IGNOU BMTC-131 Solved Assignment | 1st January, 2025 to 31st December, 2025 | CALCULUS | BSc CBCS

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IGNOU BMTC-131 Solved Assignment | 1st January, 2025 to 31st December, 2025 | CALCULUS | BSc CBCS Sample Solution

BMTC-131 Solved Assignment 2025

CALCULUS

  1. Which of the following statements are true or false? Give reasons for your answer in the form of a short proof or a counter-example, whichever is appropriate.
    a) The set { S ∈ R : x 2 − 3 x + 2 = 0 } S ∈ R : x 2 − 3 x + 2 = 0 {S inR:x^(2)-3x+2=0}\left\{S \in \mathbf{R}: \mathrm{x}^2-3 \mathrm{x}+2=0\right\}{S∈R:x2−3x+2=0} is an infinite set.
    b) The greatest interger function is continuous on R R R\mathbf{R}R.
    c) d d x [ ∫ 3 e x lnt dt ] = x e x − ln ⁡ 3 d d x ∫ 3 e x   lnt dt = x e x − ln ⁡ 3 (d)/(dx)[int_(3)^(e^(x))(lntdt)]=xe^(x)-ln 3\frac{d}{d x}\left[\int_3^{e^x} \operatorname{lnt~dt}\right]=x e^x-\ln 3ddx[∫3exlnt dt]=xex−ln⁡3.
    d) Every integrable function is monotonic.
    e) a ⊕ b = a + b a ⊕ b = a + b ao+b=sqrt(a+b)\mathrm{a} \oplus \mathrm{b}=\sqrt{\mathrm{a}+\mathrm{b}}a⊕b=a+b defines a binary operation on Q Q Q\mathbf{Q}Q, the set of rational numbers.
  2. a) Find the domain of the function f f fff given by f ( x ) = 2 − x x 2 + 1 f ( x ) = 2 − x x 2 + 1 f(x)=sqrt((2-x)/(x^(2)+1))f(x)=\sqrt{\frac{2-x}{x^2+1}}f(x)=2−xx2+1.
    b) The set R R R\mathbf{R}R of real numbers with the usual addition (+) and usual multiplication (.) is given. Define ( ∗ ∗ ^(**){ }^*∗ ) on R R R\mathbf{R}R as:
a ∗ b = a + b 2 , ∀ a , b ∈ R . a ∗ b = a + b 2 , ∀ a , b ∈ R . a**b=(a+b)/(2),AAa,binR.\mathrm{a} * \mathrm{~b}=\frac{\mathrm{a}+\mathrm{b}}{2}, \forall \mathrm{a}, \mathrm{~b} \in \mathbf{R} .a∗ b=a+b2,∀a, b∈R.
Is (*) associative in R R R\mathbf{R}R ? Is (.) distributive ( ∗ ∗ ^(**){ }^*∗ ) in R R R\mathbf{R}R ? Check.
3. a) If | z − 1 + 2 i | = 4 | z − 1 + 2 i | = 4 |z-1+2i|=4|z-1+2 i|=4|z−1+2i|=4, show that the point z + i z + i z+iz+iz+i describes a circle. Also draw this circle.
b) Express x − 1 x 3 − x 2 − 2 x x − 1 x 3 − x 2 − 2 x (x-1)/(x^(3)-x^(2)-2x)\frac{x-1}{x^3-x^2-2 x}x−1x3−x2−2x as a sum of partial fractions.
  1. a) Find the least value of a 2 sec 2 ⁡ x + b 2 cosec 2 x a 2 sec 2 ⁡ x + b 2 cosec 2 x a^(2)sec^(2)x+b^(2)cosec^(2)xa^2 \sec ^2 x+b^2 \operatorname{cosec}^2 xa2sec2⁡x+b2cosec2x, where a > 0 , b > 0 a > 0 , b > 0 a > 0,b > 0a>0, b>0a>0,b>0.
    b) Evaluate:
∫ x 2 cot − 1 ⁡ ( x 3 ) 1 + x 6 d x ∫ x 2 cot − 1 ⁡ x 3 1 + x 6 d x int(x^(2)cot^(-1)(x^(3)))/(1+x^(6))dx\int \frac{x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} d x∫x2cot−1⁡(x3)1+x6dx
c) For any two sets S and T, show that:
S ∪ T = ( S − T ) ∪ ( S ∩ T ) ∪ ( T − S ) S ∪ T = ( S − T ) ∪ ( S ∩ T ) ∪ ( T − S ) S uu T=(S-T)uu(S nn T)uu(T-S)S \cup T=(S-T) \cup(S \cap T) \cup(T-S)S∪T=(S−T)∪(S∩T)∪(T−S)
Depict this situation in the Venn diagram.
  1. a) Let f f fff and g g ggg be two functions defined on R R R\mathbf{R}R by:
f ( x ) = x 3 − x 2 − 8 x + 12 f ( x ) = x 3 − x 2 − 8 x + 12 f(x)=x^(3)-x^(2)-8x+12f(x)=x^3-x^2-8 x+12f(x)=x3−x2−8x+12
and g ( x ) = { f ( x ) x + 3 , when x ≠ − 3 α , when x = − 3 g ( x ) = f ( x ) x + 3 , when x ≠ − 3 α , when x = − 3 g(x)={[(f(x))/(x+3)”,”” when “x!=-3],[alpha”,”” when “x=-3]:}g(x)=\left\{\begin{array}{c}\frac{f(x)}{x+3}, \text { when } x \neq-3 \\ \alpha, \text { when } x=-3\end{array}\right.g(x)={f(x)x+3, when x≠−3α, when x=−3
i) Find the value of α α alpha\alphaα for which f f fff is continuous at x = − 3 x = − 3 x=-3x=-3x=−3.
ii) Find all the roots of f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0.
b) Find the area between the curve y 2 ( 4 − x ) = x 3 y 2 ( 4 − x ) = x 3 y^(2)(4-x)=x^(3)y^2(4-x)=x^3y2(4−x)=x3 and its asymptote parallel to y y yyy-axis.
6. a) If the revenue function is given by d R d x = 15 + 2 x − x 2 , x d R d x = 15 + 2 x − x 2 , x (dR)/(dx)=15+2x-x^(2),x\frac{d R}{d x}=15+2 x-x^2, xdRdx=15+2x−x2,x being the input, find the maximum revenue. Also find the revenue function R R RRR, if the initial revenue is 0 .
b) Trace the curve y 2 ( x + 1 ) = x 2 ( 3 − x ) y 2 ( x + 1 ) = x 2 ( 3 − x ) y^(2)(x+1)=x^(2)(3-x)y^2(x+1)=x^2(3-x)y2(x+1)=x2(3−x), clearly stating all the properties used for tracing it.
7. a) Find the length of the cycloid x = α ( θ − sin ⁡ θ ) , y = α ( 1 − cos ⁡ θ ) x = α ( θ − sin ⁡ θ ) , y = α ( 1 − cos ⁡ θ ) x=alpha(theta-sin theta),y=alpha(1-cos theta)x=\alpha(\theta-\sin \theta), y=\alpha(1-\cos \theta)x=α(θ−sin⁡θ),y=α(1−cos⁡θ) and show that the line θ = 2 π 3 θ = 2 π 3 theta=(2pi)/(3)\theta=\frac{2 \pi}{3}θ=2π3 divides it in the ratio 1 : 3 1 : 3 1:31: 31:3.
b) Find the condition for the curves, ax 2 + by 2 = 1 ax 2 + by 2 = 1 ax^(2)+by^(2)=1\mathrm{ax}^2+\mathrm{by}^2=1ax2+by2=1 and a ′ x 2 + b ′ y 2 = 1 a ′ x 2 + b ′ y 2 = 1 a^(‘)x^(2)+b^(‘)y^(2)=1\mathrm{a}^{\prime} \mathrm{x}^2+\mathrm{b}^{\prime} \mathrm{y}^2=1a′x2+b′y2=1 intersecting orthogonally.
  1. a) If y = e m sin − 1 ⁡ x y = e m sin − 1 ⁡ x y=e^(m)sin^(-1)xy=e^m \sin ^{-1} xy=emsin−1⁡x, then show that ( 1 − x 2 ) y 2 − x y 1 − m 2 y = 0 1 − x 2 y 2 − x y 1 − m 2 y = 0 (1-x^(2))y_(2)-xy_(1)-m^(2)y=0\left(1-x^2\right) y_2-x y_1-m^2 y=0(1−x2)y2−xy1−m2y=0. Hence using Leibnitz’s formula, find the value of ( 1 − x 2 ) y n + 2 − ( 2 n + 1 ) x y n + 1 1 − x 2 y n + 2 − ( 2 n + 1 ) x y n + 1 (1-x^(2))y_(n+2)-(2n+1)xy_(n+1)\left(1-x^2\right) y_{n+2}-(2 n+1) x y_{n+1}(1−x2)yn+2−(2n+1)xyn+1.
    b) Find the largest subset of R R R\mathbf{R}R on which the function f : R → R f : R → R f:RrarrR\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}f:R→R defined as:
f ( x ) = { 2 x , x > 5 x + 5 , 1 ≤ x ≤ 5 | x | , x < 1 f ( x ) = 2 x      , x > 5 x + 5 ,      1 ≤ x ≤ 5 | x | , x < 1 f(x)={[2x,”,”x > 5],[x+5″,”,1 <= x <= 5],[|x|quad”,”x < 1]:}f(x)= \begin{cases}2 x & , x>5 \\ x+5, & 1 \leq x \leq 5 \\ |x| \quad, x<1\end{cases}f(x)={2x,x>5x+5,1≤x≤5|x|,x<1
is continuous.
9. a) Solve the equation:
x 4 + 15 x 3 + 70 x 2 + 120 x + 64 = 0 x 4 + 15 x 3 + 70 x 2 + 120 x + 64 = 0 x^(4)+15x^(3)+70x^(2)+120 x+64=0x^4+15 x^3+70 x^2+120 x+64=0x4+15x3+70x2+120x+64=0
given that its roots are in G.P.
b) Evaluate:
lim x → 0 tan ⁡ x − sin ⁡ x x 3 lim x → 0   tan ⁡ x − sin ⁡ x x 3 lim_(x rarr0)(tan x-sin x)/(x^(3))\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3}limx→0tan⁡x−sin⁡xx3
  1. a) If I m , n = ∫ x 3 ( log ⁡ x ) n d x I m , n = ∫ x 3 ( log ⁡ x ) n d x I_(m,n)=intx^(3)(log x)^(n)dxI_{m, n}=\int x^3(\log x)^n d xIm,n=∫x3(log⁡x)ndx, show that:
( m + 1 ) I m , n = x m + 1 ( log ⁡ x ) n − n I m , n − 1 . ( m + 1 ) I m , n = x m + 1 ( log ⁡ x ) n − n I m , n − 1 . (m+1)I_(m,n)=x^(m+1)(log x)^(n)-nI_(m),n-1.(m+1) I_{m, n}=x^{m+1}(\log x)^n-n I_m, n-1 .(m+1)Im,n=xm+1(log⁡x)n−nIm,n−1.
Hence find the value of ∫ x 4 ( log ⁡ x ) 3 d x ∫ x 4 ( log ⁡ x ) 3 d x intx^(4)(log x)^(3)dx\int x^4(\log x)^3 d x∫x4(log⁡x)3dx.
b) Verigy Lagrange’s mean value theorem for the function f defined by f ( x ) = 2 x 2 − 7 x − 10 f ( x ) = 2 x 2 − 7 x − 10 f(x)=2x^(2)-7x-10\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2-7 \mathrm{x}-10f(x)=2x2−7x−10 over [ 2 , 5 ] [ 2 , 5 ] [2,5][2,5][2,5].

Answer:

Question:-1(a)

The set { S ∈ R : x 2 − 3 x + 2 = 0 } S ∈ R : x 2 − 3 x + 2 = 0 {S inR:x^(2)-3x+2=0}\left\{S \in \mathbf{R}: \mathrm{x}^2-3 \mathrm{x}+2=0\right\}{S∈R:x2−3x+2=0} is an infinite set.

Answer:

False.
The quadratic equation x 2 − 3 x + 2 = 0 x 2 − 3 x + 2 = 0 x^(2)-3x+2=0x^2 – 3x + 2 = 0x2−3x+2=0 can be factored as ( x − 1 ) ( x − 2 ) = 0 ( x − 1 ) ( x − 2 ) = 0 (x-1)(x-2)=0(x-1)(x-2) = 0(x−1)(x−2)=0, with roots x = 1 x = 1 x=1x = 1x=1 and x = 2 x = 2 x=2x = 2x=2. Thus, the set { x ∈ R : x 2 − 3 x + 2 = 0 } { x ∈ R : x 2 − 3 x + 2 = 0 } {x inR:x^(2)-3x+2=0}\{x \in \mathbf{R}: x^2 – 3x + 2 = 0\}{x∈R:x2−3x+2=0} is { 1 , 2 } { 1 , 2 } {1,2}\{1, 2\}{1,2}, which has exactly two elements and is finite, not infinite.

Question:-1(b)

The greatest integer function is continuous on R R R\mathbf{R}R.

Answer:

Statement: The greatest integer function is continuous on R R R\mathbf{R}R.
Answer: False.
Justification: The greatest integer function, often denoted as ⌊ x ⌋ ⌊ x ⌋ |__ x __|\lfloor x \rfloor⌊x⌋, gives the largest integer less than or equal to x x xxx. To be continuous on R R R\mathbf{R}R, the function must be continuous at every point in R R R\mathbf{R}R. However, at any integer n n nnn, the function has a jump discontinuity. For example, at x = 1 x = 1 x=1x = 1x=1:
  • As x → 1 − x → 1 − x rarr1^(-)x \to 1^-x→1−, ⌊ x ⌋ = 0 ⌊ x ⌋ = 0 |__ x __|=0\lfloor x \rfloor = 0⌊x⌋=0.
  • At x = 1 x = 1 x=1x = 1x=1, ⌊ x ⌋ = 1 ⌊ x ⌋ = 1 |__ x __|=1\lfloor x \rfloor = 1⌊x⌋=1.
  • As x → 1 + x → 1 + x rarr1^(+)x \to 1^+x→1+, ⌊ x ⌋ = 1 ⌊ x ⌋ = 1 |__ x __|=1\lfloor x \rfloor = 1⌊x⌋=1.
The left-hand limit ( lim x → 1 − ⌊ x ⌋ = 0 lim x → 1 −   ⌊ x ⌋ = 0 lim_(x rarr1^(-))|__ x __|=0\lim_{x \to 1^-} \lfloor x \rfloor = 0limx→1−⌊x⌋=0) does not equal the function value at x = 1 x = 1 x=1x = 1x=1 ( ⌊ 1 ⌋ = 1 ⌊ 1 ⌋ = 1 |__1__|=1\lfloor 1 \rfloor = 1⌊1⌋=1). Thus, the function is not continuous at x = 1 x = 1 x=1x = 1x=1. Since this discontinuity occurs at every integer, the greatest integer function is not continuous on R R R\mathbf{R}R.

Question:-1(c)

d d x [ ∫ 3 e x lnt dt ] = x e x − ln ⁡ 3 d d x ∫ 3 e x   lnt dt = x e x − ln ⁡ 3 (d)/(dx)[int_(3)^(e^(x))(lntdt)]=xe^(x)-ln 3\frac{d}{d x}\left[\int_3^{e^x} \operatorname{lnt~dt}\right]=x e^x-\ln 3ddx[∫3exlnt dt]=xex−ln⁡3.

Answer:

Statement: d d x [ ∫ 3 e x ln ⁡ t d t ] = x e x − ln ⁡ 3 d d x ∫ 3 e x   ln ⁡ t d t = x e x − ln ⁡ 3 (d)/(dx)[int_(3)^(e^(x))ln tdt]=xe^(x)-ln 3\frac{d}{dx}\left[\int_3^{e^x} \ln t \, dt\right] = x e^x – \ln 3ddx[∫3exln⁡tdt]=xex−ln⁡3.
Answer: False.
Justification: To evaluate d d x [ ∫ 3 e x ln ⁡ t d t ] d d x ∫ 3 e x   ln ⁡ t d t (d)/(dx)[int_(3)^(e^(x))ln tdt]\frac{d}{dx}\left[\int_3^{e^x} \ln t \, dt\right]ddx[∫3exln⁡tdt], we use the Leibniz rule for differentiating an integral with variable limits. For an integral of the form ∫ a ( x ) b ( x ) f ( t ) d t ∫ a ( x ) b ( x )   f ( t ) d t int_(a(x))^(b(x))f(t)dt\int_{a(x)}^{b(x)} f(t) \, dt∫a(x)b(x)f(t)dt, the derivative is:
d d x ∫ a ( x ) b ( x ) f ( t ) d t = f ( b ( x ) ) ⋅ b ′ ( x ) − f ( a ( x ) ) ⋅ a ′ ( x ) . d d x ∫ a ( x ) b ( x )   f ( t ) d t = f ( b ( x ) ) ⋅ b ′ ( x ) − f ( a ( x ) ) ⋅ a ′ ( x ) . (d)/(dx)int_(a(x))^(b(x))f(t)dt=f(b(x))*b^(‘)(x)-f(a(x))*a^(‘)(x).\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) – f(a(x)) \cdot a'(x).ddx∫a(x)b(x)f(t)dt=f(b(x))⋅b′(x)−f(a(x))⋅a′(x).
Here, f ( t ) = ln ⁡ t f ( t ) = ln ⁡ t f(t)=ln tf(t) = \ln tf(t)=ln⁡t, a ( x ) = 3 a ( x ) = 3 a(x)=3a(x) = 3a(x)=3, and b ( x ) = e x b ( x ) = e x b(x)=e^(x)b(x) = e^xb(x)=ex. Thus, a ′ ( x ) = 0 a ′ ( x ) = 0 a^(‘)(x)=0a'(x) = 0a′(x)=0 and b ′ ( x ) = e x b ′ ( x ) = e x b^(‘)(x)=e^(x)b'(x) = e^xb′(x)=ex. Applying the Leibniz rule:
d d x ∫ 3 e x ln ⁡ t d t = ln ⁡ ( e x ) ⋅ e x − ln ⁡ ( 3 ) ⋅ 0 = x ⋅ e x − 0 = x e x . d d x ∫ 3 e x   ln ⁡ t d t = ln ⁡ ( e x ) ⋅ e x − ln ⁡ ( 3 ) ⋅ 0 = x ⋅ e x − 0 = x e x . (d)/(dx)int_(3)^(e^(x))ln tdt=ln(e^(x))*e^(x)-ln(3)*0=x*e^(x)-0=xe^(x).\frac{d}{dx} \int_3^{e^x} \ln t \, dt = \ln(e^x) \cdot e^x – \ln(3) \cdot 0 = x \cdot e^x – 0 = x e^x.ddx∫3exln⁡tdt=ln⁡(ex)⋅ex−ln⁡(3)⋅0=x⋅ex−0=xex.
The given expression is x e x − ln ⁡ 3 x e x − ln ⁡ 3 xe^(x)-ln 3x e^x – \ln 3xex−ln⁡3. Since the correct derivative is x e x x e x xe^(x)x e^xxex, and ln ⁡ 3 ≠ 0 ln ⁡ 3 ≠ 0 ln 3!=0\ln 3 \neq 0ln⁡3≠0 (as ln ⁡ 3 ln ⁡ 3 ln 3\ln 3ln⁡3 is a positive constant), the statement x e x − ln ⁡ 3 = x e x x e x − ln ⁡ 3 = x e x xe^(x)-ln 3=xe^(x)x e^x – \ln 3 = x e^xxex−ln⁡3=xex is false. For example, at x = 0 x = 0 x=0x = 0x=0, the correct derivative is 0 ⋅ e 0 = 0 0 ⋅ e 0 = 0 0*e^(0)=00 \cdot e^0 = 00⋅e0=0, but the given expression is 0 − ln ⁡ 3 ≈ − 1.098 0 − ln ⁡ 3 ≈ − 1.098 0-ln 3~~-1.0980 – \ln 3 \approx -1.0980−ln⁡3≈−1.098, which differs. Thus, the statement is false.

Question:-1(d)

Every integrable function is monotonic.

Answer:

Statement: Every integrable function is monotonic.
Answer: False.
Justification: A function is integrable (in the Riemann sense) on an interval if it is bounded and the set of its discontinuities has measure zero. A monotonic function is either non-decreasing or non-increasing over its domain. However, not all integrable functions are monotonic.
Counterexample: Consider the function f ( x ) = sin ⁡ ( x ) f ( x ) = sin ⁡ ( x ) f(x)=sin(x)f(x) = \sin(x)f(x)=sin⁡(x) on [ 0 , 2 π ] [ 0 , 2 π ] [0,2pi][0, 2\pi][0,2π]. This function is continuous (hence integrable) because it is bounded and has no discontinuities. However, sin ⁡ ( x ) sin ⁡ ( x ) sin(x)\sin(x)sin⁡(x) is not monotonic: it increases on [ 0 , π / 2 ] [ 0 , π / 2 ] [0,pi//2][0, \pi/2][0,π/2] (e.g., sin ⁡ ( 0 ) = 0 sin ⁡ ( 0 ) = 0 sin(0)=0\sin(0) = 0sin⁡(0)=0 to sin ⁡ ( π / 2 ) = 1 sin ⁡ ( π / 2 ) = 1 sin(pi//2)=1\sin(\pi/2) = 1sin⁡(π/2)=1) and decreases on [ π / 2 , 3 π / 2 ] [ π / 2 , 3 π / 2 ] [pi//2,3pi//2][\pi/2, 3\pi/2][π/2,3π/2] (e.g., sin ⁡ ( π / 2 ) = 1 sin ⁡ ( π / 2 ) = 1 sin(pi//2)=1\sin(\pi/2) = 1sin⁡(π/2)=1 to sin ⁡ ( 3 π / 2 ) = − 1 sin ⁡ ( 3 π / 2 ) = − 1 sin(3pi//2)=-1\sin(3\pi/2) = -1sin⁡(3π/2)=−1). Since sin ⁡ ( x ) sin ⁡ ( x ) sin(x)\sin(x)sin⁡(x) is not monotonic but is integrable, the statement is false.

Question:-1(e)

a ⊕ b = a + b a ⊕ b = a + b ao+b=sqrt(a+b)\mathrm{a} \oplus \mathrm{b}=\sqrt{\mathrm{a}+\mathrm{b}}a⊕b=a+b defines a binary operation on Q Q Q\mathbf{Q}Q, the set of rational numbers.

Answer:

Statement: a ⊕ b = a + b a ⊕ b = a + b ao+b=sqrt(a+b)\mathrm{a} \oplus \mathrm{b} = \sqrt{\mathrm{a} + \mathrm{b}}a⊕b=a+b defines a binary operation on Q Q Q\mathbf{Q}Q, the set of rational numbers.
Answer: False.
Justification: A binary operation on a set Q Q Q\mathbf{Q}Q takes two elements a , b ∈ Q a , b ∈ Q a,b inQa, b \in \mathbf{Q}a,b∈Q and produces another element in Q Q Q\mathbf{Q}Q. Here, the operation is defined as a ⊕ b = a + b a ⊕ b = a + b a o+b=sqrt(a+b)a \oplus b = \sqrt{a + b}a⊕b=a+b. For this to be a binary operation on Q Q Q\mathbf{Q}Q, the result a + b a + b sqrt(a+b)\sqrt{a + b}a+b must be rational for all rational a , b a , b a,ba, ba,b.
Counterexample: Let a = 1 a = 1 a=1a = 1a=1 and b = 1 b = 1 b=1b = 1b=1, both rational numbers. Then a + b = 1 + 1 = 2 a + b = 1 + 1 = 2 a+b=1+1=2a + b = 1 + 1 = 2a+b=1+1=2, and a + b = 2 a + b = 2 sqrt(a+b)=sqrt2\sqrt{a + b} = \sqrt{2}a+b=2. Since 2 2 sqrt2\sqrt{2}2 is irrational (not in Q Q Q\mathbf{Q}Q), the operation does not always produce a rational number. Thus, ⊕ ⊕ o+\oplus⊕ is not a binary operation on Q Q Q\mathbf{Q}Q, and the statement is false.

Question:-2

a) Find the domain of the function f f fff given by f ( x ) = 2 − x x 2 + 1 f ( x ) = 2 − x x 2 + 1 f(x)=sqrt((2-x)/(x^(2)+1))f(x)=\sqrt{\frac{2-x}{x^2+1}}f(x)=2−xx2+1.

Answer:

To find the domain of the function f ( x ) = 2 − x x 2 + 1 f ( x ) = 2 − x x 2 + 1 f(x)=sqrt((2-x)/(x^(2)+1))f(x) = \sqrt{\frac{2 – x}{x^2 + 1}}f(x)=2−xx2+1, we need to determine all real numbers x x xxx for which f ( x ) f ( x ) f(x)f(x)f(x) is defined. Since the function involves a square root, the expression inside the square root must be non-negative (i.e., 2 − x x 2 + 1 ≥ 0 2 − x x 2 + 1 ≥ 0 (2-x)/(x^(2)+1) >= 0\frac{2 – x}{x^2 + 1} \geq 02−xx2+1≥0).
  1. Denominator analysis: The denominator is x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1. Since x 2 ≥ 0 x 2 ≥ 0 x^(2) >= 0x^2 \geq 0x2≥0 for all real x x xxx, we have x 2 + 1 ≥ 1 > 0 x 2 + 1 ≥ 1 > 0 x^(2)+1 >= 1 > 0x^2 + 1 \geq 1 > 0x2+1≥1>0. Thus, the denominator is always positive and never zero, so there are no restrictions from the denominator.
  2. Square root condition: For the square root to be defined, the argument must be non-negative:
    2 − x x 2 + 1 ≥ 0. 2 − x x 2 + 1 ≥ 0. (2-x)/(x^(2)+1) >= 0.\frac{2 – x}{x^2 + 1} \geq 0.2−xx2+1≥0.
    Since x 2 + 1 > 0 x 2 + 1 > 0 x^(2)+1 > 0x^2 + 1 > 0x2+1>0 for all x x xxx, we can multiply both sides by x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 without changing the inequality’s direction:
    2 − x ≥ 0. 2 − x ≥ 0. 2-x >= 0.2 – x \geq 0.2−x≥0.
    Solving this:
    2 ≥ x or x ≤ 2. 2 ≥ x or x ≤ 2. 2 >= x quad”or”quad x <= 2.2 \geq x \quad \text{or} \quad x \leq 2.2≥xorx≤2.
  3. Boundary check: At x = 2 x = 2 x=2x = 2x=2, the expression becomes:
    2 − 2 2 2 + 1 = 0 5 = 0. 2 − 2 2 2 + 1 = 0 5 = 0. (2-2)/(2^(2)+1)=(0)/(5)=0.\frac{2 – 2}{2^2 + 1} = \frac{0}{5} = 0.2−222+1=05=0.
    Since 0 = 0 0 = 0 sqrt0=0\sqrt{0} = 00=0, the function is defined at x = 2 x = 2 x=2x = 2x=2.
  4. Conclusion: The condition x ≤ 2 x ≤ 2 x <= 2x \leq 2x≤2 ensures the expression inside the square root is non-negative. There are no other restrictions (e.g., the denominator is never zero, and there are no other operations like logarithms or tangents imposing additional constraints).
Thus, the domain of f ( x ) f ( x ) f(x)f(x)f(x) is all real numbers x x xxx such that x ≤ 2 x ≤ 2 x <= 2x \leq 2x≤2.
Domain: ( − ∞ , 2 ] ( − ∞ , 2 ] (-oo,2](-\infty, 2](−∞,2] or, in interval notation, x ∈ ( − ∞ , 2 ] x ∈ ( − ∞ , 2 ] x in(-oo,2]x \in (-\infty, 2]x∈(−∞,2].
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