KVPY Sample Paper KVPY Stream-SX Model Paper-8

If \[f(x)=0\] be a quadratic equation such that \[f\,(-\,\pi )=f(\pi )=0\] and \[f\left( \frac{\pi }{2} \right)=-\frac{3{{\pi }^{2}}}{4},\] then\[\underset{n\to -\pi }{\mathop{\lim }}\,\frac{f(x)}{\sin (\sin x)}\] is equal to-

A) 0

B) \[\pi \]

C) \[2\pi \]

D) None of these

Correct Answer: C

Solution :

\[f(x)=a{{x}^{2}}+bx+c\]
\[f(\pi )=a{{\pi }^{2}}+b\pi +c=0\]	??. (1)
\[f(-\pi )=a{{\pi }^{2}}-b\pi +c=0\]	??. (2)
\[(1)-(2)\Rightarrow b=0\]
\[f(\pi /2)\,\,\Rightarrow \,\,\frac{a{{\pi }^{2}}}{4}+c\frac{-\,3{{\pi }^{2}}}{4}\]
from (1), \[\frac{a{{\pi }^{2}}}{4}-a{{\pi }^{2}}=\frac{-\,3{{\pi }^{2}}}{4}\]

\[\Rightarrow \,\,\,\frac{-\,3a{{\pi }^{2}}}{4}=\frac{-\,3{{\pi }^{2}}}{4}\]

\[\Rightarrow \,\,\,a=1,\,\,\,c=-{{\pi }^{2}}\]

\[f(x)={{x}^{2}}-{{\pi }^{2}}\]

\[\underset{n\to -\pi }{\mathop{\lim }}\,\,\,\frac{{{x}^{2}}-{{\pi }^{2}}}{\sin (\sin x)}\] (using L\[-\] H)

=\[\underset{n\to -\pi }{\mathop{\lim }}\,\,\,\frac{2x-0}{cos\,\,(\sin x)\cos x}\]

\[=\frac{-2\pi }{\cos (0)\cos (-\,\pi )}=2\pi \]