Let S be the set of positive integral values | vedclass

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

Question

(B) Given the inequality $\frac{ax^2+2(a+1)x+9a+4}{x^2-8x+32}  0$ for all $x \

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given the inequality $\frac{ax^2+2(a+1)x+9a+4}{x^2-8x+32} Since the denominator $x^2-8x+32 = (x-4)^2 + 16 > 0$ for all $x \in R$,the inequality holds if and only if the numerator $f(x) = ax^2+2(a+1)x+9a+4 For $f(x) However,the question asks for the set $S$ of positive integral values of $a$.Since $a$ must be negative for the inequality to hold for all $x$,there are no positive integral values of $a$ that satisfy the condition.Therefore,the set $S$ is empty,and the number of elements in $S$ is $0$.

Let $S$ be the set of positive integral values of $a$ for which $\frac{ax^2+2(a+1)x+9a+4}{x^2-8x+32} < 0, \forall x \in R$. Then,the number of elements in $S$ is:

Similar Questions

The set of all real numbers satisfying the inequation $x^2-|x+2|+x>0$ is

Given that $x$ is a real number satisfying $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$, then

The solution of the equation $2x^2 + 3x - 9 \le 0$ is given by

If $a < 0$,then the inequality $ax^2 - 2x + 4 > 0$ has the solution represented by:

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