# Let f(x) be a polynomial of degree 6 in x,

Question:

Let $f(x)$ be a polynomial of degree 6 in $x$, in which the coefficient of $x^{6}$ is unity and it has

extrema at $x=-1$ and $x=1$. If $\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=1$, then

$5 \cdot f(2)$ is equal to______.

Solution:

Let $f(x)=x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f$

as $\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=1$ non-zero finite

So, $d=e=f=0$

and $f(x)=x^{3}\left(x^{3}+a x^{2}+b x+c\right)$

Hence, $\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=c=1$

Now, as $f(x)=x^{6}+a x^{5}+b x^{4}+x^{3}$ and $f^{\prime}(x)=0$ at $x=1$ and $x=-1$

i.e., $f^{\prime}(x)=6 x^{5}+5 a x^{4}+4 b x^{3}+3 x^{2}$

$f^{\prime}(1)=0$

$\Rightarrow 6+5 a+4 b+3=0$

$\Rightarrow 5 a+4 b=-9$

$\& f^{\prime}(-1)=0$

$\Rightarrow-6+5 a-4 b+3=0$

$\Rightarrow 5 a-4 b=3$

Solving both we get,

$\mathrm{a}=\frac{-6}{10}=\frac{-3}{5} ; \quad \mathrm{b}=\frac{-3}{2}$

$\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^{6}-\frac{3}{5} \mathrm{x}^{5}-\frac{3}{2} \mathrm{x}^{4}+\mathrm{x}^{3}$

$\therefore 5 f(2)=5\left[64-\frac{3}{5} \cdot 32-\frac{3}{2} \cdot 16+8\right]$

$=320-96-120+40$

$=144$