The probability of selecting integers a in [- | vedclass

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(C) For the quadratic expression $x^{2}+2(a+4)x-5a+64 > 0$ to be true for all $x \in \mathbb{R}$,its discriminant $D$ must be less than $0$.$D = [2(a+

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$\frac{2}{9}$

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(C) For the quadratic expression $x^{2}+2(a+4)x-5a+64 > 0$ to be true for all $x \in \mathbb{R}$,its discriminant $D$ must be less than $0$.$D = [2(a+4)]^{2} - 4(1)(-5a+64) $4(a^{2}+8a+16) + 20a - 256 $4a^{2} + 32a + 64 + 20a - 256 $4a^{2} + 52a - 192 Dividing by $4$,we get $a^{2} + 13a - 48 Factoring the quadratic,we get $(a+16)(a-3) This implies $a \in (-16, 3)$.Since $a$ must be an integer in the range $[-5, 30]$,the possible values for $a$ are $\{-5, -4, -3, -2, -1, 0, 1, 2\}$.The number of favorable values is $8$.The total number of integers in the range $[-5, 30]$ is $30 - (-5) + 1 = 36$.Therefore,the required probability is $\frac{8}{36} = \frac{2}{9}$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

The probability of selecting integers $a \in [-5, 30]$ such that $x^{2}+2(a+4)x-5a+64 > 0$ for all $x \in \mathbb{R}$ is:

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