Question

Let $g\left(x\right)=f\left(\log x\right)+f(2-\log x)$ and $f^{\prime \prime }\left(x\right)<0\forall x\in (0,3)$. Then find the interval in which $g\left(x\right)$ increases.

• A. $(0,\frac{1}{e})$
• B. $(\frac{1}{e},e)$
• C. $(0,e)$
• D. None of these

Answer 1

The correct option is C $(0,e)$

Since $f^{\prime \prime }\left(x\right)<0,$ $f^{\prime }\left(x\right)$ is decreasing function $g^{\prime }\left(x\right),\frac{f^{\prime }\left(\log x\right)}{x}+\frac{f^{\prime }(2-\log x)}{-x}$

$\Rightarrow \frac{1}{x}(f^{\prime }\left(\log x\right)-f^{\prime }(2-\log x))$

$g^{\prime }\left(x\right)>0$ ( increasing )

$\Rightarrow (f^{\prime }\left(\log x\right)>f^{\prime }(2-\log x))$

Since $f^{\prime }\left(x\right)$ is decreasing

$\Rightarrow \log x<2-\log x$

$2\log x<1$

$\log x<\log e$

$x<e$

$x\in (0,e).$

Hence, the answer is $x\in (0,e).$