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Optimization: Local and Global Extrema PYQs for GATE DA

Solve 1+ Optimization: Local and Global Extrema previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

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2025 PYQ

Level 3: Exam Standard

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Optimization: Local and Global Extrema PYQs for GATE DA

Solve 1+ Optimization: Local and Global Extrema previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Optimization and Extrema

Chapter Roadmap

The Optimization Journey

The Terrain

Local neighborhoods vs global domain. Strict vs non-strict extrema.

Finding Flat Spots

Using the gradient to locate stationary points where slope is zero.

Reading the Curvature

Deploying the Hessian matrix. Classifying via eigenvalues and principal minors.

The Convexity Shortcut

Proving that for convex functions, every local minimum is the global minimum.

The Core Goal of Optimization

Core Concept

The Objective

Optimization is the mathematical framework for finding the best value of an objective function $f (x)$ over a feasible set.

Minimization

Find

x^{*}

such that

f (x^{*}) \leq f (x)

for all valid

x

Loss functions in ML

Maximization

Find

x^{*}

such that

f (x^{*}) \geq f (x)

Likelihood functions

Key Insight

Maximizing

f (x)

is identical to minimizing

- f (x)

. We only need tools for minimization.

Optimization: Local and Global Extrema: Solved Questions with Step-by-Step Explanations (1 Problems)

Question 1 · Calculus and Optimization · 2025 MSQ

A.
$f$ has a local minima
B.
There does not exist $x$ and $y$ , $x \neq = y$ , such that $f^{'} (x) = f^{'} (y) = 0$
C.
$f$ has at most one global minimum
D.
$f$ has at most one local minimum

Correct Answer:

["B","C","D"]

Step-by-Step Solution

Insight: Strict convexity means the function curves strictly upwards, preventing multiple flat spots or multiple dips, but not guaranteeing a minimum exists.

Exam route: Use the definition of strict convexity. (A) is false because $e^{x}$ is strictly convex but has no minimum. (B) is true because the derivative is strictly increasing. (C) and (D) are true because strict convexity forbids multiple minima.

Learning route:

Define strict convexity: $f (t x + (1 - t) y) < t f (x) + (1 - t) f (y)$ for $t \in (0, 1)$ and $x \neq = y$ .
Tempting wrong path: "Convex means bowl-shaped, so it must have a bottom." This leads to selecting (A). This breaks at the counterexample $f (x) = e^{x}$ , which is strictly convex but strictly increasing, so it never flattens out. Generalization: Strict convexity guarantees the shape, but not the existence of an extremum unless the function is coercive.
Check (A): False, as shown by $f (x) = e^{x}$ .
Check (B): If $f$ is strictly convex and differentiable at $x < y$ , then $f^{'} (x) < f^{'} (y)$ . Thus, $f^{'}$ can be zero at most once. There do not exist $x \neq = y$ with $f^{'} (x) = f^{'} (y) = 0$ . (B) is true.
Check (C): Suppose two global minima $x \neq = y$ exist with $f (x) = f (y) = m$ . Then $f ((x + y) /2) < (f (x) + f (y)) /2 = m$ , contradicting $m$ is the global minimum. (C) is true.
Check (D): For convex functions, any local minimum is a global minimum. Since there is at most one global minimum, there is at most one local minimum. (D) is true.

Verification: For $f (x) = e^{x}$ , $f^{'} (x) = e^{x} \neq = 0$ (B holds), no global min (C holds), no local min (D holds), no local min (A fails).