The maximum value of \(f(x) = \displaystyle\frac{1}{x^{2} - 6x + 13}\), \(x \in R\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{4}\)
The denominator is always positive, so f is largest where the denominator is smallest. Differentiating:
\( \displaystyle f'(x) = -\frac{2x - 6}{(x^{2} - 6x + 13)^{2}} \)
So f′(x) = 0 at x = 3. Also f′(x) > 0 for x < 3 and f′(x) < 0 for x > 3, so by the first derivative test x = 3 is the point of maximum.
\( \displaystyle f(3) = \frac{1}{9 - 18 + 13} = \frac{1}{4} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{13}\) — Incorrect. This is f(0), not the largest value.
Option (3) 4 — Incorrect. This is the least value of the denominator, not of f.
Option (4) \(\displaystyle\frac{1}{2}\) — Incorrect. The denominator is never less than 4, so f never exceeds \(\frac{1}{4}\).