JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 4)
The integer 'k', for which the inequality x2 $$-$$ 2(3k $$-$$ 1)x + 8k2 $$-$$ 7 > 0 is valid for every x in R, is :
4
2
3
0
Explanation
$${x^2} - 2(3k - 1)x + 8{k^2} - 7 > 0$$
Now, D < 0
$$ \Rightarrow 4{(3k - 1)^2} - 4 \times 1 \times (8{k^2} - 7) < 0$$
$$ \Rightarrow 9{k^2} - 6k + 1 - 8{k^2} + 7 < 0$$
$$ \Rightarrow {k^2} - 6k + 8 < 0$$
$$ \Rightarrow (k - 4)(k - 2) < 0$$
2 < k < 4
then k = 3
Now, D < 0
$$ \Rightarrow 4{(3k - 1)^2} - 4 \times 1 \times (8{k^2} - 7) < 0$$
$$ \Rightarrow 9{k^2} - 6k + 1 - 8{k^2} + 7 < 0$$
$$ \Rightarrow {k^2} - 6k + 8 < 0$$
$$ \Rightarrow (k - 4)(k - 2) < 0$$
2 < k < 4
then k = 3
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