MCQ
If $f(x) = a\sin (\log x)$, then ${x^2}f''(x) + xf'(x) = . . . $
- A$f(x)$
- ✓$ - f(x)$
- C$0$
- D$1$
Differentiating w.r.t. $x$ of $y$, we get $f(x) = a\cos (\log x) \frac{1}{x} $
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 ${x^2}f''(x) + xf'(x) = - a\sin (\log x) = - f(x)$.
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