Calculus

Application of Derivatives

Medium

core

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

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{2}{3}$

$\frac{1}{3}$

✅ Correct Option: 2

To maximize $f(x) = \frac{1}{4x^2 + 2x + 1}$ , the denominator $4x^2 + 2x + 1$ must be minimized.

The expression $4x^2 + 2x + 1$ is a parabola opening upward (since the coefficient of $x^2$ is positive), so it has a minimum value.

For $ax^2 + bx + c$ , the minimum occurs at $x = -\frac{b}{2a}$ .

With $a = 4$ and $b = 2$ :

$x = -\frac{2}{2(4)}$

$x = -\frac{2}{8}$

$x = -\frac{1}{4}$

Substituting $x = -\frac{1}{4}$ into $4x^2 + 2x + 1$ :

$4\left(-\frac{1}{4}\right)^2 + 2\left(-\frac{1}{4}\right) + 1$

$= 4 \times \frac{1}{16} - \frac{2}{4} + 1$

$= \frac{1}{4} - \frac{1}{2} + 1$

$= \frac{1}{4} - \frac{2}{4} + \frac{4}{4}$

$= \frac{3}{4}$

The maximum value of $f(x)$ is:

$f\left(-\frac{1}{4}\right) = \frac{1}{\frac{3}{4}}$

$= \frac{4}{3}$

Therefore, the maximum value is $\frac{4}{3}$ .

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CUET Mathematics Past Year Questions (Topic-Wise)

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