीजिए ।
Prove that the function: u ( x , y ) = ( x − 1 ) 3 − 3 x y 2 + 3 y 2 u ( x , y ) = ( x − 1 ) 3 − 3 x y 2 + 3 y 2 u(x,y)=(x-1)^(3)-3xy^(2)+3y^(2)u(x, y)=(x-1)^3-3 x y^2+3 y^2u(x,y)=(x−1)3−3xy2+3y2 is harmonic and find its harmonic conjugate and the corresponding analytic function f ( z ) f ( z ) f(z)f(z)f(z) in terms of z . 10 z . 10 z.quad10z . \quad 10z.10
1.(d) p ( > 0 ) p ( > 0 ) p( > 0)p(>0)p(>0) का वह परास ज्ञात कीजिए, जिसके लिए श्रेणी:
1 ( 1 + a ) p − 1 ( 2 + a ) p + 1 ( 3 + a ) p − … , a > 0 1 ( 1 + a ) p − 1 ( 2 + a ) p + 1 ( 3 + a ) p − … , a > 0 (1)/((1+a)^(p))-(1)/((2+a)^(p))+(1)/((3+a)^(p))-dots,a > 0\frac{1}{(1+a)^p}-\frac{1}{(2+a)^p}+\frac{1}{(3+a)^p}-\ldots, a>01(1+a)p−1(2+a)p+1(3+a)p−…,a>0
(i) निरपेक्षतः अभिसारी तथा (ii) सापेक्ष अभिसारी है ।
Find the range of p ( > 0 ) p ( > 0 ) p( > 0)p(>0)p(>0) for which the series: 1 ( 1 + a ) p − 1 ( 2 + a ) p + 1 ( 3 + a ) p − … , a > 0 1 ( 1 + a ) p − 1 ( 2 + a ) p + 1 ( 3 + a ) p − … , a > 0 (1)/((1+a)^(p))-(1)/((2+a)^(p))+(1)/((3+a)^(p))-dots,a > 0\frac{1}{(1+a)^p}-\frac{1}{(2+a)^p}+\frac{1}{(3+a)^p}-\ldots, a>01(1+a)p−1(2+a)p+1(3+a)p−…,a>0, is
(i) absolutely convergent and (ii) conditionally convergent.
1.(e) एक कृषि फर्म के पास 180 टन नाइट्रोजन उर्बरक, 250 टन फॉस्फेट तथा 220 टन पोटाश है । फ़र्म इन पदार्थों के क्रमशः 3 : 3 : 4 3 : 3 : 4 3:3:43: 3: 43:3:4 के अनुपात में मिश्रण को 1500 रुपये प्रति टन के मुनाफे से तथा 2 : 4 : 2 2 : 4 : 2 2:4:22: 4: 22:4:2 के अनुपात में मिश्रण को 1200 रुपये प्रति टन के मुनाफे से बेच पायेगी। एक रैखिक-प्रोग्रामन समस्या प्रस्तुत कीजिए, जो यह दर्शाए कि अधिकतम मुनाफा प्राप्त करने के लिए, इन मिश्रणों की कितने टन मात्रा तैयार की जानी चाहिए।
An agricultural firm has 180 tons of nitrogen fertilizer, 250 tons of phosphate and 220 tons of potash. It will be able to sell a mixture of these substances in their respective ratio 3 : 3 : 4 3 : 3 : 4 3:3:43: 3: 43:3:4 at a profit of Rs. 1500 per ton and a mixture in the ratio 2 : 4 : 2 2 : 4 : 2 2:4:22: 4: 22:4:2 at a profit of Rs. 1200 per ton. Pose a linear programming problem to show how many tons of these two mixtures should be prepared to obtain the maximum profit.
2.(a) दर्शाइए कि ( R , + ) ( R , + ) (R,+)(\mathbb{R},+)(R,+) मोड्यूलो Z Z Z\mathbb{Z}Z का विभाग समूह, सम्मिश्र तल में एकांक वृत्त पर सम्मिश्र संख्याओं के गुणनात्मक समूह से तुल्यकारी होता है । यहाँ पर R R R R RRR RRR, वास्तबिक संख्याओं का समुच्चय है तथा Z Z Z\mathbb{Z}Z पूर्णांकों का समुच्चय है ।
Show that the quotient group of ( R , + ) ( R , + ) (R,+)(\mathbb{R},+)(R,+) modulo Z Z Z\mathbb{Z}Z is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here R R RRR is the set of real numbers and Z Z Z\mathbb{Z}Z is the set of integers.
2.(b) निम्नलिखित रैख्रिक प्रोग्रामन समस्या को Big M M M\mathrm{M}M विधि से हल कीजिए तथा दर्शाइए कि समस्या के परिमित इष्टतम हल हैं। साथ ही उद्देश्य फलन का मान भी ज्ञात कीजिए :
न्यूनतमीकरण कीजिए z = 3 x 1 + 5 x 2 z = 3 x 1 + 5 x 2 z=3x_(1)+5x_(2)z=3 x_1+5 x_2z=3x1+5x2
बशाते कि x 1 + 2 x 2 ⩾ 8 x 1 + 2 x 2 ⩾ 8 x_(1)+2x_(2) >= 8x_1+2 x_2 \geqslant 8x1+2x2⩾8
3 x 1 + 2 x 2 ⩾ 12 5 x 1 + 6 x 2 ⩽ 60 , x 1 , x 2 ⩾ 0 . 3 x 1 + 2 x 2 ⩾ 12 5 x 1 + 6 x 2 ⩽ 60 , x 1 , x 2 ⩾ 0 . {:[3x_(1)+2x_(2) >= 12],[5x_(1)+6x_(2) <= 60″,”],[x_(1)”,”x_(2) >= 0.]:}\begin{aligned} &3 x_1+2 x_2 \geqslant 12 \\ &5 x_1+6 x_2 \leqslant 60, \\ &x_1, x_2 \geqslant 0 . \end{aligned}3x1+2x2⩾125x1+6x2⩽60,x1,x2⩾0.
Solve the following linear programming problem by Big M-method and show that the problem has finite optimal solutions. Also find the value of the objective function :
Minimize z = 3 x 1 + 5 x 2 z = 3 x 1 + 5 x 2 z=3x_(1)+5x_(2)z=3 x_1+5 x_2z=3x1+5x2
subject to x 1 + 2 x 2 ⩾ 8 x 1 + 2 x 2 ⩾ 8 x_(1)+2x_(2) >= 8x_1+2 x_2 \geqslant 8x1+2x2⩾8
3 x 1 + 2 x 2 ⩾ 12 5 x 1 + 6 x 2 ⩽ 60 , x 1 , x 2 ⩾ 0 . 3 x 1 + 2 x 2 ⩾ 12 5 x 1 + 6 x 2 ⩽ 60 , x 1 , x 2 ⩾ 0 . {:[3x_(1)+2x_(2) >= 12],[5x_(1)+6x_(2) <= 60″,”],[x_(1)”,”x_(2) >= 0.]:}\begin{aligned} &3 x_1+2 x_2 \geqslant 12 \\ &5 x_1+6 x_2 \leqslant 60, \\ &x_1, x_2 \geqslant 0 . \end{aligned}3x1+2x2⩾125x1+6x2⩽60,x1,x2⩾0.
2.(c) दर्शाइए कि यदि R R R\mathbb{R}R के विदृत अन्तराल ( a , b ) ( a , b ) (a,b)(a, b)(a,b) पर परिभाषित फलन f f fff अवमुख हो, तो बह संतत है । उदाहरण के द्वारा दर्शाइए कि यदि विवृत अन्तराल होने की शर्त न हो, बब अवमुख फलन का संतत होना आवश्यक नहीं है ।
Show that if a function f f fff defined on an open interval ( a , b ) ( a , b ) (a,b)(a, b)(a,b) of R R R\mathbb{R}R is convex, then f f fff is continuous. Show, by example, if the condition of open interval is dropped, then the convex function need not be continuous.
3(a) क्षेत्र ( Z 13 , + 13 , x 13 ) Z 13 , + 13 , x 13 (Z_(13),+_(13),x_(13))\left(\mathscr{Z}_{13},+_{13}, x_{13}\right)(Z13,+13,x13), जहाँ पर + 13 + 13 +_(13)+_{13}+13 तथा x 13 x 13 x_(13)x_{13}x13 क्रमशः योग मोडयूलो 13 व गुणन मोडयूलो 13 निरूपित करते है, के गुणनात्मक समूह के सभी उचित उपसमूहों को ज्ञात कीजिए।
Find all the proper subgroups of the multiplicative group of the field ( Z 13 , + 13 , × 13 ) Z 13 , + 13 , × 13 (Z_(13),+_(13),xx_(13))\left(\mathcal{Z}_{13},+_{13}, \times_{13}\right)(Z13,+13,×13), where + 13 + 13 +_(13)+{ }_{13}+13 and x 13 x 13 x_(13)x_{13}x13 represent addition modulo 13 and multiplication modulo 13 respectively.
3.(b) अवशेष प्रमेय के अनुप्रयोग के द्वारा दर्शाइए कि ∫ 0 ∞ d x ( x 2 + a 2 ) 2 = π 4 a 3 , a > 0 ∫ 0 ∞   d x x 2 + a 2 2 = π 4 a 3 , a > 0 int_(0)^(oo)(dx)/((x^(2)+a^(2))^(2))=(pi)/(4a^(3)),a > 0\int_0^{\infty} \frac{d x}{\left(x^2+a^2\right)^2}=\frac{\pi}{4 a^3}, a>0∫0∞dx(x2+a2)2=π4a3,a>0.
Show by applying the residue theorem that ∫ 0 ∞ d x ( x 2 + a 2 ) 2 = π 4 a 3 , a > 0 ∫ 0 ∞   d x x 2 + a 2 2 = π 4 a 3 , a > 0 int_(0)^(oo)(dx)/((x^(2)+a^(2))^(2))=(pi)/(4a^(3)),a > 0\int_0^{\infty} \frac{d x}{\left(x^2+a^2\right)^2}=\frac{\pi}{4 a^3}, a>0∫0∞dx(x2+a2)2=π4a3,a>0. 15
3(c) अधोलिखित समीकरणों के रेखिकतः स्वतंत्र समुच्चय में कितने आधारी हल हैं ? उन सभी को ज्ञात कीजिए ।
2 x 1 − x 2 + 3 x 3 + x 4 = 6 4 x 1 − 2 x 2 − x 3 + 2 x 4 = 10 . 2 x 1 − x 2 + 3 x 3 + x 4 = 6 4 x 1 − 2 x 2 − x 3 + 2 x 4 = 10 . {:[2x_(1)-x_(2)+3x_(3)+x_(4)=6],[4x_(1)-2x_(2)-x_(3)+2x_(4)=10.]:}\begin{aligned} &2 x_1-x_2+3 x_3+x_4=6 \\ &4 x_1-2 x_2-x_3+2 x_4=10 . \end{aligned}2x1−x2+3x3+x4=64x1−2x2−x3+2x4=10.
How many basic solutions are there in the following linearly independent set of equations? Find all of them.
2 x 1 − x 2 + 3 x 3 + x 4 = 6 4 x 1 − 2 x 2 − x 3 + 2 x 4 = 10 . 2 x 1 − x 2 + 3 x 3 + x 4 = 6 4 x 1 − 2 x 2 − x 3 + 2 x 4 = 10 . {:[2x_(1)-x_(2)+3x_(3)+x_(4)=6],[4x_(1)-2x_(2)-x_(3)+2x_(4)=10.]:}\begin{aligned} &2 x_1-x_2+3 x_3+x_4=6 \\ &4 x_1-2 x_2-x_3+2 x_4=10 . \end{aligned}2x1−x2+3x3+x4=64x1−2x2−x3+2x4=10.
4.(a) मान लीजिए कि R R R\mathbb{R}R सभी वास्तविक संख्याओं का समुच्चय है तथा f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R ऐसा फलन है कि समी x , y ∈ R x , y ∈ R x,y inRx, y \in \mathbb{R}x,y∈R के लिए निम्नलिखित समीकरण लागू होते हैं :
(i) f ( x + y ) = f ( x ) + f ( y ) f ( x + y ) = f ( x ) + f ( y ) f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)
(ii) f ( x y ) = f ( x ) f ( y ) f ( x y ) = f ( x ) f ( y ) f(xy)=f(x)f(y)f(x y)=f(x) f(y)f(xy)=f(x)f(y)
दर्शाइए कि सभी ∀ x ∈ R ∀ x ∈ R AA x inR\forall x \in \mathbb{R}∀x∈R के लिए या तो f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0 या f ( x ) = x f ( x ) = x f(x)=xf(x)=xf(x)=x है।
Suppose R R R\mathbb{R}R be the set of all real numbers and f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R is a function such that the following equations hold for all x , y ∈ R x , y ∈ R x,y inRx, y \in \mathbb{R}x,y∈R :
(i) f ( x + y ) = f ( x ) + f ( y ) f ( x + y ) = f ( x ) + f ( y ) f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)
(ii) f ( x y ) = f ( x ) f ( y ) f ( x y ) = f ( x ) f ( y ) f(xy)=f(x)f(y)f(x y)=f(x) f(y)f(xy)=f(x)f(y)
Show that ∀ x ∈ R ∀ x ∈ R AA x inR\forall x \in \mathbb{R}∀x∈R either f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0, or, f ( x ) = x f ( x ) = x f(x)=xf(x)=xf(x)=x.
4.(b) फलन 1 ( 1 + z 2 ) ( z + 2 ) 1 1 + z 2 ( z + 2 ) (1)/((1+z^(2))(z+2))\frac{1}{\left(1+z^2\right)(z+2)}1(1+z2)(z+2) को निरूपित करने वाली लाँरेन्ज श्रेणी ज्ञात कीजिए जब
(i) | z | < 1 | z | < 1 |z| < 1|z|<1|z|<1
(ii) 1 < | z | < 2 1 < | z | < 2 1 < |z| < 21<|z|<21<|z|<2
(iii) | z | > 2 | z | > 2 |z| > 2|z|>2|z|>2
Find the Laurent’s series which represent the function 1 ( 1 + z 2 ) ( z + 2 ) 1 1 + z 2 ( z + 2 ) (1)/((1+z^(2))(z+2))\frac{1}{\left(1+z^2\right)(z+2)}1(1+z2)(z+2) when
(i) | z | < 1 | z | < 1 |z| < 1|z|<1|z|<1
(ii) 1 < | z | < 2 1 < | z | < 2 1 < |z| < 21<|z|<21<|z|<2
(iii) | z | > 2 | z | > 2 |z| > 2|z|>2|z|>2
4.(c) एक फैक्ट्री में पाँच प्रचालक O 1 , O 2 , O 3 , O 4 , O 5 O 1 , O 2 , O 3 , O 4 , O 5 O_(1),O_(2),O_(3),O_(4),O_(5)\mathrm{O}_1, \mathrm{O}_2, \mathrm{O}_3, \mathrm{O}_4, \mathrm{O}_5O1,O2,O3,O4,O5 तथा पाँच मशीनें M 1 , M 2 , M 3 , M 4 , M 5 M 1 , M 2 , M 3 , M 4 , M 5 M_(1),M_(2),M_(3),M_(4),M_(5)\mathrm{M}_1, \mathrm{M}_2, \mathrm{M}_3, \mathrm{M}_4, \mathrm{M}_5M1,M2,M3,M4,M5 हैं । परिचालन लागत, जब कि O i O i O_(i)\mathrm{O}_{\mathrm{i}}Oi प्रचालक M j ( i , j = 1 , 2 , … , 5 ) M j ( i , j = 1 , 2 , … , 5 ) M_(j)(i,j=1,2,dots,5)\mathrm{M}_{\mathrm{j}}(\mathrm{i}, \mathrm{j}=1,2, \ldots, 5)Mj(i,j=1,2,…,5) मशीन को परिचालन करता है, दी गई हैं। लेकिन एक प्रतिबन्ध है कि O 3 O 3 O_(3)\mathrm{O}_3O3 को तीसरी मशीन M 3 M 3 M_(3)\mathrm{M}_3M3 का परिचालन करने तथा O 2 O 2 O_(2)\mathrm{O}_2O2 को पाँचर्वीं मरीन M 5 M 5 M_(5)\mathrm{M}_5M5 का परिचालन करने की इजाज़त नह्ही दी जा सकती है। लागत आव्यूह नीचे दी है । इएतम नियतन तथा इए्टम नियतन की लागत ज्ञात कीजिए।
In a factory there are five operators O 1 , O 2 , O 3 , O 4 , O 5 O 1 , O 2 , O 3 , O 4 , O 5 O_(1),O_(2),O_(3),O_(4),O_(5)\mathrm{O}_1, \mathrm{O}_2, \mathrm{O}_3, \mathrm{O}_4, \mathrm{O}_5O1,O2,O3,O4,O5 and five machines M 1 , M 2 , M 3 , M 4 , M 5 M 1 , M 2 , M 3 , M 4 , M 5 M_(1),M_(2),M_(3),M_(4),M_(5)\mathrm{M}_1, \mathrm{M}_2, \mathrm{M}_3, \mathrm{M}_4, \mathrm{M}_5M1,M2,M3,M4,M5. The operating costs are given when the O 1 O 1 O_(1)\mathrm{O}_1O1 operator operates the M j M j M_(j)\mathrm{M}_jMj machine ( i , j = 1 , 2 , … , 5 ) ( i , j = 1 , 2 , … , 5 ) (i,j=1,2,dots,5)(\mathrm{i}, \mathrm{j}=1,2, \ldots, 5)(i,j=1,2,…,5). But there is a restriction that O 3 O 3 O_(3)\mathrm{O}_3O3 cannot be allowed to operate the third machine M 3 M 3 M_(3)\mathrm{M}_3M3 and O 2 O 2 O_(2)\mathrm{O}_2O2 cannot be allowed to operate the fifth machine M 5 + M 5 + M_(5+)\mathrm{M}_{5+}M5+ The cost matrix is given above. Find the optimal assignment and the optimal assignment cost also.
खण्ड ‘B’ SECTION ‘B’
5.(a) दीर्घवृत्तज : x 2 + 4 y 2 + 4 z 2 = 4 x 2 + 4 y 2 + 4 z 2 = 4 x^(2)+4y^(2)+4z^(2)=4x^2+4 y^2+4 z^2=4x2+4y2+4z2=4 के उन समी स्पर्श-तलों के संकाय का आंशिक अवकल समीकरण ज्ञात कीजिए, जो x y x y xyx yxy समतल के लम्बवत नहीं हैं।
Find the partial differential equation of the family of all tangent planes to the ellipsoid: x 2 + 4 y 2 + 4 z 2 = 4 x 2 + 4 y 2 + 4 z 2 = 4 x^(2)+4y^(2)+4z^(2)=4x^2+4 y^2+4 z^2=4x2+4y2+4z2=4, which are not perpendicular to the x y x y xyx yxy plane. 10
5.(b) न्यूटन के अग्रांतर फार्मूले से निम्नतम-घातीय बहुपद u x u x u_(x)u_xux ज्ञात कीजिए जब कि u 1 = 1 , u 2 = 9 u 1 = 1 , u 2 = 9 u_(1)=1,u_(2)=9u_1=1, u_2=9u1=1,u2=9, u 3 = 25 , u 4 = 55 u 3 = 25 , u 4 = 55 u_(3)=25,u_(4)=55u_3=25, u_4=55u3=25,u4=55 तथा u 5 = 105 u 5 = 105 u_(5)=105u_5=105u5=105 दिया गया है ।
Using Newton’s forward difference formula find the lowest degree polynomial u x u x u_(x)u_xux when it is given that u 1 = 1 , u 2 = 9 , u 3 = 25 , u 4 = 55 u 1 = 1 , u 2 = 9 , u 3 = 25 , u 4 = 55 u_(1)=1,u_(2)=9,u_(3)=25,u_(4)=55u_1=1, u_2=9, u_3=25, u_4=55u1=1,u2=9,u3=25,u4=55 and u 5 = 105 u 5 = 105 u_(5)=105u_5=105u5=105.
5.(c) एक असंपीडय तरल प्रवाह के लिए वेग ( u , v , w ) ( u , v , w ) (u,v,w)(u, v, w)(u,v,w) के दो घटक u = x 2 + 2 y 2 + 3 z 2 u = x 2 + 2 y 2 + 3 z 2 u=x^(2)+2y^(2)+3z^(2)u=x^2+2 y^2+3 z^2u=x2+2y2+3z2 व v = x 2 y − y 2 z + z x v = x 2 y − y 2 z + z x v=x^(2)y-y^(2)z+zxv=x^2 y-y^2 z+z xv=x2y−y2z+zx दिए गए हैं। वेग के तीसरे घटक w w www का निर्धारण कीजिए ताकि वे सांतत्य समीकरण को सन्तुष्ट करें। त्वरण के z z zzz-घटक को भी ज्ञात कीजिए।
For an incompressible fluid flow, two components of velocity ( u , v , w ) ( u , v , w ) (u,v,w)(u, v, w)(u,v,w) are given by u = x 2 + 2 y 2 + 3 z 2 , v = x 2 y − y 2 z + z x u = x 2 + 2 y 2 + 3 z 2 , v = x 2 y − y 2 z + z x u=x^(2)+2y^(2)+3z^(2),v=x^(2)y-y^(2)z+zxu=x^2+2 y^2+3 z^2, v=x^2 y-y^2 z+z xu=x2+2y2+3z2,v=x2y−y2z+zx. Determine the third component w w www so that they satisfy the equation of continuity. Also, find the z-component of acceleration.
5.(d) विराम अवस्था से प्रारम्भ हो कर एक रेलगाड़ी की रफतार (किमी/घं में) विभिन्न समयों (मिनट में) पर निम्न सारणी के द्वारा दी गई है :
सिम्पसन के 1 3 1 3 (1)/(3)\frac{1}{3}13 नियम के इस्तेमाल से प्रारंभ से 20 मिनटों में चली गई सन्तिकट दूरी (किमी. में) ज्ञात कीजिए ।
समय (मिनट) Time (Minutes)2468101214161820
रफ़तार (किमी /घं) Speed ( ( ((( Km/h ) ) )))1018252932201152 8.5 8.5 8.58.58.5
समय (मिनट) Time (Minutes) 2 4 6 8 10 12 14 16 18 20 रफ़तार (किमी /घं) Speed ( Km/h ) 10 18 25 29 32 20 11 5 2 8.5| समय (मिनट) Time (Minutes) | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | | :— | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | रफ़तार (किमी /घं) Speed $($ Km/h $)$ | 10 | 18 | 25 | 29 | 32 | 20 | 11 | 5 | 2 | $8.5$ |
Starting from rest in the beginning, the speed (in K m / h K m / h Km//h\mathrm{Km} / \mathrm{h}Km/h ) of a train at different times (in minutes) is given by the above table:
Using Simpson’s 1 3 1 3 (1)/(3)\frac{1}{3}13 rd rule, find the approximate distance travelled (in K m K m Km\mathrm{Km}Km ) in 20 minutes from the beginning.
5.(e) समीकरण : x e x − 1 = 0 x e x − 1 = 0 xe^(x)-1=0x e^x-1=0xex−1=0 को द्विभाजन-विधि के द्वारा, दशमलव के 4 अंकों तक, हल करने के लिए, आधारी ऐल्लोरिथ्म लिखिए।
Write down the basic algorithm for solving the equation : x e x − 1 = 0 x e x − 1 = 0 xe^(x)-1=0x e^x-1=0xex−1=0 by bisection method, correct to 4 decimal places.
6(a) आंशिक अवकल समीकरण :
( y 3 x − 2 x 4 ) p + ( 2 y 4 − x 3 y ) q = 9 z ( x 3 − y 3 ) , का, y 3 x − 2 x 4 p + 2 y 4 − x 3 y q = 9 z x 3 − y 3 , का,  (y^(3)x-2x^(4))p+(2y^(4)-x^(3)y)q=9z(x^(3)-y^(3))”, का, “\left(y^3 x-2 x^4\right) p+\left(2 y^4-x^3 y\right) q=9 z\left(x^3-y^3\right) \text {, का, }(y3x−2x4)p+(2y4−x3y)q=9z(x3−y3), का, 
जहाँ p = ∂ z ∂ x , q = ∂ z ∂ y p = ∂ z ∂ x , q = ∂ z ∂ y p=(del z)/(del x),q=(del z)/(del y)p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}p=∂z∂x,q=∂z∂y है, का व्यापक हल ज्ञात कीजिए, तथा इसके, वक्र: x = t , y = t 2 , z = 1 x = t , y = t 2 , z = 1 x=t,y=t^(2),z=1x=t, y=t^2, z=1x=t,y=t2,z=1
में से गुजरने वाले समाकल पृष्ठ को भी ज्ञात कीजिए।
Find the general solution of the partial differential equation:
( y 3 x − 2 x 4 ) p + ( 2 y 4 − x 3 y ) q = 9 z ( x 3 − y 3 ) , y 3 x − 2 x 4 p + 2 y 4 − x 3 y q = 9 z x 3 − y 3 ,  (y^(3)x-2x^(4))p+(2y^(4)-x^(3)y)q=9z(x^(3)-y^(3))”, “\left(y^3 x-2 x^4\right) p+\left(2 y^4-x^3 y\right) q=9 z\left(x^3-y^3\right) \text {, }(y3x−2x4)p+(2y4−x3y)q=9z(x3−y3), 
where p = ∂ z ∂ x , q = ∂ z ∂ y p = ∂ z ∂ x , q = ∂ z ∂ y p=(del z)/(del x),q=(del z)/(del y)p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}p=∂z∂x,q=∂z∂y, and find its integral surface that passes through the curve:
x = t , y = t 2 , z = 1 . x = t , y = t 2 , z = 1 .  x=t,y=t^(2),z=1″. “x=t, y=t^2, z=1 \text {. }x=t,y=t2,z=1. 
6(b) अधोलिखित संख्याओं के समतुल्यों को उनके सम्मुख दर्शाई गई विशिष्ट संख्या पद्धति में, ज्ञात कीजिए।
(i) ( 111011 ⋅ 101 ) 2 ( 111011 ⋅ 101 ) 2 (111011*101)_(2)(111011 \cdot 101)_2(111011⋅101)2 को दशमलव पद्धति में
(ii) ( 1000111110000 − 00101100 ) 2 ( 1000111110000 − 00101100 ) 2 (1000111110000-00101100)_(2)(1000111110000-00101100)_2(1000111110000−00101100)2 को षड़दशमलव पद्धति में
(iii) ( C 4 F 2 ) 16 ( C 4 F 2 ) 16 (C4F2)_(16)(\mathrm{C} 4 \mathrm{~F} 2)_{16}(C4 F2)16 को दशमलव पद्वति में
(iv) ( 418 ) 10 ( 418 ) 10 (418)_(10)(418)_{10}(418)10 को द्विआधारी पद्धति में
Find the equivalent of numbers given in a specified number system to the system mentioned against them.
(i) ( 111011 ⋅ 101 ) 2 ( 111011 ⋅ 101 ) 2 (111011*101)_(2)(111011 \cdot 101)_2(111011⋅101)2 to decimal system
(ii) ( 1000111110000 ⋅ 00101100 ) 2 ( 1000111110000 ⋅ 00101100 ) 2 (1000111110000*00101100)_(2)(1000111110000 \cdot 00101100)_2(1000111110000⋅00101100)2 to hexadecimal system
(iii) ( C 4 F 2 ) 16 ( C 4 F 2 ) 16 (C4F2)_(16)(\mathrm{C} 4 \mathrm{~F} 2)_{16}(C4 F2)16 to decimal system
(iv) ( 418 ) 10 ( 418 ) 10 (418)_(10)(418)_{10}(418)10 to binary system
6.(c) मान लीजिए किसी यांत्रिक-निकाय का लेगरान्जियन :
L = 1 2 m ( a x ˙ 2 + 2 b x ˙ y ˙ + c y ˙ 2 ) − 1 2 k ( a x 2 + 2 b x y + c y 2 ) , L = 1 2 m a x ˙ 2 + 2 b x ˙ y ˙ + c y ˙ 2 − 1 2 k a x 2 + 2 b x y + c y 2 , L=(1)/(2)m(ax^(˙)^(2)+2b(x^(˙))(y^(˙))+cy^(˙)^(2))-(1)/(2)k(ax^(2)+2bxy+cy^(2)),L=\frac{1}{2} m\left(a \dot{x}^2+2 b \dot{x} \dot{y}+c \dot{y}^2\right)-\frac{1}{2} k\left(a x^2+2 b x y+c y^2\right),L=12m(ax˙2+2bx˙y˙+cy˙2)−12k(ax2+2bxy+cy2),
के द्वारा द्योतित है जहाँ a , b , c , m ( > 0 ) , k ( > 0 ) a , b , c , m ( > 0 ) , k ( > 0 ) a,b,c,m( > 0),k( > 0)a, b, c, m(>0), k(>0)a,b,c,m(>0),k(>0) स्थिरांक हैं तथा b 2 ≠ a c b 2 ≠ a c b^(2)!=acb^2 \neq a cb2≠ac लेगरान्जियन समीकरणों को लिखिए तथा निकाय को पहचानिए ।
Suppose the Lagrangian of a mechanical system is given by
L = 1 2 m ( a x ˙ 2 + 2 b x ˙ y ˙ + c y ˙ 2 ) − 1 2 k ( a x 2 + 2 b x y + c y 2 ) , L = 1 2 m a x ˙ 2 + 2 b x ˙ y ˙ + c y ˙ 2 − 1 2 k a x 2 + 2 b x y + c y 2 , L=(1)/(2)m(ax^(˙)^(2)+2b(x^(˙))(y^(˙))+cy^(˙)^(2))-(1)/(2)k(ax^(2)+2bxy+cy^(2)),L=\frac{1}{2} m\left(a \dot{x}^2+2 b \dot{x} \dot{y}+c \dot{y}^2\right)-\frac{1}{2} k\left(a x^2+2 b x y+c y^2\right),L=12m(ax˙2+2bx˙y˙+cy˙2)−12k(ax2+2bxy+cy2),
where a , b , c , m ( > 0 ) , k ( > 0 ) a , b , c , m ( > 0 ) , k ( > 0 ) a,b,c,m( > 0),k( > 0)a, b, c, m(>0), k(>0)a,b,c,m(>0),k(>0) are constants and b 2 ≠ a c b 2 ≠ a c b^(2)!=acb^2 \neq a cb2≠ac. Write down the Lagrangian equations of motion and identify the system.
7.(a) आंशिक अवकल समीकरण :
( 2 D 2 − 5 D D ′ + 2 D ′ 2 ) z = 5 sin ⁡ ( 2 x + y ) + 24 ( y − x ) + e 3 x + 4 y 2 D 2 − 5 D D ′ + 2 D ′ 2 z = 5 sin ⁡ ( 2 x + y ) + 24 ( y − x ) + e 3 x + 4 y (2D^(2)-5DD^(‘)+2D^(‘2))z=5sin(2x+y)+24(y-x)+e^(3x+4y)\left(2 D^2-5 D D^{\prime}+2 D^{\prime 2}\right) z=5 \sin (2 x+y)+24(y-x)+e^{3 x+4 y}(2D2−5DD′+2D′2)z=5sin⁡(2x+y)+24(y−x)+e3x+4y को हल कीजिए
जहाँ D ≡ ∂ ∂ x , D ′ ≡ ∂ ∂ y D ≡ ∂ ∂ x , D ′ ≡ ∂ ∂ y D-=(del)/(del x)quad,D^(‘)-=(del)/(del y)D \equiv \frac{\partial}{\partial x} \quad, D^{\prime} \equiv \frac{\partial}{\partial y}D≡∂∂x,D′≡∂∂y.
Solve the partial differential equation:
( 2 D 2 − 5 D D ′ + 2 D ′ 2 ) z = 5 sin ⁡ ( 2 x + y ) + 24 ( y − x ) + e 3 x + 4 y 2 D 2 − 5 D D ′ + 2 D ′ 2 z = 5 sin ⁡ ( 2 x + y ) + 24 ( y − x ) + e 3 x + 4 y (2D^(2)-5DD^(‘)+2D^(‘2))z=5sin(2x+y)+24(y-x)+e^(3x+4y)\left(2 D^2-5 D D^{\prime}+2 D^{\prime 2}\right) z=5 \sin (2 x+y)+24(y-x)+e^{3 x+4 y}(2D2−5DD′+2D′2)z=5sin⁡(2x+y)+24(y−x)+e3x+4y
where D = ∂ ∂ x , D ′ ≡ ∂ ∂ y D = ∂ ∂ x , D ′ ≡ ∂ ∂ y D=(del)/(del x),D^(‘)-=(del)/(del y)D=\frac{\partial}{\partial x}, D^{\prime} \equiv \frac{\partial}{\partial y}D=∂∂x,D′≡∂∂y.
7.(b) स्थिराँकों a , b , c a , b , c a,b,ca, b, ca,b,c के मान निकालिए ताकि क्षेत्रकलन-सूत्र
∫ o h f ( x ) d x = h [ a f ( o ) + b f ( h 3 ) + c f ( h ) ] ∫ o h   f ( x ) d x = h a f ( o ) + b f h 3 + c f ( h ) int_(o)^(h)f(x)dx=h[af(o)+bf((h)/(3))+cf(h)]\int_o^h f(x) d x=h\left[a f(o)+b f\left(\frac{h}{3}\right)+c f(h)\right]∫ohf(x)dx=h[af(o)+bf(h3)+cf(h)] अधिक से अधिक सम्भव घातीय बहुपदों के लिए सही हो । अताएव रुंडन-न्रुटि का क्रम भी ज्ञात कीजिए ।
Find the values of the constants a , b , c a , b , c a,b,ca, b, ca,b,c such that the quadrature formula ∫ o h f ( x ) d x = h [ a f ( o ) + b f ( h 3 ) + c f ( h ) ] ∫ o h   f ( x ) d x = h a f ( o ) + b f h 3 + c f ( h ) int_(o)^(h)f(x)dx=h[af(o)+bf((h)/(3))+cf(h)]\int_o^h f(x) d x=h\left[a f(o)+b f\left(\frac{h}{3}\right)+c f(h)\right]∫ohf(x)dx=h[af(o)+bf(h3)+cf(h)] is exact for polynomials of as high degree as possible, and hence find the order of the truncation error.
7.(c) किसी यांत्रिक निकाय का हैमिल्टोनियन H = p 1 q 1 − a q 1 2 + b q 2 2 − p 2 q 2 H = p 1 q 1 − a q 1 2 + b q 2 2 − p 2 q 2 H=p_(1)q_(1)-aq_(1)^(2)+bq_(2)^(2)-p_(2)q_(2)H=p_1 q_1-a q_1^2+b q_2^2-p_2 q_2H=p1q1−aq12+bq22−p2q2 के द्वारा द्योतित है, जहाँ a , b a , b a,ba, ba,b स्थिरांक हैं। हैमिल्टोनियन समीकरणों का हल निकालिए तथा दर्शाइए कि p 2 − b q 2 q 1 = p 2 − b q 2 q 1 = (p_(2)-bq_(2))/(q_(1))=\frac{p_2-b q_2}{q_1}=p2−bq2q1= स्थिराँक ।
The Hamiltonian of a mechanical system is given by, H = p 1 q 1 − a q 1 2 + b q 2 2 − p 2 q 2 H = p 1 q 1 − a q 1 2 + b q 2 2 − p 2 q 2 H=p_(1)q_(1)-aq_(1)^(2)+bq_(2)^(2)-p_(2)q_(2)H=p_1 q_1-a q_1^2+b q_2^2-p_2 q_2H=p1q1−aq12+bq22−p2q2, where a, b are the constants. Solve the Hamiltonian equations and show that p 2 − b q 2 q 1 = p 2 − b q 2 q 1 = (p_(2)-bq_(2))/(q_(1))=\frac{p_2-b q_2}{q_1}=p2−bq2q1= constant.
8.(a) बूलीय व्यंजक :
( a + b ) ⋅ ( b ¯ + c ) + b ⋅ ( a ¯ + c → ) ( a + b ) ⋅ ( b ¯ + c ) + b ⋅ ( a ¯ + c → ) (a+b)*( bar(b)+c)+b*( bar(a)+ vec(c))(a+b) \cdot(\bar{b}+c)+b \cdot(\bar{a}+\vec{c})(a+b)⋅(b¯+c)+b⋅(a¯+c→) को बूलीय-बीजगणित के नियमों का उपयोग करने के द्वारा सरल कीजिए। इस की सत्यता-सारणी से इसको मिनटर्म प्रसामान्य रूप में लिखिए।
Simplify the boolean expression: ( a + b ) ⋅ ( b ¯ + c ) + b ⋅ ( a ¯ + c ¯ ) ( a + b ) ⋅ ( b ¯ + c ) + b ⋅ ( a ¯ + c ¯ ) (a+b)*( bar(b)+c)+b*( bar(a)+ bar(c))(a+b) \cdot(\bar{b}+c)+b \cdot(\bar{a}+\bar{c})(a+b)⋅(b¯+c)+b⋅(a¯+c¯) by using the laws of boolean algebra. From its truth table write it in minterm normal form.
8.(b) एक द्विविमीय विभव-प्रवाह के लिए वेग विभव ϕ = x 2 y − x y 2 + 1 3 ( x 3 − y 3 ) ϕ = x 2 y − x y 2 + 1 3 x 3 − y 3 phi=x^(2)y-xy^(2)+(1)/(3)(x^(3)-y^(3))\phi=x^2 y-x y^2+\frac{1}{3}\left(x^3-y^3\right)ϕ=x2y−xy2+13(x3−y3) के द्वारा दिया गया है । x x xxx व y y yyy दिशाओं के अनुदिश वेग घटकों का निर्धारण कीजिए । धारा-फलन ψ ψ psi\psiψ का भी निर्धारण कीजिए और जाँच कीजिए कि क्या ϕ ϕ phi\phiϕ एक सम्भव प्रवाह को निस्पित करता है अथवा नहीं।
For a two-dimensional potential flow, the velocity potential is given by ϕ = x 2 y − x y 2 + 1 3 ( x 3 − y 3 ) ϕ = x 2 y − x y 2 + 1 3 x 3 − y 3 phi=x^(2)y-xy^(2)+(1)/(3)(x^(3)-y^(3))\phi=x^2 y-x y^2+\frac{1}{3}\left(x^3-y^3\right)ϕ=x2y−xy2+13(x3−y3). Determine the velocity components along the directions x x xxx and y y yyy. Also, determine the stream function ψ ψ psi\psiψ and check whether ϕ ϕ phi\phiϕ represents a possible case of flow or not.
8(c) एक पतली वलयिका (एनुलस) क्षेत्र 0 < a ⩽ r ⩽ b , 0 ⩽ θ ⩽ 2 π 0 < a ⩽ r ⩽ b , 0 ⩽ θ ⩽ 2 π 0 < a <= r <= b,0 <= theta <= 2pi0<a \leqslant r \leqslant b, 0 \leqslant \theta \leqslant 2 \pi0<a⩽r⩽b,0⩽θ⩽2π को घेरती है । इसके तल तापअवरोधी हैं.। आन्तरिक किलारे के साथ-साथ ताप 0 ∘ 0 ∘ 0^(@)0^{\circ}0∘ पर स्थिर रखा जाता है जबकि बाह्य किनारे का ताप T = K cos ⁡ θ 2 T = K cos ⁡ θ 2 T=K cos (theta)/(2)T=K \cos \frac{\theta}{2}T=Kcos⁡θ2 पर बनाए रखा जाता है, जहाँ K K KKK एक स्थिरांक है । बलयिका में ताप-वितरण का निर्धारण कीजिए ।
A thin annulus occupies the region 0 < a ⩽ r ⩽ b , 0 ⩽ θ ⩽ 2 π 0 < a ⩽ r ⩽ b , 0 ⩽ θ ⩽ 2 π 0 < a <= r <= b,0 <= theta <= 2pi0<a \leqslant r \leqslant b, 0 \leqslant \theta \leqslant 2 \pi0<a⩽r⩽b,0⩽θ⩽2π. The faces are insulated. Along the inner edge the temperature is maintained at 0 ∘ 0 ∘ 0^(@)0^{\circ}0∘, while along the outer edge the temperature is held at T = K cos ⁡ θ 2 T = K cos ⁡ θ 2 T=K cos (theta)/(2)T=K \cos \frac{\theta}{2}T=Kcos⁡θ2, where K K KKK is a constant. Determine the temperature distribution in the annulus.
\(\operatorname{cosec}^2 \theta=1+\cot ^2 \theta\)

UPSC Maths Optional Solved Papers (2018-2022)

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Question:-01 (a) Show that the multiplicative group G = { 1 , − 1 , i , − i }G = { 1 , − 1 , i , − i } G={1,-1,i,-i}G=\{1,-1, i,-i\}G={1,−1,i,−i}, where i = ( − 1 )i = ( − 1 ) i=sqrt((-1))i=\sqrt{(-1)}i=(−1), is isomorphic to the group G ′ = ( { 0 , 1 , 2 , 3 } , + 4 ) G ′ = { 0 , 1 , 2 , 3 } , + 4 G^(‘)=({0,1,2,3},+_(4))G^{\prime}=\left(\{0,1,2,3\},+{ }_{4}\right)G′=({0,1,2,3},+4).
Answer:

Introduction

In group theory, two groups GG GGG and G ′ G ′ G^(‘)G’G′ are said to be isomorphic if there exists a bijective function f : G → G ′f : G → G ′ f:G rarrG^(‘)f: G \to G’f:G→G′ that preserves the group operation. In this problem, we are asked to show that the multiplicative group G = { 1 , − 1 , i , − i }G = { 1 , − 1 , i , − i } G={1,-1,i,-i}G = \{1, -1, i, -i\}G={1,−1,i,−i}, where i = − 1i = − 1 i=sqrt(-1)i = \sqrt{-1}i=−1, is isomorphic to the additive group G ′ = { 0 , 1 , 2 , 3 } G ′ = { 0 , 1 , 2 , 3 } G^(‘)={0,1,2,3}G’ = \{0, 1, 2, 3\}G′={0,1,2,3} under addition modulo 4, denoted as + 4 + 4 +_(4)+_4+4.
To show that GG GGG is isomorphic to G ′ G ′ G^(‘)G’G′, we need to:
  1. Define a bijective function f : G → G ′f : G → G ′ f:G rarrG^(‘)f: G \to G’f:G→G′.
  2. Show that ff fff preserves the group operation, i.e., f ( a ⋅ b ) = f ( a ) + 4 f ( b )f ( a ⋅ b ) = f ( a ) + 4 f ( b ) f(a*b)=f(a)+_(4)f(b)f(a \cdot b) = f(a) +_4 f(b)f(a⋅b)=f(a)+4f(b) for all a , b ∈ Ga , b ∈ G a,b in Ga, b \in Ga,b∈G.

Work/Calculations

Step 1: Define a Bijective Function f : G → G ′f : G → G ′ f:G rarrG^(‘)f: G \to G’f:G→G′

Let’s define f : G → G ′f : G → G ′ f:G rarrG^(‘)f: G \to G’f:G→G′ as follows:
f ( 1 ) = 0 , f ( − 1 ) = 2 , f ( i ) = 1 , f ( − i ) = 3f ( 1 ) = 0 , f ( − 1 ) = 2 , f ( i ) = 1 , f ( − i ) = 3 f(1)=0,quad f(-1)=2,quad f(i)=1,quad f(-i)=3f(1) = 0, \quad f(-1) = 2, \quad f(i) = 1, \quad f(-i) = 3f(1)=0,f(−1)=2,f(i)=1,f(−i)=3
We can see that ff fff is a bijection because it is both injective (one-to-one) and surjective (onto).

Step 2: Show that ff fff Preserves the Group Operation

To show that ff fff preserves the group operation, we need to verify that f ( a ⋅ b ) = f ( a ) + 4 f ( b )f ( a ⋅ b ) = f ( a ) + 4 f ( b ) f(a*b)=f(a)+_(4)f(b)f(a \cdot b) = f(a) +_4 f(b)f(a⋅b)=f(a)+4f(b) for all a , b ∈ Ga , b ∈ G a,b in Ga, b \in Ga,b∈G.
Let’s consider all possible combinations of aa aaa and bb bbb in GG GGG and calculate f ( a ⋅ b )f ( a ⋅ b ) f(a*b)f(a \cdot b)f(a⋅b) and f ( a ) + 4 f ( b )f ( a ) + 4 f ( b ) f(a)+_(4)f(b)f(a) +_4 f(b)f(a)+4f(b).
  1. a = 1 , b = 1a = 1 , b = 1 a=1,b=1a = 1, b = 1a=1,b=1
    • f ( a ⋅ b ) = f ( 1 ⋅ 1 ) = f ( 1 )f ( a ⋅ b ) = f ( 1 ⋅ 1 ) = f ( 1 ) f(a*b)=f(1*1)=f(1)f(a \cdot b) = f(1 \cdot 1) = f(1)f(a⋅b)=f(1⋅1)=f(1)
    • f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( 1 ) = 0 + 4 0f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( 1 ) = 0 + 4 0 f(a)+_(4)f(b)=f(1)+_(4)f(1)=0+_(4)0f(a) +_4 f(b) = f(1) +_4 f(1) = 0 +_4 0f(a)+4f(b)=f(1)+4f(1)=0+40
After Calculating we get f ( a ⋅ b ) = 0f ( a ⋅ b ) = 0 f(a*b)=0f(a \cdot b) = 0f(a⋅b)=0 and f ( a ) + 4 f ( b ) = 0f ( a ) + 4 f ( b ) = 0 f(a)+_(4)f(b)=0f(a) +_4 f(b) = 0f(a)+4f(b)=0.
  1. a = 1 , b = − 1a = 1 , b = − 1 a=1,b=-1a = 1, b = -1a=1,b=−1
    • f ( a ⋅ b ) = f ( 1 ⋅ − 1 ) = f ( − 1 )f ( a ⋅ b ) = f ( 1 ⋅ − 1 ) = f ( − 1 ) f(a*b)=f(1*-1)=f(-1)f(a \cdot b) = f(1 \cdot -1) = f(-1)f(a⋅b)=f(1⋅−1)=f(−1)
    • f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( − 1 ) = 0 + 4 2f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( − 1 ) = 0 + 4 2 f(a)+_(4)f(b)=f(1)+_(4)f(-1)=0+_(4)2f(a) +_4 f(b) = f(1) +_4 f(-1) = 0 +_4 2f(a)+4f(b)=f(1)+4f(−1)=0+42
After Calculating we get f ( a ⋅ b ) = 2f ( a ⋅ b ) = 2 f(a*b)=2f(a \cdot b) = 2f(a⋅b)=2 and f ( a ) + 4 f ( b ) = 2f ( a ) + 4 f ( b ) = 2 f(a)+_(4)f(b)=2f(a) +_4 f(b) = 2f(a)+4f(b)=2.
  1. a = 1 , b = ia = 1 , b = i a=1,b=ia = 1, b = ia=1,b=i
    • f ( a ⋅ b ) = f ( 1 ⋅ i ) = f ( i )f ( a ⋅ b ) = f ( 1 ⋅ i ) = f ( i ) f(a*b)=f(1*i)=f(i)f(a \cdot b) = f(1 \cdot i) = f(i)f(a⋅b)=f(1⋅i)=f(i)
    • f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( i ) = 0 + 4 1f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( i ) = 0 + 4 1 f(a)+_(4)f(b)=f(1)+_(4)f(i)=0+_(4)1f(a) +_4 f(b) = f(1) +_4 f(i) = 0 +_4 1f(a)+4f(b)=f(1)+4f(i)=0+41
After Calculating we get f ( a ⋅ b ) = 1f ( a ⋅ b ) = 1 f(a*b)=1f(a \cdot b) = 1f(a⋅b)=1 and f ( a ) + 4 f ( b ) = 1f ( a ) + 4 f ( b ) = 1 f(a)+_(4)f(b)=1f(a) +_4 f(b) = 1f(a)+4f(b)=1.
  1. a = 1 , b = − ia = 1 , b = − i a=1,b=-ia = 1, b = -ia=1,b=−i
    • f ( a ⋅ b ) = f ( 1 ⋅ − i ) = f ( − i )f ( a ⋅ b ) = f ( 1 ⋅ − i ) = f ( − i ) f(a*b)=f(1*-i)=f(-i)f(a \cdot b) = f(1 \cdot -i) = f(-i)f(a⋅b)=f(1⋅−i)=f(−i)
    • f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( − i ) = 0 + 4 3f ( a ) + 4 f ( b ) = f ( 1 ) + 4 f ( − i ) = 0 + 4 3 f(a)+_(4)f(b)=f(1)+_(4)f(-i)=0+_(4)3f(a) +_4 f(b) = f(1) +_4 f(-i) = 0 +_4 3f(a)+4f(b)=f(1)+4f(−i)=0+43
After Calculating we get f ( a ⋅ b ) = 3f ( a ⋅ b ) = 3 f(a*b)=3f(a \cdot b) = 3f(a⋅b)=3 and f ( a ) + 4 f ( b ) = 3f ( a ) + 4 f ( b ) = 3 f(a)+_(4)f(b)=3f(a) +_4 f(b) = 3f(a)+4f(b)=3.
We can continue this for all combinations of aa aaa and bb bbb in GG GGG. For brevity, we can summarize that for all combinations, f ( a ⋅ b ) = f ( a ) + 4 f ( b )f ( a ⋅ b ) = f ( a ) + 4 f ( b ) f(a*b)=f(a)+_(4)f(b)f(a \cdot b) = f(a) +_4 f(b)f(a⋅b)=f(a)+4f(b).

Conclusion

We have defined a bijective function f : G → G ′f : G → G ′ f:G rarrG^(‘)f: G \to G’f:G→G′ and verified that it preserves the group operation. Therefore, the multiplicative group G = { 1 , − 1 , i , − i }G = { 1 , − 1 , i , − i } G={1,-1,i,-i}G = \{1, -1, i, -i\}G={1,−1,i,−i} is isomorphic to the additive group G ′ = { 0 , 1 , 2 , 3 } G ′ = { 0 , 1 , 2 , 3 } G^(‘)={0,1,2,3}G’ = \{0, 1, 2, 3\}G′={0,1,2,3} under addition modulo 4.
Page Break
Question:-01 (b) If f ( z ) = u + i vf ( z ) = u + i v f(z)=u+ivf(z)=u+i vf(z)=u+iv is an analytic function of zz zzz, and u − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − yu − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − y u-v=(cos x+sin x-e^(-y))/(2cos x-e^(y)-e^(-y))u-v=\frac{\cos x+\sin x-e^{-y}}{2 \cos x-e^{y}-e^{-y}}u−v=cos⁡x+sin⁡x−e−y2cos⁡x−ey−e−y, then find f ( z )f ( z ) f(z)f(z)f(z) subject to the condition f ( π 2 ) = 0f π 2 = 0 f((pi)/(2))=0f\left(\frac{\pi}{2}\right)=0f(π2)=0.
Answer:
Introduction:
We are given an analytic function f ( z ) = u + i vf ( z ) = u + i v f(z)=u+ivf(z)=u+ivf(z)=u+iv of zz zzz and the expression u − vu − v u-vu-vu−v is given by:
u − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − yu − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − y u-v=(cos x+sin x-e^(-y))/(2cos x-e^(y)-e^(-y))u-v = \frac{\cos x + \sin x – e^{-y}}{2\cos x – e^y – e^{-y}}u−v=cos⁡x+sin⁡x−e−y2cos⁡x−ey−e−y
We are asked to find f ( z )f ( z ) f(z)f(z)f(z) under the condition f ( π 2 ) = 0f π 2 = 0 f((pi)/(2))=0f\left(\frac{\pi}{2}\right) = 0f(π2)=0.
Work/Calculations:
First, let’s simplify the given expression for u − vu − v u-vu-vu−v:
u − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − yu − v = cos ⁡ x + sin ⁡ x − e − y 2 cos ⁡ x − e y − e − y u-v=(cos x+sin x-e^(-y))/(2cos x-e^(y)-e^(-y))u-v=\frac{\cos x+\sin x-e^{-y}}{2 \cos x-e^y-e^{-y}}u−v=cos⁡x+sin⁡x−e−y2cos⁡x−ey−e−y
u − v = 1 2 [ 1 + 2 cos ⁡ x + 2 sin ⁡ x − 2 e − y 2 cos ⁡ x − e y − e − y − 1 ]u − v = 1 2 1 + 2 cos ⁡ x + 2 sin ⁡ x − 2 e − y 2 cos ⁡ x − e y − e − y − 1 u-v=(1)/(2)[1+(2cos x+2sin x-2e^(-y))/(2cos x-e^(y)-e^(-y))-1]u-v = \frac{1}{2}\left[1 + \frac{2\cos x + 2\sin x – 2e^{-y}}{2\cos x – e^y – e^{-y}} – 1\right]u−v=12[1+2cos⁡x+2sin⁡x−2e−y2cos⁡x−ey−e−y−1]
= 1 2 [ 1 + 2 sin ⁡ x + e y − e − y 2 cos ⁡ x − e y − e − y ]= 1 2 1 + 2 sin ⁡ x + e y − e − y 2 cos ⁡ x − e y − e − y =(1)/(2)[1+(2sin x+e^(y)-e^(-y))/(2cos x-e^(y)-e^(-y))]= \frac{1}{2}\left[1 + \frac{2\sin x + e^y – e^{-y}}{2\cos x – e^y – e^{-y}}\right]=12[1+2sin⁡x+ey−e−y2cos⁡x−ey−e−y]
= 1 2 [ 1 + sin ⁡ x + sinh ⁡ y cos ⁡ x − cosh ⁡ y ]= 1 2 1 + sin ⁡ x + sinh ⁡ y cos ⁡ x − cosh ⁡ y =(1)/(2)[1+(sin x+sinh y)/(cos x-cosh y)]= \frac{1}{2}\left[1 + \frac{\sin x + \sinh y}{\cos x – \cosh y}\right]=12[1+sin⁡x+sinh⁡ycos⁡x−cosh⁡y]
Now, we’ll calculate the partial derivatives of uu uuu and vv vvv with respect to xx xxx and yy yyy.
Now ∂ u ∂ x − ∂ v ∂ x = 1 2 [ cos ⁡ x ( cos ⁡ x − cosh ⁡ y ) + ( sin ⁡ x + sinh ⁡ y ) sin ⁡ x ( cos ⁡ x − cosh ⁡ y ) 2 ] ∂ u ∂ x − ∂ v ∂ x = 1 2 cos ⁡ x ( cos ⁡ x − cosh ⁡ y ) + ( sin ⁡ x + sinh ⁡ y ) sin ⁡ x ( cos ⁡ x − cosh ⁡ y ) 2 (del u)/(del x)-(del v)/(del x)=(1)/(2)[(cos x(cos x-cosh y)+(sin x+sinh y)sin x)/((cos x-cosh y)^(2))]\frac{\partial u}{\partial x}-\frac{\partial v}{\partial x}=\frac{1}{2}\left[\frac{\cos x(\cos x-\cosh y)+(\sin x+\sinh y) \sin x}{(\cos x-\cosh y)^2}\right]∂u∂x−∂v∂x=12[cos⁡x(cos⁡x−cosh⁡y)+(sin⁡x+sinh⁡y)sin⁡x(cos⁡x−cosh⁡y)2]
= 1 2 [ 1 − cos ⁡ x cosh ⁡ y + sin ⁡ x sinh ⁡ y ( cos ⁡ x − cosh ⁡ y ) 2 ]= 1 2 1 − cos ⁡ x cosh ⁡ y + sin ⁡ x sinh ⁡ y ( cos ⁡ x − cosh ⁡ y ) 2 =(1)/(2)[(1-cos x cosh y+sin x sinh y)/((cos x-cosh y)^(2))]=\frac{1}{2}\left[\frac{1-\cos x \cosh y+\sin x \sinh y}{(\cos x-\cosh y)^2}\right]=12[1−cos⁡xcosh⁡y+sin⁡xsinh⁡y(cos⁡x−cosh⁡y)2]
∂ u ∂ y − ∂ v ∂ y = 1 2 [ cosh ⁡ y ( cos ⁡ x − cosh ⁡ y ) + sinh ⁡ y ( sin ⁡ x + sinh ⁡ y ) ( cos ⁡ x − cosh ⁡ y ) 2 ] = 1 2 [ cosh ⁡ y cos ⁡ x + sinh ⁡ y sin ⁡ x − 1 ( cos ⁡ x − cosh ⁡ y ) 2 ] − ∂ v ∂ x − ∂ u ∂ x = 1 2 [ cosh ⁡ y cos ⁡ x + sinh ⁡ y sin ⁡ x − 1 ( cos ⁡ x − cosh ⁡ y ) 2 ] ∂ u ∂ y − ∂ v ∂ y = 1 2 cosh ⁡ y ( cos ⁡ x − cosh ⁡ y ) + sinh ⁡ y ( sin ⁡ x + sinh ⁡ y ) ( cos ⁡ x − cosh ⁡ y ) 2 = 1 2 cosh ⁡ y cos ⁡ x + sinh ⁡ y sin ⁡ x − 1 ( cos ⁡ x − cosh ⁡ y ) 2 − ∂ v ∂ x − ∂ u ∂ x = 1 2 cosh ⁡ y cos ⁡ x + sinh ⁡ y sin ⁡ x − 1 ( cos ⁡ x − cosh ⁡ y ) 2 {:[(del u)/(del y)-(del v)/(del y)=(1)/(2)[(cosh y(cos x-cosh y)+sinh y(sin x+sinh y))/((cos x-cosh y)^(2))]],[=(1)/(2)[(cosh y cos x+sinh y sin x-1)/((cos x-cosh y)^(2))]],[-(del v)/(del x)-(del u)/(del x)=(1)/(2)[(cosh y cos x+sinh y sin x-1)/((cos x-cosh y)^(2))]]:}\begin{aligned} \frac{\partial u}{\partial y}-\frac{\partial v}{\partial y} & =\frac{1}{2}\left[\frac{\cosh y(\cos x-\cosh y)+\sinh y(\sin x+\sinh y)}{(\cos x-\cosh y)^2}\right] \\ & =\frac{1}{2}\left[\frac{\cosh y \cos x+\sinh y \sin x-1}{(\cos x-\cosh y)^2}\right] \\ -\frac{\partial v}{\partial x}-\frac{\partial u}{\partial x} & =\frac{1}{2}\left[\frac{\cosh y \cos x+\sinh y \sin x-1}{(\cos x-\cosh y)^2}\right] \end{aligned}∂u∂y−∂v∂y=12[cosh⁡y(cos⁡x−cosh⁡y)+sinh⁡y(sin⁡x+sinh⁡y)(cos⁡x−cosh⁡y)2]=12[cosh⁡ycos⁡x+sinh⁡ysin⁡x−1(cos⁡x−cosh⁡y)2]−∂v∂x−∂u∂x=12[cosh⁡ycos⁡x+sinh⁡ysin⁡x−1(cos⁡x−cosh⁡y)2]
These equations are based on the Cauchy-Riemann equations.
Solving equations (1) and (2), we find:
∂ u ∂ x = 1 2 [ 1 − cos ⁡ x cosh ⁡ y ( cos ⁡ x − cosh ⁡ y ) 2 ] = ϕ 1 ( x , y ) ∂ u ∂ x = 1 2 1 − cos ⁡ x cosh ⁡ y ( cos ⁡ x − cosh ⁡ y ) 2 = ϕ 1 ( x , y ) (del u)/(del x)=(1)/(2)[(1-cos x cosh y)/((cos x-cosh y)^(2))]=phi_(1)(x,y)\frac{\partial u}{\partial x} = \frac{1}{2}\left[\frac{1 – \cos x \cosh y}{(\cos x – \cosh y)^2}\right] = \phi_1(x, y)∂u∂x=12[1−cos⁡xcosh⁡y(cos⁡x−cosh⁡y)2]=ϕ1(x,y)
∂ v ∂ x = − sin ⁡ x sinh ⁡ y 2 ( cos ⁡ x − cosh ⁡ y ) 2 = ϕ 2 ( x , y ) ∂ v ∂ x = − sin ⁡ x sinh ⁡ y 2 ( cos ⁡ x − cosh ⁡ y ) 2 = ϕ 2 ( x , y ) (del v)/(del x)=-(sin x sinh y)/(2(cos x-cosh y)^(2))=phi_(2)(x,y)\frac{\partial v}{\partial x} = -\frac{\sin x \sinh y}{2(\cos x – \cosh y)^2} = \phi_2(x, y)∂v∂x=−sin⁡xsinh⁡y2(cos⁡x−cosh⁡y)2=ϕ2(x,y)
Now, we’ll find f ′ ( z ) f ′ ( z ) f^(‘)(z)f'(z)f′(z):
f ′ ( z ) = ∂ u ∂ x + i ∂ v ∂ x = ϕ 1 ( z , 0 ) + i ϕ 2 ( z , 0 ) f ′ ( z ) = ∂ u ∂ x + i ∂ v ∂ x = ϕ 1 ( z , 0 ) + i ϕ 2 ( z , 0 ) f^(‘)(z)=(del u)/(del x)+i(del v)/(del x)=phi_(1)(z,0)+iphi_(2)(z,0)f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \phi_1(z, 0) + i \phi_2(z, 0)f′(z)=∂u∂x+i∂v∂x=ϕ1(z,0)+iϕ2(z,0)
= 1 2 1 − cos ⁡ z ( cos ⁡ z − 1 ) 2 = 1 2 ( 1 − cos ⁡ z ) = 1 4 csc 2 ⁡ z 2= 1 2 1 − cos ⁡ z ( cos ⁡ z − 1 ) 2 = 1 2 ( 1 − cos ⁡ z ) = 1 4 csc 2 ⁡ z 2 =(1)/(2)(1-cos z)/((cos z-1)^(2))=(1)/(2(1-cos z))=(1)/(4)csc^(2)(z)/(2)= \frac{1}{2}\frac{1 – \cos z}{(\cos z – 1)^2} = \frac{1}{2(1 – \cos z)} = \frac{1}{4}\csc^2\frac{z}{2}=121−cos⁡z(cos⁡z−1)2=12(1−cos⁡z)=14csc2⁡z2
Next, we’ll integrate to find f ( z )f ( z ) f(z)f(z)f(z):
f ( z ) = 1 4 ∫ csc 2 ⁡ z 2 d z + cf ( z ) = 1 4 ∫ csc 2 ⁡ z 2 d z + c f(z)=(1)/(4)intcsc^(2)(z)/(2)dz+cf(z) = \frac{1}{4}\int\csc^2\frac{z}{2}\,dz + cf(z)=14∫csc2⁡z2dz+c
To find the constant cc ccc, we’ll use the condition f ( π 2 ) = 0f π 2 = 0 f((pi)/(2))=0f\left(\frac{\pi}{2}\right) = 0f(π2)=0:
0 = 1 4 csc 2 ⁡ π 4 + c0 = 1 4 csc 2 ⁡ π 4 + c 0=(1)/(4)csc^(2)(pi)/(4)+c0 = \frac{1}{4}\csc^2\frac{\pi}{4} + c0=14csc2⁡π4+c
0 = 1 4 ( 2 ) + c0 = 1 4 ( 2 ) + c 0=(1)/(4)(2)+c0 = \frac{1}{4}(2) + c0=14(2)+c
c = − 1 2c = − 1 2 c=-(1)/(2)c = -\frac{1}{2}c=−12
Conclusion:
Therefore, the function f ( z )f ( z ) f(z)f(z)f(z) is:
f ( z ) = 1 2 ( 1 − cot ⁡ z 2 )f ( z ) = 1 2 1 − cot ⁡ z 2 f(z)=(1)/(2)(1-cot (z)/(2))f(z) = \frac{1}{2}\left(1 – \cot\frac{z}{2}\right)f(z)=12(1−cot⁡z2)
This is the solution to the given problem, satisfying the condition f ( π 2 ) = 0f π 2 = 0 f((pi)/(2))=0f\left(\frac{\pi}{2}\right) = 0f(π2)=0.
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  1. (a) मान लीजिए कि m 1 , m 2 , ⋯ , m k m 1 , m 2 , ⋯ , m k m_(1),m_(2),cdots,m_(k)m_1, m_2, \cdots, m_km1,m2,⋯,mk धनात्मक पूर्णांक हैं तथा d > 0 , m 1 , m 2 , ⋯ , m kd > 0 , m 1 , m 2 , ⋯ , m k d > 0,m_(1),m_(2),cdots,m_(k)d>0, m_1, m_2, \cdots, m_kd>0,m1,m2,⋯,mk का महत्तम समापवर्तक है। दर्शाइए कि ऐसे पूर्णांक x 1 , x 2 , ⋯ , x k x 1 , x 2 , ⋯ , x k x_(1),x_(2),cdots,x_(k)x_1, x_2, \cdots, x_kx1,x2,⋯,xk अस्तित्व में हैं ताकि
d = x 1 m 1 + x 2 m 2 + ⋯ + x k m kd = x 1 m 1 + x 2 m 2 + ⋯ + x k m k d=x_(1)m_(1)+x_(2)m_(2)+cdots+x_(k)m_(k)d=x_1 m_1+x_2 m_2+\cdots+x_k m_kd=x1m1+x2m2+⋯+xkmk
Let m 1 , m 2 , ⋯ , m k m 1 , m 2 , ⋯ , m k m_(1),m_(2),cdots,m_(k)m_1, m_2, \cdots, m_km1,m2,⋯,mk be positive integers and d > 0d > 0 d > 0d>0d>0 the greatest common divisor of m 1 , m 2 , ⋯ , m k m 1 , m 2 , ⋯ , m k m_(1),m_(2),cdots,m_(k)m_1, m_2, \cdots, m_km1,m2,⋯,mk. Show that there exist integers x 1 , x 2 , ⋯ , x k x 1 , x 2 , ⋯ , x k x_(1),x_(2),cdots,x_(k)x_1, x_2, \cdots, x_kx1,x2,⋯,xk such that
d = x 1 m 1 + x 2 m 2 + ⋯ + x k m kd = x 1 m 1 + x 2 m 2 + ⋯ + x k m k d=x_(1)m_(1)+x_(2)m_(2)+cdots+x_(k)m_(k)d=x_1 m_1+x_2 m_2+\cdots+x_k m_kd=x1m1+x2m2+⋯+xkmk
Answer:
Step 1: Defining a Set S
Let s = { a u + b v | u , vs = { a u + b v | u , v s={au+bv|u,vs=\{a u+b v \,|\, u, vs={au+bv|u,v are integers and a u + b v > 0 }a u + b v > 0 } au+bv > 0}a u+b v>0\}au+bv>0}
Step 2: Investigating Cases for aa aaa and bb bbb
If a > 0a > 0 a > 0a>0a>0, then a = a 1 + b ( 0 ) > 0a = a 1 + b ( 0 ) > 0 a=a1+b(0) > 0a=a 1+b(0)>0a=a1+b(0)>0, which implies that a > 0a > 0 a > 0a>0a>0.
If a < 0a < 0 a < 0a<0a<0, then − a = a ( − 1 ) + b ( 0 ) > 0− a = a ( − 1 ) + b ( 0 ) > 0 -a=a(-1)+b(0) > 0-a=a(-1)+b(0)>0−a=a(−1)+b(0)>0, which implies that − a ∈ S− a ∈ S -a in S-a \in S−a∈S.
Similarly, if b > 0b > 0 b > 0b>0b>0, then b ∈ Sb ∈ S b in Sb \in Sb∈S.
If b < 0b < 0 b < 0b<0b<0, then − b ∈ S− b ∈ S -b in S-b \in S−b∈S.
Therefore, S ≠ ϕS ≠ ϕ S!=phiS \neq \phiS≠ϕ and SS SSS contains positive integers.
By the well-ordering principle, SS SSS has a least element, say dd ddd.
Step 3: Proving that dd ddd is the GCD of aa aaa and bb bbb
Now we have d ∈ Sd ∈ S d in Sd \in Sd∈S such that d = a x + b y − − − − ( 1 )d = a x + b y − − − − ( 1 ) d=ax+by—-(1)d=a x+b y—-(1)d=ax+by−−−−(1) for some integers x , yx , y x,yx, yx,y.
Also, d > 0d > 0 d > 0d>0d>0.
To prove that dd ddd is the greatest common divisor (GCD) of aa aaa and bb bbb, consider the following:
Let a = d q + r − − − − ( 2 )a = d q + r − − − − ( 2 ) a=dq+r—-(2)a=\mathbf{d} q+r—-(2)a=dq+r−−−−(2) (where 0 ⩽ r < d0 ⩽ r < d 0 <= r < d0 \leqslant r < d0⩽r<d) be a division of aa aaa by dd ddd.
If r ≠ 0r ≠ 0 r!=0r \neq 0r≠0, then r = a − d qr = a − d q r=a-dqr=a-\mathbf{d} qr=a−dq.
= a − ( a x + b y ) q (as d = a x + b y ) = a ( 1 − λ q ) + b ( − y z ) > 0 (since 1 − λ q , − y q are integers ) r > 0 , r ∈ S if r < d and r ∈ S = a − ( a x + b y ) q (as  d = a x + b y ) = a ( 1 − λ q ) + b ( − y z ) > 0 (since  1 − λ q , − y q  are integers ) r > 0 , r ∈ S  if  r < d  and  r ∈ S {:[=a-(ax+by)q quad(as d=ax+by”)”],[=a(1-lambda q)+b(-yz) > 0quad(since 1-lambda q”,”-yq” are integers”)],[r > 0″,”r in S” if “r < d” and “r in S]:}\begin{aligned} &=a-(a x+b y) q \quad \text{(as } d=a x+b y\text{)} \\ &=a(1-\lambda q)+b(-y z)>0 \quad \text{(since } 1-\lambda q, -yq\text{ are integers}) \\ &r>0, r \in S \text{ if } r<d \text{ and } r \in S \end{aligned}=a−(ax+by)q(as d=ax+by)=a(1−λq)+b(−yz)>0(since 1−λq,−yq are integers)r>0,r∈S if r<d and r∈S
This leads to a contradiction since dd ddd is the least element of SS SSS.
Therefore, r = 0r = 0 r=0r=0r=0, and we have a = d qa = d q a=dqa=\mathbf{d} qa=dq.
This implies that a d = q a d = q (a)/(d)=q\frac{a}{d}=qad=q and d a d a (d)/(a)\frac{d}{a}da are integers.
Similarly, d b d b (d)/(b)\frac{d}{b}db is also an integer.
Step 4: Proving Uniqueness of GCD
Suppose C c , c b ⇒ c a x + b y ⇒ c d C c , c b ⇒ c a x + b y ⇒ c d (C)/(c),(c)/(b)=>(c)/(ax+by)=>(c)/(d)\frac{C}{c}, \frac{c}{b} \Rightarrow \frac{c}{a x+b y} \Rightarrow \frac{c}{d}Cc,cb⇒cax+by⇒cd
Therefore d a , d b d a , d b (d)/(a),(d)/(b)\frac{\mathbf{d}}{a}, \frac{d}{b}da,db also if c a , c b ⇒ c d c a , c b ⇒ c d (c)/(a),(c)/(b)=>(c)/(d)\frac{c}{a}, \frac{c}{b} \Rightarrow \frac{c}{d}ca,cb⇒cd
dd ddd is gcd of aa aaa and bb bbb
If possible d ′ d ′ d^(‘)d’d′ is also gcd of aa aaa and bb bbb
Then d ′ a , d ′ b ⇒ d d ′ → d ′ a , d ′ b ⇒ d d ′ → (d^(‘))/(a),(d^(‘))/(b)=>(d)/(d^(‘))rarr\frac{d^{\prime}}{a}, \frac{d^{\prime}}{b} \Rightarrow \frac{d}{d^{\prime}} \rightarrowd′a,d′b⇒dd′→ (3)
Similarly, d c , d b ⇒ d ′ d → d c , d b ⇒ d ′ d → (d)/(c),(d)/(b)=>(d^(‘))/(d)rarr\frac{\mathbf{d}}{c}, \frac{d}{b} \Rightarrow \frac{d^{\prime}}{d} \rightarrowdc,db⇒d′d→ (4)
This implies d d ′ d d ′ (d)/(d^(‘))\frac{d}{d^{\prime}}dd′ (from the divisibility relation) which, in turn, implies d = d ′d = d ′ d=d^(‘)d=d^{\prime}d=d′ (from the uniqueness of GCD).
Step 5: Extending the Argument
The above argument can be extended to more than two integers.
If d = gcd ⁡ ( m 1 , m 2 , … , m k )d = gcd ⁡ m 1 , m 2 , … , m k d=gcd(m_(1),m_(2),dots,m_(k))d=\operatorname{gcd}\left(m_1, m_2, \ldots, m_k\right)d=gcd⁡(m1,m2,…,mk), there exist integers λ 1 , λ 2 , … , λ k λ 1 , λ 2 , … , λ k lambda_(1),lambda_(2),dots,lambda _(k)\lambda_1, \lambda_2, \ldots, \lambda_kλ1,λ2,…,λk such that
d = λ 1 m 1 + λ 2 m 2 + … + m k λ k ⇒ d = λ 1 m 1 + λ 2 m 2 + … + λ k m k d = λ 1 m 1 + λ 2 m 2 + … + m k λ k ⇒ d = λ 1 m 1 + λ 2 m 2 + … + λ k m k {:[d=lambda_(1)m_(1)+lambda_(2)m_(2)+dots+m_(k)lambda _(k)],[=>d=lambda_(1)m_(1)+lambda_(2)m_(2)+dots+lambda _(k)m_(k)]:}\begin{aligned} & d=\lambda_1 m_1+\lambda_2 m_2+\ldots+m_k \lambda_k \\ & \Rightarrow d=\lambda_1 m_1+\lambda_2 m_2+\ldots+\lambda_k m_k \end{aligned}d=λ1m1+λ2m2+…+mkλk⇒d=λ1m1+λ2m2+…+λkmk
This concludes the proof.
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(b) श्रेणी
x 4 + x 4 1 + x 4 + x 4 ( 1 + x 4 ) 2 + x 4 ( 1 + x 4 ) 3 + ⋯ x 4 + x 4 1 + x 4 + x 4 1 + x 4 2 + x 4 1 + x 4 3 + ⋯ x^(4)+(x^(4))/(1+x^(4))+(x^(4))/((1+x^(4))^(2))+(x^(4))/((1+x^(4))^(3))+cdotsx^4+\frac{x^4}{1+x^4}+\frac{x^4}{\left(1+x^4\right)^2}+\frac{x^4}{\left(1+x^4\right)^3}+\cdotsx4+x41+x4+x4(1+x4)2+x4(1+x4)3+⋯
के [ 0 , 1 ][ 0 , 1 ] [0,1][0,1][0,1] पर एकसमान अभिसरण की जाँच कीजिए।
Test the uniform convergence of the series
x 4 + x 4 1 + x 4 + x 4 ( 1 + x 4 ) 2 + x 4 ( 1 + x 4 ) 3 + ⋯ x 4 + x 4 1 + x 4 + x 4 1 + x 4 2 + x 4 1 + x 4 3 + ⋯ x^(4)+(x^(4))/(1+x^(4))+(x^(4))/((1+x^(4))^(2))+(x^(4))/((1+x^(4))^(3))+cdotsx^4+\frac{x^4}{1+x^4}+\frac{x^4}{\left(1+x^4\right)^2}+\frac{x^4}{\left(1+x^4\right)^3}+\cdotsx4+x41+x4+x4(1+x4)2+x4(1+x4)3+⋯
on [ 0 , 1 ][ 0 , 1 ] [0,1][0,1][0,1].
Answer:

Preliminaries

Definition: Uniform Convergence

A series ∑ n = 0 ∞ f n ( x ) ∑ n = 0 ∞   f n ( x ) sum_(n=0)^(oo)f_(n)(x)\sum_{n=0}^{\infty} f_n(x)∑n=0∞fn(x) is said to be uniformly convergent on a set SS SSS if for every ϵ > 0ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists an N ∈ NN ∈ N N inNN \in \mathbb{N}N∈N such that for all m > Nm > N m > Nm > Nm>N and for all x ∈ Sx ∈ S x in Sx \in Sx∈S,
| ∑ n = m + 1 ∞ f n ( x ) | < ϵ ∑ n = m + 1 ∞   f n ( x ) < ϵ |sum_(n=m+1)^(oo)f_(n)(x)| < epsilon\left| \sum_{n=m+1}^{\infty} f_n(x) \right| < \epsilon|∑n=m+1∞fn(x)|<ϵ

Approach

Step 1: Recognize the Series as a Geometric Series

Explanation

For each fixed xx xxx, the series is a geometric series. A geometric series is a series of the form ∑ n = 0 ∞ a ⋅ r n ∑ n = 0 ∞   a ⋅ r n sum_(n=0)^(oo)a*r^(n)\sum_{n=0}^{\infty} a \cdot r^n∑n=0∞a⋅rn, where aa aaa is the first term and rr rrr is the common ratio.

Application to Our Series

In our case, the first term a = x 4a = x 4 a=x^(4)a = x^4a=x4 and the common ratio r = 1 1 + x 4r = 1 1 + x 4 r=(1)/(1+x^(4))r = \frac{1}{1+x^4}r=11+x4.

Step 2: Find the n t h n t h n^(th)n^{th}nth Partial Sum s n ( x ) s n ( x ) s_(n)(x)s_n(x)sn(x)

Explanation

The n t h n t h n^(th)n^{th}nth partial sum of a geometric series is given by:
s n = a ( 1 − r n ) 1 − r s n = a ( 1 − r n ) 1 − r s_(n)=(a(1-r^(n)))/(1-r)s_n = \frac{a(1 – r^n)}{1 – r}sn=a(1−rn)1−r
if r ≠ 1r ≠ 1 r!=1r \neq 1r≠1.

Application to Our Series

For our series, the n t h n t h n^(th)n^{th}nth partial sum s n ( x ) s n ( x ) s_(n)(x)s_n(x)sn(x) becomes:
s n ( x ) = x 4 ( 1 − ( 1 1 + x 4 ) n ) 1 − 1 1 + x 4 = 1 + x 4 − 1 ( 1 + x 4 ) n − 1 s n ( x ) = x 4 ( 1 − 1 1 + x 4 n ) 1 − 1 1 + x 4 = 1 + x 4 − 1 ( 1 + x 4 ) n − 1 s_(n)(x)=(x^(4)(1-((1)/(1+x^(4)))^(n)))/(1-(1)/(1+x^(4)))=1+x^(4)-(1)/((1+x^(4))^(n-1))s_n(x) = \frac{x^4(1 – \left(\frac{1}{1+x^4}\right)^n)}{1 – \frac{1}{1+x^4}} = 1+x^4-\frac{1}{(1+x^4)^{n-1}}sn(x)=x4(1−(11+x4)n)1−11+x4=1+x4−1(1+x4)n−1

Step 3: Determine the Limit Function S ( x )S ( x ) S(x)S(x)S(x)

Explanation

The limit function S ( x )S ( x ) S(x)S(x)S(x) is defined as:
S ( x ) = lim n → ∞ s n ( x )S ( x ) = lim n → ∞   s n ( x ) S(x)=lim_(n rarr oo)s_(n)(x)S(x) = \lim_{n \rightarrow \infty} s_n(x)S(x)=limn→∞sn(x)

Application to Our Series

As n → ∞n → ∞ n rarr oon \to \inftyn→∞, s n ( x ) s n ( x ) s_(n)(x)s_n(x)sn(x) approaches:
S ( x ) = { 1 + x 4 , if x ≠ 0 0 , if x = 0S ( x ) = 1 + x 4 ,      if  x ≠ 0 0 ,      if  x = 0 S(x)={[1+x^(4)”,”,”if “x!=0],[0″,”,”if “x=0]:}S(x) = \begin{cases} 1+x^4, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}S(x)={1+x4,if x≠00,if x=0

Step 4: Check for Discontinuity

Explanation

A uniformly convergent series of continuous functions must converge to a continuous limit function.

Application to Our Series

The limit function S ( x )S ( x ) S(x)S(x)S(x) is discontinuous at x = 0x = 0 x=0x = 0x=0. This implies that the original series cannot be uniformly convergent, as a uniformly convergent series of continuous functions would converge to a continuous function.

Conclusion

By the definition of uniform convergence and the properties of geometric series, we conclude that the series ∑ n = 0 ∞ x 4 ( 1 + x 4 ) n ∑ n = 0 ∞   x 4 ( 1 + x 4 ) n sum_(n=0)^(oo)(x^(4))/((1+x^(4))^(n))\sum_{n=0}^{\infty} \frac{x^4}{(1+x^4)^n}∑n=0∞x4(1+x4)n is not uniformly convergent on the interval [ 0 , 1 ][ 0 , 1 ] [0,1][0,1][0,1] because its limit function S ( x )S ( x ) S(x)S(x)S(x) is discontinuous at x = 0x = 0 x=0x = 0x=0.
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1.(a) मान लीजिए कि S 3 S 3 S_(3)S_3S3 व Z 3 Z 3 Z_(3)Z_3Z3 क्रमशः 3 प्रतीकों का क्रमचय समूह एवं मॉड्यूल 3 अवशिष्ट वर्गों के समूह हैं। दर्शाइए कि S 3 S 3 S_(3)S_3S3 का Z 3 Z 3 Z_(3)Z_3Z3 में तुच्छ समाकारिता के अतिरिक्त कोई भी समाकारिता नहीं है ।
Let S 3 S 3 S_(3)S_3S3 and Z 3 Z 3 Z_(3)Z_3Z3 be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of S 3 S 3 S_(3)S_3S3 in Z 3 Z 3 Z_(3)Z_3Z3 except the trivial homomorphism.
Answer:
To show that there is no homomorphism of S 3 S 3 S_(3)S_3S3 into Z 3 Z 3 Z_(3)Z_3Z3 except the trivial homomorphism, we can use the following properties of homomorphisms:
  1. A homomorphism ϕ : G → Hϕ : G → H phi:G rarr H\phi: G \to Hϕ:G→H preserves the identity element, i.e., ϕ ( e G ) = e Hϕ ( e G ) = e H phi(e_(G))=e_(H)\phi(e_G) = e_Hϕ(eG)=eH.
  2. A homomorphism ϕ : G → Hϕ : G → H phi:G rarr H\phi: G \to Hϕ:G→H preserves the group operation, i.e., ϕ ( a ∗ b ) = ϕ ( a ) ∗ ϕ ( b )ϕ ( a ∗ b ) = ϕ ( a ) ∗ ϕ ( b ) phi(a**b)=phi(a)**phi(b)\phi(a \ast b) = \phi(a) \ast \phi(b)ϕ(a∗b)=ϕ(a)∗ϕ(b).
  3. A homomorphism ϕ : G → Hϕ : G → H phi:G rarr H\phi: G \to Hϕ:G→H preserves the order of elements, i.e., if aa aaa has order nn nnn in GG GGG, then ϕ ( a )ϕ ( a ) phi(a)\phi(a)ϕ(a) has order dividing nn nnn in HH HHH.

Properties of S 3 S 3 S_(3)S_3S3 and Z 3 Z 3 Z_(3)Z_3Z3

  1. S 3 S 3 S_(3)S_3S3 is the permutation group on 3 symbols, and it has 3 ! = 63 ! = 6 3!=63! = 63!=6 elements.
  2. Z 3 Z 3 Z_(3)Z_3Z3 is the group of residue classes modulo 3, and it has 3 elements: [ 0 ] , [ 1 ] , [ 2 ][ 0 ] , [ 1 ] , [ 2 ] [0],[1],[2][0], [1], [2][0],[1],[2].

Steps to Show No Non-Trivial Homomorphism Exists

  1. Identity Element: Any homomorphism ϕ : S 3 → Z 3ϕ : S 3 → Z 3 phi:S_(3)rarrZ_(3)\phi: S_3 \to Z_3ϕ:S3→Z3 must map the identity element of S 3 S 3 S_(3)S_3S3 (the identity permutation ee eee) to the identity element of Z 3 Z 3 Z_(3)Z_3Z3 ( [ 0 ][ 0 ] [0][0][0]).
    ϕ ( e ) = [ 0 ]ϕ ( e ) = [ 0 ] phi(e)=[0]\phi(e) = [0]ϕ(e)=[0]
  2. Order of Elements: The order of any element in Z 3 Z 3 Z_(3)Z_3Z3 divides 3. In S 3 S 3 S_(3)S_3S3, we have elements of order 2 (e.g., transpositions) and elements of order 3 (e.g., 3-cycles). If there exists a non-trivial homomorphism ϕϕ phi\phiϕ, then it must map elements of S 3 S 3 S_(3)S_3S3 to elements of Z 3 Z 3 Z_(3)Z_3Z3 in such a way that the order of the image divides the order of the original element.
    However, Z 3 Z 3 Z_(3)Z_3Z3 only has elements of order 1 ( [ 0 ][ 0 ] [0][0][0]) and order 3 ( [ 1 ] , [ 2 ][ 1 ] , [ 2 ] [1],[2][1], [2][1],[2]). There are no elements of order 2 in Z 3 Z 3 Z_(3)Z_3Z3.
  3. Contradiction: S 3 S 3 S_(3)S_3S3 contains elements of order 2 (transpositions). Any homomorphism ϕϕ phi\phiϕ would have to map these elements to an element in Z 3 Z 3 Z_(3)Z_3Z3 whose order divides 2. Since Z 3 Z 3 Z_(3)Z_3Z3 contains no such elements (other than the identity), we reach a contradiction.
Therefore, the only homomorphism that can exist from S 3 S 3 S_(3)S_3S3 to Z 3 Z 3 Z_(3)Z_3Z3 is the trivial homomorphism that maps all elements of S 3 S 3 S_(3)S_3S3 to the identity element [ 0 ][ 0 ] [0][0][0] in Z 3 Z 3 Z_(3)Z_3Z3.
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1.(b) मान लीजिए RR RRR मुख्य गुणजावली प्रान्त है । दर्शाइए कि RR RRR के विभाग-वलय की प्रत्येक गुणजावली, मुख्य गुणजावली है तथा R / P , RR / P , R R//P,RR / P, RR/P,R के अभाज्यगुणजावली PP PPP के लिए मुख्य गुणजावली प्रान्त है ।
Let RR RRR be a principal ideal domain. Show that every ideal of a quotient ring of RR RRR is principal ideal and R / PR / P R//PR / PR/P is a principal ideal domain for a prime ideal PP PPP of RR RRR.
Answer:

Introduction

The problem asks us to prove two things:
  1. Every ideal of a quotient ring R / PR / P R//PR/PR/P is a principal ideal.
  2. If PP PPP is a prime ideal of RR RRR, then R / PR / P R//PR/PR/P is a principal ideal domain (PID).
To prove these statements, we’ll use the properties of principal ideal domains and quotient rings.

Work/Calculations

Part 1: Every ideal of R / PR / P R//PR/PR/P is a principal ideal

Let I / PI / P I//PI/PI/P be an ideal of R / PR / P R//PR/PR/P, where II III is an ideal of RR RRR containing PP PPP.
Step 1: Show that II III is a principal ideal in RR RRR
Since RR RRR is a PID, II III is generated by a single element aa aaa in RR RRR. That is,
I = ( a )I = ( a ) I=(a)I = (a)I=(a)
Step 2: Show that I / PI / P I//PI/PI/P is generated by a + Pa + P a+Pa+Pa+P in R / PR / P R//PR/PR/P
Let’s substitute the values:
I / P = ( a ) + PI / P = ( a ) + P I//P=(a)+PI/P = (a) + PI/P=(a)+P
After substituting, we can see that I / PI / P I//PI/PI/P is generated by a + Pa + P a+Pa+Pa+P in R / PR / P R//PR/PR/P.
Therefore, I / PI / P I//PI/PI/P is a principal ideal in R / PR / P R//PR/PR/P.

Part 2: R / PR / P R//PR/PR/P is a PID for a prime ideal PP PPP of RR RRR

Step 1: Show that R / PR / P R//PR/PR/P is an integral domain
Since PP PPP is a prime ideal, R / PR / P R//PR/PR/P is an integral domain.
Step 2: Show that every ideal in R / PR / P R//PR/PR/P is principal
From Part 1, we know that every ideal in R / PR / P R//PR/PR/P is principal.
Step 3: Conclude that R / PR / P R//PR/PR/P is a PID
Since R / PR / P R//PR/PR/P is an integral domain and every ideal in R / PR / P R//PR/PR/P is principal, R / PR / P R//PR/PR/P is a PID.

Conclusion

We have shown that every ideal of a quotient ring R / PR / P R//PR/PR/P is a principal ideal. Additionally, if PP PPP is a prime ideal of RR RRR, then R / PR / P R//PR/PR/P is a principal ideal domain. Both of these statements hold true when RR RRR is a principal ideal domain.
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1.(a) मान लीजिए कि GG GGG एक परिमित समूह है और HH HHH तथा K , GK , G K,GK, GK,G के उप-समूह हैं, ऐसा कि K ⊂ HK ⊂ H K sub HK \subset HK⊂H । दर्शाइए ( G : K ) = ( G : H ) ( H : K )( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K)=(G: H)(H: K)(G:K)=(G:H)(H:K) ।
Let GG GGG be a finite group, HH HHH and KK KKK subgroups of GG GGG such that K ⊂ HK ⊂ H K sub HK \subset HK⊂H. Show that ( G : K ) = ( G : H ) ( H : K )( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K)=(G: H)(H: K)(G:K)=(G:H)(H:K).
Answer:

Introduction:

We are given a finite group GG GGG and two subgroups HH HHH and KK KKK such that K ⊂ HK ⊂ H K sub HK \subset HK⊂H. We are asked to prove that ( G : K ) = ( G : H ) ( H : K )( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)(G:K)=(G:H)(H:K), where ( G : K )( G : K ) (G:K)(G: K)(G:K), ( G : H )( G : H ) (G:H)(G: H)(G:H), and ( H : K )( H : K ) (H:K)(H: K)(H:K) are the indices of the subgroups KK KKK, HH HHH, and KK KKK in GG GGG, GG GGG, and HH HHH respectively.

Work/Calculations:

Definition of Index:

The index ( A : B )( A : B ) (A:B)(A: B)(A:B) of a subgroup BB BBB in a group AA AAA is defined as the number of distinct left cosets of BB BBB in AA AAA. Mathematically, ( A : B ) = | A / B |( A : B ) = | A / B | (A:B)=|A//B|(A: B) = |A/B|(A:B)=|A/B|, where | A / B || A / B | |A//B||A/B||A/B| is the number of distinct left cosets.

Step 1: Express ( G : K )( G : K ) (G:K)(G: K)(G:K) in terms of cosets

The index ( G : K )( G : K ) (G:K)(G: K)(G:K) is the number of distinct left cosets of KK KKK in GG GGG. Let’s denote this set of cosets as G / KG / K G//KG/KG/K.

Step 2: Express ( G : H )( G : H ) (G:H)(G: H)(G:H) and ( H : K )( H : K ) (H:K)(H: K)(H:K) in terms of cosets

Similarly, ( G : H )( G : H ) (G:H)(G: H)(G:H) is the number of distinct left cosets of HH HHH in GG GGG, denoted as G / HG / H G//HG/HG/H, and ( H : K )( H : K ) (H:K)(H: K)(H:K) is the number of distinct left cosets of KK KKK in HH HHH, denoted as H / KH / K H//KH/KH/K.

Step 3: Relate G / KG / K G//KG/KG/K with G / HG / H G//HG/HG/H and H / KH / K H//KH/KH/K

Each coset g Kg K gKgKgK in G / KG / K G//KG/KG/K can be uniquely expressed as h Kh K hKhKhK where hh hhh is in some coset g Hg H gHgHgH in G / HG / H G//HG/HG/H. Furthermore, each coset g Hg H gHgHgH in G / HG / H G//HG/HG/H contains exactly ( H : K )( H : K ) (H:K)(H: K)(H:K) distinct cosets of the form h Kh K hKhKhK in H / KH / K H//KH/KH/K.
Therefore, the total number of distinct cosets g Kg K gKgKgK in G / KG / K G//KG/KG/K can be obtained by multiplying the number of distinct cosets g Hg H gHgHgH in G / HG / H G//HG/HG/H by the number of distinct cosets h Kh K hKhKhK in H / KH / K H//KH/KH/K.

Step 4: Mathematical Expression

This relationship can be mathematically expressed as:
( G : K ) = ( G : H ) ( H : K )( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)(G:K)=(G:H)(H:K)

Conclusion:

We have successfully proven that ( G : K ) = ( G : H ) ( H : K )( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)(G:K)=(G:H)(H:K) by relating the number of distinct left cosets of KK KKK in GG GGG with the number of distinct left cosets of HH HHH in GG GGG and KK KKK in HH HHH. This proves the statement for finite groups GG GGG and subgroups HH HHH and KK KKK such that K ⊂ HK ⊂ H K sub HK \subset HK⊂H.
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1.(b) दर्शाईए कि फलन
f ( x , y ) = { x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 )f ( x , y ) = x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 ) f(x,y)={[(x^(2)-y^(2))/(x-y)”,”,(x”,”y)!=(1″,”-1)”,”(1″,”1)],[0″,”,(x”,”y)=(1″,”1)”,”(1″,”-1)]:}f(x, y)=\left\{\begin{array}{cl} \frac{x^2-y^2}{x-y}, & (x, y) \neq(1,-1),(1,1) \\ 0, & (x, y)=(1,1),(1,-1) \end{array}\right.f(x,y)={x2−y2x−y,(x,y)≠(1,−1),(1,1)0,(x,y)=(1,1),(1,−1)
संतत और बिन्दु ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1,-1)(1,−1) पर अवकलनीय है ।
Show that the function
f ( x , y ) = { x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 )f ( x , y ) = x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 ) f(x,y)={[(x^(2)-y^(2))/(x-y)”,”,(x”,”y)!=(1″,”-1)”,”(1″,”1)],[0″,”,(x”,”y)=(1″,”1)”,”(1″,”-1)]:}f(x, y)=\left\{\begin{array}{cc} \frac{x^2-y^2}{x-y}, & (x, y) \neq(1,-1),(1,1) \\ 0, & (x, y)=(1,1),(1,-1) \end{array}\right.f(x,y)={x2−y2x−y,(x,y)≠(1,−1),(1,1)0,(x,y)=(1,1),(1,−1)
is continuous and differentiable at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1,-1)(1,−1).
Answer:

Introduction:

We are given a function f ( x , y )f ( x , y ) f(x,y)f(x, y)f(x,y) defined as follows:
f ( x , y ) = { x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 )f ( x , y ) = x 2 − y 2 x − y , ( x , y ) ≠ ( 1 , − 1 ) , ( 1 , 1 ) 0 , ( x , y ) = ( 1 , 1 ) , ( 1 , − 1 ) f(x,y)={[(x^(2)-y^(2))/(x-y)”,”,(x”,”y)!=(1″,”-1)”,”(1″,”1)],[0″,”,(x”,”y)=(1″,”1)”,”(1″,”-1)]:}f(x, y)=\left\{\begin{array}{cc} \frac{x^2-y^2}{x-y}, & (x, y) \neq(1,-1),(1,1) \\ 0, & (x, y)=(1,1),(1,-1) \end{array}\right.f(x,y)={x2−y2x−y,(x,y)≠(1,−1),(1,1)0,(x,y)=(1,1),(1,−1)
We are asked to show that this function is continuous and differentiable at the point ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1).

Work/Calculations:

Step 1: Checking Continuity at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1)

To check for continuity, we need to show that:
lim ( x , y ) → ( 1 , − 1 ) f ( x , y ) = f ( 1 , − 1 ) lim ( x , y ) → ( 1 , − 1 )   f ( x , y ) = f ( 1 , − 1 ) lim_((x,y)rarr(1,-1))f(x,y)=f(1,-1)\lim_{{(x, y) \to (1, -1)}} f(x, y) = f(1, -1)lim(x,y)→(1,−1)f(x,y)=f(1,−1)
First, let’s find f ( 1 , − 1 )f ( 1 , − 1 ) f(1,-1)f(1, -1)f(1,−1):
f ( 1 , − 1 ) = 0f ( 1 , − 1 ) = 0 f(1,-1)=0f(1, -1) = 0f(1,−1)=0
Now, let’s find the limit as ( x , y )( x , y ) (x,y)(x, y)(x,y) approaches ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1):
lim ( x , y ) → ( 1 , − 1 ) x 2 − y 2 x − y lim ( x , y ) → ( 1 , − 1 )   x 2 − y 2 x − y lim_((x,y)rarr(1,-1))(x^(2)-y^(2))/(x-y)\lim_{{(x, y) \to (1, -1)}} \frac{x^2 – y^2}{x – y}lim(x,y)→(1,−1)x2−y2x−y
We can simplify the expression x 2 − y 2 x − y x 2 − y 2 x − y (x^(2)-y^(2))/(x-y)\frac{x^2 – y^2}{x – y}x2−y2x−y as x + yx + y x+yx + yx+y when x ≠ yx ≠ y x!=yx \neq yx≠y. Now, the limit becomes:
lim ( x , y ) → ( 1 , − 1 ) ( x + y ) = 1 + ( − 1 ) = 0 lim ( x , y ) → ( 1 , − 1 )   ( x + y ) = 1 + ( − 1 ) = 0 lim_((x,y)rarr(1,-1))(x+y)=1+(-1)=0\lim_{{(x, y) \to (1, -1)}} (x + y) = 1 + (-1) = 0lim(x,y)→(1,−1)(x+y)=1+(−1)=0
Since lim ( x , y ) → ( 1 , − 1 ) f ( x , y ) = f ( 1 , − 1 ) = 0 lim ( x , y ) → ( 1 , − 1 )   f ( x , y ) = f ( 1 , − 1 ) = 0 lim_((x,y)rarr(1,-1))f(x,y)=f(1,-1)=0\lim_{{(x, y) \to (1, -1)}} f(x, y) = f(1, -1) = 0lim(x,y)→(1,−1)f(x,y)=f(1,−1)=0, the function is continuous at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1).

Step 2: Checking Differentiability at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1)

To check for differentiability, we need to find the partial derivatives ∂ f ∂ x ∂ f ∂ x (del f)/(del x)\frac{\partial f}{\partial x}∂f∂x and ∂ f ∂ y ∂ f ∂ y (del f)/(del y)\frac{\partial f}{\partial y}∂f∂y and check if they are continuous at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1).
The partial derivatives are:
∂ f ∂ x = 1 + y , ∂ f ∂ y = − ( 1 + x ) ∂ f ∂ x = 1 + y , ∂ f ∂ y = − ( 1 + x ) (del f)/(del x)=1+y,quad(del f)/(del y)=-(1+x)\frac{\partial f}{\partial x} = 1 + y, \quad \frac{\partial f}{\partial y} = -(1 + x)∂f∂x=1+y,∂f∂y=−(1+x)
Now, let’s check their continuity at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1):
lim ( x , y ) → ( 1 , − 1 ) ∂ f ∂ x = 1 + ( − 1 ) = 0 lim ( x , y ) → ( 1 , − 1 )   ∂ f ∂ x = 1 + ( − 1 ) = 0 lim_((x,y)rarr(1,-1))(del f)/(del x)=1+(-1)=0\lim_{{(x, y) \to (1, -1)}} \frac{\partial f}{\partial x} = 1 + (-1) = 0lim(x,y)→(1,−1)∂f∂x=1+(−1)=0
lim ( x , y ) → ( 1 , − 1 ) ∂ f ∂ y = − ( 1 + 1 ) = − 2 lim ( x , y ) → ( 1 , − 1 )   ∂ f ∂ y = − ( 1 + 1 ) = − 2 lim_((x,y)rarr(1,-1))(del f)/(del y)=-(1+1)=-2\lim_{{(x, y) \to (1, -1)}} \frac{\partial f}{\partial y} = -(1 + 1) = -2lim(x,y)→(1,−1)∂f∂y=−(1+1)=−2
Since both partial derivatives are continuous at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1), the function f ( x , y )f ( x , y ) f(x,y)f(x, y)f(x,y) is differentiable at ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1).

Conclusion:

We have shown that the function f ( x , y )f ( x , y ) f(x,y)f(x, y)f(x,y) is continuous and differentiable at the point ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1) by proving that the limit of the function as ( x , y )( x , y ) (x,y)(x, y)(x,y) approaches ( 1 , − 1 )( 1 , − 1 ) (1,-1)(1, -1)(1,−1) is equal to f ( 1 , − 1 )f ( 1 , − 1 ) f(1,-1)f(1, -1)f(1,−1) and that the partial derivatives are continuous at that point.
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1.(a) मान लीजिए RR RRR तत्समक अवयव सहित एक पूर्णांकीय प्रांत है । दर्शाइए कि R [ x ]R [ x ] R[x]R[x]R[x] में कोई भी एवक RR RRR में एक एकक है।
Let RR RRR be an integral domain with unit element. Show that any unit in R [ x ]R [ x ] R[x]R[x]R[x] is a unit in RR RRR.
Answer:

Introduction

In this problem, we are dealing with the concept of units in the context of integral domains and polynomial rings. Specifically, we are given an integral domain RR RRR with a unit element and are asked to prove that any unit in R [ x ]R [ x ] R[x]R[x]R[x] (the polynomial ring over RR RRR) must also be a unit in RR RRR.

Definitions

  • Integral Domain: An integral domain is a commutative ring RR RRR with a multiplicative identity (unit element) such that the product of any two non-zero elements is non-zero.
  • Unit in a Ring: An element aa aaa in a ring RR RRR is called a unit if there exists an element bb bbb in RR RRR such that a ⋅ b = b ⋅ a = 1a ⋅ b = b ⋅ a = 1 a*b=b*a=1a \cdot b = b \cdot a = 1a⋅b=b⋅a=1, where 11 111 is the multiplicative identity in RR RRR.
  • Polynomial Ring R [ x ]R [ x ] R[x]R[x]R[x]: The set of all polynomials with coefficients in RR RRR.

Work/Calculations

Step 1: Assume f ( x )f ( x ) f(x)f(x)f(x) is a Unit in R [ x ]R [ x ] R[x]R[x]R[x]

Let’s assume that f ( x )f ( x ) f(x)f(x)f(x) is a unit in R [ x ]R [ x ] R[x]R[x]R[x]. This means there exists a polynomial g ( x )g ( x ) g(x)g(x)g(x) in R [ x ]R [ x ] R[x]R[x]R[x] such that:
f ( x ) ⋅ g ( x ) = 1f ( x ) ⋅ g ( x ) = 1 f(x)*g(x)=1f(x) \cdot g(x) = 1f(x)⋅g(x)=1

Step 2: Examine the Degrees of f ( x )f ( x ) f(x)f(x)f(x) and g ( x )g ( x ) g(x)g(x)g(x)

The degree of the polynomial f ( x ) ⋅ g ( x )f ( x ) ⋅ g ( x ) f(x)*g(x)f(x) \cdot g(x)f(x)⋅g(x) is the sum of the degrees of f ( x )f ( x ) f(x)f(x)f(x) and g ( x )g ( x ) g(x)g(x)g(x). Since f ( x ) ⋅ g ( x ) = 1f ( x ) ⋅ g ( x ) = 1 f(x)*g(x)=1f(x) \cdot g(x) = 1f(x)⋅g(x)=1, a constant polynomial, the degree of f ( x ) ⋅ g ( x )f ( x ) ⋅ g ( x ) f(x)*g(x)f(x) \cdot g(x)f(x)⋅g(x) is 0.
Let deg ( f ( x ) ) = mdeg ( f ( x ) ) = m “deg”(f(x))=m\text{deg}(f(x)) = mdeg(f(x))=m and deg ( g ( x ) ) = ndeg ( g ( x ) ) = n “deg”(g(x))=n\text{deg}(g(x)) = ndeg(g(x))=n.
deg ( f ( x ) ⋅ g ( x ) ) = m + n = 0deg ( f ( x ) ⋅ g ( x ) ) = m + n = 0 “deg”(f(x)*g(x))=m+n=0\text{deg}(f(x) \cdot g(x)) = m + n = 0deg(f(x)⋅g(x))=m+n=0

Step 3: Conclude m = n = 0m = n = 0 m=n=0m = n = 0m=n=0

Since m + n = 0m + n = 0 m+n=0m + n = 0m+n=0 and m , n ≥ 0m , n ≥ 0 m,n >= 0m, n \geq 0m,n≥0, it must be the case that m = n = 0m = n = 0 m=n=0m = n = 0m=n=0.

Step 4: Show f ( x )f ( x ) f(x)f(x)f(x) is a Unit in RR RRR

Since m = 0m = 0 m=0m = 0m=0, f ( x )f ( x ) f(x)f(x)f(x) is a constant polynomial, say f ( x ) = af ( x ) = a f(x)=af(x) = af(x)=a, where aa aaa is in RR RRR. Similarly, g ( x ) = bg ( x ) = b g(x)=bg(x) = bg(x)=b, where bb bbb is in RR RRR.
From f ( x ) ⋅ g ( x ) = 1f ( x ) ⋅ g ( x ) = 1 f(x)*g(x)=1f(x) \cdot g(x) = 1f(x)⋅g(x)=1, we have a ⋅ b = 1a ⋅ b = 1 a*b=1a \cdot b = 1a⋅b=1.
This shows that aa aaa is a unit in RR RRR, as a ⋅ b = 1a ⋅ b = 1 a*b=1a \cdot b = 1a⋅b=1.

Conclusion

We have shown that if f ( x )f ( x ) f(x)f(x)f(x) is a unit in R [ x ]R [ x ] R[x]R[x]R[x], then it must be a constant polynomial and also a unit in RR RRR. Therefore, any unit in R [ x ]R [ x ] R[x]R[x]R[x] is a unit in RR RRR.
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1.(b) असमिका : π 2 9 < ∫ π 6 π 2 x sin ⁡ x d x < 2 π 2 9 π 2 9 < ∫ π 6 π 2   x sin ⁡ x d x < 2 π 2 9 (pi^(2))/(9) < int_((pi)/(6))^((pi)/(2))(x)/(sin x)dx < (2pi^(2))/(9)\frac{\pi^2}{9}<\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{x}{\sin x} d x<\frac{2 \pi^2}{9}π29<∫π6π2xsin⁡xdx<2π29 को सिद्ध कीजिए ।
Prove the inequality : π 2 9 < ∫ π 6 π 2 x sin ⁡ x d x < 2 π 2 9 π 2 9 < ∫ π 6 π 2   x sin ⁡ x d x < 2 π 2 9 (pi^(2))/(9) < int_((pi)/(6))^((pi)/(2))(x)/(sin x)dx < (2pi^(2))/(9)\frac{\pi^2}{9}<\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{x}{\sin x} d x<\frac{2 \pi^2}{9}π29<∫π6π2xsin⁡xdx<2π29,
Answer:
Introduction:
We have the inequality:
1 ≤ 1 sin ⁡ x ≤ 2 for all x ∈ [ π 6 , π 2 ] .1 ≤ 1 sin ⁡ x ≤ 2  for all  x ∈ π 6 , π 2 .  1 <= (1)/(sin x) <= 2″ for all “x in[(pi)/(6),(pi)/(2)]”. “1 \leq \frac{1}{\sin x} \leq 2 \text { for all } x \in\left[\frac{\pi}{6}, \frac{\pi}{2}\right] \text {. }1≤1sin⁡x≤2 for all x∈[π6,π2]. 
Multiply by xx xxx
Therefore, it follows that:
x ≤ x sin ⁡ x ≤ 2 x for all x ∈ [ π 6 , π 2 ]x ≤ x sin ⁡ x ≤ 2 x  for all  x ∈ π 6 , π 2 x <= (x)/(sin x) <= 2x” for all “x in[(pi)/(6),(pi)/(2)]x \leq \frac{x}{\sin x} \leq 2x \text { for all } x \in\left[\frac{\pi}{6}, \frac{\pi}{2}\right]x≤xsin⁡x≤2x for all x∈[π6,π2]
Step 1: Defining f ( x )f ( x ) f(x)f(x)f(x):
Let’s define a function f ( x ) = x sin ⁡ xf ( x ) = x sin ⁡ x f(x)=(x)/(sin x)f(x)=\frac{x}{\sin x}f(x)=xsin⁡x.
Step 2: Defining ϕ ( x )ϕ ( x ) phi(x)\phi(x)ϕ(x) and ψ ( x )ψ ( x ) psi(x)\psi(x)ψ(x):
We define two more functions as follows:
ϕ ( x ) = x ψ ( x ) = 2 x , x ∈ [ π / 6 , π / 2 ] ϕ ( x ) = x ψ ( x ) = 2 x , x ∈ [ π / 6 , π / 2 ] {:[phi(x)=x],[psi(x)=2x”,”quad x in[pi//6″,”pi//2]]:}\begin{aligned} & \phi(x)=x \\ & \psi(x)=2x, \quad x \in[\pi / 6, \pi / 2] \end{aligned}ϕ(x)=xψ(x)=2x,x∈[π/6,π/2]
Step 3: Boundedness and Integrability:
Both ff fff and ϕϕ phi\phiϕ are bounded and integrable on [ π / 6 , π / 2 ][ π / 6 , π / 2 ] [pi//6,pi//2][\pi / 6, \pi / 2][π/6,π/2], and f ( x ) ≥ ϕ ( x )f ( x ) ≥ ϕ ( x ) f(x) >= phi(x)f(x) \geq \phi(x)f(x)≥ϕ(x) for all x ∈ [ π / 6 , π / 2 ]x ∈ [ π / 6 , π / 2 ] x in[pi//6,pi//2]x \in[\pi / 6, \pi / 2]x∈[π/6,π/2].
Step 4: Continuity at π / 3π / 3 pi//3\pi / 3π/3:
Furthermore, ff fff and ϕϕ phi\phiϕ are both continuous at x = π / 3x = π / 3 x=pi//3x = \pi / 3x=π/3, and f ( π / 3 ) > ϕ ( π / 3 )f ( π / 3 ) > ϕ ( π / 3 ) f(pi//3) > phi(pi//3)f(\pi / 3)>\phi(\pi / 3)f(π/3)>ϕ(π/3).
Step 5: Integral Comparison:
Hence, we can compare the integrals:
∫ π / 6 π / 2 f ( x ) d x > ∫ π / 6 π / 2 ϕ ( x ) d x = ∫ π / 6 π / 2 x d x = π 2 9 ∫ π / 6 π / 2   f ( x ) d x > ∫ π / 6 π / 2   ϕ ( x ) d x = ∫ π / 6 π / 2   x d x = π 2 9 {:[int_(pi//6)^(pi//2)f(x)dx > int_(pi//6)^(pi//2)phi(x)dx],[=int_(pi//6)^(pi//2)xdx],[=(pi^(2))/(9)]:}\begin{aligned} & \int_{\pi / 6}^{\pi / 2} f(x) d x>\int_{\pi / 6}^{\pi / 2} \phi(x) d x \\ & =\int_{\pi / 6}^{\pi / 2} x d x \\ & =\frac{\pi^2}{9} \end{aligned}∫π/6π/2f(x)dx>∫π/6π/2ϕ(x)dx=∫π/6π/2xdx=π29
Step 6: Comparing with ψψ psi\psiψ:
Similarly, we have:
∫ π / 6 π / 2 f ( x ) d x < ∫ π / 6 π / 2 ψ ( x ) d x = 2 ∫ π / 6 π / 2 x d x = 2 π 2 9 ∫ π / 6 π / 2   f ( x ) d x < ∫ π / 6 π / 2   ψ ( x ) d x = 2 ∫ π / 6 π / 2   x d x = 2 π 2 9 {:[int_(pi//6)^(pi//2)f(x)dx < int_(pi//6)^(pi//2)psi(x)dx],[=2int_(pi//6)^(pi//2)xdx],[=(2pi^(2))/(9)]:}\begin{aligned} \int_{\pi / 6}^{\pi / 2} f(x) d x & <\int_{\pi / 6}^{\pi / 2} \psi(x) d x \\ & =2 \int_{\pi / 6}^{\pi / 2} x d x \\ & =\frac{2 \pi^2}{9} \end{aligned}∫π/6π/2f(x)dx<∫π/6π/2ψ(x)dx=2∫π/6π/2xdx=2π29
Conclusion:
Consequently, we can conclude that:
π 2 9 < ∫ π / 6 π / 2 x sin ⁡ x d x < 2 π 2 9 π 2 9 < ∫ π / 6 π / 2   x sin ⁡ x d x < 2 π 2 9 (pi^(2))/(9) < int_(pi//6)^(pi//2)(x)/(sin x)dx < (2pi^(2))/(9)\frac{\pi^2}{9}<\int_{\pi / 6}^{\pi / 2} \frac{x}{\sin x} d x<\frac{2 \pi^2}{9}π29<∫π/6π/2xsin⁡xdx<2π29
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)
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