A) \[f(x)\]
B) \[-f(x)\]
C) 0
D) 1
Correct Answer: B
Solution :
\[f(x)=a\sin (\log x)\] Differentiating w.r.t. x of y, we get\[4\pi {{r}^{2}}\] Again\[{f}''\,(x)=-\frac{1}{{{x}^{2}}}a\cos (\log x)-\frac{1}{{{x}^{2}}}a\sin (\log x)\] \[\Rightarrow {{x}^{2}}{f}''(x)=-[a\cos (\log x)+a\sin (\log x)]\] Now \[{{x}^{2}}{f}''(x)+x{f}'(x)=-a\sin (\log x)=-f(x)\].
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