JEE Main 2021 — Application Of Derivatives Question with Solution

Let f(x) = ax³ + bx² + cx + d

f'(x) = 3ax² + 2bx + c $$\Rightarrow$$ f''(x) = 6ax + 2b

f'(x) has local minima at x = $$-$$1, so

$$\because$$ f''($$-$$1) = 0 $$\Rightarrow$$ $$-$$6a + 2b = 0 $$\Rightarrow$$ b = 3a ..... (i)

f(x) has local minima at x = 1

f'(1) = 0

$$\Rightarrow$$ 3a + 6a + c = 0

$$\Rightarrow$$ c = $$-$$9a ..... (ii)

f(1) = $$-$$10

$$\Rightarrow$$ $$-$$5a + d = $$-$$10 ..... (iii)

f($$-$$1) = 6

$$\Rightarrow$$ 11a + d = 6 ..... (iv)

Solving Eqs. (iii) and (iv)

a = 1, d = $$-$$5

From Eqs. (i) and (ii),

b = 3, c = $$-$$9

$$\therefore$$ f(x) = x³ + 3x² $$-$$ 9x $$-$$ 5

So, f(3) = 27 + 27 $$-$$ 27 $$-$$ 5 = 22