12) Which of the following functions are decreasing on 0 , \pi /2


(A) \cos x (B) \cos 2x (C) \cos 3x (D) \tan x

Answers (1)

(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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