IGNOU MMT-004 Solved Assignment 2024 for M.Sc. MACS

IGNOU MMT-004 Solved Assignment 2024 | M.Sc. MACS

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

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IGNOU MMT-004 Assignment Question Paper 2024

mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e

mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e

MMT-003 Solved Assignment 2024 QP
  1. a) Let X = C [ 0 , 1 ] X = C [ 0 , 1 ] X=C[0,1]\mathrm{X}=\mathrm{C}[0,1]X=C[0,1]. Define d : X × X R d : X × X R d:X xx X rarrRd: X \times X \rightarrow \mathbf{R}d:X×XR by d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginX\mathrm{d}(\mathrm{f}, \mathrm{g})=\int_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})| \mathrm{dt}, \mathrm{f}, \mathrm{g} \in \mathrm{X}d(f,g)=01|f(t)g(t)|dt,f,gX where the integral is the Riemann integral. Show that d d ddd is a metric on X X XXX. Find d ( f , g ) d ( f , g ) d(f,g)d(f, g)d(f,g) where f ( x ) = 4 x f ( x ) = 4 x f(x)=4xf(x)=4 xf(x)=4x and g ( x ) = x 3 , x [ 0 , 1 ] g ( x ) = x 3 , x [ 0 , 1 ] g(x)=x^(3),x in[0,1]g(x)=x^3, x \in[0,1]g(x)=x3,x[0,1].
b) Let ( X , d ) X , d ) X,d)X, d)X,d) be a metric space and a X a X a in Xa \in XaX be a fixed point of X X XXX. Show that the function f a : X R f a : X R f_(a):X rarrRf_a: X \rightarrow \mathbf{R}fa:XR given by f a ( x ) = d ( x , a ) f a ( x ) = d ( x , a ) f_(a)(x)=d(x,a)\mathrm{f}_{\mathrm{a}}(\mathrm{x})=\mathrm{d}(\mathrm{x}, \mathrm{a})fa(x)=d(x,a) is continuous. Is it uniformly continuous? Justify you answer.
2. a) Let A A AAA and B B BBB be any two subsets of a metric space (X, d), then show that
i) int A = { E A = { E A=uu{E\mathrm{A}=\cup\{\mathrm{E}A={E : is open and E A } E A } EsubeA}\mathrm{E} \subseteq \mathrm{A}\}EA}
ii) int ( A B ) = int A int B int ( A B ) = int A int B int(AnnB)=int Ann int B\operatorname{int}(\mathrm{A} \cap \mathrm{B})=\operatorname{int} \mathrm{A} \cap \operatorname{int} \mathrm{B}int(AB)=intAintB
iii) int ( A B ) int ( A B ) quad int(A uu B)supe\quad \operatorname{int}(A \cup B) \supseteqint(AB) int A A A nnA \capA int B B BBB
iv) A B ¯ A ¯ B ¯ A B ¯ A ¯ B ¯ bar(AnnB)sube bar(A)nn bar(B)\overline{\mathrm{A} \cap \mathrm{B}} \subseteq \overline{\mathrm{A}} \cap \overline{\mathrm{B}}AB¯A¯B¯.
b) Find the interior, boundary and closure of the following sets A A A\mathbf{A}A in R R R\mathbf{R}R with the usual metric and discrete metric.
i) A = Q A = Q A=QA=\mathbf{Q}A=Q, the set of rationals in R R R\mathbf{R}R
ii) A = ] 1 , 2 ] ] 2 , 4 [ A = ] 1 , 2 ] ] 2 , 4 [ A=]1,2]uu]2,4[\mathrm{A}=] 1,2] \cup] 2,4[A=]1,2]]2,4[
3. a) Let ( X , d 1 ) X , d 1 (X,d_(1))\left(X, d_1\right)(X,d1) and ( Y , d 2 ) Y , d 2 (Y,d_(2))\left(Y, d_2\right)(Y,d2) be metric spaces. Show that f : X Y f : X Y f:X rarr Yf: X \rightarrow Yf:XY is continuous if and only if f ( A ¯ ) f ( A ) ¯ f ( A ¯ ) f ( A ) ¯ f( bar(A))sube bar(f(A))f(\bar{A}) \subseteq \overline{f(A)}f(A¯)f(A)¯ where A A AAA is any subset of X X XXX
b) Let ( X 1 , d 1 ) X 1 , d 1 (X_(1),d_(1))\left(\mathrm{X}_1, \mathrm{~d}_1\right)(X1, d1) and ( X 2 , d 2 ) X 2 , d 2 (X_(2),d_(2))\left(\mathrm{X}_2, \mathrm{~d}_2\right)(X2, d2) be two discrete metric spaces. Verify that the product metric on X 1 × X 2 X 1 × X 2 X_(1)xxX_(2)\mathrm{X}_1 \times \mathrm{X}_2X1×X2 is discrete.
c) Show that an infinite discrete metric space X X X\mathrm{X}X is bounded but not totally bounded.
4. a) Find the first derivative f ( a ) f ( a ) f^(‘)(a)\mathrm{f}^{\prime}(\mathbf{a})f(a) of the function f f f\mathrm{f}f defined by f : R 3 R 2 f : R 3 R 2 f:R^(3)rarrR^(2)f: \mathbf{R}^3 \rightarrow \mathbf{R}^2f:R3R2 given by f ( x , y , z ) = ( x y z , x + y + z 2 ) f ( x , y , z ) = x y z , x + y + z 2 f(x,y,z)=(xyz,x+y+z^(2))f(x, y, z)=\left(x y z, x+y+z^2\right)f(x,y,z)=(xyz,x+y+z2) where a = ( 1 . 1 , 2 ) a = ( 1 . 1 , 2 ) a=(1.-1,2)\mathbf{a}=(1 .-1,2)a=(1.1,2).
b) Let E E E\mathrm{E}E be an open subset of R n R n R^(n)\mathbf{R}^nRn and f : E R m f : E R m f:E rarrR^(m)f: E \rightarrow \mathbf{R}^mf:ERm be a function such that each of its components function f i f i f_(i)f_ifi are differentiable, then show that f f fff is differentiable. Is the converse of this result true? Justify your answer.
c) Near what points may the surface z 2 + x z + y = 0 z 2 + x z + y = 0 z^(2)+xz+y=0z^2+x z+y=0z2+xz+y=0 be represented uniquely as a graph of a differentiable function z = k ( x , y ) z = k ( x , y ) z=k(x,y)\mathrm{z}=\mathrm{k}(\mathrm{x}, \mathrm{y})z=k(x,y) ? Locate such a point.
5. a) Use the method of Lagrange’s multiplier method to find the shortest possible distance from the ellipse x 2 + 2 y 2 = 2 x 2 + 2 y 2 = 2 x^(2)+2y^(2)=2x^2+2 y^2=2x2+2y2=2 to the line x + y = 2 x + y = 2 x+y=2x+y=2x+y=2.
b) Find the directional derivative of the function f : R 4 R 3 f : R 4 R 3 f:R^(4)rarrR^(3)f: \mathbf{R}^4 \rightarrow \mathbf{R}^3f:R4R3 defined by
f ( x , y , z , w ) = ( x 2 y , x y z , x 2 + y 2 + z 2 ) f ( x , y , z , w ) = x 2 y , x y z , x 2 + y 2 + z 2 f(x,y,z,w)=(x^(2)y,xyz,x^(2)+y^(2)+z^(2))f(x, y, z, w)=\left(x^2 y, x y z, x^2+y^2+z^2\right)f(x,y,z,w)=(x2y,xyz,x2+y2+z2)
at a = ( 1 , 2 , 1 , 2 ) a = ( 1 , 2 , 1 , 2 ) a=(1,2,-1,-2)\mathrm{a}=(1,2,-1,-2)a=(1,2,1,2) in the direction v = ( 0 , 1 , 2 , 2 ) v = ( 0 , 1 , 2 , 2 ) v=(0,1,2,-2)\mathrm{v}=(0,1,2,-2)v=(0,1,2,2).
6. a) Let A be a compact non-empty subset of a metric space (X, d) and let F be a closed subset of X X XXX such that A F = ϕ A F = ϕ A nn F=phiA \cap F=\phiAF=ϕ, then show that d ( A , F ) > 0 d ( A , F ) > 0 d(A,F) > 0d(A, F)>0d(A,F)>0 where d ( A , F ) = inf { d ( a , b ) : a A , b F } d ( A , F ) = inf { d ( a , b ) : a A , b F } d(A,F)=i n f{d(a,b):a in A,b in F}d(A, F)=\inf \{d(a, b): a \in A, b \in F\}d(A,F)=inf{d(a,b):aA,bF}.
b) Give an example of the following with justification
i) A vector-valued function f : R 3 R 3 f : R 3 R 3 f:R^(3)rarrR^(3)f: \mathbf{R}^3 \rightarrow \mathbf{R}^3f:R3R3 which is not differentiable at ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0).
ii) A function which is Legesgue measurable on R R R\mathbf{R}R.
c) Show that the components of a metric space is either identical or pairwise disjoint.
7. a) Let Q Q Q\mathbf{Q}Q be the set of rationals with the metric defined on Q Q Q\mathbf{Q}Q by d : Q × Q R d : Q × Q R d:QxxQrarrRd: \mathbf{Q} \times \mathbf{Q} \rightarrow \mathbf{R}d:Q×QR, defined by d ( x , y ) = | x y | , x , y R d ( x , y ) = | x y | , x , y R d(x,y)=|x-y|,AA x,y inRd(x, y)=|x-y|, \forall x, y \in \mathbf{R}d(x,y)=|xy|,x,yR.
Show that { ( 1 + 1 n ) n } 1 + 1 n n {(1+(1)/(n))^(n)}\left\{\left(1+\frac{1}{\mathrm{n}}\right)^{\mathrm{n}}\right\}{(1+1n)n} is Cauchy sequence in Q Q Q\mathbf{Q}Q, but does not converge in Q Q Q\mathbf{Q}Q and { 1 3 n } 1 3 n {(1)/(3^(n))}\left\{\frac{1}{3^n}\right\}{13n} is a Cauchy sequence Q Q Q\mathbf{Q}Q which converges in Q Q Q\mathbf{Q}Q to the limit 0 .
b) Which of the following sets are totally bounded? Give reasons for your answer. Are they compact?
i) 2 N 2 N quad2N\quad 2 \mathbf{N}2N in ( N , d ) ( N , d ) (N,d)(\mathbf{N}, d)(N,d) where d d ddd is the discrete metric.
ii) [ 0 , 2 ] [ 5 , 10 ] [ 0 , 2 ] [ 5 , 10 ] quad[0,2]uu[5,10]\quad[0,2] \cup[5,10][0,2][5,10] in ( R , d ) ( R , d ) (R,d)(\mathbf{R}, d)(R,d) where d d ddd is the Euclidean metric.
c) Which of the following sets are connected sets in R 2 R 2 R^(2)\mathbf{R}^2R2 with the metric given against it? Justify your answer.
i) A = { ( x , y ) : 0 x 1 , 0 y 2 } A = { ( x , y ) : 0 x 1 , 0 y 2 } quadA={(x,y):0 <= x <= 1,0 <= y <= 2}\quad \mathrm{A}=\{(\mathrm{x}, \mathrm{y}): 0 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 2\}A={(x,y):0x1,0y2} under the standard metric.
ii) A = { ( x , y ) : x 2 + y 2 = 1 } A = ( x , y ) : x 2 + y 2 = 1 A={(x,y):x^(2)+y^(2)=1}\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^2+\mathrm{y}^2=1\right\}A={(x,y):x2+y2=1} under the discrete metric.
8. a) Consider Z Z Z\mathbf{Z}Z and let F 1 F 1 F_(1)\mathcal{F}_1F1 denote the class of subsets of Z Z Z\mathbf{Z}Z, given by F 1 = { A Z F 1 = { A Z F_(1)={AsubZ\mathcal{F}_1=\{\mathrm{A} \subset \mathbf{Z}F1={AZ : either A A A\mathrm{A}A is finite or A c A c A^(c)\mathrm{A}^{\mathrm{c}}Ac is finite }. Check whether F 1 F 1 F_(1)\mathcal{F}_1F1 is a σ σ sigma\sigmaσ algebra or not.
b) Let A be any set in R R R\mathbf{R}R, show that m ( A ) = m ( A + x ) m ( A ) = m ( A + x ) m^(**)(A)=m^(**)(A+x)m^*(A)=m^*(A+x)m(A)=m(A+x) where m m m^(**)m^*m denotes the outer measure.
c) Find the measure of the following sets.
i) E = n = 1 ( a 1 n , b ) E = n = 1 a 1 n , b quad E=nnn_(n=1)^(oo)(a-(1)/(n),b)\quad E=\bigcap_{n=1}^{\infty}\left(a-\frac{1}{n}, b\right)E=n=1(a1n,b)
ii) E = Q { 1 , 2 , 3 , 4 } E = Q { 1 , 2 , 3 , 4 } E=Quu{1,2,3,4}\mathrm{E}=\mathbf{Q} \cup\{1,2,3,4\}E=Q{1,2,3,4}
iii) E = ] 5 , 7 [ [ 7 , 7.5 ] E = ] 5 , 7 [ [ 7 , 7.5 ] E=]5,7[uu[7,7.5]\mathrm{E}=] 5,7[\cup[7,7.5]E=]5,7[[7,7.5].
9. a) Show that if f f fff is measurable, then the function f a ( x ) f a ( x ) f^(a)(x)f^a(x)fa(x) given by
f a ( x ) = { a if f ( x ) > a f ( x ) if f ( x ) a f a ( x ) = a if f ( x ) > a f ( x ) if f ( x ) a f^(a)(x)={[a,” if “f(x) > a],[f(x),” if “f(x) <= a]:}f^a(x)=\left\{\begin{array}{cc} a & \text { if } f(x)>a \\ f(x) & \text { if } f(x) \leq a \end{array}\right.fa(x)={a if f(x)>af(x) if f(x)a
is also measurable.
b) Verify Bounded Convergence Theorem for the sequence of functions { f n } f n {f_(n)}\left\{f_n\right\}{fn} where
f n ( x ) = 1 ( 1 + x / n ) n , 0 x 1 , n N f n ( x ) = 1 ( 1 + x / n ) n , 0 x 1 , n N f_(n)(x)=(1)/((1+x//n)^(n)),0 <= x <= 1,ninN\mathrm{f}_{\mathrm{n}}(\mathrm{x})=\frac{1}{(1+\mathrm{x} / \mathrm{n})^{\mathrm{n}}}, 0 \leq \mathrm{x} \leq 1, \mathrm{n} \in \mathbf{N}fn(x)=1(1+x/n)n,0x1,nN
c) Find the fourier series of the function f f fff defined by
f ( x ) = { x 2 , π < x 0 x 2 , 0 < x < π f ( x ) = x 2 , π < x 0 x 2 , 0 < x < π f(x)={[-x^(2)”,”-pi < x <= 0],[x^(2)”,”0 < x < pi]:}f(x)=\left\{\begin{array}{c} -x^2,-\pi<x \leq 0 \\ x^2, 0<x<\pi \end{array}\right.f(x)={x2,π<x0x2,0<x<π
  1. State whether the following statements are True or False. Justify your answers.
    a) The sequence { ( 1 n , 1 n ) : n N } 1 n , 1 n : n N {((1)/(n),(1)/(n)):ninN}\left\{\left(\frac{1}{\mathrm{n}}, \frac{1}{\mathrm{n}}\right): \mathrm{n} \in \mathbf{N}\right\}{(1n,1n):nN} is convergent in R 2 R 2 R^(2)\mathbf{R}^2R2 under the discrete metric on R 2 R 2 R^(2)\mathbf{R}^2R2.
    b) A subset in a metric space is compact if it is closed.
    c) Continuous image of a path connected space is path connected.
    d) The second derivative of a linear map from R n R n R^(n)\mathbf{R}^nRn to R m R m R^(m)\mathbf{R}^mRm never vanishes.
    e) If A f d m = A g d m A f d m = A g d m int_(A)fdm=int_(A)gdm\int_{\mathrm{A}} \mathrm{fdm}=\int_{\mathrm{A}} \mathrm{gdm}Afdm=Agdm for all A M A M Ain M\mathrm{A} \in \boldsymbol{M}AM, then f = g f = g f=g\mathrm{f}=\mathrm{g}f=g.
\(cos\:3\theta =4\:cos^3\:\theta -3\:cos\:\theta \)

MMT-004 Sample Solution 2024

mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

MMT-004 Solved Assignment 2024 SS
  1. a) Let X = C [ 0 , 1 ] X = C [ 0 , 1 ] X=C[0,1]\mathrm{X}=\mathrm{C}[0,1]X=C[0,1]. Define d : X × X R d : X × X R d:X xx X rarrRd: X \times X \rightarrow \mathbf{R}d:X×XR by d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginX\mathrm{d}(\mathrm{f}, \mathrm{g})=\int_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})| \mathrm{dt}, \mathrm{f}, \mathrm{g} \in \mathrm{X}d(f,g)=01|f(t)g(t)|dt,f,gX where the integral is the Riemann integral. Show that d d ddd is a metric on X X XXX. Find d ( f , g ) d ( f , g ) d(f,g)d(f, g)d(f,g) where f ( x ) = 4 x f ( x ) = 4 x f(x)=4xf(x)=4 xf(x)=4x and g ( x ) = x 3 , x [ 0 , 1 ] g ( x ) = x 3 , x [ 0 , 1 ] g(x)=x^(3),x in[0,1]g(x)=x^3, x \in[0,1]g(x)=x3,x[0,1].
Answer:
To show that d d ddd is a metric on X X XXX, we need to verify the following properties for all f , g , h X f , g , h X f,g,h in Xf, g, h \in Xf,g,hX:
  1. Non-negativity: d ( f , g ) 0 d ( f , g ) 0 d(f,g) >= 0d(f, g) \geq 0d(f,g)0
  2. Identity of indiscernibles: d ( f , g ) = 0 d ( f , g ) = 0 d(f,g)=0d(f, g) = 0d(f,g)=0 if and only if f = g f = g f=gf = gf=g
  3. Symmetry: d ( f , g ) = d ( g , f ) d ( f , g ) = d ( g , f ) d(f,g)=d(g,f)d(f, g) = d(g, f)d(f,g)=d(g,f)
  4. Triangle inequality: d ( f , h ) d ( f , g ) + d ( g , h ) d ( f , h ) d ( f , g ) + d ( g , h ) d(f,h) <= d(f,g)+d(g,h)d(f, h) \leq d(f, g) + d(g, h)d(f,h)d(f,g)+d(g,h)
Let’s verify each property:
  1. Non-negativity:
    For any f , g X f , g X f,g in Xf, g \in Xf,gX, the absolute value function | | | | |*||\cdot||| ensures that | f ( t ) g ( t ) | 0 | f ( t ) g ( t ) | 0 |f(t)-g(t)| >= 0|f(t) – g(t)| \geq 0|f(t)g(t)|0 for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1]. Therefore, the integral 0 1 | f ( t ) g ( t ) | d t 0 1 | f ( t ) g ( t ) | d t int_(0)^(1)|f(t)-g(t)|dt\int_0^1 |f(t) – g(t)| \, dt01|f(t)g(t)|dt is also non-negative. Hence, d ( f , g ) 0 d ( f , g ) 0 d(f,g) >= 0d(f, g) \geq 0d(f,g)0.
  2. Identity of indiscernibles:
    If f = g f = g f=gf = gf=g, then f ( t ) g ( t ) = 0 f ( t ) g ( t ) = 0 f(t)-g(t)=0f(t) – g(t) = 0f(t)g(t)=0 for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1], so d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 0 d t = 0 d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 0 d t = 0 d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)0dt=0d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 0 \, dt = 0d(f,g)=01|f(t)g(t)|dt=010dt=0.
    Conversely, if d ( f , g ) = 0 d ( f , g ) = 0 d(f,g)=0d(f, g) = 0d(f,g)=0, then 0 1 | f ( t ) g ( t ) | d t = 0 0 1 | f ( t ) g ( t ) | d t = 0 int_(0)^(1)|f(t)-g(t)|dt=0\int_0^1 |f(t) – g(t)| \, dt = 001|f(t)g(t)|dt=0. Since the integrand is non-negative, it must be zero almost everywhere, implying that f ( t ) = g ( t ) f ( t ) = g ( t ) f(t)=g(t)f(t) = g(t)f(t)=g(t) for almost all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1]. Since f f fff and g g ggg are continuous, they must be equal everywhere on [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1], so f = g f = g f=gf = gf=g.
  3. Symmetry:
    By the properties of the absolute value function, | f ( t ) g ( t ) | = | g ( t ) f ( t ) | | f ( t ) g ( t ) | = | g ( t ) f ( t ) | |f(t)-g(t)|=|g(t)-f(t)||f(t) – g(t)| = |g(t) – f(t)||f(t)g(t)|=|g(t)f(t)| for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1]. Therefore, d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 | g ( t ) f ( t ) | d t = d ( g , f ) d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 | g ( t ) f ( t ) | d t = d ( g , f ) d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)|g(t)-f(t)|dt=d(g,f)d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 |g(t) – f(t)| \, dt = d(g, f)d(f,g)=01|f(t)g(t)|dt=01|g(t)f(t)|dt=d(g,f).
  4. Triangle inequality:
    For any f , g , h X f , g , h X f,g,h in Xf, g, h \in Xf,g,hX and for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1], we have | f ( t ) h ( t ) | = | f ( t ) g ( t ) + g ( t ) h ( t ) | | f ( t ) g ( t ) | + | g ( t ) h ( t ) | | f ( t ) h ( t ) | = | f ( t ) g ( t ) + g ( t ) h ( t ) | | f ( t ) g ( t ) | + | g ( t ) h ( t ) | |f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)| <= |f(t)-g(t)|+|g(t)-h(t)||f(t) – h(t)| = |f(t) – g(t) + g(t) – h(t)| \leq |f(t) – g(t)| + |g(t) – h(t)|