Applications Of Derivatives Question 300
Question: The function $ x\sqrt{1-x^{2}},(x>0) $ has
Options:
A) A local maxima
B) A local minima
C) Neither a local maxima nor a local minima
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ f(x)=x\sqrt{1-x^{2}} $
therefore $ f’(x)=\frac{1-2x^{2}}{\sqrt{1-x^{2}}}=0\Rightarrow x=\pm \frac{1}{\sqrt{2}} $
But as $ x>0 $ , we have $ x=\frac{1}{\sqrt{2}} $
Now, again $ {f}’’(x)=\frac{\sqrt{1-x^{2}}(-4x)-(1-2x^{2})\frac{-x}{\sqrt{1-x^{2}}}}{(1-x^{2})} $
$ =\frac{2x^{3}-3x}{{{(1-x^{2})}^{3/2}}} $
$ \Rightarrow {f}’’( \frac{1}{\sqrt{2}} )=-ve $ . Then $ f(x) $ is maximum at $ x=\frac{1}{\sqrt{2}} $ .
BETA
