JEE Main2003MathematicsContinuity and DifferentiabilityActual
If f(x)= cases x e^ - ( 1 |x| + 1 x ) , & x 0 then f(x) is 0 & , x=0 cases
Options
- Adiscontinuous every where
- Bcontinuous as well as differentiable for all x
- Ccontinuous for all x but not differentiable at x=0
- Dneither differentiable nor continuous at x=0
Correct answer
C. continuous for all x but not differentiable at x=0
Step-by-step solution
f(0)=0 ; f(x)=x e^ - ( 1 |x| + 1 x ) R.H.L. _ h 0 (0+h) e ^ -2 / h = _ h 0 h e ^ 2 / h =0 L.H.L Lim _ h 0 (0-h) e^ - ( 1 h - 1 h ) =0 Therefore, f(x) is continuous R.H.D. _ h 0 (0+h) e^ - ( 1 h + 1 h ) -h e^ - ( 1 h + 1 h ) h =0 L.H.D. Lim _ h 0 (0-h) e ^ - ( 1 h - 1 h ) -h e^ - ( 1 h + 1 h ) -h =1 Therefore, L.H.D. R.H.D. f ( x ) is not differentiable at x =0 .