Let a be an integer such that all the real roots of the polynomial
2x⁵ + 5x⁴ + 10x³ + 10x² + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to ___________.

Solution

Maths sir · 4 steps

Differentiate first: f' is a perfect square, so f only rises and has exactly one real root.

1. Look at the derivative. Let f(x) = 2x⁵ + 5x⁴ + 10x³ + 10x² + 10x + 10. Then f'(x) = 10x⁴ + 20x³ + 30x² + 20x + 10, and the bracket inside factors neatly.

   f'(x) = 10(x⁴ + 2x³ + 3x² + 2x + 1) = 10(x² + x + 1)²

2. Only one root. x² + x + 1 has discriminant 1 - 4 = -3 < 0, so it is never zero and f'(x) > 0 for every x. So f is strictly increasing. An odd-degree polynomial that always rises crosses the x-axis exactly once.
3. Trap the root. f(-1) = -2 + 5 - 10 + 10 - 10 + 10 = 3 > 0 and f(-2) = -64 + 80 - 80 + 40 - 20 + 10 = -34 < 0. The sign changes, so the root lies in (-2, -1).
4. So. The interval (a, a+1) is (-2, -1), so a = -2 and |a| = 2.

The common slip: 1. Taking a as the right endpoint -1 of the interval (-2, -1) gives 1, but a is the left endpoint, so a = -2.

The idea to keep. For an odd-degree polynomial, check f' first. If it never changes sign, there is exactly one real root, and sign checks at consecutive integers pin it down.

Answer from the NTA final answer key. Solution by PrepWiser.