Question

Let $f(x)$ be a polynomial satisfying $\underset{x\to \infty }{\mathrm{Lim}}\frac{x^{2}f(x)}{2x^5+3}=6$, also $f(1)=3,f(3)=7$ and $f(5)=11$, then find the value of $\left(\frac{f(6)+5f(4)}{29}\right)$.

Answer 1

Answer: 2

$f(x)$ must be polynomial of degree '$3$'
$f(x)=\lambda (x-1)(x-3)(x-5)+(2x+1)$
$\underset{x\to \infty }{\mathrm{Lim}}\frac{x^2(\lambda (x-1)(x-3)(x-5)+2x+1)}{2x^5+3}=6$
$\frac{\lambda }{2}=6\Rightarrow \lambda =12$
Hence, $f(x)=12(x-1)(x-3)(x-5)+(2x+1)$
$f(6)=193$
$f(4)=-27$
$\left(\frac{f(6)+5f(4)}{29}\right)=2$