GATE 2023 EC GA Q.7: solution and answer

GATE 2023 · ECGA Q.7 · 2 marks · MCQ

Which one of the following options represents the given graph?

GATE 2023 EC GA question 7: A graph of f(x) against x. — Figure from the GATE 2023 EC 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 $f (x) = x^{2} 2^{- ∣ x ∣}$ , option (A).

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

Read the graph's features: it is symmetric about the $f (x)$ -axis (an even function), $f (0) = 0$ , $f (x) \geq 0$ for every $x$ , and $f (x) \to 0$ both as $x \to + \infty$ and as $x \to - \infty$ .
Test each option for evenness, i.e. $f (- x) = f (x)$ . (B) $f (- x) = - x 2^{- ∣ - x ∣} = - x 2^{- ∣ x ∣} = - f (x)$ , so it is odd, not even. (D) $f (- x) = - x 2^{x}$ , which is not equal to $f (x)$ . (C) $f (- x) = ∣ - x ∣ 2^{x} = ∣ x ∣ 2^{x} \neq = ∣ x ∣ 2^{- x}$ for $x \neq = 0$ . Only (A) is even: $(- x)^{2} 2^{- ∣ - x ∣} = x^{2} 2^{- ∣ x ∣}$ .
Check the sign at a negative point: at $x = - 1$ , (B) gives $- 1 \cdot 2^{- 1} = - 0.5$ and (D) gives $- 1 \cdot 2^{1} = - 2$ . Both go below the $x$ -axis, but the graph never does.
Check the left tail: at $x = - 4$ , (C) gives $4 \cdot 2^{4} = 64$ and (D) gives $- 4 \cdot 2^{4} = - 64$ . For negative $x$ the factor $2^{- x}$ grows without bound, so neither (C) nor (D) can decay to 0 on the left the way the graph does.
Confirm that (A) has every feature: $f (0) = 0$ ; $x^{2} \geq 0$ and $2^{- ∣ x ∣} > 0$ make $f (x) \geq 0$ ; and exponential decay beats polynomial growth, so $x^{2} 2^{- ∣ x ∣} \to 0$ as $∣ x ∣ \to \infty$ .
(A) also has the two humps. For $x > 0$ , $f (x) = x^{2} 2^{- x}$ and $f^{'} (x) = 2^{- x} (2 x - x^{2} ln 2) = x 2^{- x} (2 - x ln 2)$ , which is zero at $x = \frac{2}{ln 2} \approx 2.89$ . By symmetry the other maximum is at $x \approx - 2.89$ , with $f (0) = 0$ as the minimum between them. The figure is a sketch, not drawn to scale: the axis ticks are unlabelled, so the peak positions cannot be checked against a scale, and near the origin $f (x) \approx x^{2}$ , so the true minimum is flat ( $f^{'} (0) = 0$ ) rather than the sharp point drawn. The question is decided by symmetry, sign and decay, and only (A) has all three.
Hence the graph is $f (x) = x^{2} 2^{- ∣ x ∣}$ , option (A).

KEY IDEA

Match a sketch to a formula by its qualitative features, not by plotting. Here there are three: symmetry (an even function needs $f (- x) = f (x)$ ), sign (the curve is never negative) and end behaviour (it decays to 0 on both sides). Only $x^{2} 2^{- ∣ x ∣}$ is even, never negative and decaying at both ends, because $2^{- ∣ x ∣}$ decays whichever way $x$ goes and beats the polynomial factor.

WATCH OUT FOR

Looking only at the right half of the graph, where all four options behave alike (zero at 0, a rise, then decay), and then picking (B) or (C). Test a negative value of $x$ : (B) and (D) go negative there, and (C) and (D) blow up because $2^{- x}$ grows as $x \to - \infty$ . A second trap is rejecting (A) because the sketch shows a sharp point at the origin while $x^{2} 2^{- ∣ x ∣}$ is actually flat there. The figure is only a qualitative sketch.

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

GATE 2023 EC GA Q.7: solution and answer

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