QJEEM20Mad505: JEE Main 2020 Mathematics - Class 12 JEEM20Mad505 Solution: Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its…

let f(x) = ax⁵ + bx⁴ + cx³ + dx² + ex + f

Given $\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$

$\Rightarrow$ $\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f} \over {{x^3}}}} \right)$ = 4

As this limit exists so d = e = f = 0

$\therefore$ f(x) = ax⁵ + bx⁴ + cx³

$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3}} \over {{x^3}}}} \right)$ = 4

$\Rightarrow$ 2 + c = 4

$\Rightarrow$ c = 2

$\therefore$ f(x) = ax⁵ + bx⁴ + 2x³

$\Rightarrow$ f'(x) = 5ax⁴ + 4bx³ + 6x²

As x = ±1 are its critical points so f'(x) = 0 at x = ±1.

f'(1) = 5a + 4b + 6 = 0 ....(1)

and f'(-1) = 5a - 4b + 6 = 0 .....(2)

Solving (1) and (2),

a = $- {6 \over 5}$ and b = 0

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

So f'(x) = -6x⁴ + 6x² = 6x²(1 + x)(1 - x)

Sign scheme for f'(x)



It is clear that maxima at x = 1 and minima at x = –1.