The largest interval for which is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
Let f(x) = x¹²- x⁹ + x⁴ - x + 1. Consider three cases:
x ≥ 1: Group as (x¹²- x⁹) + (x⁴ - x) + 1 = x⁹(x³-1) + x(x³-1) + 1. Both grouped terms are ≥ 0 for x ≥ 1, so f(x) ≥ 1 > 0.
0 ≤ x < 1: Group as x¹² + (x⁴ - x⁹) + (1 - x) = x¹² + x⁴(1-x⁵) + (1-x). Since x < 1, all terms are ≥ 0 and the last term (1-x) > 0, so f(x) > 0.
x < 0: Let x = -t where t > 0. Then f(-t) = t¹² + t⁹ + t⁴ + t + 1 > 0 always.
Since f(x) > 0 for all real x, the answer is -∞ < x < ∞.
x ≥ 1: Group as (x¹²- x⁹) + (x⁴ - x) + 1 = x⁹(x³-1) + x(x³-1) + 1. Both grouped terms are ≥ 0 for x ≥ 1, so f(x) ≥ 1 > 0.
0 ≤ x < 1: Group as x¹² + (x⁴ - x⁹) + (1 - x) = x¹² + x⁴(1-x⁵) + (1-x). Since x < 1, all terms are ≥ 0 and the last term (1-x) > 0, so f(x) > 0.
x < 0: Let x = -t where t > 0. Then f(-t) = t¹² + t⁹ + t⁴ + t + 1 > 0 always.
Since f(x) > 0 for all real x, the answer is -∞ < x < ∞.
Previous-year question, JEE Main.