Question

Let f(x) be twice differentiable function such that f "(x) > 0 in [0, 2]. Then which of the following option(s) is/are correct?

• A. $f\left(0\right)+f\left(2\right)=2f\left(c\right)$, for atlease one $c,cϵ(0,2)$
  
• B. $f\left(0\right)+f\left(2\right)<2f\left(1\right)$
  
• C. $f\left(0\right)+f\left(2\right)>2f\left(1\right)$
  
• D. $2f\left(0\right)+f\left(2\right)>3f\left(\frac{2}{3}\right)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(0\right)+f\left(2\right)=2f\left(c\right)$, for atlease one $c,cϵ(0,2)$

C $f\left(0\right)+f\left(2\right)>2f\left(1\right)$

D $2f\left(0\right)+f\left(2\right)>3f\left(\frac{2}{3}\right)$

$f"\left(x\right)>0\forall xϵ[0,2]$

$\Rightarrow$ f' (x) is increasing function

$\Rightarrow f^{\prime }\left(c_1\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0},c_1ϵ(0,1)$

$f^{\prime }\left(c_2\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1},c_2ϵ(1,2)$

$\therefore f^{\prime }\left(c_1\right)<f^{\prime }\left(c_2\right)\Rightarrow f\left(0\right)+f\left(2\right)>2f\left(1\right)$

similarly applying LMVT between

$[0,\frac{2}{3}]\text{and}[\frac{2}{3},2]$; we get

$\frac{f\left(2\right)-f\left(\frac{2}{3}\right)}{\frac{4}{3}}>\frac{f\left(\frac{2}{3}\right)-f\left(0\right)}{\frac{2}{3}}$

$\Rightarrow 2f\left(0\right)+f\left(2\right)>3f\left(\frac{2}{3}\right)$

$f\left(0\right)+f\left(2\right)=2f\left(c\right)$, for atlease one $c,cϵ(0,2)\text{By intermediate Value Theorem}$