Let the set of all positive values of λ, for which the point of local minimum of the function f(x)=1+x(λ2−x2) satisfies x2+5x+6x2+x+2<0, be (α,β). Then α2+β2 is equal to ______.
Answer: 39
Solution
f′(x)=λ2−3x2=0⇒x=±3λ; since f′′=−6x, the local minimum is at x=−3λ (λ>0). The inequality (numerator always positive) needs x2+5x+6<0, i.e. x∈(−3,−2). So −3λ∈(−3,−2)⇒λ∈(23,33), giving α2+β2=12+27=39.