Mathematics · Trigonometry Ratios and Identities

JEE Advanced 2019 — Paper 2 — Question 38

Let $f(x)=\sin (\pi \cos x)$ and $g(x)=\cos (2 \pi \sin x)$ be two functions defined for $x>0$ . Define the

following sets whose elements are written in the increasing order:

\begin{aligned} & X=\{x: f(x)=0\}, Y=\left\{x: f^{\prime}(x)=0\right\} \\& Z=\{x: g(x)=0\}, W=\left\{x: g^{\prime}(x)=0\right\} \end{aligned}

List - I contains the set X, Y, Z and W. List - II contains some information regarding these sets.

List - I	List - II
(I) X	(P) $\supseteq\left\{\frac{\pi}{2}, \frac{3 \pi}{2}, 4 \pi, 7 \pi\right\}$
(II) Y	(Q) an arithmetic progression
(III) Z	(R) NOT an arithmetic progression
(IV) W	(S) $\supseteq\left\{\frac{\pi}{6}, \frac{7 \pi}{6}, \frac{13 \pi}{6}\right\}$
	(T) $\supseteq\left\{\frac{\pi}{3}, \frac{2 \pi}{3}, \pi\right\}$
	(U) $\supseteq\left\{\frac{\pi}{6}, \frac{3 \pi}{4}\right\}$

Which of the following is the only CORRECT combination?

Answer: C

f(x) = \sin(\pi \cos x) = 0

\Rightarrow \pi \cos x = n_1 \pi, n_1 \in \mathbb{I}

\Rightarrow \cos x = -1, \cos x = 0, \cos x = 1

\Rightarrow x = n_2 \pi, \frac{(2n_3 + 1)\pi}{2}, n_2, n_3 \in \mathbb{I}

\Rightarrow X = \{ \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \ldots \}

f'(x) = -\cos(\pi \cos x) \cdot \pi (-\sin x) = 0

\Rightarrow \sin x = 0 \text{ or } \cos(\pi \cos x) = 0

\Rightarrow x = n_4 \pi \text{ or } \pi \cos x = (2n_5 + 1) \frac{\pi}{2}

\Rightarrow x = n_4 \pi \text{ or } \cos x = \frac{2n_5 + 1}{2}

\Rightarrow \cos x = \pm \frac{1}{2}

\Rightarrow x = n_4 \pi, \pm \frac{\pi}{3} + 2n_6 \pi

Y = \{ \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi, \ldots \}

g(x) = \cos(2\pi \sin x) = 0

\Rightarrow 2\pi \sin x = (2n_6 + 1) \frac{\pi}{2}

\Rightarrow \sin x = \frac{2n_6 + 1}{4} = \pm \frac{1}{4}, \pm \frac{3}{4}

Z = \{ x | \sin x = \pm \frac{1}{4}, \pm \frac{3}{4}, x > 0 \}

g'(x) = -2\pi \cos x \sin(2\pi \sin x) = 0

\Rightarrow \cos x = 0 \text{ or } \sin(2\pi \sin x) = 0

\Rightarrow x = (2n_7 + 1) \frac{\pi}{2} \text{ or } 2\pi \sin x = n_8 \pi

\Rightarrow \sin x = \frac{n_8}{2} = -1, -\frac{1}{2}, 0, \frac{1}{2}, 1

W = \{ x | \sin x = \pm 1, \pm \frac{1}{2}, 0, x > 0 \}

\Rightarrow \text{I - (P), (Q)}

\Rightarrow \text{II - (Q), (T)}

\Rightarrow \text{III - (R)}

\Rightarrow \text{IV - (P), (R), (S)}

Answer key and solution verified before publishing.

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