Question

If $\varphi \left(x\right)=f\left(x\right)+f(2a-x)$ and $f^{\prime \prime }\left(x\right)>0,a>0,0\le x\le 2a$ then

• A. $\varphi \left(x\right)$ increases in $(a,2a)$
• B. $\varphi \left(x\right)$ increases in $(0,a)$
• C. $\varphi \left(x\right)$ decreases in $(0,a)$
• D. $\varphi \left(x\right)$ decreases in $(a,2a)$

Answer 1

Answer: (A), (C)

The correct options are
A $\varphi \left(x\right)$ increases in $(a,2a)$
C $\varphi \left(x\right)$ decreases in $(0,a)$

given that $f^{\prime \prime }\left(x\right)>0$ where $0\le x\le 2a$
Therefore, $f^{\prime }\left(x\right)$ is an increasing function.
$\varphi \left(x\right)=f\left(x\right)+f(2a-x)$
${\varphi }^{\prime }\left(x\right)=f^{\prime }\left(x\right)-f^{\prime }(2a-x)$
If $\varphi \left(x\right)$ increases, then ${\varphi }^{\prime }\left(x\right)>0$
$f^{\prime }\left(x\right)>f^{\prime }(2a-x)$
Since, $f^{\prime }\left(x\right)$ is an increasing function. Therefore, $x>2a-x$
$\Rightarrow x>a$
and if $\varphi \left(x\right)$ decreases, then ${\varphi }^{\prime }\left(x\right)<0$
$f^{\prime }\left(x\right)<f^{\prime }(2a-x)$
Since, $f^{\prime }\left(x\right)$ is an increasing function. Therefore, $x<2a-x$
$\Rightarrow x<a$
$\Rightarrow 0<x<a$
Ans: A,C