Let $f(x)$ be a polynomial of degree 5 such that $x=\pm 1$ are its critical points. If $\lim _{x \rightarrow 0}\left(2+\frac{f(x)}{x^{3}}\right)=4$, then which one of the following is not true?
Correct Option: , 4
$f(x)=a x^{5}+b x^{4}+c x^{3}$
$\lim _{x \rightarrow 0}\left(2+\frac{a x^{5}+b x^{4}+c x^{3}}{x^{3}}\right)=4$
$\Rightarrow 2+c=4 \Rightarrow c=2$
$f^{\prime}(x)=5 a x^{4}+4 b x^{3}+6 x^{2}$
$=x^{2}\left(5 a x^{2}+4 b x+6\right)$
Since, $x=\pm 1$ are the critical points,
$\therefore \quad f^{\prime}(1)=0 \Rightarrow 5 a+4 b+6=0$ ...(1)
$f^{\prime}(-1)=0 \Rightarrow 5 a-4 b+6=0$......(2)
From eqns. (1) and (2),
$b=0$ and $a=-\frac{6}{5}$
$f(x)=\frac{-6}{5} x^{5}+2 x^{3}$
$f^{\prime}(x)=-6 x^{4}+6 x^{2}=6 x^{2}\left(-x^{2}+1\right)$
$=-6 x^{2}(x+1)(x-1)$
$\therefore f(x)$ has minima at $x=-1$ and maxima at $x=1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.