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← JEE Advanced 2019 · Paper 2
MathematicsList match+3 / -1Question 51 of 54

JEE Advanced 2019 · Paper 2, Mathematics question 15

Passage

Answer Q.15 and Q.16 by appropriately matching the lists based on the information given in the paragraph.

Let f(x)=sin⁡(πcos⁡x)f(x) = \sin(\pi\cos x) and g(x)=cos⁡(2πsin⁡x)g(x) = \cos(2\pi\sin x) be two functions defined for x>0x > 0. Define the following sets whose elements are written in the increasing order:

X={x:f(x)=0},Y={x:f′(x)=0},X = \{x : f(x) = 0\},\qquad Y = \{x : f'(x) = 0\},

Z={x:g(x)=0},W={x:g′(x)=0}.Z = \{x : g(x) = 0\},\qquad W = \{x : g'(x) = 0\}.

List I contains the sets XX, YY, ZZ and WW. List II contains some information regarding these sets.

Which of the following is the only CORRECT combination?

List-I

  • I
    XX
  • II
    YY
  • III
    ZZ
  • IV
    WW

List-II

  • P
    ⊇{π2,3π2,4π,7π}\supseteq\left\{\frac\pi2, \frac{3\pi}2, 4\pi, 7\pi\right\}
  • Q
    an arithmetic progression
  • R
    NOT an arithmetic progression
  • S
    ⊇{π6,7π6,13π6}\supseteq\left\{\frac\pi6, \frac{7\pi}6, \frac{13\pi}6\right\}
  • T
    ⊇{π3,2π3,π}\supseteq\left\{\frac\pi3, \frac{2\pi}3, \pi\right\}
  • U
    ⊇{π6,3π4}\supseteq\left\{\frac\pi6, \frac{3\pi}4\right\}
Choose one option
Answer and worked solution

Official answer: B

X. sin⁡(πcos⁡x)=0\sin(\pi\cos x)=0 needs cos⁡x∈{−1,0,1}\cos x\in\{-1,0,1\}, so X={kπ2}X=\{\frac{k\pi}2\}. That is an arithmetic progression and contains π2,3π2,4π,7π\frac\pi2,\frac{3\pi}2,4\pi,7\pi.

Y.
f′(x)=−πsin⁡xcos⁡(πcos⁡x)=0  ⟺  sin⁡x=0  or  cos⁡x=±12.f'(x)=-\pi\sin x\cos(\pi\cos x)=0\iff\sin x=0\ \text{ or }\ \cos x=\pm\tfrac12.
So Y={kπ3}Y=\{\frac{k\pi}3\}. That is an arithmetic progression and contains π3,2π3,π\frac\pi3,\frac{2\pi}3,\pi.

The options.

  • (A) fails because X is an AP, so not (R).
  • (B) (II), (Q), (T) is correct.
  • (C) fails: X does not contain π6\frac\pi6.
  • (D) fails: Y is an AP.

(B)