If the function f(x)= cases 2x \ (k1+1)x+ (k2-1)x \, & x<0\\[2mm] 4, & x=0\\[2mm] 2x e ( 2+k1x2+k2x ), & x>0 cases is continuous at x=0, then k12+k22 is equal to :
- 5
- 10
- 20
- 8
Answer
(2) 10
Solution
Continuity: LHL = RHL =f(0)=4.
LHL, using axx=a: 2 [(k1+1)+(k2-1) ]=4 k1+k2=2
RHL, using (1+t) t: 2x (k1-k2)x2=4 k1-k2=4
k12+k22= (k1+k2)2+(k1-k2)22= 4+162=10
Answer: (2)