GATE 2020 EC GA Q.5: solution and answer

GATE 2020 · ECGA Q.5 · 1 mark · MCQ

A superadditive function $f (\cdot)$ satisfies the following property

f (x_{1} + x_{2}) \geq f (x_{1}) + f (x_{2})

Which of the following functions is a superadditive function for $x > 1$ ?

(A)
$e^{x}$
(B)
$x$
(C)
$1/ x$
(D)
$e^{- x}$

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Answer and solution

ANSWER

(A) $e^{x}$ . For $x_{1}, x_{2} > 1$ , $e^{x_{1} + x_{2}} = e^{x_{1}} e^{x_{2}} > e^{x_{1}} + e^{x_{2}}$ , because $(e^{x_{1}} - 1) (e^{x_{2}} - 1) > (e - 1)^{2} > 1$ .

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

Take any $x_{1}, x_{2} > 1$ . An option qualifies only if $f (x_{1} + x_{2}) \geq f (x_{1}) + f (x_{2})$ holds for every such pair.
(A) $e^{x}$ : by the law of exponents, $e^{x_{1} + x_{2}} = e^{x_{1}} e^{x_{2}}$ . Put $a = e^{x_{1}}$ and $b = e^{x_{2}}$ . Then $ab \geq a + b ⟺ ab - a - b + 1 \geq 1 ⟺ (a - 1) (b - 1) \geq 1$ .
For $x > 1$ we have $a, b > e \approx 2.718$ , so $(a - 1) (b - 1) > (e - 1)^{2} \approx 2.95 > 1$ . So $e^{x_{1} + x_{2}} > e^{x_{1}} + e^{x_{2}}$ always holds, and $e^{x}$ is superadditive.
(B) $x$ : both sides are positive, so compare their squares. $(x_{1} + x_{2})^{2} = x_{1} + x_{2} + 2 x_{1} x_{2} > x_{1} + x_{2}$ , so $x_{1} + x_{2} < x_{1} + x_{2}$ . The inequality goes the wrong way (the function is subadditive), so (B) fails.
(C) $1/ x$ : $\frac{1}{x _{1} + x _{2}} < \frac{1}{x _{1}} < \frac{1}{x _{1}} + \frac{1}{x _{2}}$ , because a larger denominator gives a smaller positive fraction and $\frac{1}{x _{2}} > 0$ . So (C) fails.
(D) $e^{- x}$ : $e^{- (x_{1} + x_{2})} = e^{- x_{1}} e^{- x_{2}} < e^{- x_{1}}$ because $0 < e^{- x_{2}} < 1$ . That is already less than $e^{- x_{1}} + e^{- x_{2}}$ , so (D) fails.
Quick check with $x_{1} = x_{2} = 2$ : $e^{4} \approx 54.6 > 2 e^{2} \approx 14.8$ , while $4 = 2 < 22 \approx 2.83$ , $\frac{1}{4} < \frac{1}{2} + \frac{1}{2} = 1$ and $e^{- 4} \approx 0.018 < 2 e^{- 2} \approx 0.27$ . Only $e^{x}$ works, so the answer is (A).

KEY IDEA

Superadditive means the value at a sum is at least the sum of the values. For $e^{x}$ the value at a sum is a product, $e^{x_{1}} e^{x_{2}}$ , and a product of two numbers larger than 2 beats their sum. Written exactly: $ab \geq a + b ⟺ (a - 1) (b - 1) \geq 1$ . The other three functions are concave or decreasing on $x > 1$ , so for them the value at the sum comes out smaller than the sum of the values.

WATCH OUT FOR

Reading the inequality backwards and choosing $x$ because it "grows". For $x$ , $x_{1} + x_{2} < x_{1} + x_{2}$ always holds, so it is subadditive. A second trap is to forget the condition $x > 1$ . On $(0, 1)$ , $e^{x_{1}} e^{x_{2}} \geq e^{x_{1}} + e^{x_{2}}$ can fail: at $x_{1} = x_{2} = 0.1$ the left side is $e^{0.2} \approx 1.22$ and the right side is $2 e^{0.1} \approx 2.21$ . The claim for (A) rests on $e^{x} > e > 2$ .

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GATE 2020 EC GA Q.5: solution and answer

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