The number of integral values of m for which | vedclass

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

Question

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

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) For the quadratic expression $f(x) = ax^2 + bx + c$ to be always positive for all $x \in R$,we must have $a > 0$ and the discriminant $D Step $1$: Condition $a > 0$$1 + 2m > 0 \Rightarrow 2m > -1 \Rightarrow m > -\frac{1}{2}$.Step $2$: Condition $D $D = [-2(1 + 3m)]^2 - 4(1 + 2m)(4(1 + m)) $4(1 + 6m + 9m^2) - 16(1 + 3m + 2m^2) Divide by $4$: $(1 + 6m + 9m^2) - 4(1 + 3m + 2m^2) $1 + 6m + 9m^2 - 4 - 12m - 8m^2 $m^2 - 6m - 3 Step $3$: Solve the inequality $m^2 - 6m - 3 The roots of $m^2 - 6m - 3 = 0$ are $m = \frac{6 \pm \sqrt{36 - 4(1)(-3)}}{2} = \frac{6 \pm \sqrt{48}}{2} = 3 \pm 2\sqrt{3}$.Since $\sqrt{3} \approx 1.732$,$2\sqrt{3} \approx 3.464$.So,$3 - 3.464 Step $4$: Intersection with $m > -0.5$The interval is $(-0.464, 6.464)$.The integral values of $m$ are ${0, 1, 2, 3, 4, 5, 6}$.There are $7$ such values.

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is:

Similar Questions

Number of integral values of $a$ for which the smaller root of the quadratic equation $x^2 - 2ax + a^2 - 4 = 0$ is smaller than $1$ and the bigger root is greater than $6$ is:

If $f(x) = \frac{1}{4x^2 + 2x + 1}$,then its maximum value is

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$ and $k$ is a real number,then the condition for $\alpha < k < \beta$ is given by:

If $f(x) = x^2 - 2(4K - 1)x + g(K) > 0$ for all $x \in R$ and for $K \in (a, b)$. If $g(K) = 15K^2 - 2K - 7$,then:

The values of $a$ for which one root of the equation $x^2 - (a + 1)x + a^2 + a - 8 = 0$ exceeds $2$ and the other is lesser than $2$,are given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module