Match List-I with List-II
| List-I | List-II |
| (A) \( \int_{\pi/6}^{\pi/3}\displaystyle\cos x\,dx \) | (I) 0 |
| (B) \( \int_{-\pi}^{\pi}\displaystyle(\sin^{61}x+x^{123})\,dx \) | (II) 4 |
| (C) \( \int_0^{\pi/2}\displaystyle\frac{1}{1+\tan^3x}\,dx \) | (III) \( \displaystyle\frac{\sqrt{3}-1}{2} \) |
| (D) \( \int_{-2}^{2}\displaystyle(2-|x|)\,dx \) | (IV) \( \displaystyle\frac{\pi}{4} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
(A) \( \int_{\pi/6}^{\pi/3}\displaystyle\cos x\,dx \)
This matches (III).
\( \displaystyle \Big[\sin x\Big]_{\pi/6}^{\pi/3}=\frac{\sqrt{3}}{2}-\frac{1}{2} \)
\( \displaystyle =\frac{\sqrt{3}-1}{2} \)
(B) \( \int_{-\pi}^{\pi}\displaystyle(\sin^{61}x+x^{123})\,dx \)
This matches (I) 0. Both terms are odd functions, and the interval is symmetric about 0, so the integral is 0.
(C) \( \int_0^{\pi/2}\displaystyle\frac{1}{1+\tan^3x}\,dx \)
This matches (IV) π/4. The integrand is \( \frac{\cos^3x}{\cos^3x+\sin^3x} \). Replacing x by \( \frac{\pi}{2}-x \) turns it into \( \frac{\sin^3x}{\sin^3x+\cos^3x} \). Adding the two forms of the integral:
\( \displaystyle 2I=\int_0^{\pi/2}1\,dx=\frac{\pi}{2} \)
\( \displaystyle I=\frac{\pi}{4} \)
(D) \( \int_{-2}^{2}\displaystyle(2-|x|)\,dx \)
This matches (II) 4. The integrand is even, so
\( \displaystyle \int_{-2}^{2}(2-|x|)\,dx=2\int_0^2(2-x)\,dx \)
\( \displaystyle =2\left[2x-\frac{x^2}{2}\right]_0^2 \)
\( \displaystyle =2\times 2=4 \)
So the matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).