Question

Let $f:R\to R$ be defined by $f\left(x\right)=x^3-3\lambda x^2+({\lambda }^2+8)x+24,$ where $\lambda$ is the largest number for which $f\left(x\right)$ is bijective. Then the value of $\left(\frac{f\left(1\right)+f^{-1}\left(31\right)}{4}\right)$ is

Answer 1

$f\left(x\right)=x^3-3\lambda x^2+({\lambda }^2+8)x+24$
Clearly, $f\left(x\right)$ is onto.
For $f\left(x\right)$ to be one-one,
$f^{\prime }\left(x\right)\ge 0\forall x\in R$
$f^{\prime }\left(x\right)=3x^2-6\lambda x+({\lambda }^2+8)D\le 0\Rightarrow 36{\lambda }^2-12({\lambda }^2+8)\le 0\Rightarrow 2{\lambda }^2-8\le 0\Rightarrow \lambda \in [-2,2]$
Largest value of $\lambda$ is $2.$
Now, $f\left(x\right)=x^3-6x^2+12x+24$
$f\left(1\right)=31$
$\therefore \frac{f\left(1\right)+f^{-1}\left(31\right)}{4}=\frac{31+1}{4}=8$