Option : (C)
f(x) = x logex
Differentiating f(x) with respect to x, we get
f'(x) = x \(\times\) \(\frac{1}{x}\)+ logex \(\times\)1
= 1 + logex
Differentiating f’(x) with respect to x, we get
f"(x) = \(\frac{1}{x}\)
For minima at x = c,
f’(c) = 0 and f’’(c) > 0
f’(x) = 0
⇒ x = \(\frac{1}{e}\)
f"(\(\frac{1}{e}\)) = e > 0
Hence,
x = \(\frac{1}{e}\) is a point of minima for f(x) and f(\(\frac{1}{e}\)) = \(-\frac{1}{e}\) is the minimum value of f(x).
