JEE Main 2026 — 4 April, Evening Shift — Question 29

$f(x)=(\lambda+2) x^{2}-3 \lambda x+4 \lambda$ C-1 $\operatorname{af}(0)>0$ $(\lambda+2) 4 \lambda>0$ $\Rightarrow \lambda<-2$ or $\lambda>0$ C-2 $-\frac{\mathrm{b}}{2 \mathrm{a}}>0$ $\frac{3 \lambda}{2(\lambda+2)}>0$ $\Rightarrow \lambda<-2$ or $\lambda>0$ C-3 $\mathrm{D} \geq 0$ $(-3 \lambda)^{2}-4(\lambda+2) \times 4 \lambda \geq 0$ $\lambda(7 \lambda+32) \leq 0$ $\Rightarrow \lambda \in\left[\frac{-32}{7}, 0\right]$ Intersection of $\mathrm{C}-1, \mathrm{C}-2$ and $\mathrm{C}-3$ $\Rightarrow \lambda \in\left[\frac{-32}{7},-2\right)$ $\lambda \in[-4.57,-2)$ $\Rightarrow \lambda=-4,-3$ ∴ Number of values of $\lambda=2$

Given $(\lambda+2)x^2 - 3\lambda x + 4\lambda = 0$, $\lambda eq -2$, has two positive roots.

Conditions:

1. $a f(0) > 0$: $(\lambda+2)(4\lambda) > 0$

$\Rightarrow \lambda < -2$ or $\lambda > 0$. 2. Sum of roots $> 0$: $\frac{3\lambda}{2(\lambda+2)} > 0$

$\Rightarrow \lambda < -2$ or $\lambda > 0$. 3. Discriminant $\ge 0$: $(-3\lambda)^2 - 4(\lambda+2)(4\lambda) \ge 0$

$\Rightarrow 9\lambda^2 - 16\lambda(\lambda+2) \ge 0$

$\Rightarrow 9\lambda^2 - 16\lambda^2 - 32\lambda \ge 0$

$\Rightarrow -7\lambda^2 - 32\lambda \ge 0$

$\Rightarrow \lambda(7\lambda+32) \le 0$

$\Rightarrow -\frac{32}{7} \le \lambda \le 0$.

Intersection of conditions: $\lambda \in [-\frac{32}{7}, -2)$.

Integral values: $\lambda = -4, -3$.

Thus, number of possible integral values $= 2.$