Question

Let f: [0,2] →R be a twice differentiable function such that $f''\left( x \right)>0$ for all $x\in (0,2)$If $~\phi (x)=f(x)+f(2-x)$then $\phi$ is:
a) Decreasing on (0,2)
b) increasing on (0,1) and decreasing on (1,2)
c) increasing on (0,2)
d) decreasing on (0,1) and increasing on (1,2)

Answer 1

We know that if y=f(x) is a differentiable function on interval (a,b), if for any two point ${{x}_{1}}$ and ${{x}_{2}}$ lies in domain (a, b) such that${{x}_{1}} < {{x}_{2}}$, if there holds a inequality $f({{x}_{1}})\le f({{x}_{2}})$, then the function is called increasing function. If the inequality is strict i.e, $f({{x}_{1}}) < f({{x}_{2}})$ then the function is called strictly increasing on the interval (a,b).

Complete step-by-step answer:
We have $\phi (x)=f(x)+f(2-x)$
We will differentiate the above equation w.r.t x
$\Rightarrow \phi '(x)=f'(x)-f'(2-x)$
It is given that $f''\left( x \right)>0$ which is only possible if $f'\left( x \right)$ is strictly increasing.
So, for increasing of function $\phi$

& \Rightarrow \phi '(x)>0 \\\ & \Rightarrow f'(x)-f'(2-x)>0 \\\ & \Rightarrow f'(x)>f'(2-x) \\\ & \Rightarrow x>2-x \\\ & \Rightarrow 2x>2 \\\ & \Rightarrow x>1 \\\ \end{aligned}$$ So, function $$\phi $$ is increasing in domain where x>1 i.e. (1,2) So, for decreasing of function $$\phi $$ $$\begin{aligned} & \Rightarrow \phi '(x)<0 \\\ & \Rightarrow f'(x)-f'(2-x)<0 \\\ & \Rightarrow f'(x)>f'(2-x) \\\ & \Rightarrow x<2-x \\\ & \Rightarrow 2x<2 \\\ & \Rightarrow x<1 \\\ \end{aligned}$$ So, function $$\phi $$ is decreasing in domain where x<1 i.e. (0,1) **So, the correct answer is “Option d”.** **Note:** We know that if y=f(x) is a differentiable function on interval (a , b), if for any two point ${{x}_{1}}$ and ${{x}_{2}}$ lies in domain (a, b) such that ${{x}_{1}}<{{x}_{2}}$, if there holds a inequality $f({{x}_{1}})\le f({{x}_{2}})$, then the function is called increasing function. If the inequality is strict i.e., $f({{x}_{1}}) < f({{x}_{2}})$ then the function is called strictly increasing on the interval (a, b). If for any two point ${{x}_{1}}$ and ${{x}_{2}}$ lies in domain (a, b) such that${{x}_{1}} > {{x}_{2}}$, if there holds a inequality $f({{x}_{1}})\ge f({{x}_{2}})$, then the function is called decreasing function. If the inequality is strict i.e., $f({{x}_{1}})>f({{x}_{2}})$ then the function is called strictly decreasing on the interval (a, b).