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QJEEM24Mac603: JEE Main 2024 Mathematics - Class 12 JEEM24Mac603 Solution: Let the set of all positive values of , for which the point of…

Let the set of all positive values of λ\lambda , for which the point of local minimum of the function (1+x(λ2−x2))(1+x(\lambda^2-x^2)) satisfies \frac{x^2+x+2}{x^2+5 x+6}<0 , be (α,β)(\alpha, \beta) . Then α2+β2\alpha^2+\beta^2 is equal to _________.

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JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Application of Derivatives Question 19 English Explanation 1

amp;f(x)=1+x(λ2−x2)amp;f(x)=−x3+(λ2x+1)amp;f′(x)=−3x2+λ2amp;x=±λ3\begin{aligned}& f(x)=1+x\left(\lambda^2-x^2\right) \\& f(x)=-x^3+\left(\lambda^2 x+1\right) \\& f^{\prime}(x)=-3 x^2+\lambda^2 \\& x= \pm \frac{\lambda}{\sqrt{3}}\end{aligned}

−λ3-\frac{\lambda}{\sqrt{3}} should satisfy the given condition

amp;x2+x+2x2+5x+6lt;6amp;1(x+2)(x+3)lt;0\begin{aligned}& \frac{x^2+x+2}{x^2+5 x+6}<6 \\& \frac{1}{(x+2)(x+3)}<0\end{aligned}

JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Application of Derivatives Question 19 English Explanation 2

amp;amp;x∈(−3,−2)amp;−3lt;−λ3lt;−2amp;−33lt;−λlt;−23amp;23lt;λlt;33amp;α=23β=23}amp;(23)2+(32)2amp;12+27amp;39\begin{aligned}&\begin{aligned}& x \in(-3,-2) \\& -3<-\frac{\lambda}{\sqrt{3}}<-2 \\& -3 \sqrt{3}<-\lambda<-2 \sqrt{3} \\& 2 \sqrt{3}<\lambda<3 \sqrt{3} \\& \left.\begin{array}{l}\alpha=2 \sqrt{3} \\\beta=2 \sqrt{3}\end{array}\right\} \\& (2 \sqrt{3})^2+(3 \sqrt{2})^2 \\& 12+27 \\\end{aligned}\\&39\end{aligned}

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