JEE Mathematics Definite Integration MCQs Set 35

Answer: (c) \( \frac1{60} \)

Explanation:
1. This is the beta integral \( B(3, 4) = 2!\,3!/6! = 12/720 = 1/60 \).

Teacher's Note:
1. Beta function with exponents \( 2 \) and \( 3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 110 \)

Explanation:
1. The integrand is \( 11^{\{x\}} \), which has period \( 1 \), so the integral is \( 11\int_0^111^x dx \).
2. = \( 11 \cdot 10/\ln 11 = 110/\ln 11 \).

Teacher's Note:
1. Periodicity of \( 11^{x}/11^{[x]} \) gives eleven copies of one period.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \pi \)

Explanation:
1. \( g(a) = B(1-a, a) = \pi/\sin(\pi a) \).
2. \( g(1/2) = \pi \).

Teacher's Note:
1. The integral is a beta function.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac{2}{e^2}e^{\pi/2} \)

Explanation:
1. Taking logarithms, the limit \( L \) satisfies \( \ln L = \int_0^1\ln(1+x^2)dx \).
2. \( \int_0^1\ln(1+x^2)dx = [x\ln(1+x^2)]_0^1 - \int_0^1\frac{2x^2}{1+x^2}dx = \ln2 - 2 + \pi/2 \).
3. \( L = 2e^{\pi/2}/e^2 \).

Teacher's Note:
1. Log of the product turns into a Riemann sum.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \left(\frac34,3\right) \)

Explanation:
1. Area = \( \int_0^t f = t + t^2 + t^3 + t^4 \), which is increasing in \( t \).
2. Since \( 1/2 \lt t \lt 3/4 \), the area lies between \( 15/16 \) and about \( 2.05 \), which is inside \( (3/4, 3) \).

Teacher's Note:
1. The area function is increasing in \( t \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 0 \)

Explanation:
1. Replace \( x \) by a - \( x \) in the h-equation and solve the two equations: \( h(x) = h(a - x) = -5 \).
2. So \( fgh(a - x) = f(x)(-g(x))(-5) = 5f(x)g(x) \), while \( fgh(x) = -5f(x)g(x) \); thus \( fgh(x) + fgh(a-x) = 0 \) and the integral is \( 0 \).

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

Answer: (a) \( \frac14\ln\frac32 \)

Explanation:
1. Put \( u = x^2 \): I = \( \frac12\int_{\ln2}^{\ln3}\frac{\sin u}{\sin u+\sin(\ln6-u)}du \).
2. The limits add to \( \ln6 \), so the integral of the fraction is half the interval length \( (\ln3 - \ln2)/2 \).
3. I = \( \frac12\cdot\frac12\ln\frac32 = \frac14\ln\frac32 \).

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

Answer: (b) m = 1, M = 12

Explanation:
1. On \( [1/2, 1], 3 \ge 2 + \sin^4\pi x \ge 2 \), so \( 64x^3 \le f'(x) \le 96x^3 \) and \( 16(x^4 - 1/16) \le f(x) \le 24(x^4 - 1/16) \).
2. \( \int_{1/2}^1(x^4 - 1/16)dx = 13/80 \), so \( 2.6 \le \int_{1/2}^1f \le 3.9 \).
3. A valid pair needs \( m \le 2.6 \) and \( M \ge 3.9 \), which only \( (m, M) = (1, 12) \) satisfies.

Teacher's Note:
1. Bound \( f' \) above and below, then integrate twice.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 2 \)

Explanation:
1. \( \int_6^{f(x)}4t^3dt = f(x)^4 - 6^4 \), so \( g(x) = \frac{f(x)^4 - 1296}{9(x-2)} \).
2. By L'Hospital, \( \lim g = \frac{4f(2)^3f'(2)}9 = \frac{4\cdot216/48}9 = \frac{18}9 = 2 \).

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

Answer: (d) \( 21 \)

Explanation:
1. Integrating \( p'(x) = p'(1-x): p(x) = -p(1-x) + C \), so \( p(x) + p(1-x) = C = p(0) + p(1) = 42 \).
2. \( \int_0^1p = \frac12\int_0^1[p(x) + p(1-x)]dx = 21 \).

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

Answer: (a) \( 3 \)

Explanation:
1. \( f'(x) = (x-1)(x-2)^2(x-3)^3(x-4)^5 \); the sign changes only at roots with odd power: \( x = 1, 3, 4 \).
2. The root \( x = 2 \) has an even power, so no sign change there. There are \( 3 \) extrema.

Teacher's Note:
1. Count odd-multiplicity roots of \( f' \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac\pi2 \)

Explanation:
1. \( \frac{\sin2kx}{\sin \) x} = \( 2[\cos x + \cos3x + \cdots + \cos(2k-1)x] \), so \( \sin2kx\cot x = 2\cos x[\cos x + \cos3x + \cdots + \cos(2k-1)x] \).
2. For the term \( 2\cos x\cos x: \int_0^{\pi/2}2\cos^2x\,dx = \pi/2 \). For \( j \ge 2: 2\cos x\cos(2j-1)x = \cos2jx + \cos(2j-2)x \), and both integrate to \( 0 \) over \( [0, \pi/2] \).
3. So the integral equals \( \pi/2 \).

Teacher's Note:
1. Only the \( j = 1 \) term survives.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac{27}{e^2} \)

Explanation:
1. \( \ln L = \lim\frac1n\sum_{k=1}^{2n}\ln\left(1+\frac kn\right) = \int_0^2\ln(1+x)dx = 3\ln3 - 2 \).
2. So \( L = 27/e^2 \).

Teacher's Note:
1. Logarithm turns the product into a Riemann sum.
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. The numerator is about \( n^{k+1}/(k+1) \); the denominator is about \( (n^3/3)(n^4/4) = n^7/12 \).
2. For \( k \le 6 \) the limit is finite \( (zero \) for \( k \lt 6) \), and for \( k = 6 \) it equals \( (1/7)/(1/12) = 12/7 \).

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

Answer: (a) TFTT

Explanation:
1. \( S1: \cos^{-1}(\cos x) = x \) on \( [0, \pi] \) and \( 2\pi - x \) on \( [\pi, 2\pi] \), so the integral is \( \pi^2/2 + \pi^2/2 = \pi^2 (true) \).
2. \( S2: |x-2| + |y+1| = 1 \) is a square with diagonals \( 2 \), so its area is \( 2 \), not \( 3 (false) \).
3. \( S3: \int \) ff' = \( [f^2/2]_a^b (true). S4: \int_0^1(x - x^2)dx = 1/6 (true) \).

Teacher's Note:
1. The only false statement is \( S2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) (i), (ii) and (iii) only

Explanation:
1. Because \( f \) is odd, \( g \) is even \( (g(-x) = \int_0^{-x}f = g(x)) \), so \( (iv) \) is false and \( (ii) \) is true.
2. \( \int_0^2f = \int_{-1}^1f = 0 \), so \( g(x + 2) = g(x) + 0 (period 2) \) and \( g(2n) = 0 \).

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

Answer: (c) \( \frac34 \)

Explanation:
1. \( [x^2] = 0 \) on \( [0, 1), 1 \) on \( [1, \sqrt2), 2 \) on \( [\sqrt2, 1.5] \).
2. \( \int_1^{\sqrt2}x\,dx = \frac12 \) and \( \int_{\sqrt2}^{1.5}2x\,dx = 2.25 - 2 = \frac14 \). Total \( 3/4 \).

Teacher's Note:
1. Find where \( x^2 \) crosses \( 1 \) and \( 2 \).
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. By parts, \( \int_1^3x^2F'dx = [x^2F]_1^3 - 2\int_1^3xF\,dx = -36 - 2\int_1^3f \), which equals \( -12 \), so \( \int_1^3f = -12 \).
2. \( \int_1^3x^3F''dx = [x^3F']_1^3 - 3\int x^2F' = 27F'(3) - F'(1) + 36 = 40 \), so \( 27F'(3) - F'(1) = 4 \).
3. \( f'(3) = -4 + 3F'(3) \) and \( f'(1) = F'(1) \), so \( 9f'(3) - f'(1) + 32 = 27F'(3) - F'(1) - 4 = 0 \).

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. \( f'(x) = 2e^{-(x+1/x)}/x \gt 0 \), so \( f \) is increasing on all of \( (0, \infty) \), which makes \( (i) \) true and \( (ii) \) false.
2. \( f(1/x) = \int_x^{1/x} = -f(x) \), so \( (iii) \) holds, and then \( f(2^{-x}) = -f(2^x) \), so \( f(2^x) \) is odd.

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

Answer: (b) \( \frac7{12} \)

Explanation:
1. The curves meet at \( x = 0 (limit) \) and at \( x = 1 \), where \( \ln x = 2 - 2x \).
2. Area = \( \int_0^1(2x - 2x^2 - x\ln x)dx = 1 - 2/3 + 1/4 = 7/12 \).

Teacher's Note:
1. Use \( \int_0^1 x ln x dx = -1/4 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( (-\sqrt2,\sqrt2) \)

Explanation:
1. The integrand \( x^4 - 2x^2 \) is negative exactly for \( |x| \lt \sqrt2 \).
2. I is least when \( [a, b] \) is exactly that interval, so \( (a, b) = (-\sqrt2, \sqrt2) (with \) a < \( b) \).

Teacher's Note:
1. Include all the region where the integrand is negative and nothing else.
2. Verify the final value with a quick numerical or limiting check.

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

Explanation:
1. \( (i) e^x \ge 1 \) while \( \int_0^xf\sin t\,dt \lt \int_0^1\sin t\,dt \lt 1 \), so it is never zero. \( (ii) \) Both \( f(x) \) and the integral are positive, so it is never zero.
2. \( (iii) \) At \( x = 0 \) it is \( -\int_0^{\pi/2}f\cos t\,dt \lt 0 \); at \( x = 1 \) it is \( 1 - \int_0^{\pi/2-1}f\cos t\,dt \gt 1 - \sin(0.57) \gt 0 \). By the intermediate value theorem it has a zero in \( (0, 1) \).
3. \( (iv) \) At \( 0 \) it is \( -f(0) \lt 0 \) and at \( 1 \) it is \( 1 - f(1) \gt 0 \), so it has a zero.

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

Answer: (a) \( 4 \)

Explanation:
1. Area of the triangle = \( 1 \). Area between \( y = x \) and \( y = x^n \) is \( \int_0^1(x - x^n)dx = \frac12 - \frac1{n+1} \).
2. \( \frac12 - \frac1{n+1} = 0.3 \) gives \( n + 1 = 5 \), so \( n = 4 \).

Teacher's Note:
1. Set \( 1/2 - 1/(n+1) = 0.3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \left(-\frac34,-\frac12\right) \)

Explanation:
1. \( f'(x) = 2 + 6x + 12x^2 \) has negative discriminant, so \( f \) is increasing and has exactly one real root \( s \).
2. \( f(-1/2) = 1/4 \gt 0 \) and \( f(-3/4) = -1/2 \lt 0 \), so \( s \) lies in \( (-3/4, -1/2) \).

Teacher's Note:
1. Sign change of \( f \) between \( -3/4 \) and \( -1/2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac38 \)

Explanation:
1. \( f \) is even, so the integral is \( 2\int_0^1 f \).
2. On \( [0, 1/4] f = x (area 1/32) \), on \( [1/4, 3/4] f = 1/4 (area 1/8) \), on \( [3/4, 1] f = 1 - x (area 1/32) \).
3. \( \int_0^1 f = 3/16 \), so the total is \( 3/8 \).

Teacher's Note:
1. Sketch the three functions and find where the minimum switches.

Answer: (d) P is false and Q is true

Explanation:
1. \( P \) reduces to \( (1-x)^2\sin^2x = (x-1)^2 + 1 \), i.e. \( -(1-x)^2\cos^2x = 1 \), which is impossible, so \( P \) is false.
2. \( Q \) reduces to \( 2(1-x)^2\sin^2x = 2x - 1 \). At \( x = 1/2 \) the left side is positive while the right is \( 0 \); at \( x = 0.6 \) the left side \( (about 0.10) \) is below the right side \( (0.2) \). By the intermediate value theorem a solution exists, so \( Q \) is true.

Teacher's Note:
1. Test \( P \) algebraically and \( Q \) with the intermediate value theorem.
2. Verify the final value with a quick numerical or limiting check.