Let $f(x)$ be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at $x=-1\;and\; x=1.$ If $\lim_{x \to0 }\frac{f(x)}{x^{3}}=1\;$ then 5f(2) is equal to________ Option: 1 12 Option: 2 36 Option: 3 72 Option: 4 144

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

Now, as $f(x)=x^{6}+a x^{5}+b x^{4}+x^{3}$

and $\mathrm{f}^{\prime}(\mathrm{x})=0\text{ at }\mathrm{x}=1\text{ and }\mathrm{x}=-1$

\begin{aligned} &\text {i.e., } f^{\prime}(x)=6 x^{5}+5 a x^{4}+4 b x^{3}+3 x^{2}\\ &\;\;\;\;\;\;\;\mathrm{f}^{\prime}(1)=0\\ &\Rightarrow 6+5 a+4 b+3=0\\ &\Rightarrow 5 a+4 b=-9 \end{aligned}

$\& \\\mathrm{f}^{\prime}(-1)=0 \\ \Rightarrow-6+5 \mathrm{a}-4 \mathrm{~b}+3=0 \\ \Rightarrow 5 \mathrm{a}-4 \mathrm{~b}=3$

\begin{aligned} &\text {Solving both we get, }\\ & a=\frac{-6}{10}=\frac{-3}{5} ; \quad b=\frac{-3}{2} \\ & \therefore \quad 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 \end{aligned}

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