CUET UG 2025 — Mathematics Calculus

To maximize $f(x) = \frac{1}{4x^2 + 2x + 1}$f(x)=4x2+2x+11​, the denominator $4x^2 + 2x + 1$4x2+2x+1 must be minimized.

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The expression $4x^2 + 2x + 1$4x2+2x+1 is a parabola opening upward (since the coefficient of $x^2$x2 is positive), so it has a minimum value.

For $ax^2 + bx + c$ax2+bx+c, the minimum occurs at $x = -\frac{b}{2a}$x=−2ab​.

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

$$x = -\frac{2}{2(4)}$$x=−2(4)2​

$$x = -\frac{2}{8}$$x=−82​

$$x = -\frac{1}{4}$$x=−41​

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Substituting $x = -\frac{1}{4}$x=−41​ into $4x^2 + 2x + 1$4x2+2x+1:

$$4\left(-\frac{1}{4}\right)^2 + 2\left(-\frac{1}{4}\right) + 1$$4(−41​)2+2(−41​)+1

$$= 4 \times \frac{1}{16} - \frac{2}{4} + 1$$=4×161​−42​+1

$$= \frac{1}{4} - \frac{1}{2} + 1$$=41​−21​+1

$$= \frac{1}{4} - \frac{2}{4} + \frac{4}{4}$$=41​−42​+44​

$$= \frac{3}{4}$$=43​

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

$$f\left(-\frac{1}{4}\right) = \frac{1}{\frac{3}{4}}$$f(−41​)=43​1​

$$= \frac{4}{3}$$=34​

Therefore, the maximum value is $\frac{4}{3}$34​.