# 12) Which of the following functions are decreasing on $0 , \pi /2$ $(A) \cos x (B) \cos 2x (C) \cos 3x (D) \tan x$

(A)
$f(x) = \cos x \\ f^{'}(x) = -\sin x$
$f^{'}(x) < 0$   for x in $(0,\frac{\pi}{2})$
Hence, $f(x) = \cos x$ is decreasing function in  $(0,\frac{\pi}{2})$

(B)
$f(x) = \cos 2x \\ f^{'}(x) = -2\sin2 x$
Now, as
$0 < x < \frac{\pi}{2}\\ 0 < 2x < \pi$
$f^{'}(x) < 0$   for 2x in $(0,\pi)$
Hence, $f(x) = \cos 2x$ is decreasing function in  $(0,\frac{\pi}{2})$

(C)
$f(x) = \cos 3x \\ f^{'}(x) = -3\sin3 x$
Now, as
$0 < x < \frac{\pi}{2}\\ 0 < 3x < \frac{3\pi}{2}$
$f^{'}(x) < 0$  for $x \ \epsilon \ \left ( 0,\frac{\pi}{3} \right )$   and    $f^{'}(x) > 0 \ x \ \epsilon \ \left ( \frac{\pi}{3} , \frac{\pi}{2}\right )$
Hence, it is clear that  $f(x) = \cos 3x$ is neither increasing nor decreasing in  $(0,\frac{\pi}{2})$

(D)
$f(x) = \tan x\\ f^{'}(x) = \sec^{2}x$
$f^{'}(x) > 0$   for x in $(0,\frac{\pi}{2})$
Hence, $f(x) = \tan x$ is strictly increasing function in the interval  $(0,\frac{\pi}{2})$

So, only (A) and (B) are decreasing functions in  $(0,\frac{\pi}{2})$

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