CUET · MATHS · PYQ PAPER 2025
The maximum value of \(f(x)=\frac{1}{4 x^2+2 x+1}\) is
- A \(\frac{3}{4}\)
- B \(\frac{4}{3}\)
- C \(\frac{2}{3}\)
- D \(\frac{1}{3}\)
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 \(f(x) = \frac{1}{\frac{3}{4}} = \frac{4}{3}\).
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