For the function \(f(x) = (\log x)^{2}\), \(x \gt 0\):
(A) \(f'(x) = \displaystyle\frac{2\log x}{x}\).
(B) f is strictly decreasing in (0, 1).
(C) f has a local maximum at x = 1.
(D) The minimum value of f is 0.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
By the chain rule:
\( \displaystyle f'(x) = 2\log x \cdot \frac{1}{x} = \frac{2\log x}{x} \)
Since x > 0, the sign of f′(x) is the sign of log x:
\(f'(x) \lt 0\) for \(0 \lt x \lt 1\)
\(f'(x) \gt 0\) for \(x \gt 1\)
So f falls up to x = 1 and rises after it, and \(f(1) = (\log 1)^{2} = 0\).
So f′(x) = 2 log x / x, f decreases in (0, 1), increases in (1, ∞), and has its minimum value 0 at x = 1.
Checking the statements
(A) \(f'(x) = \displaystyle\frac{2\log x}{x}\). — Correct. Chain rule with \((\log x)' = \frac{1}{x}\).
(B) f is strictly decreasing in (0, 1). — Correct. For 0 < x < 1, log x < 0, so f′(x) < 0.
(C) f has a local maximum at x = 1. — Incorrect. f′ changes from − to + at x = 1, so it is a local minimum.
(D) The minimum value of f is 0. — Correct. A square is never negative, and f(1) = 0.