JEE Mathematics Definite Integration MCQs Set 14

Answer: (d) 1/4

Explanation:
1. \( x^2f-xf^2=\frac{x^3}{4}-x\left(f-\frac x2\right)^2 \).
2. \( J-K=\frac1{16}-\int_0^1x\left(f-\frac x2\right)^2dx\le\frac1{16} \).
3. So J − K ≤ 1/16; the value 1/4 is impossible, while 1/16 (f = x/2), 0 (f = 0) and −1 (f a large constant) are all attainable.

Teacher's Note:
1. Complete the square in f.
2. The correction term is always non-negative.

Answer: (b) 1 and 4 only

Explanation:
1. Pair x with −x: \( \frac1{1+e^x}+\frac1{1+e^{-x}}=1 \), so \( I=\int_0^\pi\frac{\sin nx}{\sin x}dx \).
2. This equals π for odd n and 0 for even n (including n = 0).
3. So I can be 0 or π only.

Teacher's Note:
1. Use the same pairing trick as in 1/(1 + e^x) integrals.
2. Evaluate ∫ sin nx / sin x over [0, π] for odd and even n.

Answer: (c) 1, 2, 3 and 4 only

Explanation:
1. Since f is odd, with t = −u: \( g(-x)=\int_0^{-x}f(t)dt=-\int_0^{x}f(-u)\,du=\int_0^xf(u)\,du=g(x) \), so g is even.
2. Over one full period an odd function integrates to 0: ∫₋₁¹ f = 0, so ∫₀² f = 0, hence g(2) = 0.
3. Then g(x + 2) = g(x) + ∫ₓ^{x+2} f = g(x) + 0, so g has period 2 and g(2n) = 0.
4. All four statements are correct.

Teacher's Note:
1. Use that an odd function integrates to zero over a symmetric interval.
2. A zero integral over a period makes the antiderivative periodic.

Answer: (b) 1/2

Explanation:
1. Expand: \( (\sin x+\sin ax)^2=\sin^2x+\sin^2ax+2\sin x\sin ax \). The time average of sin²x is 1/2.
2. For a = 0: sin ax = 0, so L = average of sin²x = 1/2.

Teacher's Note:
1. Average of sin² over a long interval is 1/2.
2. For a = 0 the second term vanishes.

Answer: (d) 2

Explanation:
1. Expand: \( (\sin x+\sin ax)^2=\sin^2x+\sin^2ax+2\sin x\sin ax \). The time average of sin²x is 1/2.
2. For a = 1: (2 sin x)² = 4 sin²x, whose average is 2.

Teacher's Note:
1. Combine the two equal terms first.
2. Then use the average of sin².

Answer: (a) 0

Explanation:
1. Expand: \( (\sin x+\sin ax)^2=\sin^2x+\sin^2ax+2\sin x\sin ax \). The time average of sin²x is 1/2.
2. For a = −1: sin ax = −sin x, so the integrand is 0 and L = 0.

Teacher's Note:
1. Notice that sin(−x) = −sin x.
2. The integrand then vanishes identically.

Answer: (d) 1

Explanation:
1. Expand: \( (\sin x+\sin ax)^2=\sin^2x+\sin^2ax+2\sin x\sin ax \). The time average of sin²x is 1/2.
2. For other a the cross term averages to 0 and sin²ax averages to 1/2, so L = 1/2 + 1/2 = 1.

Teacher's Note:
1. The product sin x sin ax is a sum of two cosines with non-zero frequency.
2. Their average over a long interval is 0.

Answer: (a) −1

Explanation:
1. Put x = t⁶: \( \int_0^1\frac{6t^5dt}{t^3+t^2}=6\int_0^1\frac{t^3}{t+1}dt \).
2. \( 6\int_0^1\left(t^2-t+1-\frac1{t+1}\right)dt=6\left(\frac56-\ln2\right)=5-6\ln2 \).
3. So a = 5, b = −6 and a + b = −1.

Teacher's Note:
1. Use x = t⁶ to clear both roots.
2. Divide t³ by t + 1 before integrating.

Answer: (d) 1

Explanation:
1. Since \( \frac{d}{d\theta}\left(e^{-\theta}\tan\theta\right)=e^{-\theta}(\sec^2\theta-\tan\theta) \), the integral equals e⁻ⁿ tan n.
2. So \( e^n\cdot e^{-n}\tan n=1 \), giving tan n = 1.

Teacher's Note:
1. Look for a derivative of e^(−θ) tan θ.
2. Then the factors e^n cancel.

Answer: (b) 1/2

Explanation:
1. For x between 1/(n+1) and 1/n, nx lies between n/(n+1) and 1, and tends to 1.
2. So tan⁻¹(nx) → π/4 and sin⁻¹(nx) → π/2 uniformly on the interval.
3. Both integrals have the same interval length, so the ratio tends to (π/4)/(π/2) = 1/2.

Teacher's Note:
1. Observe that nx → 1 on the whole interval.
2. The common interval length cancels.

Answer: (a) 1

Explanation:
1. Write as \( \frac{\int_{e^3}^x dt/\ln t}{x/\ln x} \) (form ∞/∞) and apply L'Hôpital's rule.
2. Numerator derivative: 1/ln x. Denominator derivative: \( \frac{\ln x-1}{\ln^2x} \).
3. Ratio: \( \frac{\ln x}{\ln x-1}\to1 \).

Teacher's Note:
1. Rewrite so that L'Hôpital applies.
2. Differentiate numerator and denominator.

Answer: (d) not existent (the expression tends to −∞)

Explanation:
1. \( \sqrt{x^4+1}=x^2+\frac{1}{2x^2}+\ldots \), so \( e^{\sqrt{x^4+1}}\approx e^{x^2} \) while \( e^{x^2+1}=e\cdot e^{x^2} \).
2. The difference is about \( (1-e)\,e^{x^2}\to-\infty \), so the limit does not exist as a finite number.

Teacher's Note:
1. Expand √(x⁴ + 1) for large x.
2. Compare the growth of e^(x²) and e·e^(x²).

Answer: (c) 1/2

Explanation:
1. Let N(x) = tan⁻¹(x/(x + 1)) and D(x) = tan⁻¹((1 + 2x − 2x²)/2). Then D(1 − x) = D(x).
2. By the tangent addition formula, \( \tan\left(N(x)+N(1-x)\right)=\frac{\frac{x}{x+1}+\frac{1-x}{2-x}}{1-\frac{x(1-x)}{(x+1)(2-x)}}=\frac{1+2x-2x^2}{2} \), so N(x) + N(1 − x) = D(x).
3. Hence the integrand r(x) = N/D satisfies r(x) + r(1 − x) = 1, and \( 2\int_0^1r(x)dx=\int_0^1[r(x)+r(1-x)]dx=1 \). The integral is 1/2.

Teacher's Note:
1. Use the property ∫f(x)dx = ∫f(1 − x)dx.
2. The pair r(x) + r(1 − x) is constant.

Answer: (b) positive

Explanation:
1. Direct integration gives \( f(\theta)-g(\theta)=(\sin\theta-\cos\theta)(1+\sin\theta+\cos\theta) \).
2. Here \( \sin\theta-\cos\theta>0 \) for θ in (π/4, 5π/4), and \( 1+\sin\theta+\cos\theta\ge0 \) exactly for \( \theta\notin(\pi,\tfrac{3\pi}{2}) \) (it is 0 at π and 3π/2).
3. On (π/4, 3π/4) both factors are positive, so the quantity is positive.

Teacher's Note:
1. Factor f(θ) − g(θ) into a product and study the sign of each factor.
2. Check the end points separately.

Answer: (c) non-negative

Explanation:
1. Direct integration gives \( f(\theta)-g(\theta)=(\sin\theta-\cos\theta)(1+\sin\theta+\cos\theta) \).
2. Here \( \sin\theta-\cos\theta>0 \) for θ in (π/4, 5π/4), and \( 1+\sin\theta+\cos\theta\ge0 \) exactly for \( \theta\notin(\pi,\tfrac{3\pi}{2}) \) (it is 0 at π and 3π/2).
3. On (3π/4, π) both factors are positive and at θ = π the second factor is 0, so the quantity is non-negative.

Teacher's Note:
1. Factor f(θ) − g(θ) into a product and study the sign of each factor.
2. Check the end points separately.

Answer: (d) non-positive

Explanation:
1. Direct integration gives \( f(\theta)-g(\theta)=(\sin\theta-\cos\theta)(1+\sin\theta+\cos\theta) \).
2. Here \( \sin\theta-\cos\theta>0 \) for θ in (π/4, 5π/4), and \( 1+\sin\theta+\cos\theta\ge0 \) exactly for \( \theta\notin(\pi,\tfrac{3\pi}{2}) \) (it is 0 at π and 3π/2).
3. At 3π/2 the product is 0; on (3π/2, 7π/4) the first factor is negative and the second is positive, so the quantity is non-positive.

Teacher's Note:
1. Factor f(θ) − g(θ) into a product and study the sign of each factor.
2. Check the end points separately.

Answer: (d) 2008

Explanation:
1. Using \( \int_a^b\frac{d\theta}{1+\tan\theta} \) with the substitution θ → a + b − θ, and since \( \theta_1+\theta_2=\frac{1004\pi}{2008}=\frac\pi2 \), the integrand pairs to 1: \( \frac1{1+\tan\theta}+\frac{1}{1+\cot\theta}=1 \).
2. So the integral equals \( \frac{\theta_2-\theta_1}{2}=\frac{1002\pi}{2\cdot2008}=\frac{501\pi}{2008} \), hence k = 2008.

Teacher's Note:
1. Notice θ₁ + θ₂ = π/2.
2. Pair the integrand with θ → π/2 − θ.

Answer: (d) 2009

Explanation:
1. The integrand equals \( \frac{d}{dx}\left[f(x)\left(g(x)+\frac1{g(x)}\right)\right] \): indeed the derivative is \( f'\left(g+\frac1g\right)+f\,g'\left(1-\frac1{g^2}\right) \).
2. Value = \( f(1)\left(1+1\right)-f(0)\left(\pi+\frac1\pi\right)=2009 \).

Teacher's Note:
1. Guess the antiderivative f(x)(g + 1/g).
2. Evaluate with the boundary values.

Answer: (a) 0

Explanation:
1. The curve \( y=(1-x^{2008})^{1/2007} \) is equivalent to \( x^{2008}+y^{2007}=1 \), and the second curve to \( x^{2007}+y^{2008}=1 \).
2. Interchanging x and y maps one curve onto the other, so both integrals equal the same area.
3. Hence the difference is 0.

Teacher's Note:
1. Rewrite each integrand as a curve in the x–y plane.
2. The areas under inverse curves in the unit square are equal.

Answer: (d) \( \frac{[x]}{2} \)

Explanation:
1. With n = [x], {t} has period 1 and integrates to 1/2 over each unit interval.
2. So the integral over [0, n] is n/2 = [x]/2.

Teacher's Note:
1. Use the period of {t}.
2. Count n unit intervals.

Answer: (b) \( \frac{[x]}2+\frac{\{x\}^2}2 \)

Explanation:
1. Split at n = [x]: \( \int_0^n\{t\}dt=\frac n2 \) and \( \int_n^x\{t\}dt=\int_0^{\{x\}}u\,du=\frac{\{x\}^2}2 \).
2. Total: [x]/2 + {x}²/2.

Teacher's Note:
1. Split the range at the integer part of x.
2. On [n, x], {t} = t − n.

Answer: (a) \( \frac{[x]^2}{2}-\frac{[x]}{2} \)

Explanation:
1. With n = [x], on [k, k+1) we have [t] = k, so the integral is \( \sum_{k=0}^{n-1}k=\frac{n(n-1)}2 \).
2. That equals [x]²/2 − [x]/2.

Teacher's Note:
1. Integrate the step function unit by unit.
2. Sum 0 + 1 + … + (n − 1).

Answer: (c) \( \frac{[x]}{2}\left(x-1+\{x\}\right) \)

Explanation:
1. With n = [x]: \( \int_0^x[t]dt=\frac{n(n-1)}2+n\{x\} \) (the last piece has [t] = n over length {x}).
2. Since n = x − {x}, this equals \( \frac n2(n-1+2\{x\})=\frac n2(x-1+\{x\}) \).

Teacher's Note:
1. Add the full unit intervals and the last partial piece.
2. Replace n by x − {x}.

Answer: (b) 10

Explanation:
1. Differentiate with respect to x: \( f(x)-f'''(3)=x^3+f'(1)x^2+f''(2)x \).
2. So f is a cubic with leading coefficient 1, hence f‴ = 6 and f(x) = x³ + px² + qx + 6 where p = f′(1), q = f″(2).
3. f′(1) = 3 + 2p + q = p gives p + q = −3; f″(2) = 12 + 2p = q. Solving: p = −5, q = 2.
4. f′(x) = 3x² − 10x + 2, so f′(4) = 48 − 40 + 2 = 10.

Teacher's Note:
1. Differentiate the identity once to get f in terms of three constants.
2. Use the definitions of the constants to solve for them.

Answer: (c) 11

Explanation:
1. \( \int_0^x\cos u^2du=x-\frac{x^5}{10}+\ldots \), so the numerator is \( \frac{x^5}{10} \).
2. \( \sin^{-1}x=x+\frac{x^3}6+\frac{3x^5}{40}+\ldots \), so the denominator is \( \frac{9x^5}{20} \).
3. Limit = (1/10)/(9/20) = 2/9, so p + q = 11.

Teacher's Note:
1. Use Maclaurin series for both numerator and denominator.
2. Compare the x⁵ coefficients.

Answer: (a) 5

Explanation:
1. Take logs: \( \ln L=-\frac12\left(1+\frac1n\right)\ln n+\frac1{n^2}\sum_{k=1}^nk\ln k \).
2. By Euler–Maclaurin, \( \sum_{k=1}^nk\ln k=\frac{n^2}{2}\ln n-\frac{n^2}4+\frac n2\ln n+O(\ln n) \).
3. Dividing by n²: \( \frac12\ln n-\frac14+\frac{\ln n}{2n} \), which cancels the first term except for −1/4.
4. So L = e^{−1/4}, p = 1, q = 4 and p + q = 5.

Teacher's Note:
1. Take logarithms and use the asymptotic of Σ k ln k.
2. The ln n terms cancel exactly.

Answer: (d) 4

Explanation:
1. A differentiable additive function is f(x) = kx, with k = f(1).
2. \( I_n=n\cdot\frac{kn^2}{2}=\frac{kn^3}{2} \).
3. Σ n³ from 1 to 5 = 225, so 225k/2 = 450 and k = 4.
4. f(1) = 4.

Teacher's Note:
1. Additive and differentiable means linear.
2. Use the sum of the first five cubes.