Consider the quadratic equation, . Let be the set of all integral values of for which one root of the equation lies in the interval and its other root lies in the interval . Then, the number of elements in is
- A.11
- B.10
- C.12
- D.18
Show answer and solution
Correct option: A
Solution
Let f(x) = (c-5)x² - 2cx + (c-4). For one root in (0,2) and the other in (2,3), we need f(0), f(2), f(3) to satisfy: f(0)·f(2) < 0 and f(2)·f(3) < 0, which means f(0), f(2), f(3) alternate in sign — specifically f(0) and f(3) have the same sign, and f(2) has the opposite sign.
Compute: f(0) = c-4, f(2) = (c-5)·4 - 4c + (c-4) = 4c-20-4c+c-4 = c-24, f(3) = (c-5)·9 - 6c + (c-4) = 9c-45-6c+c-4 = 4c-49.
Conditions: f(0)·f(2) < 0 → (c-4)(c-24) < 0 → 4 < c < 24. f(2)·f(3) < 0 → (c-24)(4c-49) < 0 → (c-24)(c-49/4) < 0 → 49/4 < c < 24, i.e., 12.25 < c < 24.
Intersection: 12.25 < c < 24. Integral values: c = 13, 14, 15, ..., 23 → that's 11 values.
The number of elements in S is 11.
Compute: f(0) = c-4, f(2) = (c-5)·4 - 4c + (c-4) = 4c-20-4c+c-4 = c-24, f(3) = (c-5)·9 - 6c + (c-4) = 9c-45-6c+c-4 = 4c-49.
Conditions: f(0)·f(2) < 0 → (c-4)(c-24) < 0 → 4 < c < 24. f(2)·f(3) < 0 → (c-24)(4c-49) < 0 → (c-24)(c-49/4) < 0 → 49/4 < c < 24, i.e., 12.25 < c < 24.
Intersection: 12.25 < c < 24. Integral values: c = 13, 14, 15, ..., 23 → that's 11 values.
The number of elements in S is 11.
[JEE Main 2019 · 10 Jan · Shift I]