Le tf(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 $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1,$  then 5 f(2) is equal to……………….

Let f (x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$    Non zero finite

So, d = e = f = 0

f (x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$ Non zero finite

f' (x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$

f' (1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 . (i)

f' (-1) = 0

-6 + 5a – 4b + 3 = 0 . (ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

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

$\therefore 5.f\left(2\right)=144$