GATE 2023 EE GA Q.7: solution and answer

GATE 2023 · EEGA Q.7 · 2 marks · MCQ

Which one of the following options represents the given graph?

GATE 2023 EE GA question 7: A plot of f(x) against x. — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
$f (x) = x^{2} 2^{- ∣ x ∣}$
(B)
$f (x) = x 2^{- ∣ x ∣}$
(C)
$f (x) = ∣ x ∣ 2^{- x}$
(D)
$f (x) = x 2^{- x}$

Answer and solution

ANSWER

The graph is $f (x) = x 2^{- ∣ x ∣}$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2023 EE (IIT Kanpur)

Read the features off the graph: $f (0) = 0$ ; $f (x) < 0$ for $x < 0$ and $f (x) > 0$ for $x > 0$ ; the curve is symmetric about the origin (odd, so $f (- x) = - f (x)$ ); and $f (x) \to 0$ as $x \to \pm \infty$ .
Sign test on (A): $x^{2} \geq 0$ and $2^{- ∣ x ∣} > 0$ , so $f (x) \geq 0$ for every $x$ . The graph goes negative for $x < 0$ , so (A) is out. (It is also even, not odd.)
Sign test on (C): $∣ x ∣ \geq 0$ and $2^{- x} > 0$ , so $f (x) \geq 0$ everywhere. Rejected for the same reason.
Behaviour of (D) for large negative $x$ : put $x = - t$ with $t > 0$ , so $f (- t) = - t 2^{t} \to - \infty$ as $t \to \infty$ . The graph flattens to 0 on the left instead of falling away, so (D) is out. Check: $f (- 5) = - 5 \times 2^{5} = - 160$ .
Check (B): $f (- x) = (- x) 2^{- ∣ - x ∣} = - x 2^{- ∣ x ∣} = - f (x)$ , so it is odd. Since $2^{- ∣ x ∣} > 0$ , $f$ has the same sign as $x$ . And $x 2^{- ∣ x ∣} = \frac{x}{2 ^{∣ x ∣}}$ , where the exponential in the denominator grows faster than the linear numerator, so $f (x) \to 0$ on both sides. Every feature matches.
The turning points agree too. For $x > 0$ , $f (x) = x 2^{- x}$ gives $f^{'} (x) = 2^{- x} (1 - x ln 2)$ , which is zero only at $x = \frac{1}{ln 2} \approx 1.44$ . It is positive before that point and negative after it, so this is a single maximum. By oddness there is a matching minimum at $x \approx - 1.44$ with the same depth. That gives the single peak and single trough in the figure. The axes carry no numbers, so only the shape is compared, not where the peak sits.
Therefore the graph is $f (x) = x 2^{- ∣ x ∣}$ , option (B).

KEY IDEA

Match a graph to a formula by testing its visible features one at a time, each of which rules out options: the sign on each side of the origin, symmetry (odd or even), and the limits as $x \to \pm \infty$ . A factor $∣ x ∣$ or $x^{2}$ can never be negative. A factor $2^{- x}$ with no modulus blows up as $x \to - \infty$ . Only $x 2^{- ∣ x ∣}$ is odd, has the sign of $x$ , and decays on both sides.

WATCH OUT FOR

Picking (D), $x 2^{- x}$ , because it matches the graph exactly for $x > 0$ (there it is the same function as (B)). For negative $x$ , $2^{- x}$ grows without bound, so $x 2^{- x} \to - \infty$ instead of returning to zero. Another slip is to see the modulus in $2^{- ∣ x ∣}$ and choose (A) without noticing that $x^{2} \geq 0$ can never produce the negative lobe.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2023 EE GA Q.7: solution and answer

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