GATE 2026 IN Q.42: solution and answer

GATE 2026 · INQ.42 · 2 marks · MCQ

In the table shown below, List-I are the system responses to input $x (t)$ and List-II are the system properties. The correct matching is ____.

| List-I | List-II |
|---|---|
| (M) $x^{2} (t)$ | (I) Linear and Time-invariant |
| (N) $x (- t)$ | (II) Non-linear and Time-varying |
| (O) $∣ x (3 - t)∣$ | (III) Linear and Time-varying |
| (P) $x (t - 3)$ | (IV) Non-linear and Time-invariant |

(A)
M-IV, N-III, O-II, P-I
(B)
M-II, N-I, O-IV, P-III
(C)
M-III, N-IV, O-II, P-I
(D)
M-I, N-IV, O-III, P-II

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Answer and solution

ANSWER

The correct matching is M-IV, N-III, O-II, P-I, which is option (A).

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

Tests used. Linearity: $x_{1} \to y_{1}$ and $x_{2} \to y_{2}$ must imply $a x_{1} + b x_{2} \to a y_{1} + b y_{2}$ for all constants $a, b$ . Time invariance: feed in the delayed input $x (t - t_{0})$ and compare the output with the delayed output $y (t - t_{0})$ . The two properties are tested separately. What the system does to the amplitude decides linearity, and what it does to the time argument decides time invariance.
(M) $y (t) = x^{2} (t)$ . Scaling fails: $a x (t) \to a^{2} x^{2} (t) \neq = a x^{2} (t)$ for $a \neq = 0, 1$ and any $x$ that is not identically zero, so the system is non-linear. The delayed input gives $x^{2} (t - t_{0}) = y (t - t_{0})$ , so it is time-invariant. This is (IV).
(N) $y (t) = x (- t)$ . Superposition holds: the input $a x_{1} + b x_{2}$ gives $a x_{1} (- t) + b x_{2} (- t) = a y_{1} (t) + b y_{2} (t)$ , so the system is linear. With the delayed input $x_{1} (t) = x (t - t_{0})$ the output is $y_{1} (t) = x_{1} (- t) = x (- t - t_{0})$ , while $y (t - t_{0}) = x (- (t - t_{0})) = x (- t + t_{0})$ . The input is shifted one way and the output the other, so for $t_{0} \neq = 0$ they differ in general. The system is time-varying. This is (III).
(O) $y (t) = ∣ x (3 - t)∣$ . Take $a = - 1$ : the input $- x$ gives $∣ - x (3 - t)∣ = ∣ x (3 - t)∣$ . That is not $- y (t)$ unless $x (3 - t) = 0$ , so the system is non-linear. With the delayed input $x_{1} (t) = x (t - t_{0})$ the output is $∣ x_{1} (3 - t)∣ = ∣ x (3 - t - t_{0})∣$ , while $y (t - t_{0}) = ∣ x (3 - t + t_{0})∣$ . As in (N), the time reversal inside the argument turns a delay of the input into an advance of the output, so these differ for $t_{0} \neq = 0$ . The system is time-varying. This is (II).
(P) $y (t) = x (t - 3)$ . A pure delay obeys superposition, $a x_{1} (t - 3) + b x_{2} (t - 3) = a y_{1} (t) + b y_{2} (t)$ , so the system is linear. The delayed input gives $x (t - t_{0} - 3)$ , which equals $y (t - t_{0})$ , so it is time-invariant. This is (I).
Matching: M-IV, N-III, O-II, P-I, which is option (A).

KEY IDEA

Test the amplitude and the time argument separately. A non-linear operation on the amplitude, such as squaring or taking the modulus, makes a system non-linear. Anything done to the time argument other than a pure shift, such as the reversal $t \to - t$ , makes it time-varying. $x (t - 3)$ is only a shift, so it is LTI. $x (- t)$ and $∣ x (3 - t)∣$ both reverse time, so both are time-varying. $x^{2} (t)$ and $∣ \cdot ∣$ fail scaling, so both are non-linear.

WATCH OUT FOR

Getting the time-invariance test wrong for $x (- t)$ by substituting $t \to t - t_{0}$ everywhere in $y (t)$ . That only produces $y (t - t_{0})$ and compares it with itself. The response to the delayed input must be found by delaying $x$ first, which gives $x (- t - t_{0})$ . A second mistake is to read $x (3 - t)$ as a shift by 3 and so call (O) time-invariant, when $x (3 - t) = x (- (t - 3))$ contains a reversal. A third is to treat the modulus as linear because it keeps the shape of a positive signal.

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