Let the set of all positive values of λ, for which the point of local minimum of the function (1+x(λ2-x2)) satisfies x2+x+2x2+5x+6<0, be (α,β). Then α2+β2 is equal to _____________ . [2024]
(39)
Let f(x)=1+x(λ2-x2)
f(x)=-x3+(λ2x+1)
⇒f'(x)=-3x2+λ2
Put f'(x)=0
⇒λ2-3x2=0⇒(λ-3x)(λ+3x)=0
x=±λ3,f''(x)=-6x
So, x=-λ3 is point of minima.
Now, -λ3 should satisfy the given condition x2+x+2x2+5x+6<0
i.e., 1(x+2)(x+3)<0 [∵ x2+x+2>0]
⇒x∈(-3,-2)
⇒-3<-λ3<-2⇒-33<-λ<-23
23<λ<33⇒λ∈(23,33)≡(α,β) (∵ Given)
∴ (23)2+(33)2=12+27=39