Question

Let $f:[0,2]\to R$ be a twice differentiable function such that $f^{\prime \prime }\left(x\right)>0$, for all $x\in (0,2)$. If $\varphi \left(x\right)=f\left(x\right)+f(2-x),$ then $\varphi$ is:

• A. increasing on $(0,1)$ and decreasing on $(1,2)$
• B. decreasing on $(0,1)$ and increasing on $(1,2)$
• C. increasing on $(0,2)$
• D. decreasing on $(0,2)$

Answer 1

The correct option is B decreasing on $(0,1)$ and increasing on $(1,2)$

Given,
$f^{\prime \prime }\left(x\right)>0$
$f^{\prime }\left(x\right)$ is increasing function
$\varphi \left(x\right)=f\left(x\right)+f(2-x)$
${\varphi }^{\prime }\left(x\right)=f^{\prime }\left(x\right)-f^{\prime }(2-x)$

Case (i)
$\varphi \left(x\right)$ is increasing
i.e. ${\varphi }^{\prime }\left(x\right)>0$
$f^{\prime }\left(x\right)-f^{\prime }(2-x)>0$
$f^{\prime }\left(x\right)>f^{\prime }(2-x)$
$x>2-x$
$2x>2$
$x>1$
i.e. $x\in (1,2)$

Case (ii)
$\varphi \left(x\right)$ is decreasing
i.e. ${\varphi }^{\prime }\left(x\right)<0$
$f^{\prime }\left(x\right)-f^{\prime }(2-x)<0$
$f^{\prime }\left(x\right)<f^{\prime }(2-x)$
$x<2-x$
$x<1$
i.e. $x\in (0,1)$