JEE Mathematics Definite Integration MCQs Set 13

Answer: (b) 1/6

Explanation:
1. Near 0, \( \sin x-x\cos x\approx\frac{x^3}{3} \), so \( f(x)\approx\frac x3 \).
2. Hence \( \int_0^t f\,dx\approx\frac{t^2}{6} \).
3. The required limit is 1/6.

Teacher's Note:
1. Expand sin x and cos x as series to get the leading behaviour of f.
2. Integrate that leading term and compare with t².

Answer: (c) 2

Explanation:
1. Since 1 − cos 2θ = 2 sin²θ, \( f(n)=\int_0^{\pi/2}\frac{\sin^2nx}{\sin^2x}dx \).
2. \( f(n+1)-f(n)=\int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2 \) for every n, and f(0) = 0.
3. So \( f(n)=\frac{n\pi}{2} \) and \( \frac{f(n+1)+f(n-1)}{f(n)}=\frac{(n+1)+(n-1)}{n}=2 \).

Teacher's Note:
1. The difference f(n+1) − f(n) is a standard integral equal to π/2.
2. This makes f(n) a linear function of n.

Answer: (d) 0

Explanation:
1. With f(k) = kπ/2 the determinant equals \( \left(\frac\pi2\right)^3\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix} \).
2. In each row the entries are in arithmetic progression, so C₃ − C₂ = C₂ − C₁ and the columns are linearly dependent.
3. The determinant is 0.

Teacher's Note:
1. Factor out π/2 from every entry.
2. Rows or columns forming an arithmetic progression give a zero determinant.

Answer: (a) 8

Explanation:
1. For ℓ non-zero, b = 1 and \( \ell=\frac{x^3/(3a^{1/p})}{x^3/6}=\frac{2}{a^{1/p}} \).
2. With p = 3 and ℓ = 1: a^{1/3} = 2, so a = 8.

Teacher's Note:
1. Find ℓ in general as 2 / a^(1/p).
2. Then substitute p = 3 and ℓ = 1.

Answer: (b) 2/3

Explanation:
1. ℓ is non-zero only for b = 1, and then \( \ell=\frac{2}{a^{1/p}}=\frac{2}{9^{1/2}} \).
2. So ℓ = 2/3.

Teacher's Note:
1. Use the general formula ℓ = 2 / a^(1/p).
2. Remember that a non-zero limit needs b = 1.

Answer: (b) 9

Explanation:
1. Let \( h(x)=f(x)f(-x) \). Then \( h'(x)=f'(x)f(-x)-f(x)f'(-x)=0 \).
2. So h is constant, equal to h(0) = f(0)² = 9.

Teacher's Note:
1. Differentiate the product f(x)f(−x) and use the given relation.
2. A zero derivative means the product is constant.

Answer: (a) 17

Explanation:
1. Since f(x)f(−x) = 9, \( \frac1{3+f(x)}+\frac1{3+f(-x)}=\frac1{3+f}+\frac{f}{3(f+3)}=\frac13 \).
2. Hence \( \int_{-51}^{51}g\,dx=\int_0^{51}[g(x)+g(-x)]dx=\frac{51}{3}=17 \).

Teacher's Note:
1. Pair x with −x using the integral property over a symmetric interval.
2. The pair sum of the integrand turns out to be constant.

Answer: (c) \( \frac{e^2-1}{2}\left(a^2+\frac{b^2}{e^2}\right) \)

Explanation:
1. \( D(t)=\frac{1}{\sqrt{a^2e^{2t}+b^2e^{-2t}}} \), so \( \frac1{D^2}=a^2e^{2t}+b^2e^{-2t} \).
2. \( I=a^2\frac{e^2-1}{2}+b^2\frac{1-e^{-2}}{2}=\frac{e^2-1}{2}\left(a^2+\frac{b^2}{e^2}\right) \).

Teacher's Note:
1. The distance from the origin to px + qy = 1 is 1/√(p² + q²).
2. Integrate the two exponentials separately.

Answer: (d) √e

Explanation:
1. Put a = 1/b: \( I=\frac{e^2-1}{2}\left(\frac1{b^2}+\frac{b^2}{e^2}\right) \).
2. By AM–GM, \( \frac1{b^2}+\frac{b^2}{e^2}\ge\frac2e \), with equality when \( b^4=e^2 \), i.e. b = √e.

Teacher's Note:
1. Eliminate a using ab = 1.
2. Apply the AM–GM inequality to the two positive terms.

Answer: (b) e − 1/e

Explanation:
1. The minimum of \( \frac1{b^2}+\frac{b^2}{e^2} \) is 2/e.
2. So \( I_{min}=\frac{e^2-1}{2}\cdot\frac2e=e-\frac1e \).

Teacher's Note:
1. Use the minimum found by AM–GM.
2. Multiply by the factor (e² − 1)/2.

Answer: (b) 1, 3 and 4 only

Explanation:
1. Statement 1 is true: replacing x by π − x and adding gives \( 2I=\pi\int f(\sin x)dx \).
2. Statement 2 is false: it needs f(x)² to be an even function, which is not true for every f (for example f(x) = x + 1).
3. Statement 3 is true because cos²x has period π, so n periods give n times one period.
4. Statement 4 is true by the substitution x + c = u.

Teacher's Note:
1. Check each property against its usual conditions.
2. Property 2 needs an even integrand, which is not given.

Answer: (a) 1, 3 and 4 only

Explanation:
1. Partial fractions: \( \frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}=\frac{2}{x+1}-\frac{1}{(x+1)^2+1} \).
2. \( I=\left[2\ln(x+1)-\tan^{-1}(x+1)\right]_0^1=2\ln2-\tan^{-1}2+\frac\pi4 \), about 1.0645, so statement 1 is true.
3. Since \( \tan\left(\frac\pi4-\tan^{-1}2\right)=-\frac13 \), we get \( \frac\pi4-\tan^{-1}2=-\cot^{-1}3 \) (statement 3 true) and \( -\frac\pi4+\cot^{-1}2=-\cot^{-1}3 \) (statement 4 true). Statement 2 gives about 1.85, so it is false.

Teacher's Note:
1. Reduce all options to a common form using inverse-tangent identities.
2. Compare each with the value of the integral.

Answer: (d) 1, 2 and 3 only

Explanation:
1. In I₁ replace x by π/2 − x: cos(π cos²x) = cos(π − π sin²x) = −cos(π sin²x), so I₁ = −I₁ and I₁ = 0.
2. Write \( \cos(2\pi\sin^2x)=\cos(\pi-\pi\cos2x)=-\cos(\pi\cos2x) \). With y = 2x and symmetry about π/2, \( I_2=-\int_0^{\pi/2}\cos(\pi\cos y)\,dy=-\int_0^{\pi/2}\cos(\pi\sin y)\,dy=-I_3 \).
3. So statements 1, 2 and 3 are true. Statement 4 would need I₃ = 0, but I₃ ≈ −0.478, so it is false.

Teacher's Note:
1. Use x → π/2 − x on I₁ to see that it equals its own negative.
2. Reduce I₂ to cos(π cos 2x) with the identity for cos(π − θ).

Answer: (d) 15

Explanation:
1. This is a Riemann sum with x = 3k/n and dx = 3/n over [0, 3].
2. \( \int_0^3(x^2+2)\,dx=9+6=15 \).

Teacher's Note:
1. Identify x = 3k/n and the width 3/n to convert the sum to an integral.
2. The upper limit is 3, not 1.

Answer: (b) 1 and 2 only

Explanation:
1. Integrating by parts, \( I_n=\frac{1}{2^n}+2n\int_0^1\frac{x^2}{(1+x^2)^{n+1}}dx=\frac1{2^n}+2n(I_n-I_{n+1}) \), which rearranges to statement 1.
2. For n = 1 (I₁ = π/4): \( 2I_2=\frac12+\frac\pi4 \), so \( I_2=\frac\pi8+\frac14 \). Statement 2 is true and 3 is false.
3. For n = 2: \( 4I_3=\frac14+3I_2=1+\frac{3\pi}{8} \), so \( I_3=\frac14+\frac{3\pi}{32}\approx0.5445 \), not \( \frac\pi{16}-\frac5{48} \). Statement 4 is false.

Teacher's Note:
1. Integration by parts gives a reduction formula for I_n.
2. Test each closed form with n = 1 and n = 2.

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

Explanation:
1. Put x → 1/x: \( I=\int_0^\infty\frac{dx}{(1+x)(1+x^2)} \).
2. Adding the two forms: \( 2I=\int_0^\infty\frac{(x+1)\,dx}{(1+x)(1+x^2)}=\int_0^\infty\frac{dx}{1+x^2}=\frac\pi2 \).
3. So I = π/4 (it is also equal to the integral without x in the numerator).

Teacher's Note:
1. The substitution x → 1/x maps one integral onto the other.
2. The sum of the two forms is a basic arctangent integral.

Answer: (d) 2, 3 and 4 only

Explanation:
1. With x → 1/x, \( u=\int_0^\infty\frac{x^2dx}{1+7x^2+x^4}=v \), so u = v and statement 1 is false.
2. \( u+v=\int_0^\infty\frac{(1+x^2)dx}{x^4+7x^2+1}=\int_0^\infty\frac{(1+1/x^2)dx}{x^2+1/x^2+7} \); put t = x − 1/x to get \( \int_{-\infty}^{\infty}\frac{dt}{t^2+9}=\frac\pi3 \).
3. So u + v = π/3 (true), u = v = π/6, hence 6v = π (true) and 3u + 2v = 5π/6 (true).

Teacher's Note:
1. Show u = v with x → 1/x.
2. Use the substitution t = x − 1/x for u + v.

Answer: (d) \( e^{2}\text{ or }e^{-2} \)

Explanation:
1. \( f(1/x)=\int_1^{1/x}\frac{\ln t}{1+t}dt \); put t = 1/s to get \( f(1/x)=\int_1^x\frac{\ln s}{s(1+s)}ds \).
2. So \( f(x)+f(1/x)=\int_1^x\frac{\ln t}{t}dt=\frac{\ln^2x}{2} \).
3. Setting this equal to 2 gives ln x = ±2, i.e. x = e² or x = e⁻².

Teacher's Note:
1. Substitute t = 1/s in f(1/x) and add.
2. The sum becomes the integral of (ln t)/t.

Answer: (b) 2 and 4 only

Explanation:
1. Let \( K=\int_0^1f(t)e^{-t}dt \). Putting x = 0 gives 0 = 1 − cK, so cK = 1.
2. Differentiating: \( f(x)=e^x-2cKe^{2x}=e^x-2e^{2x} \).
3. \( K=\int_0^1(1-2e^{t})dt=3-2e \), so c = 1/K = 1/(3 − 2e).
4. Statements 2 and 4 are true; 1 and 3 are false.

Teacher's Note:
1. Put x = 0 to find cK, then differentiate.
2. Recompute K from the f you found.

Answer: (c) 3 and 4 only

Explanation:
1. f(0) = ∫(1 − |t|) dt over [−1, 1] = 1, so statement 1 is false.
2. The integrand is continuous in (x, t), so f is continuous and \( \lim_{x\to0}f(x)=f(0)=1 \).
3. So statements 3 and 4 are true; statement 2 is false.

Teacher's Note:
1. Evaluate f(0) directly.
2. An integral of a jointly continuous integrand is continuous in the parameter.

Answer: (d) 1, 3 and 4 only

Explanation:
1. Differentiating: \( f'(x)=\int_0^xe^t\cos(x-t)dt \) (the boundary term is 0), and \( f''(x)=e^x-\int_0^xe^t\sin(x-t)dt=e^x-f(x) \).
2. So g(x) = f + f″ = eˣ. It is positive, not constant, g(1) = e, and g′ = g.
3. Statements 1, 3 and 4 are true.

Teacher's Note:
1. Differentiate under the integral sign twice.
2. The result simplifies to g(x) = e^x.

Answer: (a) 1 and 2 only

Explanation:
1. Differentiate under the integral sign: \( f'(x)=\int_0^{\pi/2}\frac{d\theta}{1+x\sin^2\theta}=\frac{\pi}{2\sqrt{1+x}} \).
2. Since f(0) = 0, integrating gives \( f(x)=\pi(\sqrt{x+1}-1) \).
3. Statements 1 and 2 are both true.

Teacher's Note:
1. Use differentiation under the integral sign (Leibniz rule).
2. Use the standard integral of 1/(1 + x sin²θ) over [0, π/2].

Answer: (b) 1, 2 and 4 only

Explanation:
1. Applying f to both sides: f(1 − x) = f(f(f(x))) = 1 − f(x), so f(x) + f(1 − x) = 1. Statement 1 holds.
2. Then \( J=\int_0^1f(x)dx=\int_0^1f(1-x)dx \) and adding gives 2J = 1, so J = 1/2 (statement 2).
3. Statement 3: if f(1/3) = a then f(2/3) = 1 − a and a(1 − a) ≤ 1/4 < 1, so it is false.
4. Statement 4: by symmetry \( \int_0^{\pi/2}\frac{\sin x}{(\sin x+\cos x)^3}dx=\frac12\int_0^{\pi/2}\frac{dx}{(\sin x+\cos x)^2}=\frac12\int_0^{\infty}\frac{dt}{(1+t)^2}=\frac12 \) (t = tan x), equal to J.

Teacher's Note:
1. Show f(x) + f(1 − x) = 1 by applying f again.
2. Use the sin x ↔ cos x symmetry in the last integral.

Answer: (b) 3 and 4 only

Explanation:
1. Differentiate: \( f'(x)+g(x)=\cos2x-\sin2x \).
2. With \( f'^2+g^2=1 \) we get \( 2f'g=(\cos2x-\sin2x)^2-1=-2\sin2x\cos2x \), so {f′, g} = {cos 2x, −sin 2x}.
3. Option 3: f′ = cos 2x and g = −sin 2x works. Option 4: f′ = −sin 2x and g = cos 2x also works (and f(0) = 0 in both cases). Options 1 and 2 fail the first equation.

Teacher's Note:
1. Differentiate the given integral equation.
2. Use the second condition to separate f′ and g.

Answer: (a) positive

Explanation:
1. Let h(x) = g(1 − x) − g(x). Since g decreases, h < 0 for x < 1/2 and h > 0 for x > 1/2, and h(1 − x) = −h(x).
2. Then \( I=\int_0^1fh\,dx=\int_{1/2}^1[f(x)-f(1-x)]\,h(x)\,dx \).
3. For x > 1/2, f(x) > f(1 − x) and h(x) > 0, so I > 0.

Teacher's Note:
1. Pair x with 1 − x to rewrite I over [1/2, 1].
2. Both factors are positive there.