Question

If $f\left(x\right)=-4e^{\frac{1-x}{2}}+1+x+\frac{x^2}{2}+\frac{x^3}{3}$ and $g\left(x\right)=f^{-1}\left(x\right)$ then the reciprocal of $g^{\prime }\left(\frac{-7}{6}\right)$ is:

• A. $4$
• B. $5$
• C. $10$
• D. $20$

Answer 1

The correct option is B $5$

Given :

$f\left(x\right)=4e^{\frac{(1-x)}{2}}+1+x+\frac{x^2}{2}+\frac{x^3}{3}$

Now , $g\left(x\right)=f^{-1}\left(x\right)$

We know that

$f\left(f^{-1}\left(x\right)\right)=x\Rightarrow f^{-1}\left(f\right(x\left)\right)=x\Rightarrow f\left(f\left(x\right)\right)=x\Rightarrow g^{{}^{\prime }}\left(f\right(x\left)\right).f^1\left(x\right)=1$

Replacing $f\left(x\right)=\frac{-7}{6}$, we have

$4e^{\frac{(1-x)}{2}}+1+x+\frac{x^2}{2}+\frac{x^3}{3}=\frac{-7}{6}4e^{\frac{(1-x)}{2}}+x+\frac{x^2}{2}+\frac{x^3}{3}=\frac{-13}{6}$

We see that the LHS contains an exponential term,but RHS does not have any exponential term.

$e^{\frac{(1-x)}{2}}=1\Rightarrow \frac{1-x}{2}=0\Rightarrow x=1$

$\therefore f\left(x\right)=\frac{7}{6}$ we have $x=1$

Now , $f^{{}^{\prime }}\left(x\right)=-4e^{\frac{(1-x)}{2}}\left(\frac{-1}{2}\right)+0+1+x+x^2\therefore f^{{}^{\prime }}\left(1\right)=2+0+1+1+1=5\therefore g^{{}^{\prime }}\left(f\right(x\left)\right).f^{{}^{\prime }}\left(x\right)=1$

for $x=1f\left(x\right)=\frac{-7}{6}$ and $f^{{}^{\prime }}\left(1\right)=5$

$\therefore g^{{}^{\prime }}\left(\frac{-7}{6}\right).5=1\Rightarrow g^{{}^{\prime }}\left(\frac{-7}{6}\right)=\frac{1}{5}$