JEE Main 2020 — Application Of Derivatives Question with Solution

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

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

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

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

$$\therefore$$ f(x) = ax⁵ + bx⁴ + cx³

$$\underset{x\to 0}{\lim }\left(2+\frac{a{x}^5+b{x}^4+c{x}^3}{x^3}\right)$$ = 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 = $$-\frac{6}{5}$$ and b = 0

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

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.