JEE Mathematics Definite Integration MCQs Set 11

Answer: (d) \( l(x)=\begin{cases}x\sin x & 0<x\le\frac\pi2\\ \frac\pi2\sin(x+\pi) & \frac\pi2<x<\pi\end{cases} \)

Explanation:
1. (a) sin x is continuous everywhere. (b) g is an integral of a bounded function, so it is continuous.
2. (c) At x = 3π/4: the left value is 1 and the right limit is 2 sin(π/6) = 1, so h is continuous.
3. (d) At x = π/2: the left value is (π/2) sin(π/2) = π/2, while the right limit is (π/2) sin(3π/2) = −π/2. They differ, so l is not continuous.

Teacher's Note:
1. Check the join points of piecewise functions by comparing left and right limits.
2. An integral of a bounded function is always continuous.

Answer: (a) A > 0 ; B < 0

Explanation:
1. \( A=\left[-\frac{\cos x^2}2\right]_{-1}^2=\frac{\cos1-\cos4}2 \) = (0.540 + 0.654)/2 > 0.
2. \( B=\left[\frac{\sin x^2}2\right]_{-1}^2=\frac{\sin4-\sin1}2 \) = (−0.757 − 0.841)/2 < 0.

Teacher's Note:
1. Both integrands are exact derivatives because of the factor x.
2. Use cos 4 < 0 and sin 4 < 0 (4 radians is in the third quadrant).

Answer: (c) 4

Explanation:
1. The integrand is even, so the integral is \( 2\int_0^1x^{-1/2}dx \). The singularity at 0 is integrable.
2. = 2 × 2 = 4.

Teacher's Note:
1. Improper integrals with x^(−1/2) converge at 0.
2. Use symmetry to double the integral over [0, 1].

Answer: (b) a = 2 ; b = 0

Explanation:
1. The term x³/(x² + 2|x| + 1) is odd and integrates to 0.
2. The rest is even: \( 2\int_0^1\frac{x+1}{(x+1)^2}dx=2\ln2 \).
3. So a = 2, b = 0.

Teacher's Note:
1. Split into odd and even parts over a symmetric interval.
2. For x ≥ 0 the denominator is (x + 1)².

Answer: (a) \( \frac14\ln^2\frac13 \)

Explanation:
1. Let u = ln((1 + x)/(1 − x)); then du = 2dx/(1 − x²).
2. The integral is \( \frac12\int_0^{\ln3}u\,du=\frac{\ln^23}{4} \).
3. Since ln²(1/3) = ln²3, this equals (1/4) ln²(1/3), option (a).

Teacher's Note:
1. The derivative of ln((1 + x)/(1 − x)) is 2/(1 − x²).
2. This turns the integral into ∫u du.

Answer: (d) 17

Explanation:
1. The sum is \( \frac1n\sum\frac kn\sqrt{1+(k/n)^2}\to\int_0^1x\sqrt{1+x^2}dx=\frac{2\sqrt2-1}{3} \).
2. Write \( \frac{2\sqrt2-1}{3}=\frac{\sqrt8-1}{\sqrt9} \), so a = 8, b = 9 and a + b = 17.

Teacher's Note:
1. Rewrite as a Riemann sum with step 1/n.
2. Match the final form √a − 1 over √b.

Answer: (c) \( \frac\pi4-\frac12\ln2 \)

Explanation:
1. By x → π/2 − x, the integrals of sin x and cos x over the same denominator are equal. Call the common value S.
2. 2S = \( \int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sin x+\cos x}dx=\int_0^{\pi/2}\left(1-\frac{1}{1+\sin x+\cos x}\right)dx=\frac\pi2-\ln2 \).
3. S = π/4 − (1/2) ln 2.

Teacher's Note:
1. Use the symmetry x → π/2 − x to relate the sine and cosine integrals.
2. Write (sin + cos)/(1 + sin + cos) = 1 − 1/(1 + sin + cos).

Answer: (b) π

Explanation:
1. cos⁶x + sin⁶x = 1 − 3sin²x cos²x = 1 − (3/4) sin²2x.
2. With u = 2x: \( \int_0^{\pi/2}\frac{dx}{1-\frac34\sin^22x}=\frac12\int_0^\pi\frac{du}{1-\frac34\sin^2u} \).
3. Using \( \int_0^\pi\frac{du}{1-k\sin^2u}=\frac{\pi}{\sqrt{1-k}} \) with k = 3/4: \( \frac12\cdot\frac{\pi}{1/2}=\pi \).

Teacher's Note:
1. Reduce cos⁶ + sin⁶ to a function of sin 2x.
2. Use ∫₀^π du/(1 − k sin²u) = π/√(1 − k).

Answer: (d) 2/3

Explanation:
1. The function F(x) = 1/(1 + 2^{1/x}) is discontinuous at 0, so split the integral at 0 and use limits: F(0⁻) = 1 (2^{1/x} → 0) and F(0⁺) = 0 (2^{1/x} → ∞).
2. Total = [F(0⁻) − F(−1)] + [F(1) − F(0⁺)] = (1 − 2/3) + (1/3 − 0) = 2/3.

Teacher's Note:
1. The Fundamental Theorem needs continuity, so treat the break at x = 0 with one-sided limits.
2. F(−1) = 1/(1 + 1/2) = 2/3 and F(1) = 1/3.

Answer: (b) 4u + v = 0

Explanation:
1. With x = tan θ: \( u=\int_0^{\pi/4}\ln(1+\tan\theta)d\theta=\frac\pi8\ln2 \) (standard result).
2. \( v=\int_0^{\pi/2}\ln\sin2x\,dx=-\frac\pi2\ln2 \).
3. So 4u + v = (π/2) ln 2 − (π/2) ln 2 = 0.

Teacher's Note:
1. Both integrals are classical results involving π ln 2.
2. Use ∫₀^{π/4} ln(1 + tan θ)dθ = (π/8) ln 2.

Answer: (a) \( \sqrt{\frac{3c}{2}} \)

Explanation:
1. The integral is \( \frac{ac}{b+2}+\frac{a^3b}{3b+6}=\frac{1}{b+2}\left(ac+\frac{a^3b}{3}\right) \).
2. For it to be independent of b, the bracket must be proportional to (b + 2): \( ac+\frac{a^3}3b=\frac{a^3}3(b+2)\Rightarrow ac=\frac{2a^3}3 \).
3. So a² = 3c/2.

Teacher's Note:
1. Integrate term by term and factor out 1/(b + 2).
2. Match the coefficient of b with the constant term.

Answer: (b) \( x_1f(x_1) \)

Explanation:
1. \( \lim\frac{1}{x-x_1}\int_{x_1}^xf(t)dt=f(x_1) \) and x → x₁.
2. So the limit is x₁ f(x₁).

Teacher's Note:
1. The quotient is the average value of f over a shrinking interval.
2. The factor x tends to x₁.

Answer: (d) 1/6

Explanation:
1. sin x sin 3x = (cos 2x − cos 4x)/2, so the integrand is \( \frac12\left(\sin2x\cos2x-\sin2x\cos4x\right)=\frac14\sin4x-\frac14(\sin6x-\sin2x) \).
2. Integrating over [0, π/2]: ∫sin 4x = 0, ∫sin 6x = 1/3, ∫sin 2x = 1.
3. Value = (1/4)(0) − (1/4)(1/3 − 1) = 1/6.

Teacher's Note:
1. Convert products of sines into sums with product-to-sum formulas.
2. Integrate each sine over [0, π/2].

Answer: (c) \( U_n=2^{2n}V_n \)

Explanation:
1. x(2 − x) = 1 − (1 − x)², so \( U_n=\int_0^1(1-u^2)^ndu \).
2. Put x = (1 + s)/2 in V: x(1 − x) = (1 − s²)/4, so \( V_n=4^{-n}\int_0^1(1-s^2)^nds=2^{-2n}U_n \).
3. Hence Uₙ = 2²ⁿ Vₙ.

Teacher's Note:
1. Complete the square: x(2 − x) = 1 − (x − 1)² and x(1 − x) = 1/4 − (x − 1/2)².
2. Compare the two resulting integrals.

Answer: (b) \( e^{\frac1h\int_a^{a+h}\ln\phi(x)dx} \)

Explanation:
1. Take the logarithm: \( \frac1n\sum_{k}\ln\phi\!\left(a+\frac{kh}{n}\right)=\frac1h\cdot\frac hn\sum\ln\phi\!\left(a+\frac{kh}n\right)\to\frac1h\int_a^{a+h}\ln\phi(x)dx \).
2. Exponentiating gives option (b).

Teacher's Note:
1. The log of a product is a sum, which is a Riemann sum with step h/n.
2. The variable runs from a to a + h.

Answer: (b) derivable and continuous in its domain

Explanation:
1. \( S'(x)=3x^2\ln x^3-2x\ln x^2=9x^2\ln x-4x\ln x \).
2. \( H(x)=(9x-4)\ln x \), which is a product of differentiable functions on x > 0.
3. So H is both derivable and continuous.

Teacher's Note:
1. Differentiate both limits with the chain rule.
2. Every derivable function is continuous.

Answer: (c) a = 2, b = 14

Explanation:
1. Using x → a + b − x with a + b = 16, the integral equals half the interval length: \( \frac{b-a}{2} \) (valid for any n when a + b = 16).
2. So b − a = 12 and a + b = 16, giving a = 2, b = 14.

Teacher's Note:
1. When a + b = 16 the fraction and its partner add to 1, so I = (b − a)/2.
2. Only option (c) satisfies both a + b = 16 and b − a = 12.

Answer: (c) 2

Explanation:
1. The derivative is \( \frac{\cos x}{1-\sin^2x}+\frac{\sin x}{1-\cos^2x}=\frac1{\cos x}+\frac1{\sin x} \).
2. Set sin x + cos x = 2√2 sin x cos x. With s = sin x + cos x, sin x cos x = (s² − 1)/2, so √2 s² − s − √2 = 0 and s = √2 or s = −1/√2.
3. s = √2 gives x = π/4. s = −1/√2 gives sin(x + π/4) = −1/2, so x = 11π/12 in [0, π]. Two solutions.

Teacher's Note:
1. Differentiate the integral with the Leibniz rule.
2. Solve the resulting equation using s = sin x + cos x.

Answer: (d) indeterminate (does not exist)

Explanation:
1. The integrand is not defined at x = 1 and \( \int\frac{dx}{(1-x)^2} \) diverges near x = 1.
2. The formal value −2 obtained from the antiderivative 1/(1 − x) is not valid because the integrand is not bounded on [0, 2]. The integral does not exist.

Teacher's Note:
1. The Fundamental Theorem requires a continuous integrand on the whole interval.
2. 1/(1 − x)² behaves like 1/(x − 1)² near the break, which is not integrable.

Answer: (a) \( \frac2\pi\int_0^\pi f(x)dx \)

Explanation:
1. \( f(x)f\!\left(\frac\pi2-x\right)=\frac{\sin x\cos x}{x(\pi/2-x)}=\frac{\sin2x}{\pi}\left(\frac1x+\frac1{\pi/2-x}\right) \).
2. The two parts give equal integrals over [0, π/2], so the integral is \( \frac2\pi\int_0^{\pi/2}\frac{\sin2x}{x}dx=\frac2\pi\int_0^\pi\frac{\sin u}{u}du \).

Teacher's Note:
1. Use partial fractions 1/(x(π/2 − x)) = (2/π)(1/x + 1/(π/2 − x)).
2. The substitution x → π/2 − x shows the two integrals are equal.

Answer: (a) 1

Explanation:
1. Split at x = α/2: \( \int_0^{\alpha/2}\sin(\alpha-2x)dx=\frac{1-\cos\alpha}2 \).
2. \( \int_{\alpha/2}^{\pi/2}\sin(2x-\alpha)dx=\frac{1+\cos\alpha}2 \).
3. Sum = 1.

Teacher's Note:
1. Remove the modulus by splitting the interval where 2x − α changes sign.
2. The two pieces add up to a constant independent of α.

Answer: (a) 1/2

Explanation:
1. Put u = 1 − sin t: \( a_n=2\int_0^1u^n(1-u)du=\frac{2}{(n+1)(n+2)} \).
2. \( \sum_{n=1}^\infty\frac{2}{n(n+1)(n+2)}=2\cdot\frac14=\frac12 \) (telescoping).

Teacher's Note:
1. sin 2t dt = 2 sin t cos t dt, and cos t dt = −du.
2. 1/(n(n + 1)(n + 2)) = ½[1/(n(n + 1)) − 1/((n + 1)(n + 2))].