Question

If f and g are inverse of each other then which of the following statement is CORRECT ?
$I.$ if $f\left(x\right)=x^3+2e^x$ then $g^{\prime }\left(2\right)=\frac{1}{2}.$
$II.$ if $f\left(x\right)=e^x-e^{-x}$ then $g^{\prime }\left(2\right)=\frac{1}{2\sqrt{2}}.$
$III.$ if $f\left(x\right)=x^3+2x+2$ then $g^{\prime }\left(2\right)=\frac{1}{2}$

• A. I and II only
• B. II and III only
• C. III and I only
• D. I , II and III

Answer 1

The correct option is D I , II and III

As $g\left(x\right)=f^{-1}\left(x\right)\Rightarrow f\left(x\right)=g^{-1}\left(x\right)$
$\Rightarrow g\left(f\right(x\left)\right)=x\Rightarrow g^{\prime }\left(f\right(x\left)\right).f^{\prime }\left(x\right)=1$
$\Rightarrow g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}$
I. put $f\left(x\right)=x^3+2e^x=2\Rightarrow x=0$
$\therefore f^{\prime }\left(x\right)=3x^2+2e^x{|}_{x=0}=2$
$\therefore g^{\prime }\left(f\right(x\left)\right)=g^{\prime }\left(2\right)=\frac{1}{f^{\prime }\left(x\right)}=\frac{1}{2}$
II. $f^{\prime }\left(x\right)=e^x+e^{-x}$
and $f\left(x\right)=2\Rightarrow e^x-e^{-x}=2$
$(e^x{)}^2-2e^x-1=0$
$e^x=\frac{2\pm \sqrt{4+4}}{2}=1\pm \sqrt{2}$
$e^x=1+\sqrt{2}as1-\sqrt{2}<0$
$e^{-x}=\sqrt{2}-1$
$e^x+e^{-x}=2\sqrt{2}\therefore g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(x\right)}=\frac{1}{2\sqrt{2}}$
III. $f\left(x\right)=x^3+2x+2=2\Rightarrow x=0$
$g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}=\frac{1}{3x^2+2}=\frac{1}{2}$