JEE Mathematics Definite Integration MCQs Set 31

Answer: (c) (i), (ii) and (iv) only

Explanation:
1. \( F'(x) = f(x) \) changes sign from - to + at \( 1 \), from + to - at \( 2 \), and from - to + at \( 5 \). So \( F \) has local minima at \( 1 \) and \( 5 \) and a local maximum at \( 2 \) only; \( (iii) \) is false.
2. \( F(x) = x^4/4 - 8x^3/3 + 17x^2/2 - 10x \) is negative on \( (0, 5): F(1) \approx -3.92, F(2) \approx -3.33 \), and \( F \) decreases further to \( F(5) \lt 0 \), so \( F \ne 0 \) on \( (0, 5) \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (d) \( \sqrt2-1 \)

Explanation:
1. \( cos(tan^{-1} x) = 1/\sqrt{1+x^2} \).
2. The integral becomes \( \int_0^1 x/\sqrt{1+x^2}\,dx = [\sqrt{1+x^2}]_0^1 \).
3. = \( \sqrt2 - 1 \).

Teacher's Note:
1. Convert the trigonometric-inverse composite to an algebraic expression with a right triangle.

Answer: (a) \( 0 \)

Explanation:
1. \( \int_0^1\sin\alpha x\sin\beta x\,dx = \frac{\beta\sin\alpha\cos\beta - \alpha\cos\alpha\sin\beta}{\alpha^2 - \beta^2} \).
2. With \( \sin\alpha = 2\alpha\cos\alpha \) and \( \sin\beta = 2\beta\cos\beta \) the numerator becomes \( 2\alpha\beta\cos\alpha\cos\beta - 2\alpha\beta\cos\alpha\cos\beta = 0 \).

Teacher's Note:
1. Use \( tan(alpha) = 2 \) alpha and \( tan(beta) = 2 \) beta.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 1+e^2 \)

Explanation:
1. The curves meet at \( x = 2 \) and \( 1/(4(x-1)) \ge 1/x^2 \) for \( x \gt 1 (since (x-2)^2 \ge 0) \). Area = \( \int_2^a\left[\frac1{4(x-1)} - \frac1{x^2}\right]dx = \frac14\ln(a-1) + \frac1a - \frac12 \).
2. Area = \( 1/a \) means \( \frac14\ln(a-1) = \frac12 \), so \( \ln(a-1) = 2 \) and a = \( 1 + e^2 \).

Teacher's Note:
1. Set area = \( 1/a \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac{\pi}{2}-\ln 2 \)

Explanation:
1. For \( 0 \le x \le 1, \sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x \).
2. \( \int_0^1 2\tan^{-1}x\,dx = 2[x\tan^{-1}x - \tfrac12\ln(1+x^2)]_0^1 \).
3. = \( 2(\pi/4 - \tfrac12\ln 2) = \pi/2 - \ln 2 \).

Teacher's Note:
1. Put \( x = tan t \) to see the identity, valid for \( x \) in \( [0,1] \).

Answer: (d) \( \frac43 \)

Explanation:
1. \( y = -(x-1)(x-3) \) is non-negative on \( [1, 3] \).
2. Area = \( \int_1^3(4x - x^2 - 3)dx = [2x^2 - x^3/3 - 3x]_1^3 = 4/3 \).

Teacher's Note:
1. The parabola cuts the x-axis at \( 1 \) and \( 3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) (ii) and (iv) only

Explanation:
1. For \( x \) in \( [k, k+1], (k+1)/(x+1) \le 1 \), so \( \frac{k+1}{x(x+1)} \le \frac1x \) and I < \( \int_1^{99}\frac{dx}x = \ln99 \).
2. Also, for \( x \le k + 1, \frac{k+1}{x(x+1)} \ge \frac{1}{k+2} \), so I > \( \sum_{k=1}^{98}\frac1{k+2} \approx 3.69 \), which exceeds \( 49/50 \).
3. Numerically I \( \approx 4.29 \), which lies between \( 49/50 \) and \( \ln99 \approx 4.60 \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (b) \( \frac13 \)

Explanation:
1. At \( x = 0, e^0f(0) = 2 \), so \( f(0) = 2 \) and \( (f^{-1})'(2) = 1/f'(0) \).
2. Differentiating: \( e^{-x}(f' - f) = \sqrt{x^4+1} \); at \( x = 0, f'(0) - 2 = 1 \), so \( f'(0) = 3 \).
3. \( (f^{-1})'(2) = 1/3 \).

Teacher's Note:
1. Use \( f(0) = 2 \) and differentiate the defining equation.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) (ii) and (iii) only

Explanation:
1. \( I_1 = \frac\pi2\int_0^\pi\frac{\sin x}{1+\cos^2x}dx = \frac\pi2[-\tan^{-1}\cos x]_0^\pi = \frac\pi2\cdot\frac\pi2 = \pi^2/4 \).
2. Since \( x^3 + (\pi - x)^3 = \pi(\pi^2 - 3\pi x + 3x^2) \), adding \( I_2 \) to its \( x \to \pi - x \) image gives \( 2I_2 = \pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx = \pi^2/2 \), so \( I_2 = \pi^2/4 = I_1 \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (d) (i) and (iv) only

Explanation:
1. Differentiating \( \int_0^xf = x^2 + x^3 \) gives \( f(x) = 2x + 3x^2 \) for \( x \ge 0 \), so \( f'(r) = 2 + 6r \) and \( \sum_{r=1}^nf'(r) = 2n + 3n(n+1) = 3n^2 + 5n \).
2. \( f \) is even, so for \( x \le 0, f(x) = f(-x) = -2x + 3x^2 = 3x^2 - 2x \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. The numerator is about \( \int_0^{x^2}\sqrt t\,dt = \frac23|x|^3 \) and the denominator is about \( x^3/6 \).
2. For \( x \to 0^+ \) the ratio tends to \( 4 \); for \( x \to 0^- \) the denominator is negative, so \( L_2 = -4 \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (b) \( 1 \)

Explanation:
1. \( \ln y_n = \frac1n\sum_{k=1}^n\ln\frac{n+k}n \to \int_0^1\ln(1+x)dx = 2\ln2 - 1 \).
2. \( L = 4/e \approx 1.47 \), so \( [L] = 1 \).

Teacher's Note:
1. Take logarithms and form a Riemann sum.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 6\pi \)

Explanation:
1. \( \int_0^\pi f''\sin x\,dx = [f'\sin x - f\cos x]_0^\pi - \int_0^\pi f\sin x\,dx = f(\pi) + f(0) - \int_0^\pi f\sin x\,dx \).
2. So the whole integral equals \( f(\pi) + f(0) = 7\pi \), giving \( f(0) = 6\pi \).

Teacher's Note:
1. The integral of \( f sin x \) cancels out.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 3(\sqrt3-1) \)

Explanation:
1. With \( x = \cos\theta, 4x^3 - 3x = \cos3\theta \). For \( x \ge 1/2 (\theta \le \pi/3), f = 3\cos^{-1}x \); for \( 0 \le x \le 1/2, f = 2\pi - 3\cos^{-1}x \).
2. \( \int_{1/2}^13\cos^{-1}x\,dx = \frac{3\sqrt3}2 - \frac\pi2 \) and \( \int_0^{1/2}(2\pi - 3\cos^{-1}x)dx = \pi - 3\left(\frac\pi6 - \frac{\sqrt3}2 + 1\right) \).
3. The total is \( 3\sqrt3 - 3 \).

Teacher's Note:
1. Use \( x = cos(theta) \) and the branches of \( cos^{-1} \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 1 \)

Explanation:
1. Interchange the order of integration in \( A_1: \int_0^xf(t)\left(\int_t^xdu\right)dt = \int_0^xf(t)(x - t)dt = A_2 \).
2. Hence the ratio is \( 1 \).

Teacher's Note:
1. \( A_1 \) and \( A_2 \) are the same double integral.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac32 \)

Explanation:
1. The line meets the x-axis at \( x = -1 \) and meets the cosine curve at \( (0, 1) \); the cosine curve meets the x-axis at \( x = \pi/2 \).
2. Area = \( (triangle \) under the line, \( 1/2) + \int_0^{\pi/2}\cos x\,dx = 1/2 + 1 = 3/2 \).

Teacher's Note:
1. A triangle of area \( 1/2 \) plus the cosine arch of area \( 1 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac{16}3 \)

Explanation:
1. The curves meet at \( (0, 0) \) and \( (4, 4) \).
2. Area = \( \int_0^4\left(2\sqrt x - \frac{x^2}4\right)dx = \frac{32}3 - \frac{16}3 = \frac{16}3 \).

Teacher's Note:
1. Subtract the lower curve from the upper.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac{125}3 \)

Explanation:
1. The identity \( 4t^2f(-1) + 4tf(1) + f(2) = 3t^2 + 3t \) holds for three values of \( t \), so it holds for all \( t: f(-1) = 3/4, f(1) = 3/4, f(2) = 0 \). Hence \( f(x) = 1 - x^2/4 \), with \( V = (0, 1) \).
2. Take A = \( (2, 0). B = (x, 1 - x^2/4) \) with \( \vec{VA}\cdot\vec{VB} = 0: 2x + x^2/4 = 0 \), so \( B = (-8, -15) \).
3. The area between a parabola \( y = -x^2/4 + \ldots \) and its chord over an x-range of \( 10 \) is \( \frac14\cdot\frac{10^3}6 = \frac{125}3 \).

Teacher's Note:
1. Compare coefficients of the polynomial identity in \( t \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac{64}{3} \)

Explanation:
1. \( \sqrt3\sin x \ge \cos x \) for \( x \ge \pi/6 \), so \( U = \int_{\pi/6}^{\pi/2}\sqrt3\sin x\,dx = \sqrt3\cos(\pi/6) = 3/2 \).
2. \( V = \int_1^5x^2dx - \int_{-3}^1x^2dx = 124/3 - 28/3 = 32 \).
3. \( \lambda = 32/(3/2) = 64/3 \).

Teacher's Note:
1. \( U = 3/2 \) and \( V = 32 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 2 \)

Explanation:
1. The integrand is \( \frac{1+\sqrt3}{(x+1)^{1/2}(1-x)^{3/2}} \). Put \( t = \frac{1+x}{1-x}, dt = \frac{2\,dx}{(1-x)^2} \).
2. Then \( \frac{dx}{(1+x)^{1/2}(1-x)^{3/2}} = \frac{dt}{2\sqrt \) t}, and \( t \) runs from \( 1 \) to \( 3 \).
3. The value is \( (1+\sqrt3)[\sqrt t]_1^3 = (1+\sqrt3)(\sqrt3 - 1) = 2 \).

Teacher's Note:
1. The substitution \( t = (1+x)/(1-x) \) removes both factors.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 6-4\ln2 \)

Explanation:
1. The region is inside the square \( [0,2] x [0,2]. e^x \) reaches \( 2 \) at \( x = \ln2 \), and \( \ln x \) is positive only for \( x \gt 1 \).
2. Area = \( \int_0^{\ln2}e^xdx + 2(2 - \ln2) - \int_1^2\ln x\,dx = 1 + (4 - 2\ln2) - (2\ln2 - 1) = 6 - 4\ln2 \).

Teacher's Note:
1. Cut off at \( y = 2 \) and subtract the area below \( ln x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) (i) and (iii) only

Explanation:
1. Area of \( R = \int_0^1(x - x^3)dx = 1/4 \). Half of it: \( \frac{\alpha^2}2 - \frac{\alpha^4}4 = \frac18 \), i.e. \( 2\alpha^4 - 4\alpha^2 + 1 = 0 \).
2. \( \alpha^2 = 1 - \frac{\sqrt2}2 \approx 0.293 \), so \( \alpha \approx 0.54 \), which lies in \( (1/2, 1) \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (a) \( \frac12(\sqrt3-1) \)

Explanation:
1. The roots are \( \alpha = \pi/6 \) and \( \beta = \pi/3. (g o f)(x) = \cos((\sqrt x)^2) = \cos x \).
2. Area = \( \int_{\pi/6}^{\pi/3}\cos x\,dx = \sin\frac\pi3 - \sin\frac\pi6 = \frac{\sqrt3 - 1}2 \).

Teacher's Note:
1. The composition simplifies to \( cos x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 1 \)

Explanation:
1. \( \log(36 - 12x + x^2) = \log(6 - x)^2 \). The sum of the limits is \( 6 \), and \( x \to 6 - x \) swaps the two logarithms.
2. So the integral is half the interval length: \( (4 - 2)/2 = 1 \).

Teacher's Note:
1. Use the a + \( b - x \) property.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) (ii) and (iii) only

Explanation:
1. Taking logarithms and using a Riemann sum: \( \ln f(x) = x\int_0^1[\ln(1+xs) - \ln(1+x^2s^2)]ds = (1+x)\ln(1+x) - x\ln(1+x^2) + x - 2\tan^{-1}x \).
2. Differentiating, \( (\ln f)' = \ln\frac{1+x}{1+x^2} \), which is positive for \( 0 \lt x \lt 1 \) and negative for \( x \gt 1 \). So \( f \) increases on \( (0, 1) \) and decreases on \( (1, \infty) \).
3. Thus \( f(1/2) \lt f(1) (i false), f(1/3) \lt f(2/3) (ii true), f'(2) \lt 0 (iii true) \), and \( (\ln f)'(3) = \ln0.4 \lt (\ln f)'(2) = \ln0.6 (iv false) \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (d) \( y=x-1 \)

Explanation:
1. \( y(1) = \int_1^1t^2dt = 0 \).
2. \( dy/dx = 2x(x^2)^2 - x^2 = 2x^5 - x^2 \), which is \( 1 \) at \( x = 1 \).
3. The tangent is \( y = 1\cdot(x - 1) \).

Teacher's Note:
1. Evaluate \( y \) and the slope at \( x = 1 \).
2. Verify the final value with a quick numerical or limiting check.