Applications-Of-Derivatives Question 438
Question: Find the greatest value of the function $ f(x)=\frac{\sin 2x}{\sin ( x+\frac{\pi }{4} )} $ on the interval $ [ 0,\frac{\pi }{2} ] $
Options:
A) 1
B) 2
C) 3
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ f(x)=\frac{\sin 2x}{\sin ( x+\frac{\pi }{4} )}= $ $ \sqrt{2}{ \frac{{{(\sin x+\cos )}^{2}}-1}{\sin x+\cos x} } $ $ =\sqrt{2}( \frac{y^{2}-1}{y} ) $ , where $ y=\sin x+\cos x $ Let $ \phi (y)=\sqrt{2}( \frac{y^{2}-1}{y} ), $ and $ g(x) $ $ =\sin x+\cos x $ We have, $ g’(x)=\cos x-\sin x $ . For max or min. $ g’(x)=0\Rightarrow \tan x=1 $
$ \Rightarrow x=\pi /4 $ . For this value of x. $ g’’(x)<0 $ . Thus, $ g(x) $ is max at $ x=\pi /4 $ and hence the domain of $ g(x) $ is $ [1,\sqrt{2}] $ i.e. y lies between 1 and $ \sqrt{2} $ Now, $ \phi ‘(y)=\sqrt{2}( 1+\frac{1}{y^{2}} )>0 $ for all $ y\in [1,\sqrt{2}] $ . That is $ \phi (y) $ is increasing for all $ y\in [1,\sqrt{2}] $ Thus it attains the greatest value at $ \sqrt{2} $ and is equal to $ \sqrt{2}( \frac{{{(\sqrt{2})}^{2}}-1}{\sqrt{2}} )=1 $ Hence, greatest value of f(x) on $ [0,\pi /2]= $ greatest value of $ \phi (y) $ on $ [1,\sqrt{2}]=1 $ .
BETA
