f"(x)=λ(x−1)
Integrating both the side with respect to x
⇒f′(x)=λ2(x−1)2+c
As f′(1)=0⇒c=−2λ
f′(x)=λ2(x−1)2−2λ
Integrating both the side with respect to x
f(x)=6λ(x−1)3−2λx+u
f(1)=−6⟹−2λ+u=−6
f(−1)=10⟹32λ+u=10
∴λ=6 and u=6
f′(x)=0⟹x−1 or x=3
f′′(3)>0
∴x=3, is a point of local minima.