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A superadditive function $f(\cdot)$ satisfies the following property $$f\left ( x_{1} +x_{2}\right )\geq f\left ( x_{1} \right ) + f\left ( x_{2} \right )$$

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

  1. $e^{x}$
  2. $\sqrt{x}$
  3. $1/x$
  4. $e^{-x}$
in Numerical Ability by (1.9k points)
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2 Answers

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Best answer

Given that, a superadditive function $f(\cdot):$ $$f\left ( x_{1} +x_{2}\right )\geq f\left ( x_{1} \right ) + f\left ( x_{2} \right )$$

Since $x >1,$ a superadditive function can never be a decreasing function. So, optionc C and D can staright away be ruled out.

We can check options A and B by taking the value of $x_{1} = 2$ and $x_{2} = 3.$ 

  1. $f(x) = \sqrt{x}$
    $f(2+3) \geq f(2) + f(3)$
    $\implies f(5) \geq f(2) + f(3)$
    $\implies \sqrt{5} \geq \sqrt{2} + \sqrt{3}$
    $\implies 2.236 \geq 1.414 + 1.732$
    $\implies 2.236 \geq 3.146 \;{\color{Red} {\textbf{(False)}}}$

So, correct answer should be $(A).$

Reference:https://en.wikipedia.org/wiki/Superadditivity

by (3.8k points)
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Solve it using exploring the options

Let $X_{1}$ = 2 and $X_{2}$ = 3

A. f(x) = $e^{x}$

      $e^{5}$ > $e^{2}$ + $e^{3}$ ( using the calculator)

it satisfies the condition so option A is correct.

by (1.4k points)
0
Just one example working does not prove anything. A counter-example can prove a statement wrong but an example cannot prove a statement right.
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