Question

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

Answer 1

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