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

Hint: 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 \[\phi \]
\[\begin{align}
  & \Rightarrow \phi '(x)>0 \\
 & \Rightarrow f'(x)-f'(2-x)>0 \\
 & \Rightarrow f'(x)>f'(2-x) \\
 & \Rightarrow x>2-x \\
 & \Rightarrow 2x>2 \\
 & \Rightarrow x>1 \\
\end{align}\]
So, function \[\phi \] is increasing in domain where x>1 i.e. (1,2)
So, for decreasing of function \[\phi \]
\[\begin{align}
  & \Rightarrow \phi '(x)<0 \\
 & \Rightarrow f'(x)-f'(2-x)<0 \\
 & \Rightarrow f'(x)>f'(2-x) \\
 & \Rightarrow x<2-x \\
 & \Rightarrow 2x<2 \\
 & \Rightarrow x<1 \\
\end{align}\]
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).