Option : (A)
f(x) = \(\frac{x}{log x}\)
f'(x) = \(\frac{log x-1}{(logx)^2}\)
f’(x) = 0
\(\frac{log x-1}{(logx)^2}\) = 0
log x -1 = 0
⇒ x = e
for second derivative we find f’(x)
Hence by second derivative test
f’’(x) > 0
So it’s a point of minimum.
Therefore,
f''(e) = \(\frac{-1}{e(loge)^2}+\)\(\frac{2}{e(loge)^2}\)
= \(-(\frac{1}{e})+(\frac{2}{e})\)
⇒ \(\frac{1}{e}\) > 0
x = e is a point of minimum
So, minimum value is f(e) = e
