If is a non-zero polynomial of degree four, having local extreme points at , then the set contains exactly
- A.four rational numbers
- B.two irrational and two rational numbers
- C.four irrational numbers
- D.two irrational and one rational number
Show answer and solution
Correct option: D
Solution
Since f'(x) has roots at x = -1, 0, 1, we write f'(x) = a·x(x+1)(x-1) = a(x³ - x).
Integrating: f(x) = a(x⁴/4 - x²/2) + C, so f(0) = C.
Setting f(x) = f(0) = C:
a(x⁴/4 - x²/2) = 0
x²(x² - 2) = 0
x = 0, 0, ±√2
So S = {0, √2, -√2}, which contains exactly one rational number (0) and two irrational numbers (±√2) — two irrational and one rational number.
Integrating: f(x) = a(x⁴/4 - x²/2) + C, so f(0) = C.
Setting f(x) = f(0) = C:
a(x⁴/4 - x²/2) = 0
x²(x² - 2) = 0
x = 0, 0, ±√2
So S = {0, √2, -√2}, which contains exactly one rational number (0) and two irrational numbers (±√2) — two irrational and one rational number.
[JEE Main 2019 · 9 Jan · Shift I]