The maximum value of \(f(x) = \displaystyle\frac{1}{4x^2 + 2x + 1}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{4}{3}\)
The denominator is always positive, so f is largest where the denominator is smallest.
The quadratic \(4x^2 + 2x + 1\) opens upwards, so its minimum is at the vertex:
\( \displaystyle x = -\frac{b}{2a} = -\frac{2}{8} = -\frac{1}{4} \)
\( \displaystyle 4\left(\frac{1}{16}\right) + 2\left(-\frac{1}{4}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4} \)
So the maximum value of f is:
\( \displaystyle f\left(-\frac{1}{4}\right) = \frac{1}{3/4} = \frac{4}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{4}\) — Incorrect. This is the minimum value of the denominator, not the maximum value of f, which is its reciprocal.
Option (3) \(\displaystyle\frac{2}{3}\) — Incorrect. f reaches \( \frac{4}{3} \), which is larger, so \( \frac{2}{3} \) is not the maximum.
Option (4) \(\displaystyle\frac{1}{3}\) — Incorrect. f reaches \( \frac{4}{3} \), which is larger, so \( \frac{1}{3} \) is not the maximum.
NCERT Exemplar:
Application of Derivatives
