JEE Mathematics Definite Integration MCQs Set 09

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

Explanation:
1. For |x| ≤ 1/2, sin⁻¹(3x − 4x³) = 3 sin⁻¹x.
2. With x = cos θ, 4x³ − 3x = cos 3θ, and for x ∈ [−1/2, 1/2] we have 3θ ∈ [π, 2π], so cos⁻¹(cos 3θ) = 2π − 3θ = 2π − 3cos⁻¹x.
3. The integrand is 3sin⁻¹x + 3cos⁻¹x − 2π = 3π/2 − 2π = −π/2, a constant. Over an interval of length 1 the integral is −π/2.

Teacher's Note:
1. Reduce the inverse functions to principal values on the given interval before integrating.
2. If the integrand becomes constant the integral is that constant times the length.

Answer: (b) \( \frac\pi2\ln2 \)

Explanation:
1. \( (\sin x)^{1/n}=e^{\frac1n\ln\sin x}\approx1+\frac{\ln\sin x}{n} \) for large n.
2. So n(1 − (sin x)^{1/n}) → −ln sin x.
3. \( -\int_0^{\pi/2}\ln\sin x\,dx=\frac\pi2\ln2 \).

Teacher's Note:
1. Expand the n-th root as an exponential and keep the first-order term.
2. Use the standard result ∫₀^{π/2} ln sin x dx = −(π/2) ln 2.

Answer: (d) \( \frac12 \)

Explanation:
1. Put u = nx: \( C_n=\frac1n\int_{n/(n+1)}^{1}\frac{\tan^{-1}u}{\sin^{-1}u}du \).
2. The interval has length 1/(n+1) and u → 1, where the integrand tends to (π/4)/(π/2) = 1/2.
3. So \( n^2C_n\approx n^2\cdot\frac1n\cdot\frac1{n+1}\cdot\frac12\to\frac12 \).

Teacher's Note:
1. For a very short interval, the integral is approximately the length times the value at the end point.
2. Evaluate the integrand at u = 1.

Answer: (a) 2/17

Explanation:
1. f′(x) = e^{g(x)}·g′(x), with g(2) = 0 and \( g'(x)=\frac{x}{1+x^4} \).
2. f′(2) = e⁰ · 2/17 = 2/17.

Teacher's Note:
1. Apply the chain rule and the Fundamental Theorem of Calculus.
2. g(2) = 0 because both limits are 2.

Answer: (b) \( b_n=0,\ a_n=-\frac{3}{2n+3} \)

Explanation:
1. Take p = 0, q = 1: ∫g dx = 0. The odd terms vanish, leaving 2bₙ = 0, so bₙ = 0.
2. Take p = 1, q = 0: \( \int_{-1}^1x\,g_n\,dx=\frac{2}{2n+3}+\frac{2a_n}{3}=0 \).
3. So \( a_n=-\frac{3}{2n+3} \).

Teacher's Note:
1. The condition must hold for every p and q, so apply it separately to 1 and to x.
2. Odd functions integrate to zero over [−1, 1].

Answer: (c) \( \frac1{10} \)

Explanation:
1. Write the term as \( \frac1n\cdot\frac{1}{\sqrt{r/n}\,(3\sqrt{r/n}+4)^2} \). The sum becomes \( \int_0^4\frac{dx}{\sqrt x(3\sqrt x+4)^2} \).
2. Put u = √x: \( 2\int_0^2\frac{du}{(3u+4)^2}=\frac23\left[\frac14-\frac1{10}\right]=\frac1{10} \).

Teacher's Note:
1. Factor out 1/n to see the Riemann sum for the function of r/n.
2. The upper limit is 4n/n = 4.

Answer: (c) 12

Explanation:
1. For an invertible increasing f: \( \int_a^bf+\int_{f(a)}^{f(b)}f^{-1}=b\,f(b)-a\,f(a) \).
2. So \( \int_3^7f^{-1}=5\cdot7-2\cdot3-17=12 \).

Teacher's Note:
1. The area under f and the area under its inverse together fill a rectangle minus a corner.
2. Plug a = 2, b = 5, f(a) = 3, f(b) = 7.

Answer: (d) 4

Explanation:
1. As t → 0, \( \frac{|\sin(x+t)-\sin x|}{|t|}\to|\cos x| \).
2. \( \int_0^{2\pi}|\cos x|dx=4 \).

Teacher's Note:
1. The difference quotient tends to the derivative; the absolute value stays outside.
2. |cos x| has integral 1 over each quarter period... that is, 4 over [0, 2π].

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

Explanation:
1. Put x = sin θ: \( \int_{-\pi/2}^{\pi/2}\frac{d\theta}{2-\sin\theta} \).
2. Pair θ and −θ: \( \frac1{2-\sin\theta}+\frac1{2+\sin\theta}=\frac{4}{4-\sin^2\theta} \), so the integral is \( \int_0^{\pi/2}\frac{4\,d\theta}{4-\sin^2\theta} \).
3. \( \int_0^{\pi/2}\frac{d\theta}{4-\sin^2\theta}=\frac{\pi}{2\sqrt{4\cdot3}}=\frac\pi{4\sqrt3} \), so the value is π/√3.

Teacher's Note:
1. The integral is improper but convergent, because 1/√(1 − x²) is integrable at ±1.
2. Use the standard result ∫₀^{π/2} dθ/(a − sin²θ) = π/(2√(a(a − 1))).

Answer: (a) \( \frac{\sqrt3}{3} \)

Explanation:
1. The bracket is the sum of sec²(kπ/6n) for k = 1 to n (the last term sec²(π/6) = 4/3).
2. This is a Riemann sum: \( \int_0^{\pi/6}\sec^2x\,dx=\tan\frac\pi6=\frac1{\sqrt3} \).

Teacher's Note:
1. Recognise the pattern of a Riemann sum with step π/(6n) on [0, π/6].
2. The extra term 4/3 is just the k = n term.

Answer: (c) 2

Explanation:
1. f is even, so f(cos x) is symmetric about x = π/2, and \( I_1=2\int_0^{\pi/2}f(\cos x)dx \).
2. Since \( \int_0^{\pi/2}f(\cos x)dx=\int_0^{\pi/2}f(\sin x)dx \), I₁ = 2I₂.
3. The ratio is 2.

Teacher's Note:
1. Use ∫₀^{π/2} g(sin x)dx = ∫₀^{π/2} g(cos x)dx.
2. Because f is even, f(cos x) takes the same value at x and π − x.

Answer: (a) 1

Explanation:
1. It is a Riemann sum with step π/(2n) for cos x on [0, π/2].
2. \( \int_0^{\pi/2}\cos x\,dx=1 \).

Teacher's Note:
1. The terms are cos(kπ/2n) for k = 0 to n − 1.
2. The sum of function values times the step approaches the integral.

Answer: (c) 16

Explanation:
1. Divide the numerator by a: the terms x² + 1 contribute \( \frac1a\int_0^\infty\frac{x^2+1}{1+x^4}\tan^{-1}\frac1x\,dx\to0 \). So the limit is \( I=\int_0^\infty\frac{x\cot^{-1}x}{1+x^4}dx \).
2. Replace x by 1/x: \( I=\int_0^\infty\frac{x\tan^{-1}x}{1+x^4}dx \). Adding the two forms: \( 2I=\frac\pi2\int_0^\infty\frac{x}{1+x^4}dx=\frac\pi2\cdot\frac\pi4 \).
3. So \( I=\frac{\pi^2}{16} \) and k = 16.

Teacher's Note:
1. Only the term linear in a survives after dividing by a.
2. tan⁻¹x + cot⁻¹x = π/2 combines the two forms.

Answer: (d) 8f′(1)

Explanation:
1. \( \int_4^{f(x)}2t\,dt=f(x)^2-16 \).
2. The limit is \( \lim_{x\to1}\frac{f(x)^2-f(1)^2}{x-1}=2f(1)f'(1)=8f'(1) \).

Teacher's Note:
1. Evaluate the integral first; the limit then becomes a derivative.
2. f(1) = 4 gives the factor 2 × 4.

Answer: (a) 2u = v

Explanation:
1. Use 2sin²x = 1 − cos 2x and put θ = 2x: \( u=\frac12\int_0^\pi\cos\left(\frac\pi3-\frac\pi3\cos\theta\right)d\theta \).
2. Expand cos(A − B): the sin(π/3)·sin((π/3)cos θ) part integrates to 0 over [0, π] (it is odd about θ = π/2), so \( u=\frac12\cos\frac\pi3\int_0^\pi\cos\left(\frac\pi3\cos\theta\right)d\theta=\frac14\int_0^\pi\cos\left(\frac\pi3\cos\theta\right)d\theta \).
3. The integral over [0, π] is twice that over [0, π/2], and by θ → π/2 − θ it equals \( 2\int_0^{\pi/2}\cos\left(\frac\pi3\sin x\right)dx=2v \). So u = v/2, i.e. 2u = v.

Teacher's Note:
1. Look for a substitution that turns sin²x into a first power of a sine or cosine.
2. cos(π/3) = 1/2 supplies the second factor of 1/2.

Answer: (c) \( \frac12 \)

Explanation:
1. By the Leibniz rule: \( \frac{x^{3/2}}{x-1}\cdot\frac{1}{2\sqrt x}=\frac{x}{2(x-1)} \).
2. As x → ∞ this tends to 1/2.

Teacher's Note:
1. The upper limit is √x, so multiply the integrand at r = √x by its derivative 1/(2√x).
2. Divide numerator and denominator by x to find the limit.

Answer: (b) \( \frac32-\frac1e \)

Explanation:
1. Integrate by parts: \( \int_1^ef''\ln x=[f'\ln x]_1^e-\int_1^e\frac{f'}{x}dx=1-\int_1^e\frac{f'}{x}dx \).
2. Again by parts: \( \int_1^e\frac{f'}{x}dx=\left[\frac fx\right]_1^e+\int_1^e\frac f{x^2}dx=\frac1e-1+\frac12 \).
3. Result: \( 1-\left(\frac1e-\frac12\right)=\frac32-\frac1e \).

Teacher's Note:
1. Integrate by parts twice, moving derivatives from f″ to ln x.
2. Substitute the given values f(1) = f(e) = f′(e) = 1.

Answer: (c) \( \frac12\int_0^{\pi/2}\frac{x}{\sin x}dx \)

Explanation:
1. Put x = tan θ: \( \int_0^{\pi/4}\frac{\theta\sec^2\theta}{\tan\theta}d\theta=\int_0^{\pi/4}\frac{\theta}{\sin\theta\cos\theta}d\theta=2\int_0^{\pi/4}\frac{\theta}{\sin2\theta}d\theta \).
2. Let u = 2θ: \( 2\cdot\frac14\int_0^{\pi/2}\frac{u}{\sin u}du=\frac12\int_0^{\pi/2}\frac{u}{\sin u}du \).

Teacher's Note:
1. Substituting x = tan θ makes the inverse tangent simple.
2. Use sin 2θ = 2 sin θ cos θ and then rescale the variable.

Answer: (b) a ∈ R⁺

Explanation:
1. Left side: \( 2-2\log_2a \). Right side: \( \log_24-\log_2a^2=2-2\log_2a \).
2. The two sides are equal for every a for which the logarithm is defined, i.e. a > 0.

Teacher's Note:
1. Both sides simplify to the same expression, so the equation is an identity.
2. The only restriction comes from the domain of log₂ a.

Answer: (a) 1/3

Explanation:
1. For small t, ln(1 + t²) ≈ t² and 1/(1 + eᵗ) ≈ 1/2.
2. \( \int_{-1/x}^{1/x}\frac{t^2}{2}dt=\frac{1}{3x^3} \).
3. Multiplying by x³ gives 1/3.

Teacher's Note:
1. Replace the integrand by its leading behaviour near t = 0.
2. The interval shrinks as x grows, so only small t matters.

Answer: (b) y

Explanation:
1. x = sinh⁻¹y, so y = sinh x.
2. Differentiating twice: d²y/dx² = sinh x = y.

Teacher's Note:
1. The integral is the inverse hyperbolic sine.
2. Alternatively, dx/dy = 1/√(1 + y²) gives dy/dx = √(1 + y²) and d²y/dx² = y.

Answer: (a) \( 2-\sqrt2 \)

Explanation:
1. The integrand is \( e^{-x}(\operatorname{cosec}x+\operatorname{cosec}x\cot x) \), which is \( -\frac{d}{dx}\left(e^{-x}\operatorname{cosec}x\right) \).
2. Value = \( e^{-\pi/6}\operatorname{cosec}\frac\pi6-e^{-\pi/4}\operatorname{cosec}\frac\pi4=2e^{-\pi/6}-\sqrt2e^{-\pi/4} \).
3. a = 2, b = −√2, so a + b = 2 − √2.

Teacher's Note:
1. Spot that the numerator is exactly the derivative of e⁻ˣ cosec x with a negative sign.
2. Evaluate at the two limits.

Answer: (c) 1/2

Explanation:
1. \( \int_0^\infty e^{-ax}\sin bx\,dx=\frac{b}{a^2+b^2} \) and \( \int_0^\infty e^{-ax}\cos bx\,dx=\frac{a}{a^2+b^2} \).
2. With a = b = 2 each integral is 2/8 = 1/4, so the total is 1/2.

Teacher's Note:
1. Use the standard Laplace-type integrals for sine and cosine times an exponential.
2. Both have the same value when a = b.