The given function fis f(x)={λ(x2−2x),4x+1,​ if x≤0 if x>0​ If f is continuous at x=0, then​x→0−lim​f(x)=x→0∘lim​f(x)=f(0)⇒x→0−lim​λ(x2−2x)=x→0+lim​(4x+1)=λ(02−2×0)⇒λ(02−2×0)=4×0+1=0​ ⇒0=1=0 , which is not possibleTherefore, there is no value of A~Â»for which fis continuous at x=0 At x=1, Therefore, for any values of ​f(1)=4x+1=4×1+1=5x→1lim​(4x+1)=4×1+1=5∴x→1lim​f(x)=f(1)​ is continuous at x=1