Match List I with List II

Choose the correct answer from the options given below :

A-III, B-IV, C-I, D-II

The correct answer is Option (2) → A-III, B-IV, C-I, D-II

(A) $y=\sin 2x$

$y'=2\cos 2x=0$

$x=\frac{π}{4}$

so $\sin 2x$ increases from 0 → 1 in $(0,\frac{π}{4})$ (III)

(B) $y=\tan x$

$y'=\sec^2x=0⇒x=\frac{π}{2}$

but $\sec^2x≥0$ always of $\tan x$ increasing in each quadrant (IV)

(C) $y=\cos x$

$y'=-\sin x=0⇒x=0,π,2π,....$

$⇒\cos x$ is decreasing from 1 to 0 in $(0,\frac{π}{6})$ (I)

(D) $y=\cos 3x$

= $\cos z$ decreasing in $z∈(0,\frac{π}{2})$

$⇒x∈(0,\frac{π}{6})$ (II)