Applications Of Derivatives Question 282
Question: The function $ f(x)={{\tan }^{-1}}(\sin x+\cos x) $ , $ x>0 $ is always an increasing function on the interval
[Kerala (Engg.) 2005]
Options:
A) $ (0,,\pi ) $
B) $ (0,,\pi /2) $
C) $ (0,\pi /4) $
D) $ (0,,3\pi /4) $
E) $ (0,,5\pi /4) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=y={{\tan }^{-1}}( \sqrt{2}\sin ( x+\frac{\pi }{4} ) ) $
$ \Rightarrow \tan y=\sqrt{2}\sin ( x+\frac{\pi }{4} )\Rightarrow {{\sec }^{2}}y\frac{dy}{dx}=\sqrt{2}\cos ( x+\frac{\pi }{4} ) $
$ \frac{dy}{dx}>0\Rightarrow \cos ( x+\frac{\pi }{4} )>0 $ .
$ \therefore ,x\in ( 0,\frac{\pi }{4} ) $ .
BETA
