PrepRung
Sign inSign up free

JEE Advanced 2024 Paper 1 · Mathematics · Limits · medium

Match the entries of List-I with those of List-II. List-I P. lim x→0(1-cos xcos 2x)/(x^2 Q. ∫0^π(xsin x)/(1+cos ^2x) dx R. Area of the region bounded…

Single correct option

Match the entries of List-I with those of List-II.

List-I

P. lim⁡x→01−cos⁡xcos⁡2xx2\displaystyle\lim_{x\to0}\dfrac{1-\cos x\cos2x}{x^2}

Q. ∫0πxsin⁡x1+cos⁡2x dx\displaystyle\int_0^{\pi}\dfrac{x\sin x}{1+\cos^2x}\,dx

R. Area of the region bounded by y=x2y=x^2 and y=2−∣x∣y=2-|x|

S. (Local maximum value) −- (local minimum value) of x3−3x+1x^3-3x+1

List-II

(1) 73\dfrac73

(2) 44

(3) π24\dfrac{\pi^2}{4}

(4) 52\dfrac52

(5) π22\dfrac{\pi^2}{2}
  1. A. P→4, Q→3, R→2, S→1
  2. B. P→3, Q→4, R→1, S→2
  3. C. P→4, Q→3, R→1, S→2
  4. D. P→4, Q→5, R→1, S→2
Show answer and solution

Answer: (C) P→4, Q→3, R→1, S→2

P: 1−cos⁡xcos⁡2x≈1−(1−x22)(1−2x2)≈52x21-\cos x\cos2x\approx1-\left(1-\tfrac{x^2}{2}\right)\left(1-2x^2\right)\approx\tfrac52x^2, so the limit is 52\tfrac52, i.e. (4). Q: with I=∫0πxsin⁡x1+cos⁡2xdxI=\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx the substitution x→π−xx\to\pi-x gives I=π2∫0πsin⁡x1+cos⁡2xdx=π2[−tan⁡−1(cos⁡x)]0π=π2⋅π2=π24I=\frac\pi2\int_0^\pi\frac{\sin x}{1+\cos^2x}dx=\frac\pi2\left[-\tan^{-1}(\cos x)\right]_0^\pi=\frac\pi2\cdot\frac\pi2=\frac{\pi^2}{4}, i.e. (3). R: the curves meet at x=±1x=\pm1, and by symmetry the area is 2∫01(2−x−x2) dx=2(2−12−13)=732\int_0^1(2-x-x^2)\,dx=2\left(2-\tfrac12-\tfrac13\right)=\tfrac73, i.e. (1). S: f′=3x2−3f'=3x^2-3 gives a local maximum at x=−1x=-1 with value 33 and a local minimum at x=1x=1 with value −1-1; the difference is 44, i.e. (2). So P→4, Q→3, R→1, S→2.
Report a mistake

Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake