JEE Main 2024 — 9 April, Shift 1 — Question 23

$\frac{x^{2}+x+2}{x^{2}+5 x+6}<0$

$\Rightarrow \frac{1}{(x+2)(x+3)}<0$

$f(x)=1+x\left(\lambda^{2}-x^{2}\right)$

Finding local minima $f^{\prime}(x)=\left(\lambda^{2}-x^{2}\right)+(-2 x) \cdot x$

Put $\mathrm{f}^{\prime}(\mathrm{x})=0$

$\Rightarrow \lambda^{2}=3 \mathrm{x}^{2}$

$\Rightarrow \mathrm{x}= \pm \frac{\lambda}{\sqrt{3}}$

Local min Local max We want local min $\Rightarrow x=\frac{-\lambda}{\sqrt{3}}$

from (1) $x \in(-3,-2)$

$-3<\frac{-\lambda}{\sqrt{3}}<-2$

$\Rightarrow-\int_{\infty}^{2} \frac{\mathrm{dt}}{\mathrm{t} \sqrt{\mathrm{t}^{2}-2}}$

$3 \sqrt{3}>\lambda>2 \sqrt{3}$

$\alpha=2 \sqrt{3}, \beta=3 \sqrt{3}$

$\Rightarrow-\int_{\infty}^{2} \frac{\mathrm{tdt}}{\mathrm{t}^{2} \sqrt{\mathrm{t}^{2}-2}}$

$\alpha^{2}+\beta^{2}=12+27=39$