Differentiation Question 23
Question: If $ f(x)=0 $ is a quadratic equation such that $ f(-\pi )=f(\pi )=0 $ and $ f( \frac{\pi }{2} )=-\frac{3{{\pi }^{2}}}{4} $ , then $ \underset{x\to -\pi }{\mathop{\lim }}\frac{f(x)}{\sin (sinx)} $ is equal to
Options:
A) 0
B) $ \pi $
C) $ 2\pi $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ f(x)=x^{2}-{{\pi }^{2}} $
$ \underset{x\to -\pi }{\mathop{\lim }}\frac{x^{2}-{{\pi }^{2}}}{\sin (sinx)}=\underset{h\to 0}{\mathop{\lim }}\frac{{{(-\pi +h)}^{2}}-{{\pi }^{2}}}{\sin (sin(-\pi +h))} $
$ \underset{h\to 0}{\mathop{\lim }}\frac{-2h\pi +h^{2}}{-\sin (sin+h)} $
$ \underset{h\to 0}{\mathop{\lim }}\frac{h-2\pi }{\frac{-\sin (sinh)}{\sinh }\times \frac{\sinh }{h}}=2\pi $
BETA
