If (2k-1)x^2 - 2(3k-2)x + 4k 0 for every x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For the quadratic expression $f(x) = ax^2 + bx + c$ to be positive for all $x \in R$,we must have $a > 0$ and the discriminant $D < 0$.Here,$a = 2

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) For the quadratic expression $f(x) = ax^2 + bx + c$ to be positive for all $x \in R$,we must have $a > 0$ and the discriminant $D Here,$a = 2k - 1$,$b = -2(3k - 2)$,and $c = 4k$.Condition $1$: $a > 0 \implies 2k - 1 > 0 \implies k > \frac{1}{2}$.Condition $2$: $D $[-2(3k - 2)]^2 - 4(2k - 1)(4k) $4(9k^2 - 12k + 4) - 16k(2k - 1) Divide by $4$: $(9k^2 - 12k + 4) - 4k(2k - 1) $9k^2 - 12k + 4 - 8k^2 + 4k $k^2 - 8k + 4 Roots of $k^2 - 8k + 4 = 0$ are $k = \frac{8 \pm \sqrt{64 - 16}}{2} = \frac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3}$.Since $2\sqrt{3} \approx 3.46$,the roots are $4 - 3.46 = 0.54$ and $4 + 3.46 = 7.46$.So,$0.54 Combining with $k > 0.5$,we get $0.54 The integral values of $k$ are $1, 2, 3, 4, 5, 6, 7$.The sum is $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$.

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

Similar Questions

The smallest negative integer satisfying both the quadratic inequalities $x^2 < 4x + 77$ and $x^2 > 4$ is

For what values of $x$ is $\frac{8x^2 + 16x - 51}{(2x - 3)(x + 4)} > 3$?

The set of all values of $k$ for which the inequality $x^2 - (3k + 1)x + 4k^2 + 3k - 3 > 0$ is true for all real values of $x$ is

The set of all values of $x$ for which the inequalities $x^2-7x+10 \geq 0$ and $2x+3-x^2 > 0$ hold simultaneously is

Let $f(x) = x^2 + 4x + 1$. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module