JEE Main 2024 (Online) 31st January Morning ShiftQuestion 1
Let $S$ be the set of positive integral values of $a$ for which $\frac{ax^{2}+2(a+1)x+9a+4}{x^{2}-8x+32}<0$, $\forall x\in\mathbb{R}$. Then, the number of elements in $S$ is :
Quadratic Equation · Sign, Range and Inequalities of Quadratic Expressions
Solution
Concept. The denominator is always positive, so the numerator must be negative for every $x$, which needs a negative leading coefficient. Formulas used. (1) $x^2-8x+32=(x-4)^2+16>0$(2) $ax^2+bx+c<0$ for all $x$ $\iff a<0$ and $D<0$Step 1 - Reduce. By (1), we need $ax^2+2(a+1)x+9a+4<0$ for all real $x$. Step 2 - Use (2). That needs $a<0$, but $a$ is a positive integer. (For $a>0$ the left side is positive for large $x$.) Step 3 - Count. $S$ is empty: $0$ elements. Answer: (C)