The maximum value of \( f(x)=\displaystyle\frac{1}{3(x^2-x+1)} \) is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{4}{9} \)
f(x) is largest when the denominator is smallest. Completing the square,
\( \displaystyle x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4} \)
\( \displaystyle x^2-x+1\ge\frac{3}{4} \)
The least value of the denominator is \( 3\times\frac{3}{4}=\frac{9}{4} \), at x = 1/2.
\( \displaystyle f_{\max}=\frac{1}{9/4} \)
\( \displaystyle =\frac{4}{9} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{9}{4} \) — Incorrect. This is the least value of the denominator, not the maximum value of f.
Option (3) \( \displaystyle\frac{1}{2} \) — Incorrect. The maximum value is 4/9, not 1/2.
Option (4) \( -\displaystyle\frac{1}{2} \) — Incorrect. The denominator is always positive, so f(x) is never negative.