Question

A function $f:R\to R$ is defined as $f\left(x\right)=x^3-2x^2+5x-1$ and $g:R\to R$ be its inverse function. If $k{g}^{\prime }\left(3\right)=1,$ then value(s) of $k$ is/are

• A. $-3$
• B. $17$
• C. $27$
• D. $7$

Answer 1

The correct option is D $7$

Since, $g$ and $f$ are inverse function
$\therefore g\circ f=I\Rightarrow g\left[f\right(x\left)\right]=x$
Differentiating both the sides,
$g^{\prime }\left[f\right(x\left)\right]f^{\prime }\left(x\right)=1$
If $f\left(x\right)=3,$ then
$g^{\prime }\left(3\right)\cdot f^{\prime }\left(x\right)=1...\left(1\right)$
Given, $k{g}^{\prime }\left(3\right)=1...\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we have
$k=f^{\prime }\left(x\right)...\left(3\right)$
Now, $f\left(x\right)=3$
$\Rightarrow x^3-2x^2+5x-1=3$
$\Rightarrow x^3-2x^2+5x-4=0$
$\Rightarrow (x-1)(x^2-x+4)=0$
$\Rightarrow x=1$

$\Rightarrow f^{\prime }\left(x\right)=6x^2-4x+5$
$\Rightarrow f^{\prime }\left(1\right)=6-4+5=7$
$\therefore k=f^{\prime }\left(1\right)=7$