JEE Main 2020 — Application Of Derivatives Question with Solution

Let f(x) = ax³ + bx² + cx + d

Given f(-1) = 10, f(1) = -6

$$\therefore$$ -a + b - c + d = 10 ....(i)

and a + b + c + d = -6 ......(ii)

adding (i) + (ii)

2(b + d) = 4

$$\Rightarrow$$ b + d = 2 ....(iii)

f'(x) = 3ax² + 2bx + c

Given f'(-1) = 0

$$\Rightarrow$$ 3a - 2b + c = 0 .....(iv)

f"(x) = 6ax + 2b

Given f"(1) = 0

$$\therefore$$ 6a + 2b = 0 ....(v)

$$\Rightarrow$$ b = -3a

adding (iv) + (v), we get

9a + c = 0 ....(vi)

$$\Rightarrow$$ $$9\left(\frac{-b}{3}\right)$$ + c = 0

$$\Rightarrow$$ c = 3b

f(x) = $$\frac{-b}{3}{x}^3$$ + bx² + 3bx + (2 - b)

$$\Rightarrow$$ f'(x) = -bx² + 2bx + 3b

= -b(x² - 2x - 3)

At maxima and minima f'(x) = 0

$$\therefore$$ (x² - 2x - 3) = 0

$$\Rightarrow$$ (x - 3) (x + 1) = 0

x = 3, -1

As a + b + c + d = -6

$$\Rightarrow$$ $$\frac{-b}{3}$$ + b + 3b + 2 - b = -6

$$\Rightarrow$$ b = -3

$$\therefore$$ f'(x) = 3(x² - 2x - 3)

$$\Rightarrow$$ f''(x) = 3(2x - 2)

At x = 3, f''(x) = 3(2.3 - 2) = 12 > 0

$$\therefore$$ Minima at x = 3.