JEE Mathematics Definite Integration MCQs Set 32

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

Explanation:
1. For \( 0 \lt x \lt 1/2 \) and \( n \gt 2, x^n \lt x^2 \), so \( \frac1{\sqrt{1-x^n}} \lt \frac1{\sqrt{1-x^2}} \); hence \( I_n \lt \sin^{-1}\frac12 = \pi/6 \).
2. Also \( 1/\sqrt{1-x^n} \gt 1 \), so \( I_n \gt 1/2 \).

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

Answer: (b) \( 2\sqrt2\lt I\lt 6\sqrt{10} \)

Explanation:
1. \( \sqrt{x^4+x^2} = x\sqrt{x^2+1} \), whose exact integral is \( [(x^2+1)^{3/2}/3]_1^3 = (10\sqrt{10} - 2\sqrt2)/3 \approx 9.6 \).
2. \( 2\sqrt2 \approx 2.83 \lt 9.6 \lt 6\sqrt{10} \approx 18.97 \).

Teacher's Note:
1. The exact value is about \( 9.6 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac{4-\pi}{4\sqrt2} \)

Explanation:
1. Take \( \tan^{-1}x \) as the first function and \( x(1+x^2)^{-3/2} (integral -(1+x^2)^{-1/2}) \) as the second.
2. I = \( [-\tan^{-1}x/\sqrt{1+x^2}]_0^1 + \int_0^1 (1+x^2)^{-3/2}dx = -\frac{\pi}{4\sqrt2} + [\frac{x}{\sqrt{1+x^2}}]_0^1 \).
3. = \( \frac{1}{\sqrt2} - \frac{\pi}{4\sqrt2} = \frac{4-\pi}{4\sqrt2} \).

Teacher's Note:
1. Integrate by parts and use \( d(x/sqrt(1+x^2)) = (1+x^2)^{-3/2} dx \).

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

Explanation:
1. By \( x \to \pi - x \) and \( f(\sin(\pi - x)) = f(\sin x), \int_0^\pi xf(\sin x)dx = \frac\pi2\int_0^\pi f(\sin x)dx \).
2. Since \( \sin x \) is symmetric about \( \pi/2, \int_0^\pi f(\sin x)dx = 2\int_0^{\pi/2}f(\sin x)dx \), giving \( (ii) \).

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

Answer: (a) \( \frac2{35} \)

Explanation:
1. Put \( u = sin x: \int_0^1u^4(1-u^2)du = 1/5 - 1/7 = 2/35 \).

Teacher's Note:
1. Substitute \( u = sin x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac{\pi}{4(\pi-1)} \)

Explanation:
1. \( \int_0^{1/2}\cos^2\pi x\,dx = \frac14 \) and \( \int_0^{1/2}\cos\pi x\,dx = \frac1\pi \), so \( a_n = \frac14 + \frac{a_{n-1}}\pi \).
2. The limit \( L \) satisfies \( L = \frac14 + \frac L\pi \), so \( L = \frac{1/4}{1 - 1/\pi} = \frac{\pi}{4(\pi-1)} \).

Teacher's Note:
1. Turn the integral recurrence into a linear recurrence.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac\pi4-\frac12 \)

Explanation:
1. Since \( \sin x/\cos^3x = d(1/(2\cos^2x)) \), integrate by parts: \( [x/(2\cos^2x)]_0^{\pi/4} - (1/2)\int_0^{\pi/4}\sec^2x\,dx \).
2. = \( \pi/4 - 1/2 \).

Teacher's Note:
1. By parts with dv = \( sin x / cos^3 x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac12[x] \)

Explanation:
1. Let \( [x] = n \). Then \( \int_0^n\{x\}dx = n\int_0^1 x\,dx = n/2 \).

Teacher's Note:
1. Each unit interval contributes \( 1/2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 5 \)

Explanation:
1. \( f'(y) = (ay + \) a + \( b)e^y \), so \( \int_0^xe^{x-y}f'(y)dy = e^x(ax^2/2 + (a+b)x) \).
2. Comparing \( (ax + b) = ax^2/2 + (a+b)x - x^2 + x - 1: x^2 \) gives a = \( 2, x \) gives \( b = -1 \), and the constant term agrees \( (b = -1) \).
3. \( a^2 + b^2 = 5 \).

Teacher's Note:
1. Compare coefficients after dividing by \( e^x \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) a = 4, b = 1

Explanation:
1. The numerator behaves like \( x^3/(3\sqrt a) \). The denominator is of order \( x \) unless \( b = 1 \); for a finite non-zero limit we need \( b = 1 \), when it is about \( x^3/6 \).
2. The limit is \( \frac{x^3/(3\sqrt a)}{x^3/6} = \frac2{\sqrt \) a} = \( 1 \), so a = \( 4 \).

Teacher's Note:
1. Match orders of \( x \): both must be \( x^3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \int_{x+5}^5g(t)dt \)

Explanation:
1. \( g \) even makes \( f \) odd. Then \( f(5 - x) = f(5 + x) (from g(-x) = g(x)) \) and \( f(5 + x) = -f(x - 5) \), so \( f(x + 10) = -f(x) \), i.e. \( g(x + 5) = f(x + 10) = -f(x) \).
2. Hence \( \frac d{dx}\int_{x+5}^5g(t)dt = -g(x + 5) = f(x) \), and both sides vanish at \( x = 0 \), so \( \int_0^xf = \int_{x+5}^5g \).

Teacher's Note:
1. Check by differentiating both sides.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac3{20}(4\pi-3) \)

Explanation:
1. On \( [-\pi/2, -1): [x] + [\sin x] + 4 = -2 - 1 + 4 = 1 \). On \( [-1, 0): -1 - 1 + 4 = 2 \). On \( [0, 1): 0 + 0 + 4 = 4 \). On \( [1, \pi/2]: 1 + 0 + 4 = 5 \).
2. Integral = \( (\frac\pi2 - 1) + \frac12 + \frac14 + \frac{\pi/2 - 1}5 = \frac{3\pi}5 - \frac9{20} = \frac3{20}(4\pi - 3) \).

Teacher's Note:
1. Split at \( -1, 0 \) and \( 1 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac53-2\ln2 \)

Explanation:
1. Put \( \sqrt x = t \), so \( dx = 2t\,dt: \int_0^1\frac{2t^3}{1+t}dt \).
2. \( \frac{t^3}{1+t} = t^2 - t + 1 - \frac{1}{1+t} \).
3. \( 2(\frac13 - \frac12 + 1 - \ln 2) = \frac53 - 2\ln 2 \).

Teacher's Note:
1. Substitute \( t = sqrt(x) \) to remove the root, then divide polynomials.

Answer: (b) \( \frac{2}{\sqrt3} \)

Explanation:
1. For \( x \gt 1 \) the integral equals \( sec^{-1}x - sec^{-1}1 = sec^{-1}x \).
2. \( sec^{-1}x = \pi/6 \) means \( cos(\pi/6)=1/x \), so \( x = 2/\sqrt3 \).

Teacher's Note:
1. The integral equals \( sec^{-1}x \), so \( x = sec(pi/6) = 2/sqrt3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac{4-3\ln3}{2} \)

Explanation:
1. For \( x \gt 0: y = x \) for \( x \le 2 \) and \( y = 4 - x \) for \( x \gt 2. y = 3/x \) meets these at \( x = \sqrt3 \) and \( x = 3 \).
2. Area = \( \int_{\sqrt3}^2(x - 3/x)dx + \int_2^3(4 - x - 3/x)dx = (1/2 - 3\ln2 + \tfrac32\ln3) + (3/2 + 3\ln2 - 3\ln3) \).
3. = \( 2 - (3/2)\ln3 \).

Teacher's Note:
1. Two pieces of the tent are needed.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac{\pi\ln2}{2} \)

Explanation:
1. Put \( t = \sqrt x \): I = \( \int_0^{\pi/2}\frac{2t\,dt}{1+\sin t+\cos \) t}. Using \( t \to \pi/2 - t \), I = \( (\pi/2)\int_0^{\pi/2}\frac{dt}{1+\sin t+\cos \) t}.
2. With \( u = \tan(t/2) \) the last integral is \( \int_0^1\frac{du}{1+u} = \ln 2 \).
3. So I = \( (\pi/2)\ln 2 \).

Teacher's Note:
1. The integral of \( 1/(1+sin+cos) \) over \( [0, pi/2] \) is \( ln 2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac\pi3+2\sqrt3 \)

Explanation:
1. \( 1 + 2cos x = 0 \) at \( x = 2\pi/3 \); it is positive before and negative after.
2. \( \int_0^{2\pi/3}(1+2\cos x)dx = 2\pi/3 + \sqrt3 \) and \( \int_{2\pi/3}^{\pi}-(1+2\cos x)dx = -\pi/3 + \sqrt3 \).
3. Sum = \( \pi/3 + 2\sqrt3 \).

Teacher's Note:
1. Split at \( 2pi/3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) continuous at x = 0

Explanation:
1. \( f(x) = \pi\sin^{-1}x \) is continuous \( (indeed differentiable) \) at \( x = 0 \).
2. At \( x = 1 \) the derivative \( \pi/\sqrt{1-x^2} \) is infinite, so \( f \) is not differentiable there.

Teacher's Note:
1. \( f \) is the inverse sine times pi.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \pi-\frac83 \)

Explanation:
1. The circle \( (x-2)^2 + y^2 = 4 \) and the parabola \( y^2 = 2x \) meet at \( (2, 2) \); for \( 0 \le x \le 2 \) the circle lies above the parabola.
2. Area = \( \int_0^2\left(\sqrt{4x - x^2} - \sqrt{2x}\right)dx = \pi - \frac83 (the \) first integral is a quarter disc of radius \( 2) \).

Teacher's Note:
1. The circle part is a quarter circle.
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. The point \( r/n \) runs over \( [0, 1] \) for \( r = 1..n \), so \( (i) \) gives \( \int_0^1f \), and \( r = 1..2n \) gives \( \int_0^2f \).
2. For \( r = n + 1..2n (or \) equivalently \( (r + n)/n \) with \( r = 1..n) \) the points run over \( (1, 2] \), giving \( \int_1^2f \).

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

Answer: (a) \( \pi\ln 2 \)

Explanation:
1. \( \tan x + \cot x = 2/\sin2x \), so the integrand is \( \ln2 - \ln\sin2x \).
2. \( \int_0^{\pi/2}\ln\sin2x\,dx = -(\pi/2)\ln2 \), so I = \( (\pi/2)\ln2 + (\pi/2)\ln2 = \pi\ln2 \).

Teacher's Note:
1. The standard integral of \( ln sin x \) gives \( -(pi/2) ln 2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac{32}{3} \)

Explanation:
1. The parabola \( x = 4 - y^2 \) meets \( x = 0 \) at \( y = \pm2 \).
2. Area = \( \int_{-2}^{2}(4 - y^2)dy = 16 - 16/3 = 32/3 \).

Teacher's Note:
1. Integrate along \( y \) from \( -2 \) to \( 2 \).
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. For \( 0 \le x \le 1: e^{-x^2} \ge e^{-1} \), so \( S \ge 1/e \); and \( e^{-x^2} \ge e^{-x} \), so \( S \ge 1 - 1/e \).
2. \( S \approx 0.7468 \gt \frac14(1 + e^{-1/2}) \approx 0.40 \), so \( (iii) \) fails.
3. Splitting \( [0, 1] \) at \( 1/\sqrt2 \) and using the maximum on \( [0, 1/\sqrt2] \) and on \( [1/\sqrt2, 1] \) gives \( S \le 1/\sqrt2 + (1/\sqrt e)(1 - 1/\sqrt2) \approx 0.885 \), so \( (iv) \) holds.

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

Answer: (d) \( \frac52 \)

Explanation:
1. \( g'(x) = f(x) - f(5 - x) = (2x - 5)(2x^2 - 10x - 1) \).
2. On \( (0, 5) \) the quadratic factor is negative \( (its \) roots are outside \( (0, 5) \) apart from \( x \gt 5.09) \). So \( g' \gt 0 \) exactly when \( 2x - 5 \lt 0 \), i.e. \( x \lt 5/2 \).
3. \( g \) is increasing on \( (0, 5/2] \), so \( p = 5/2 \).

Teacher's Note:
1. Factor \( g'(x) \) completely.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac43 \)

Explanation:
1. The parabola lies above the x-axis for \( 1 \lt x \lt 3 \), and the area is \( \int_1^3(4x - x^2 - 3)dx = 4/3 \).

Teacher's Note:
1. Same area as the strip under the arch.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. \( a_k + a_{k-1} = \frac1{2k-1} \), so the sum is \( 1 - \frac13 + \frac15 - \cdots \).
2. This is the Leibniz series, which converges to \( \frac\pi4 (also \) because the sum telescopes to \( a_0 + (-1)^{n-1}a_n \) with \( a_n \to 0) \).

Teacher's Note:
1. The alternating sum telescopes to \( a_0 \) in the limit.
2. Verify the final value with a quick numerical or limiting check.