Question

Let $f$ be a twice differentiable function in $(-\infty ,\infty )$ such that $f^{\prime \prime }\left(x\right)\le 0\forall x\in R.$ If $g\left(x\right)=f\left(x\right)+f(1-x)$ and $g^{\prime }\left(\frac{1}{4}\right)=0,$ then

• A. $g\left(x\right)$ is increasing in $(-\infty ,0)$
• B. $g\left(x\right)$ attains local minimum at $x=\frac{1}{2}$
• C. $g^{\prime \prime }\left(x\right)=0$ has at least two roots in $[0,1]$
• D. $g^{\prime }\left(x\right)<0$ for all $x\in (\frac{1}{2},\infty )$

Answer 1

Answer: (A), (C), (D)

$g^{\prime }\left(x\right)<0$ for all $x\in (\frac{1}{2},\infty )$

$f^{\prime \prime }\left(x\right)\le 0\forall x\in R$
$\Rightarrow f^{\prime }$ is decreasing in $R$
$g^{\prime }\left(x\right)=f^{\prime }\left(x\right)-f^{\prime }(1-x)\cdots \left(1\right)$

If $x>1-x$ i.e., $x>\frac{1}{2}$
then $f^{\prime }\left(x\right)\le f^{\prime }(1-x)$
$\Rightarrow f^{\prime }\left(x\right)-f^{\prime }(1-x)\le 0$
$\Rightarrow g^{\prime }\left(x\right)\le 0$ for all $x\in (\frac{1}{2},\infty )$

If $x<1-x$ i.e., $x<\frac{1}{2}$
then $f^{\prime }\left(x\right)\ge f^{\prime }(1-x)$
$\Rightarrow g^{\prime }\left(x\right)\ge 0$ for all $x\in (-\infty ,\frac{1}{2})$
By first derivative test,
$g\left(x\right)$ has local maximum at $x=\frac{1}{2}$

$g\left(x\right)=f\left(x\right)+f(1-x)$
Replacing $x$ by $1-x$, we get
$g\left(x\right)=g(1-x)$
Again, replacing $x$ by $x+\frac{1}{2}$
$g(\frac{1}{2}+x)=g(\frac{1}{2}-x)$
$\Rightarrow$ Graph of $y=g\left(x\right)$ is symmetric about $x=\frac{1}{2}$
Given, $g^{\prime }\left(\frac{1}{4}\right)=0$
By symmetricity, $g^{\prime }\left(\frac{3}{4}\right)=0$
Also, from $\left(1\right),$ $g^{\prime }\left(\frac{1}{2}\right)=0$
By Rolle's theorem, $g^{\prime \prime }\left(x\right)$ vanishes for at least two values in $[0,1]$