The number of real roots of the equation is
- A.1
- B.3
- C.4
- D.2
Show answer and solution
Correct option: A
Solution
Let t = 2x (t > 0). The equation becomes 5 + |t - 1| = t(t - 2).
Case 1: t ≥ 1, so |t-1| = t-1. Equation: 5 + t - 1 = t² - 2t → t² - 3t - 4 = 0 → (t-4)(t+1) = 0 → t = 4 (since t > 0). This gives 2x = 4, so x = 2. ✓
Case 2: t < 1, so |t-1| = 1-t. Equation: 5 + 1 - t = t² - 2t → t² - t - 6 = 0 → (t-3)(t+2) = 0 → t = 3 or t = -2, both invalid since t < 1.
Only one real solution exists: x = 2.
Case 1: t ≥ 1, so |t-1| = t-1. Equation: 5 + t - 1 = t² - 2t → t² - 3t - 4 = 0 → (t-4)(t+1) = 0 → t = 4 (since t > 0). This gives 2x = 4, so x = 2. ✓
Case 2: t < 1, so |t-1| = 1-t. Equation: 5 + 1 - t = t² - 2t → t² - t - 6 = 0 → (t-3)(t+2) = 0 → t = 3 or t = -2, both invalid since t < 1.
Only one real solution exists: x = 2.
[JEE Main 2019 · 10 April · Shift II]