For the function \(f(x) = \displaystyle\frac{2}{1 + x^{2}}\), \(x \in R\):
(A) \(f'(x) = \displaystyle\frac{-4x}{(1 + x^{2})^{2}}\).
(B) f is strictly increasing in \((0, \infty)\).
(C) The maximum value of f is 2.
(D) The least value of f on R is 0.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
Writing \(f(x) = 2(1 + x^{2})^{-1}\) and using the chain rule:
\( \displaystyle f'(x) = -2(1 + x^{2})^{-2} \cdot 2x = \frac{-4x}{(1 + x^{2})^{2}} \)
The sign of f′(x) is the sign of −x:
\(f'(x) \gt 0\) for \(x \lt 0\)
\(f'(x) \lt 0\) for \(x \gt 0\)
So f has its maximum f(0) = 2. Also \(f(x) \gt 0\) for every x, and f(x) gets as close to 0 as we like for large |x|, but never reaches 0.
So f increases in (−∞, 0), decreases in (0, ∞), has maximum value 2 and has no least value.
Checking the statements
(A) \(f'(x) = \displaystyle\frac{-4x}{(1 + x^{2})^{2}}\). — Correct. Chain rule on \(2(1 + x^{2})^{-1}\).
(B) f is strictly increasing in \((0, \infty)\). — Incorrect. For x > 0, f′(x) < 0, so f is decreasing; for example f(1) = 1 > f(2) = 0.4.
(C) The maximum value of f is 2. — Correct. f′ changes from + to − at x = 0, and f(0) = 2.
(D) The least value of f on R is 0. — Incorrect. \(1 + x^{2}\) is finite, so f(x) > 0 for every x; the value 0 is never attained.