JEE Main 2022 · 27 June · morning · Single correct · +4 / −1
The number of distinct real roots of x⁴ - 4x + 1 = 0 is :
- (a) 4
- (b) 2
- (c) 1
- (d) 0
Answer and solution
Answer (b) 2
Solution
Maths sir · 4 steps
One turning point and a negative minimum: the graph dips below the x-axis once, so it crosses it twice.
- Find the turning point. Let f(x) = x⁴ - 4x + 1. Then f'(x) = 4x³ - 4 = 4(x³ - 1), which is zero only at x = 1.
- Read the shape. For x < 1, f'(x) < 0, so f decreases. For x > 1, f'(x) > 0, so f increases. The graph has one minimum at x = 1, and f(x) → +∞ at both ends.
f(1) = 1 - 4 + 1 = -2 < 0
- Locate the roots. The minimum value is negative, so the graph comes down, crosses the x-axis, and goes back up to cross it again. The signs confirm this: f(0) = 1 > 0, f(1) = -2 < 0 and f(2) = 9 > 0. That puts one root in (0, 1) and one in (1, 2).
- So. The equation has exactly 2 distinct real roots, option (b).
The idea to keep. To count real roots of a polynomial, find where f'(x) = 0 and check the sign of f there. Each change in sign between turning points and the ends gives exactly one root.
Answer from the NTA final answer key. Solution by PrepWiser.