Let f:R be defined as : f(x)= cases a-b 2xx2; & x<0\ x2+cx+2; & 0 x 1\ 2x+1; & x>1 cases If f is continuous everywhere in R and m is the number of points where f is NOT differentiable then m+a+b+c equals :
- 1
- 4
- 3
- 2
Answer
(4) 2
Solution
Continuity at x=1: 1+c+2=3 c=0
Continuity at x=0: x 0- a-b 2xx2=f(0)=2
∵ denominator 0, numerator must 0 a=b
x 0- 2b 2xx2=2b=2 a=b=1
At x=1: LHD =2x+c=2, RHD =2 ⇒ differentiable
At x=0: for x<0, f(x)= 2 2xx2=2- 2x23+ ⇒ LHD =0; RHD =2x+c=0 ⇒ differentiable
m=0, m+a+b+c=0+1+1+0=2
Answer: (4)