JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 21)
Let f(x) be a cubic polynomial with f(1) = $$-$$10, f($$-$$1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = $$-$$1. Then f(3) is equal to ____________.
Answer
22
Explanation
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
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