If x^2-4ax+5+a0 for all x in R whenever a in | vedclass

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

Question

(B) For the quadratic expression $f(x) = x^2 - 4ax + (5+a)$ to be greater than $0$ for all $x \in R$,the discriminant $D$ must be less than $0$.$D = (

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) For the quadratic expression $f(x) = x^2 - 4ax + (5+a)$ to be greater than $0$ for all $x \in R$,the discriminant $D$ must be less than $0$.$D = (-4a)^2 - 4(1)(5+a) $16a^2 - 20 - 4a $4a^2 - a - 5 Factoring the quadratic: $(4a - 5)(a + 1) This inequality holds when $a \in (-1, 5/4)$.Comparing this with $a \in (\alpha, \beta)$,we get $\alpha = -1$ and $\beta = 5/4$.Therefore,$4\beta + \alpha = 4(5/4) + (-1) = 5 - 1 = 4$.

If $x^2-4ax+5+a>0$ for all $x \in R$ whenever $a \in (\alpha, \beta)$,then $4\beta+\alpha=$

Similar Questions

For the inequality $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$,the set of real values of $x$ is:

The set of values of $x$ for which the inequalities $x^2-3x-10 < 0$ and $10x-x^2-16 > 0$ hold simultaneously is:

If $x^2+2px-2p+8>0$ for all real values of $x$,then the set of all possible values of $p$ 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module