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CUET · MATHS · PYQ PAPER 2025

The maximum value of \(f(x)=\frac{1}{4 x^2+2 x+1}\) is

  1. A \(\frac{3}{4}\)
  2. B \(\frac{4}{3}\)
  3. C \(\frac{2}{3}\)
  4. D \(\frac{1}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{4}{3}\)

Step-by-step Solution

Detailed explanation

Min value of denominator \(4x^2+2x+1 = 1 - \frac{2^2}{4(4)} = 1 - \frac{4}{16} = \frac{3}{4}\). Maximum value of \(f(x) = \frac{1}{\frac{3}{4}} = \frac{4}{3}\).