TIFR CSE 2019 | Part A | Question: 6

Question

A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is said to be $\textit{convex}$ if for all $x,y \in \mathbb{R}$ and $\lambda$ such that $0 \leq \lambda \leq1,$

      $f(\lambda x+ (1-\lambda)y) \leq \lambda f (x) + (1-\lambda) f(y)$.

Let  $f:$$\mathbb{R}$ $→$ $\mathbb{R}$ be a convex function , and define the following functions:

             $p(x) = f(-x) , \: \: \: q(x) = -f(-x), \text{ and } r(x) = f(1-x)$.

Which of the functions $p,q$ and $r$ must be convex?

• A. Only $p$
• B. Only $q$
• C. Only $r$
• D. Only $p$ and $r$
• E. Only $q$ and $r$

Answer 1

A twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative.

For eg :- $f(x) = x^2 \implies f''(x) = 2 >0$  so $f$ is a convex function.

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Now , we know $f(x)=x^2$ is a convex function.

$p(x) = f(-x) = x^2 \implies p''(x) = 2 >0$

$q(x) =- f(-x) = -x^2 \implies q''(x) = -2 <0$

$ r(x) =- f(1-x) = (1-x)^2\implies r''(x) = 2(1-x) = 2 >0$

Hence $p(x)$ and $r(x)$ are convex functions.

 

$\therefore$ Option $D$ is the correct choice.

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References:-

https://en.wikipedia.org/wiki/Convex_function

Comments

r(x) is given as f(1-x) then only you can get the positive value. I think it was just a typo.