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