MCQ
If $f(x) = a\sin (\log x)$, then ${x^2}f''(x) + xf'(x) =  . . . $
  • A
    $f(x)$
  • $ - f(x)$
  • C
    $0$
  • D
    $1$

Answer

Correct option: B.
$ - f(x)$
b
(b) $f(x) = a\sin (\log x)$

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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