JEE Mathematics Definite Integration MCQs Set 08

Answer: (d) none of these

Explanation:
1. The integrand equals the sign of \( g=\frac{2[x]}{3x-[x]} \). On [k, k+1) with k ≤ −2, \( 3x-k=2k+3\{x\}<0 \) and 2k < 0, so g > 0 and the integrand is +1: total 9 over [−10, −1).
2. On [−1, 0): \( 3x-[x]=3x+1 \) is negative for x < −1/3 (integrand +1) and positive for x > −1/3 (integrand −1).
3. Contribution: \( \left(\tfrac23\right)(1)+\left(\tfrac13\right)(-1)=\tfrac13 \). Total \( =9+\tfrac13=\tfrac{28}{3} \), which is not among (a)-(c).

Teacher's Note:
1. |g|/g is the sign function: +1 where g > 0, −1 where g < 0.
2. Treat the interval [−1, 0) separately because the denominator changes sign there.

Answer: (a) \( \pi \)

Explanation:
1. Use the identity \( \frac{\sin(n+\frac12)x}{\sin\frac x2}=1+2\sum_{k=1}^n\cos kx \).
2. Integrate over [0, π]: \( \int_0^\pi1\,dx=\pi \) and \( \int_0^\pi\cos kx\,dx=0 \) for every positive integer k.
3. So the value is π.

Teacher's Note:
1. This is the Dirichlet kernel; its expansion is a sum of cosines.
2. Each cosine term integrates to zero over [0, π].

Answer: (b) \( 1-\frac1{\sqrt2} \)

Explanation:
1. For 0 < x < π/4 < 1, [x] = 0 so \( d(x-[x])=dx \).
2. \( \int_0^{\pi/4}\sin x\,dx=1-\cos\frac\pi4=1-\frac{1}{\sqrt2} \).

Teacher's Note:
1. Inside the interval of integration [x] is constant, so d[x] = 0.
2. The integral reduces to an ordinary integral of sin x.

Answer: (a) \( -\cos x \)

Explanation:
1. Differentiate: \( f'(x)=-\sin x-\int_0^xf(t)\,dt \) (the boundary term (x − x)f(x) vanishes).
2. Differentiate again: \( f''(x)=-\cos x-f(x) \).
3. Hence \( f''(x)+f(x)=-\cos x \).

Teacher's Note:
1. Use the Leibniz rule; the term from the variable integrand cancels with the upper limit.
2. Two differentiations remove the integral completely.

Answer: (c) \( 21/2 \)

Explanation:
1. \( \frac{16\pi}3=5\pi+\frac\pi3 \). Each interval of length π contributes 2, so the first 5π give 10.
2. The remaining part: \( \int_0^{\pi/3}\sin x\,dx=1-\cos\frac\pi3=\frac12 \).
3. Total \( 10+\frac12=\frac{21}2 \).

Teacher's Note:
1. |sin x| is periodic with period π.
2. Split the interval into whole periods and a remainder.

Answer: (c) −5

Explanation:
1. \( \int_2^43\,dx=6 \), so \( 6-\int_2^4f=7\Rightarrow\int_2^4f=-1 \).
2. \( \int_{-1}^2f=\int_{-1}^4f-\int_2^4f=4+1=5 \).
3. Therefore \( \int_2^{-1}f=-5 \).

Teacher's Note:
1. Use additivity of integrals over intervals.
2. Reversing the limits changes the sign.

Answer: (d) none of these

Explanation:
1. \( zx-z^2=\frac{x^2}4-\left(z-\frac x2\right)^2 \), so \( I_1=e^{x^2/4}\int_{-x/2}^{x/2}e^{-u^2}du=e^{x^2/4}\cdot2\int_0^{x/2}e^{-u^2}du \).
2. In \( I_2 \) put z = 2u: \( I_2=2\int_0^{x/2}e^{-u^2}du \).
3. Hence \( I_1=e^{x^2/4}I_2 \), which is not given in (a), (b) or (c).

Teacher's Note:
1. Complete the square in the exponent.
2. Substitute u = z − x/2 in I₁ and z = 2u in I₂ to compare them.

Answer: (a) 2/5

Explanation:
1. Differentiate with respect to t: \( t^2f(t^2)\cdot2t=2t^4 \), so \( f(t^2)=t \) for t > 0.
2. Take \( t^2=\frac4{25} \), t = 2/5: \( f\left(\frac4{25}\right)=\frac25 \).

Teacher's Note:
1. Differentiate both sides with respect to t using the chain rule on the upper limit.
2. Pick the positive t since the upper limit t² is the argument.

Answer: (a) x = 1, −1

Explanation:
1. \( \phi'(x)=e^{-x^2/2}(1-x^2) \).
2. It vanishes at x = ±1 and changes sign at both (negative for x < −1, positive between, negative for x > 1), so both are extrema.

Teacher's Note:
1. The derivative of the integral is the integrand.
2. Check the sign change of φ′ at each root.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Strict inequality means g takes both positive and negative values, so by the intermediate value theorem g(x) = 0 has a root in (a, b): R is true.
2. Because g changes sign and f > 0 is continuous, the integral \( \int fg \) is a continuous function of how the positive and negative parts are weighted and can equal zero: A is true, and it rests on the sign change of g.

Teacher's Note:
1. The strict triangle inequality for integrals forces a sign change.
2. A continuous function that changes sign has a zero.

Answer: (c) A is true, but R is false.

Explanation:
1. On (0, π/4) we have x² < x, so \( e^{x^2}\lt e^x \) and \( I_2>I_1 \); cos x < 1 gives \( I_1>I_3 \); and sin x < cos x on (0, π/4) gives \( I_3>I_4 \). So A is true.
2. R is false: sin x > cos x only for x > π/4, so it fails on (0, π/4) (although x > x² holds on (0, 1)).

Teacher's Note:
1. Compare the integrands pointwise on (0, π/4).
2. On (0, π/4), cos x > sin x, the opposite of R.