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 $\displaystyle\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=1$, then $5 \cdot f (2)$ is equal to _____

Answer 1

Let $f(x)=x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+$
as $\displaystyle\lim _{x \rightarrow 0} \frac{ f ( x )}{ x ^{3}}=1$ non-zero finite
So, $d = e = f =0$
and $f(x)=x^{3}\left(x^{3}+a x^{2}+b x+c\right)$
Hence, $\displaystyle\lim_{x \rightarrow 0} \frac{ f ( x )}{ x ^{3}}= c =1$
Now, as $f(x)=x^{6}+a x^{5}+b x^{4}+x^{3}$
and $f'( x )=0$ at $x =1$ and $x =-1$
i.e., $f'(x)=6 x^{5}+5 a x^{4}+4 b x^{3}+3 x^{2}$
$f '(1)=0$
$\Rightarrow 6+5 a+4 b+3=0$
$\Rightarrow 5 a+4 b=-9$
&$ f '(-1)=0$
$\Rightarrow -6+5 a -4 b +3=0$
$\Rightarrow 5 a -4 b =-9$
Solving both we get,
$a=\frac{-6}{10}=\frac{-3}{5} ; b=\frac{-3}{2}$
$ \therefore f ( x )= x ^{6}-\frac{3}{5} x ^{5}-\frac{3}{2} x ^{4}+ 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 $
$=144$