Question

Let $y=f\left(x\right)$ be a polynomial function whose degree is greater than zero such that $f\left(\alpha \right)$ and $f\left(\frac{1}{\alpha }\right)$ satisfy the equation $x^3-(1-a)x^2-2ax+a=0$ $\forall \alpha \in R-\left\{0\right\}$ where $a\in R$. If ${\begin{array}{c}\frac{d^{4}y}{d{x}^4}\end{array}|}_{x=2}=0$ and ${\begin{array}{c}\frac{d^{3}y}{d{x}^3}\end{array}|}_{x=2}=-6,$ then for $\alpha =2,$ the value of $8a$ is

Answer 1

$x^3-(1-a)x^2-2ax+a=0$
Roots of above equation are $1,f\left(\alpha \right),f\left(\frac{1}{\alpha }\right)$
$1+f\left(\alpha \right)+f\left(\frac{1}{\alpha }\right)=1-a\dots \left(1\right)$
$1\cdot f\left(\alpha \right)\cdot f\left(\frac{1}{\alpha }\right)=-a\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$f\left(\alpha \right)\cdot f\left(\frac{1}{\alpha }\right)=f\left(\alpha \right)+f\left(\frac{1}{\alpha }\right)\Rightarrow f\left(x\right)=1\pm x^n{f}^{\prime }\left(x\right)=\pm n{x}^{n-1}{f}^{\prime \prime }\left(x\right)=\pm n(n-1)x^{n-2}{f}^{\prime \prime \prime }\left(x\right)=\pm n(n-1(n-2)x^{n-3}\Rightarrow f^{\prime \prime \prime }(2)=\pm n(n-1\left)\right(n-2)2^{n-3}=-6$
$n=3$ and negative sign is to be taken.
Again, $f^{\prime \prime \prime \prime }\left(x\right)=\pm n(n-1)(n-2)(n-3)x^{n-4}$
$f^{\prime \prime \prime \prime }\left(2\right)=0\Rightarrow n=0,1,2,3$

$\therefore n=3$
Now, $f\left(x\right)=1-x^3$
$f\left(2\right)=-7,f\left(\frac{1}{2}\right)=\frac{7}{8}$
From $\left(2\right),$ $a=\frac{49}{8}$