The values of a for which the expression (a^2 | vedclass

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

Question

(D) For a quadratic expression $Ax^2 + Bx + C$ to be positive for all $x$,we must have $A > 0$ and the discriminant $D = B^2 - 4AC < 0$.Here,$A = a^2

Vedclass · Accepted Answer

$a  1$

Vedclass · Answer

(D) For a quadratic expression $Ax^2 + Bx + C$ to be positive for all $x$,we must have $A > 0$ and the discriminant $D = B^2 - 4AC Here,$A = a^2 - 1$,$B = 2(a - 1)$,and $C = 2$.Condition $1$: $A > 0 \implies a^2 - 1 > 0 \implies a^2 > 1 \implies a  1$.Condition $2$: $D $4(a - 1)^2 - 8(a - 1)(a + 1) Dividing by $4(a - 1)$ (assuming $a 
eq 1$): $(a - 1) - 2(a + 1) $a - 1 - 2a - 2  -3$.Combining the conditions: $(a  1)$ $AND$ $(a > -3)$.This gives $a \in (-3, -1) \cup (1, \infty)$.Note: If $a = 1$,the expression becomes $2$,which is positive for all $x$. If $a = -1$,the expression becomes $-4x + 2$,which is not positive for all $x$.Re-evaluating the discriminant condition: $4(a-1)^2 - 8(a^2-1) $4(a-1)[(a-1) - 2(a+1)]  0$.This holds for $a  1$.Intersection with $a^2 - 1 > 0$ ($a  1$) gives $a  1$.

The values of $a$ for which the expression $(a^2 - 1)x^2 + 2(a - 1)x + 2$ is positive for all $x \in \mathbb{R}$ are:

Similar Questions

If $a, b, c$ are distinct odd natural numbers, then the number of rational roots of the equation $ax^2 + bx + c = 0$ is:

If $-1$ is a twice repeated root of the equation $a(x^3+x^2)+bx+c=0$,then $a:b:c=$

Let $\alpha, \beta, \gamma$ be the three roots of the equation $x^3+bx+c=0$. If $\beta \gamma=1=-\alpha$,then $b^3+2c^3-3\alpha^3-6\beta^3-8\gamma^3$ is equal to $......$.

The number of real roots of the equation $\frac{(x^2+1)^3}{x^3} + \frac{x^2+1}{3x} = 0, (x \neq 0)$ is

The number of real values of $x$ for which the equality $|3x^2 + 12x + 6| = 5x + 16$ holds is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module