Let S be the set of positive integral values of a for which \fraca x^2+2(a+1) x+9 a+4x^2-8 x+32 \lt 0, \forall x \in \mathbbR . Then, the number of elements in S is :

Quadratic Equation and Inequalities PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q33 with Solution

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Correct answer: A

We need

\quad \forall x\in\mathbb R.$$ First simplify the denominator: $$x^2-8x+32=(x-4)^2+16.$$ Since $$(x-4)^2+16>0 \quad \forall x\in\mathbb R,$$ the denominator is always positive. Hence the given fraction is negative for all real $x$ **iff** the numerator is negative for all real $x$: $$a x^2+2(a+1)x+9a+4<0 \quad \forall x\in\mathbb R.$$ --- 2. **Condition for a quadratic to be negative for all real $x$** Let $$N(x)=a x^2+2(a+1)x+9a+4.$$ For $N(x)<0$ for all real $x$, we must have: - leading coefficient $a<0$, and - discriminant $<0$. But the problem asks for **positive integral values of $a$**. So $a>0$. That already makes it impossible, because if $a>0$, then as $|x|\to\infty$, $$N(x)\sim a x^2>0,$$ so $N(x)$ cannot be negative for all real $x$. Thus no positive integer $a$ satisfies the condition. --- 3. **Optional discriminant check** For completeness, $$\Delta=[2(a+1)]^2-4(a)(9a+4).$$ So $$\Delta=4(a+1)^2-4a(9a+4) =4\big(a^2+2a+1-9a^2-4a\big) =4(1-2a-8a^2).$$ But this is irrelevant once $a>0$, because the upward-opening parabola cannot stay below $0$ for all real $x$. --- 4. **Conclusion** So, $$S=\varnothing,$$ and the number of elements in $S$ is $$0.$$ Therefore the correct option is **A**.

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Correct answer: A

We need

\quad \forall x\in\mathbb R.$$ First simplify the denominator: $$x^2-8x+32=(x-4)^2+16.$$ Since $$(x-4)^2+16>0 \quad \forall x\in\mathbb R,$$ the denominator is always positive. Hence the given fraction is negative for all real $x$ **iff** the numerator is negative for all real $x$: $$a x^2+2(a+1)x+9a+4<0 \quad \forall x\in\mathbb R.$$ --- 2. **Condition for a quadratic to be negative for all real $x$** Let $$N(x)=a x^2+2(a+1)x+9a+4.$$ For $N(x)<0$ for all real $x$, we must have: - leading coefficient $a<0$, and - discriminant $<0$. But the problem asks for **positive integral values of $a$**. So $a>0$. That already makes it impossible, because if $a>0$, then as $|x|\to\infty$, $$N(x)\sim a x^2>0,$$ so $N(x)$ cannot be negative for all real $x$. Thus no positive integer $a$ satisfies the condition. --- 3. **Optional discriminant check** For completeness, $$\Delta=[2(a+1)]^2-4(a)(9a+4).$$ So $$\Delta=4(a+1)^2-4a(9a+4) =4\big(a^2+2a+1-9a^2-4a\big) =4(1-2a-8a^2).$$ But this is irrelevant once $a>0$, because the upward-opening parabola cannot stay below $0$ for all real $x$. --- 4. **Conclusion** So, $$S=\varnothing,$$ and the number of elements in $S$ is $$0.$$ Therefore the correct option is **A**.