JEE Mathematics Definite Integration MCQs Set 12

Answer: (a) \( 2\sqrt{e^{\pi/2}} \)

Explanation:
1. The integrand is \( e^{x/2}\left(\sqrt{\sin x}+\frac{\cos x}{\sqrt{\sin x}}\right)=\frac{d}{dx}\left(2e^{x/2}\sqrt{\sin x}\right) \).
2. Value = \( 2e^{\pi/4}\cdot1-0=2\sqrt{e^{\pi/2}} \).

Teacher's Note:
1. Look for a derivative of a product e^{x/2}·√(sin x).
2. Evaluate at π/2 and 0.

Answer: (c) \( -\frac{\ln^22}2 \)

Explanation:
1. Since \( \frac d{dx}\ln^2(\cos x)=-2\tan x\ln\cos x \), the integrand is \( -\frac12\frac d{dx}\ln^2\cos x \).
2. Value = \( -\frac12\ln^2\left(\frac12\right)=-\frac{\ln^22}{2} \).

Teacher's Note:
1. Recognise tan x as the negative derivative of ln cos x.
2. ln²(1/2) = ln²2.

Answer: (c) \( \frac{27}{4e} \)

Explanation:
1. (3n)!/(2n)! = (2n + 1)(2n + 2)…(3n). So \( \ln\frac{P_n}n=\frac1n\sum_{k=2n+1}^{3n}\ln\frac kn\to\int_2^3\ln x\,dx \).
2. \( \int_2^3\ln x\,dx=3\ln3-2\ln2-1 \), so the limit is \( e^{3\ln3-2\ln2-1}=\frac{27}{4e} \).

Teacher's Note:
1. Take logarithms to make a Riemann sum.
2. ∫ ln x dx = x ln x − x.

Answer: (a) \( n^2\pi\ln2 \)

Explanation:
1. Using x → nπ − x: \( I=\frac{n\pi}2\int_0^{n\pi}\frac{|\sin x|}{1+|\cos x|}dx \).
2. The integrand has period π: \( \int_0^{n\pi}=n\cdot2\int_0^{\pi/2}\frac{\sin x}{1+\cos x}dx=2n\ln2 \).
3. I = (nπ/2)(2n ln 2) = n²π ln 2.

Teacher's Note:
1. Use the property for g(nπ − x) = g(x).
2. |sin x|/(1 + |cos x|) repeats every π and is symmetric about π/2.

Answer: (d) \( e^2+e^{-1}-3 \)

Explanation:
1. eˣ − 1 changes sign at x = 0.
2. \( \int_{-1}^0(1-e^x)dx=1-(1-e^{-1})=e^{-1} \) and \( \int_0^2(e^x-1)dx=e^2-3 \).
3. Total = e² + e⁻¹ − 3.

Teacher's Note:
1. Split the integral where the expression inside the modulus changes sign.
2. e⁰ = 1 gives the split point x = 0.

Answer: (b) \( e^{\sin^2y} \)

Explanation:
1. The bracket equals \( \int_y^{x+y}e^{\sin^2t}dt \).
2. Divided by x, as x → 0 this tends to the integrand at t = y: e^{sin²y}.

Teacher's Note:
1. Combine the two integrals into one over [y, x + y].
2. The average over a shrinking interval is the value at its left end point.

Answer: (b) 0

Explanation:
1. Put x = 1/u: \( I=\int_{\cot\theta}^{\tan\theta}f\!\left(\frac1u\right)\left(-\frac{du}{u^2}\right)=\int_{\tan\theta}^{\cot\theta}\frac{f(1/u)}{u^2}du \).
2. Since f(1/u) = −u² f(u), this is −I. So I = 0.

Teacher's Note:
1. The limits tan θ and cot θ are reciprocals, which suggests the substitution x → 1/x.
2. An integral equal to its own negative is zero.

Answer: (b) \( \ln\left|\tan^{-1}(\sec x+\cos x)\right|+c \)

Explanation:
1. Put t = cos x; then sin²x = 1 − t² and sin x dx = −dt.
2. Divide numerator and denominator by t²: with w = t + 1/t (so dw = (1 − 1/t²)dt and t² + 1/t² + 3 = w² + 1), the integral becomes \( \int\frac{dw}{(w^2+1)\tan^{-1}w} \).
3. With z = tan⁻¹w this is ∫dz/z = ln|tan⁻¹(sec x + cos x)| + c.

Teacher's Note:
1. The combination t + 1/t = sec x + cos x suggests w = t + 1/t.
2. Then tan⁻¹w has derivative 1/(1 + w²), which appears in the integrand.

Answer: (c) 0

Explanation:
1. Treat f as a probability density: mean = α and E[x²] = α², so the variance is α² − α² = 0.
2. \( \int_0^1f(x)(x-\alpha)^2dx=\alpha^2-2\alpha^2+\alpha^2=0 \).
3. A positive continuous f would make this integral positive. So no such function exists: 0.

Teacher's Note:
1. Combine the three conditions into ∫ f(x)(x − α)² dx = 0.
2. The integrand of a positive continuous function times a square is non-negative and not identically zero.

Answer: (a) \( \frac1{1+\ln2} \)

Explanation:
1. Let c = 2e, so the integrand is t cᵗ. An antiderivative is \( c^t\left(\frac t{\ln c}-\frac1{\ln^2c}\right) \), which tends to 0 as t → −∞.
2. f(x) = 0 when x/ln c = 1/ln²c, i.e. x = 1/ln c = 1/(1 + ln 2).

Teacher's Note:
1. Combine 2ᵗ eᵗ = (2e)ᵗ and integrate by parts.
2. The lower limit −∞ contributes 0.

Answer: (d) None of these

Explanation:
1. max(|x + y|, |x − y|) = |x| + |y|, so the curve is |x| + |y| = 2.
2. The part above the x-axis is f(x) = 2 − |x|.
3. \( \int_{-1/2}^{1/2}(2-|x|)dx=2-\frac14=\frac74 \), which is not among (a) to (c).

Teacher's Note:
1. max(|a|, |b|) with a = x + y and b = x − y equals |x| + |y|.
2. The curve is a square tilted at 45°.

Answer: (c) (2, ∞)

Explanation:
1. For a > 0: \( \int_0^{2a}|x-a|dx=a^2 \) and \( \int_0^{2a}5\,dx=10a \), so we need a² + 10a > 24, i.e. (a + 12)(a − 2) > 0, giving a > 2.
2. For a < 0 the integral equals −a² + 10a (negative), which is never greater than 24. For a = 0 it is 0.
3. Hence a ∈ (2, ∞).

Teacher's Note:
1. The integral over [0, 2a] is symmetric about x = a when a > 0.
2. Check separately that negative a gives a negative integral.

Answer: (c) \( \frac3{\sqrt2} \)

Explanation:
1. The integer part is 2 for 3/(x² + 1) ∈ [2, 3), i.e. 0 < x ≤ 1/√2; it is 1 for 1/√2 < x ≤ √2; and 0 for x > √2.
2. Integral = 2 × (1/√2) + 1 × (√2 − 1/√2) = √2 + √2/2 = 3/√2.

Teacher's Note:
1. Find where 3/(x² + 1) crosses the integers 2 and 1.
2. Then add the areas of the rectangular pieces.

Answer: (b) 4

Explanation:
1. tan⁻¹(eˣ) + tan⁻¹(e⁻ˣ) = π/2.
2. With u = tan⁻¹p + tan⁻¹x, du = dx/(1 + x²): \( \frac\pi2\int\frac{du}{u}=\frac\pi2\ln\frac{2\tan^{-1}p}{\tan^{-1}p}=\frac\pi2\ln2 \).
3. So a = 2 and b = 2 (the only pair among the given sums), giving a + b = 4.

Teacher's Note:
1. tan⁻¹ y + tan⁻¹(1/y) = π/2 for y > 0.
2. The substitution u = tan⁻¹p + tan⁻¹x gives a logarithm.

Answer: (a) monotonic increasing

Explanation:
1. As k increases, 1 − k cos²x decreases for every x, so 1/√(1 − k cos²x) increases for every x.
2. Hence the integral increases with k: f is monotonic increasing.

Teacher's Note:
1. Compare the integrand pointwise for two values of k.
2. No need to evaluate the elliptic integral.

Answer: (a) \( \frac\pi{2ab(a+b)} \)

Explanation:
1. Partial fractions: \( \frac1{(x^2+a^2)(x^2+b^2)}=\frac1{b^2-a^2}\left(\frac1{x^2+a^2}-\frac1{x^2+b^2}\right) \).
2. \( \int_0^\infty\frac{dx}{x^2+a^2}=\frac\pi{2a} \), so the value is \( \frac1{b^2-a^2}\cdot\frac\pi2\left(\frac1a-\frac1b\right)=\frac{\pi}{2ab(a+b)} \).

Teacher's Note:
1. Use partial fractions in x².
2. ∫₀^∞ dx/(x² + a²) = π/(2a).

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

Explanation:
1. d(x²) = 2x dx. By parts: \( I_n=\left[x^2(\ln x)^n\right]_1^e-n\int_1^ex(\ln x)^{n-1}dx=e^2-\frac n2I_{n-1} \), since \( \int_1^e(\ln x)^{n-1}\,d(x^2)=I_{n-1} \) equals 2∫x(ln x)ⁿ⁻¹dx.
2. So 2Iₙ + nIₙ₋₁ = 2e².

Teacher's Note:
1. Integrate by parts with u = (ln x)ⁿ and dv = d(x²).
2. The remaining integral is a multiple of Iₙ₋₁.

Answer: (c) 5/2

Explanation:
1. \( I(n)=\frac{1-\cos n\pi}{n} \), which is 2/n for odd n.
2. 5ⁿ is always odd, so I(5ⁿ) = 2/5ⁿ.
3. \( \sum_{n=0}^\infty\frac2{5^n}=\frac{2}{1-1/5}=\frac52 \).

Teacher's Note:
1. cos(nπ) = (−1)ⁿ, so only odd n contribute.
2. The sum is a geometric series with ratio 1/5.

Answer: (c) 2/3

Explanation:
1. For 0 ≤ x ≤ 1: x² ≥ x³, so f = x² and g = x³; f − g = x² − x³.
2. For −1 ≤ x < 0: x|x| = −x² < 0 and x²|x| = −x³ > 0, so f = −x³ and g = −x²; f − g = −x³ + x².
3. Integral = (1/3 − 1/4) + (1/4 + 1/3) = 2/3.

Teacher's Note:
1. Compare the two expressions separately for x > 0 and x < 0.
2. Over [−1, 0] substitute x → −x to evaluate.

Answer: (c) Statement-1 is true, Statement-2 is false

Explanation:
1. Statement-1: √(1 + t²) is even, so its integral from 0 is odd, and f′(x) = √(1 + x²) is even. True.
2. Statement-2: if f′ is even then f(x) = (odd function) + constant; f is odd only if the constant is 0 (for example f(x) = sin x + 1 has even derivative cos x but is not odd). False.

Teacher's Note:
1. The integral of an even function from 0 is odd.
2. Anti-derivatives are determined only up to a constant.

Answer: (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1

Explanation:
1. By parts twice: \( \int_0^{2\pi}Pf''dx=[Pf'-P'f]_0^{2\pi}+\int_0^{2\pi}P''f\,dx \).
2. The boundary term is 0 because f(0) = f(2π) = 0 and f′(0) = f′(2π) = 0. P″ = 2, so the integral is 2∫f dx.
3. ∫₀^{2π} sin³x dx = 0 (Statement-2), and this is precisely why Statement-1 holds.

Teacher's Note:
1. Integrate by parts twice and use the periodicity of sin³x.
2. The second derivative of a monic quadratic is the constant 2.

Answer: (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1

Explanation:
1. 1/x² is unbounded near 0 and \( \int_0^3\frac{dx}{x^2} \) diverges, so the integral over [−1, 3] does not exist (Statement-1 is true).
2. The integrand is discontinuous at 0, so the Fundamental Theorem cannot be applied across 0 (Statement-2 is true) and the 'value' −4/3 from the antiderivative −1/x is invalid. Statement-2 explains Statement-1.

Teacher's Note:
1. Check continuity of the integrand on the whole interval before using an antiderivative.
2. An infinite discontinuity of order 2 is not integrable.