CUET UG 2025 — Mathematics Calculus

Given: $f(x) = x + \frac{a^2}{x}$f(x)=x+xa2​, where $a > 0$a>0 and $x \neq 0$x=0

To find local maxima, find the critical points where $f'(x) = 0$f′(x)=0 and test their nature.

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Finding the first derivative:

$$f'(x) = 1 - \frac{a^2}{x^2}$$f′(x)=1−x2a2​

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Setting $f'(x) = 0$f′(x)=0 to find critical points:

$$1 - \frac{a^2}{x^2} = 0$$1−x2a2​=0

$$1 = \frac{a^2}{x^2}$$1=x2a2​

$$x^2 = a^2$$x2=a2

$$x = \pm a$$x=±a

The critical points are $x = a$x=a and $x = -a$x=−a.

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Finding the second derivative:

$$f''(x) = \frac{2a^2}{x^3}$$f′′(x)=x32a2​

For $f''(x) < 0$f′′(x)<0, the critical point is a local maxima.

For $f''(x) > 0$f′′(x)>0, the critical point is a local minima.

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At $x = a$x=a:

$$f''(a) = \frac{2a^2}{a^3}$$f′′(a)=a32a2​

$$f''(a) = \frac{2}{a}$$f′′(a)=a2​

Since $a > 0$a>0, $f''(a) > 0$f′′(a)>0

This is a local minima.

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At $x = -a$x=−a:

$$f''(-a) = \frac{2a^2}{(-a)^3}$$f′′(−a)=(−a)32a2​

$$f''(-a) = \frac{2a^2}{-a^3}$$f′′(−a)=−a32a2​

$$f''(-a) = -\frac{2}{a}$$f′′(−a)=−a2​

Since $a > 0$a>0, $f''(-a) < 0$f′′(−a)<0

This is a local maxima.

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Therefore, the function has a local maxima at $x = -a$x=−a.