If the set of all , for which the equation has no real root, is the interval , and , then is equal to : [2025]

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Solution

Q.
If the set of all $a \in R$ , for which the equation $2 x^{2} + (a - 5) x + 15 = 3 a$ has no real root, is the interval $(α, β)$ , and $X = {x \in Z : α < x < β}$ , then $\sum_{x \in X} x^{2}$ is equal to : [2025]

1 2129

2 2119

3 2109

4 2139

Ans.
(4)

Since, the given equation has no real root

$∴ b^{2} - 4 a c < 0 \Rightarrow {(a - 5)}^{2} - 4 (2) (15 - 3 a) < 0$

$\Rightarrow {(a - 5)}^{2} - 8 (15 - 3 a) < 0$

$\Rightarrow a^{2} - 10 a + 25 - 120 + 24 a < 0$

$\Rightarrow a^{2} - 14 a - 95 < 0 \Rightarrow a \in (- 5, 19)$

$∴ \sum_{x \in X} x^{2} = (1^{2} + 2^{2} + 3^{2} + 4^{2}) + 0^{2} + {(1^{2} + 2^{2} + . . . + 18)}^{2}$

$= \frac{4 (5) (9)}{6} + \frac{18 (19) (37)}{6} = \frac{12834}{6} = 2139$ .

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