NDA 2026 — Mathematics PYQ

The given expression is a quadratic function in $x$:

$$f(x)=x^2+kx+k^2$$

To find the minimum value, we can complete the square. The general form $x^2+kx$ can be completed by adding and subtracting $(\frac{k}{2}{)}^2$:

$$f(x)=\left(x^2+kx+{\left(\frac{k}{2}\right)}^2\right)-{\left(\frac{k}{2}\right)}^2+k^2$$

Now, simplify the expression:

$$f(x)={\left(x+\frac{k}{2}\right)}^2-\frac{k^2}{4}+k^2$$

$$f(x)={\left(x+\frac{k}{2}\right)}^2+\left(\frac{-k^2+4k^2}{4}\right)$$

$$f(x)={\left(x+\frac{k}{2}\right)}^2+\frac{3k^2}{4}$$

Since the square of any real number ${\left(x+\frac{k}{2}\right)}^2$ is always greater than or equal to $0$, the minimum value of the expression occurs when the squared term is $0$:

$$\text{Minimum value}=0+\frac{3k^2}{4}=\frac{3k^2}{4}$$

Correct Option: (c) $\frac{3k^2}{4}$