Question

If \( (2k - 1)x^2 - 2(3k - 2)x + 4k>0 \) for every \( x \in \mathbb{R} \), then the sum of all possible integral values of \( k \) is:

• A. \(21\)
• B. \(27\)
• C. \(36\)
• D. \(28\)

Hint:

For quadratic expressions to be positive for all real \( x \), ensure the leading coefficient is positive and the discriminant is negative. Use inequalities to filter valid ranges and count integer solutions carefully.

Answer 1

The Correct Option is D

We are given the inequality: \[ (2k - 1)x^2 - 2(3k - 2)x + 4k>0 \quad \text{for all} \quad x \in \mathbb{R}. \]
Step 1: For a quadratic expression \( ax^2 + bx + c>0 \) to be positive for all real \( x \), the following conditions must be satisfied:
- The leading coefficient \( a>0 \),
- The discriminant \( D = b^2 - 4ac<0 \).
Step 2: Let us identify \( a, b, c \):
\[ a = 2k - 1, \quad b = -2(3k - 2) = -6k + 4, \quad c = 4k. \]
Now compute the discriminant: \[ D = b^2 - 4ac = (-6k + 4)^2 - 4(2k - 1)(4k). \]
Step 3: Simplify the expression: \[ D = (36k^2 - 48k + 16) - 32k(2k - 1). \]
\[ = 36k^2 - 48k + 16 - (64k^2 - 32k). \]
\[ = (36k^2 - 64k^2) + (-48k + 32k) + 16 = -28k^2 - 16k + 16. \]
So, \[ D<0 \quad \Rightarrow \quad -28k^2 - 16k + 16<0. \]
Step 4: Solve the inequality: \[ 28k^2 + 16k - 16>0. \]
Divide throughout by 4: \[ 7k^2 + 4k - 4>0. \]
Find the roots of the corresponding quadratic: \[ k = \frac{-4 \pm \sqrt{16 + 112}}{14} = \frac{-4 \pm \sqrt{128}}{14} = \frac{-4 \pm 8\sqrt{2}}{14}. \]
\[ k = \frac{-2 \pm 4\sqrt{2}}{7}. \]
So the inequality is satisfied when: \[ k<\frac{-2 - 4\sqrt{2}}{7} \quad \text{or} \quad k>\frac{-2 + 4\sqrt{2}}{7}. \]
Approximate \( \frac{-2 + 4\sqrt{2}}{7} \approx \frac{-2 + 5.656}{7} = \frac{3.656}{7} \approx 0.522 \).
Step 5: Also need \( a = 2k - 1>0 \Rightarrow k>\frac{1}{2} \).
So overall condition: \[ k>\max\left(\frac{1}{2}, \frac{-2 + 4\sqrt{2}}{7}\right) \approx 0.522. \]
Hence, the integral values of \( k \) satisfying all conditions are: \[ k = 1, 2, 3, 4, 5, 6, 7. \]
\[ \text{Sum} = 1 + 2 + 3 + 4 + 5 + 6 + 7 = \boxed{28}. \]