GATE 2025 EE Q.11: solution and answer

GATE 2025 · EEQ.11 · 1 mark · MCQ

Consider the set $S$ of points $(x, y) \in ℜ^{2}$ which minimize the real valued function
$f (x, y) = (x + y - 1)^{2} + (x + y)^{2}$
Which of the following statements is true about the set $S$ ?

(A)
The number of elements in the set $S$ is finite and more than one.
(B)
The number of elements in the set $S$ is infinite.
(C)
The set $S$ is empty.
(D)
The number of elements in the set $S$ is exactly one.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The minimizers are every point on the line $x + y = \frac{1}{2}$ , which is an infinite set. The answer is (B).

Official answer: BSource: Official answer key, GATE 2025 EE (IIT Roorkee)

Stationary points need $\nabla f = 0$ : $\frac{\partial f}{\partial x} = 2 (x + y - 1) + 2 (x + y) = 0$ and $\frac{\partial f}{\partial y} = 2 (x + y - 1) + 2 (x + y) = 0$ .
Both conditions are the same single equation, $4 (x + y) - 2 = 0$ . So the stationary points are every point on the line $x + y = \frac{1}{2}$ , not one isolated point.
Check that these points are minima. The Hessian is $H = [4444]$ and $det H = 0$ , so the second-derivative test cannot decide. Use the structure of $f$ instead.
$f$ depends only on $u = x + y$ : $f = g (u) = (u - 1)^{2} + u^{2} = 2 u^{2} - 2 u + 1 = 2 (u - \frac{1}{2})^{2} + \frac{1}{2}$ (completing the square).
So $f (x, y) \geq \frac{1}{2}$ for every $(x, y)$ , with equality exactly when $u = x + y = \frac{1}{2}$ . The global minimum value is $\frac{1}{2}$ .
Hence $S = {(x, y) \in ℜ^{2} : x + y = \frac{1}{2}}$ , a straight line (for example $(0, \frac{1}{2})$ , $(\frac{1}{2}, 0)$ , $(1, - \frac{1}{2})$ , and so on). It has infinitely many elements, so option (B) is correct.

KEY IDEA

Here $f$ depends on $x$ and $y$ only through $u = x + y$ . Minimise the one-variable function $g (u) = (u - 1)^{2} + u^{2}$ to get $u = \frac{1}{2}$ . Every $(x, y)$ with $x + y = \frac{1}{2}$ gives that same minimum, so the minimizers make up a whole line.

WATCH OUT FOR

Setting $\partial f / \partial x = 0$ and $\partial f / \partial y = 0$ and expecting two independent equations with one solution, which leads to (D). The two equations are identical, so they fix only $x + y$ . Another trap is reading the singular Hessian ( $det H = 0$ ) as "no minimum exists" and picking (C). A zero determinant only means the test cannot decide, and completing the square shows $f \geq \frac{1}{2}$ with equality all along the line.

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 39 Calculus questions, with notes

GATE 2025 EE Q.11: solution and answer

Answer and solution

More Calculus questions