GATE EC 2020 Set 1 — Question 5

A function $f(x)$ is superadditive if $f(x_1+x_2) \geq f(x_1) + f(x_2)$ for all $x_1, x_2$ in its domain.
Let's test each option for $x > 1$:(A) $f(x) = e^x$
We need to check if $e^{x_1+x_2} \geq e^{x_1} + e^{x_2}$ for $x_1, x_2 > 1$.
We can rewrite the inequality as $e^{x_1}e^{x_2} - e^{x_1} - e^{x_2} \geq 0$.
Adding 1 to both sides, we get $e^{x_1}e^{x_2} - e^{x_1} - e^{x_2} + 1 \geq 1$.
This can be factored as $(e^{x_1}-1)(e^{x_2}-1) \geq 1$.
Since $x_1 > 1$, $e^{x_1} > e \approx 2.718$, so $e^{x_1}-1 > 1.718$.
Similarly, since $x_2 > 1$, $e^{x_2}-1 > 1.718$.
Therefore, $(e^{x_1}-1)(e^{x_2}-1) > 1.718 \times 1.718 \approx 2.95 > 1$.
The inequality holds true. So, $f(x) = e^x$ is a superadditive function for $x > 1$.(B) $f(x) = \sqrt{x}$
We need to check if $\sqrt{x_1+x_2} \geq \sqrt{x_1} + \sqrt{x_2}$.
Squaring both sides (since both sides are positive for $x_1, x_2 > 1$):
$x_1+x_2 \geq (\sqrt{x_1} + \sqrt{x_2})^2$
$x_1+x_2 \geq x_1 + x_2 + 2\sqrt{x_1x_2}$
$0 \geq 2\sqrt{x_1x_2}$
This is false for $x_1, x_2 > 1$ (as $2\sqrt{x_1x_2}$ would be positive). Thus, $f(x) = \sqrt{x}$ is not superadditive; it is subadditive.(C) $f(x) = 1/x$
We need to check if $1/(x_1+x_2) \geq 1/x_1 + 1/x_2$.
$1/(x_1+x_2) \geq (x_2+x_1)/(x_1x_2)$
Since $x_1, x_2 > 1$, $x_1x_2$ and $x_1+x_2$ are positive. We can cross-multiply:
$x_1x_2 \geq (x_1+x_2)^2$
$x_1x_2 \geq x_1^2 + 2x_1x_2 + x_2^2$
$0 \geq x_1^2 + x_1x_2 + x_2^2$
This is false for $x_1, x_2 > 1$ (as $x_1^2 + x_1x_2 + x_2^2$ would be positive). Thus, $f(x) = 1/x$ is not superadditive; it is subadditive.(D) $f(x) = e^{-x}$
We need to check if $e^{-(x_1+x_2)} \geq e^{-x_1} + e^{-x_2}$.
This is equivalent to $1/e^{x_1+x_2} \geq 1/e^{x_1} + 1/e^{x_2}$.
Let $y_1 = e^{x_1}$ and $y_2 = e^{x_2}$. Since $x_1, x_2 > 1$, $y_1, y_2 > e > 1$.
The inequality becomes $1/(y_1y_2) \geq 1/y_1 + 1/y_2$.
This is the same form as $1/x$ and will also be false for $y_1, y_2 > 1$. Thus, $f(x) = e^{-x}$ is not superadditive; it is subadditive.The final answer is $\boxed{A}$