JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 13)
Consider the quadratic equation (c – 5)x2 – 2cx + (c – 4) = 0, c $$ \ne $$ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is -
12
18
10
11
Explanation
_10th_January_Morning_Slot_en_13_1.png)
Let f(x) = (c $$-$$ 5)x2 $$-$$ 2cx + c $$-$$ 4
$$ \therefore $$ f(0)f(2) < 0 . . . . .(1)
& f(2)f(3) < 0 . . . . .(2)
from (1) and (2)
(c $$-$$ 4)(c $$-$$ 24) < 0
& (c $$-$$ 24)(4c $$-$$ 49) < 0
$$ \Rightarrow $$ $${{49} \over 4}$$ < c < 24
$$ \therefore $$ s = {113, 14, 15, . . . . . 23}
Number of elements in set S = 11
Comments (0)
