Exam

JEE Main

Year

2024

Date

9 April

Shift

Morning Shift

Subject

Mathematics

Chapter

Application of Derivatives

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Application of Derivatives

Let the set of all positive values of λ , for which the point of...

Let the set of all positive values of $\lambda$, for which the point of local minimum of the function \left(1+x\left(\lambda^2-x^2\right)\right) satisfies \frac{x^2+x+2}{x^2+5 x+6} < 0$, be$(\alpha, \beta)$. Then$\alpha^2+\beta^2 is equal to _________

Step-by-step solution

$$\begin{aligned} & \frac{x^2+x+2}{x^2+5 x+6} < 0 \\ & \Rightarrow \frac{1}{(x+2)(x+3)} < 0\end{aligned}$$
$$\begin{aligned} & x \in(-3,-2) \ldots \ldots \ldots . .(1) \\ & f(x)=1+x\left(\lambda^2-x^2\right) \end{aligned}$$
Finding local minima $f^{\prime}(x)=\left(\lambda^2-x^2\right)+(-2 x) \cdot x$ Put $\mathrm{f}^{\prime}(\mathrm{x})=0$$\begin{aligned} & \Rightarrow \lambda^2=3 x^2 \\ & \Rightarrow x= \pm \frac{\lambda}{\sqrt{3}} \end{aligned}$<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/QaJ6UNnAmRG1F2UfM0ZfbS3oldd9AeNxUxM7pM44vTc.original.fullsize.png"> We want local min $\Rightarrow \mathrm{x}=\frac{-\lambda}{\sqrt{3}}$ from (1) $x \in(-3,-2)\begin{aligned} & -3 < \frac{-\lambda}{\sqrt{3}} < -2 \\ & 3 \sqrt{3}>\lambda>2 \sqrt{3} \\ & \alpha=2 \sqrt{3}, \beta=3 \sqrt{3} \\ & \alpha^2+\beta^2=12+27=39\end{aligned}$$$