Question

If $f(-3x-5x^3-2x^5)=x,$ $x\in R,$ then which of the following statements is(are) CORRECT ?

• A. $f\left(x\right)$ is increasing for all $x\in R.$
• B. origin is the point of inflection for $y=f\left(x\right).$
• C. Area enclosed by $f^{-1}\left(x\right)$ and $x-$axis from $x=-1$ to $x=1$ is $\frac{37}{6}$ sq. units.
• D. $f^{-1}\left(x\right)$ is concave upward in $(-\infty ,0)$ and concave downward in $(0,\infty ).$

Answer 1

Answer: (B), (C), (D)

$f^{-1}\left(x\right)$ is concave upward in $(-\infty ,0)$ and concave downward in $(0,\infty ).$

$f(-3x-5x^3-2x^5)=x$
Let $g\left(x\right)=f^{-1}\left(x\right)=-3x-5x^3-2x^5$
$\Rightarrow g^{\prime }\left(x\right)=-3-15x^2-10x^4$
$\Rightarrow g^{\prime }\left(x\right)<0$
$\Rightarrow g\left(x\right)$ is decreasing.
$\Rightarrow f\left(x\right)$ is also decreasing.

$g^{\prime \prime }\left(x\right)=-30x-40x^3=0$
$\Rightarrow -10x(4x^2+3)=0\Rightarrow x=0$
$g^{\prime \prime }\left(x\right)$ changes sign about $x=0$
$(0,0)$ is the point of inflection of $y=f^{-1}\left(x\right).$
Hence, it is also point of inflection of $y=f\left(x\right).$

Required area $=2\left|\underset{-1}{\overset{0}{\int }}{f}^{-1}\right(x\left)dx\right|$
$=2\left|\underset{-1}{\overset{0}{\int }}\right(-3x-5x^3-2x^5\left)dx\right|$
$=2\left|{[\frac{-3x^2}{2}-\frac{5x^4}{4}-\frac{x^6}{3}]}_{-1}^0\right|$
$=\frac{37}{6}$ sq.units

$g^{\prime \prime }\left(x\right)>0$ in $(-\infty ,0)$
$g^{\prime \prime }\left(x\right)<0$ in $(0,\infty )$