The function f(x) = x + (a^2)/x, a gt 0, x ≠ 0 has a local maxima at

The correct answer is Option (1) → \(x = -a\)

A local maximum occurs at a critical point where the second derivative is negative (second derivative test).

    \( \displaystyle f'(x) = 1 - \frac{a^2}{x^2} \)

Putting \(f'(x) = 0\):

    \(x^2 = a^2\)
\(x = a \text{ or } x = -a\)

    \( \displaystyle f''(x) = \frac{2a^2}{x^3} \)

At \(x = -a\):

    \( \displaystyle f''(-a) = \frac{2a^2}{-a^3} = -\frac{2}{a} \lt 0 \)

So f has a local maximum at \(x = -a\).

At \(x = a\):

    \( \displaystyle f''(a) = \frac{2a^2}{a^3} = \frac{2}{a} \gt 0 \)

So \(x = a\) gives a local minimum, not a maximum.