Exams Joint Entrance Examination Mathematics Application of Derivatives Question 58 Question 58 of 212 Medium NUMERICAL
Let the set of all positive values of λ \lambda λ , for which the point of local minimum of the function ( 1 + x ( λ 2 − x 2 ) ) (1+x({\lambda }^{2}-{x}^{2})) ( 1 + x ( λ 2 − x 2 )) satisfies x 2 + x + 2 x 2 + 5 x + 6 < 0 \frac{{x}^{2}+x+2}{{x}^{2}+5x+6}<0 x 2 + 5 x + 6 x 2 + x + 2 < 0 , be ( α , β ) (\alpha ,\beta ) ( α , β ) . Then α 2 + β 2 {\alpha }^{2}+{\beta }^{2} α 2 + β 2 is equal to \_\_\_\_\_\_\_\_\_.
Answer: 39
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f ( x ) = 1 + x ( λ 2 − x 2 ) f ( x ) = − x 3 + ( λ 2 x + 1 ) f ′ ( x ) = − 3 x 2 + λ 2 x = ± λ 3 \begin{gathered} f(x)=1+x({\lambda }^{2}-{x}^{2}) \\ f(x)=-{x}^{3}+({\lambda }^{2}x+1) \\ {f}^{'}(x)=-3{x}^{2}+{\lambda }^{2} \\ x=\pm \frac{\lambda }{\sqrt{3}} \end{gathered} f ( x ) = 1 + x ( λ 2 − x 2 ) f ( x ) = − x 3 + ( λ 2 x + 1 ) f ′ ( x ) = − 3 x 2 + λ 2 x = ± 3 λ
− λ 3 -\frac{\lambda }{\sqrt{3}} − 3 λ should satisfy the given condition
\[\& \frac{{x}^{2}+x+2}{{x}^{2}+5x+6}\]
\[\begin{gathered} x\in (-3,-2) \\ -3<-\frac{\lambda }{\sqrt{3}}<-2 \\ -3\sqrt{3}<-\lambda <-2\sqrt{3} \\ 2\sqrt{3}<\lambda <3\sqrt{3} \\ \alpha =2\sqrt{3} \\ \beta =2\sqrt{3} \\ (2\sqrt{3}{)}^{2}+(3\sqrt{2}{)}^{2} \\ 12+27 \\ 39 \end{gathered}\] Answer the question to unlock