Question

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

• A. 2
• B. 3
• C. 4
• D. 0

Hint:

If a quadratic expression is always positive, its graph is an upward-opening parabola that never touches the x-axis ($D<0$).

Answer 1

The Correct Option is B

Step 1: For $ax^2 + bx + c>0$ to be true for all $x \in \mathbb{R}$, we need $a>0$ and $D<0$.
Step 2: Here $a=1>0$. Calculate $D = [-2(3k-1)]^2 - 4(1)(8k^2 - 7)$.
Step 3: $D = 4(9k^2 - 6k + 1) - 32k^2 + 28 = 36k^2 - 24k + 4 - 32k^2 + 28 = 4k^2 - 24k + 32$.
Step 4: $D<0 \Rightarrow 4k^2 - 24k + 32<0 \Rightarrow k^2 - 6k + 8<0$.
Step 5: $(k-2)(k-4)<0 \Rightarrow 2<k<4$.
Step 6: The integer in the interval $(2, 4)$ is $k=3$.