Question

Let $f^{\prime \prime }\left(x\right)>0\forall x\in R$ and $g\left(x\right)=f(2-x)+f(4+x)$. Then, $g\left(x\right)$ is increasing in

• A. $(-\infty ,-1)$
• B. $(-\infty ,0)$
• C. $(-1,\infty )$
• D. none of these

Answer 1

The correct option is C $(-1,\infty )$

$f^{\prime \prime }\left(x\right)>0\forall x\in R\Rightarrow f^{\prime }\left(x\right)$ is increasing $\forall x\in R$

Now $g^{\prime }\left(x\right)=-f^{\prime }(2-x)+f^{\prime }(4+x)$

If $g^{\prime }\left(x\right)>0$, then $f^{\prime }(4+x)>f^{\prime }(2-x)$

$\Rightarrow 4+x>2-x$ $[\because f^{\prime }(x)$ is increasing $\forall x\in R]$

$\Rightarrow 2x>-2$ or $x>-1$

$\therefore g\left(x\right)$ is increasing in $(-1,\infty )$.