QJEEAdv19Mae97e: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Mae97e Solution: Let f(x) = sin( cos x) and g(x) = cos(2 sin x) be two functions …

For, X = {x : f(x) = 0}, x > 0

Now, f(x) = 0

$\Rightarrow$ sin( $\pi$ cos x) = 0, x > 0

$\Rightarrow$ $\pi$ cos x = n $\pi$ , n $\in$ Integer.

$\Rightarrow$ cos x = n

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

{ $\because$ cos x $\in$ [ $-$ 1, 1]}

When cos x = ± 1 $\Rightarrow$ x = n $\pi$

When cos x = 0 $\Rightarrow$ x = (2n + 1) ${{\pi \over 2}}$

Hence, (i) $\to$ (P), (Q)

For, Y = {x : f'(x) = 0}, x > 0

Now, f'(x) = 0

$\Rightarrow$ $-$ $\pi$ sin x cos( $\pi$ cos x) = 0

$\Rightarrow$ either sin x = 0 $\Rightarrow$ x = n $\pi$ , n is an integer,

or cos( $\pi$ cos x) = 0

$\Rightarrow$ $\pi$ cos x = (2n + 1) ${{\pi \over 2}}$ , n is an integer

$\Rightarrow$ cos x = ${{{2n + 1} \over 2}}$

$\Rightarrow$ $\cos x = \pm {1 \over 2}$ { $\because$ cos x $\in$ [ $-$ 1, 1]}

$\Rightarrow$ x = $2n\pi \pm {\pi \over 3}$ or $2n\pi \pm {{2\pi } \over 3}$ , n is an integer.

So, (ii) $\to$ (Q), (T)

Hence, option (a) is correct.