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

3

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2

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4

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0

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Solution

The correct option is A $3$
For $x^{2} - 2 (3 k - 1) x + 8 k^{2} - 7 > 0,$
$D < 0$
$\Rightarrow (2 (3 k - 1))^{2} - 4 (8 k^{2} - 7) < 0$
$\Rightarrow 9 k^{2} + 1 - 6 k - (8 k^{2} - 7) < 0$
$\Rightarrow k^{2} - 6 k + 8 < 0$
$\Rightarrow (k - 4) (k - 2) < 0$
$\Rightarrow 2 < k < 4$
Then $k = 3$

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