Question

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

Answer 1

Step- 1: Determine $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$:

$\underset{x\to \infty }{\lim }\frac{x^{2}f\left(x\right)}{(2x^5+3)}=6$ is finite and non-zero. $f\left(x\right)$ will be a polynomial of degree $3$.

$\therefore f\left(x\right)=\lambda \left[(x-1)(x-3)(x-5)+2x+1\right]$

Now,

$$\begin{gathered} \Rightarrow \underset{x\to \infty }{\lim }\frac{x^2\lambda \left[(x-1)(x-3)(x-5)+2x+1\right]}{(2x^5+3)}=6 \\ \Rightarrow \frac{\lambda }{2}=6 \\ \Rightarrow \lambda =12 \end{gathered}$$

$f\left(x\right)=12\left[(x-1)(x-3)(x-5)+2x+1\right]$

Step- 2:Find the value of $\frac{\mathbf{\left[}\mathbf{f}\mathbf{\right(}\mathbf{6}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{5}\mathbf{f}\mathbf{(}\mathbf{4}\mathbf{\left)}\mathbf{\right]}}{\mathbf{29}}$

$$\begin{gathered} f\left(6\right)=12\left[\right(5\left)\right(3\left)\right(1\left)\right]+12+1 \\ \Rightarrow =193 \\ andf\left(4\right)=12\left[\right(3\left)\right(1\left)\right(-1\left)\right]+9 \\ \Rightarrow =-27 \\ \frac{\left[f\right(6)+5f(4\left)\right]}{29}=\frac{193+5(-27)}{29} \\ \Rightarrow =\frac{58}{29} \\ \Rightarrow =2 \end{gathered}$$

Hence, the value of $\frac{\mathbf{\left[}\mathbf{f}\mathbf{\right(}\mathbf{6}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{5}\mathbf{f}\mathbf{(}\mathbf{4}\mathbf{\left)}\mathbf{\right]}}{\mathbf{29}}$ is $2$