Question

$S_1:$ If $f\left(x\right)$ is an increasing function with downward concavity, then concavity of $f^{-1}\left(x\right)$ is upwards.
$S_2:$ If $f\left(x\right)$ is decreasing function with downward concavity, then concavity of $f^{-1}\left(x\right)$ is upwards.

• A. $S_1$ and $S_2$ both are true
• B. $S_1$ is true and $S_2$ is false
• C. $S_1$ is false and $S_2$ is true
• D. $S_1$ and $S_2$ both are false

Answer 1

The correct option is B $S_1$ is true and $S_2$ is false

Let $g\left(x\right)=f^{-1}\left(x\right)$
$\Rightarrow f\left(g\right(x\left)\right)=x$
$\Rightarrow f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)=1$
$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\right(x\left)\right)}$
$\Rightarrow g^{\prime \prime }\left(x\right)=-\frac{f^{\prime \prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)}{\left[f^{\prime }\right(g\left(x\right)){]}^2}\cdots \left(i\right)$
$S_1:$ Since, $f\left(x\right)$ is increasing, so $g\left(x\right)$ will also be increasing
$\Rightarrow g^{\prime }\left(x\right)>0$
$\Rightarrow$ if $f\left(x\right)$ is downward concaving $(i.e.f^{\prime \prime }(x)<0)$
$\therefore g^{\prime \prime }\left(x\right)>0(i.e.\text{upward concaving})$

$S_2:$ if $f\left(x\right)$ is decreasing, then $g\left(x\right)$ will also be decreasing, $g^{\prime }\left(x\right)<0$
and if $f\left(x\right)$ is downward concave, $f^{\prime \prime }\left(x\right)<0$
$\Rightarrow$ from $\left(i\right),g^{\prime \prime }\left(x\right)<0$
$\therefore$ downward concaving.
So, $S_1$ is true and $S_2$ is flase