The integer k,for which the inequality x^{2}- | vedclass

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(A) For the quadratic inequality $ax^{2}+bx+c > 0$ to be valid for all $x \in \mathbb{R}$,the discriminant $D$ must be less than $0$ and $a > 0$.Here,

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$3$

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(A) For the quadratic inequality $ax^{2}+bx+c > 0$ to be valid for all $x \in \mathbb{R}$,the discriminant $D$ must be less than $0$ and $a > 0$.Here,$a = 1 > 0$,so we only need $D $D = [-2(3k-1)]^{2} - 4(1)(8k^{2}-7) $4(9k^{2}-6k+1) - 4(8k^{2}-7) Dividing by $4$:$9k^{2}-6k+1 - 8k^{2}+7 $k^{2}-6k+8 $(k-2)(k-4) This implies $k \in (2, 4)$.Since $k$ is an integer,the only integer value in the interval $(2, 4)$ is $k = 3$.

The integer $k$,for which the inequality $x^{2}-2(3k-1)x+8k^{2}-7>0$ is valid for every $x \in \mathbb{R}$,is

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