Let \(f(x)=\frac{1}{4x^2+2x+1}\). then i

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

Let \(f(x)=\frac{1}{4x^2+2x+1}\). then its maximum value is

Options:

\(\frac{1}{4}\)

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

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

\(\frac{5}{4}\)

Correct Answer:

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

Explanation:

Find minimum value of \(4x^2+2x+1=4(x+\frac{1}{4})^2+\frac{3}{4}\)