(D) Given inequalities are:$1$) $x^2 - 3x - 10 0$Multiply by $-1$ (inequal

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(D) Given inequalities are:$1$) $x^2 - 3x - 10 $(x - 5)(x + 2) This implies $x \in (-2, 5)$.$2$) $10x - x^2 - 16 > 0$Multiply by $-1$ (inequality sign reverses):$x^2 - 10x + 16 $(x - 2)(x - 8) This implies $x \in (2, 8)$.To find the values of $x$ that satisfy both inequalities simultaneously, we take the intersection of the two sets:$x \in (-2, 5) \cap (2, 8) = (2, 5)$.Therefore, the correct option is $D$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic inequations and Newton Formula

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The set of values of $x$ for which the inequalities $x^2-3x-10 < 0$ and $10x-x^2-16 > 0$ hold simultaneously is:

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A
$(-2, 5)$
B
$(2, 8)$
C
$(-2, 8)$
D
$(2, 5)$

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4-2.Quadratic Equations and Inequations

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The set of values of $x$ for which the inequalities $x^2-3x-10 < 0$ and $10x-x^2-16 > 0$ hold simultaneously is:

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