The integer 'k', for which the inequality x² - 2(3k - 1)x + 8k² - 7 > 0 is valid for every x in R, is :

Solution

Maths sir · 3 steps

The parabola opens upward, so it stays above zero for every x only if it never touches the x-axis: D < 0.

1. Condition needed. The coefficient of x² is 1 > 0, so the graph opens upward. It is positive for every real x only when it has no real root, that is, when the discriminant is strictly negative.

   D/4 = (3k-1)² - (8k²-7) = k² - 6k + 8 = (k-2)(k-4)

2. Solve the inequality. (k-2)(k-4) < 0 holds only for 2 < k < 4. The only integer in this open interval is k = 3.

   Option	D/4 = k² - 6k + 8	Expression is
   k = 4	0	(x-11)², equals 0 at x = 11
   k = 2	0	(x-5)², equals 0 at x = 5
   k = 3	-1, negative	x² - 16x + 65, always positive
   k = 0	8, positive	x² + 2x - 7, negative between its roots

3. So. Only k = 3 makes the expression positive for every real x, so the answer is (c).

The idea to keep. For ax² + bx + c > 0 for all real x, you need a > 0 and D < 0. If the inequality were ≥ 0, then D ≤ 0 would be enough, so check the sign of the inequality.

Answer from the NTA final answer key. Solution by PrepWiser.