MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 1983Moderate
Visualized Solution (English)
Visualizing the Piecewise Function
- Given function: f(x)={2x2,2x2−3x+23,0≤x<11≤x≤2
- The critical point for continuity is x=1.
- We need to examine f(x), f′(x), and f′′(x) at this point.
Continuity of f(x) at x=1
- Left Hand Limit (LHL): limx→1−f(x)=limx→1−2x2=21
- Right Hand Limit (RHL): limx→1+f(x)=limx→1+(2x2−3x+23)=2(1)2−3(1)+23=21
- Since LHL=RHL=f(1), f(x) is continuous at x=1.
Finding the First Derivative f′(x)
- Differentiating f(x) piecewise:
- For 0≤x<1: f′(x)=dxd(2x2)=x
- For 1<x≤2: f′(x)=dxd(2x2−3x+23)=4x−3
Continuity of f′(x) at x=1
- Left Hand Derivative (LHD): f′(1−)=1
- Right Hand Derivative (RHD): f′(1+)=4(1)−3=1
- Since LHD=RHD, f′(x) is continuous at x=1.
Finding the Second Derivative f′′(x)
- Differentiating f′(x) piecewise:
- For 0≤x<1: f′′(x)=dxd(x)=1
- For 1<x≤2: f′′(x)=dxd(4x−3)=4
Discontinuity of f′′(x) at x=1
- Left Hand Limit of f′′(x): limx→1−f′′(x)=1
- Right Hand Limit of f′′(x): limx→1+f′′(x)=4
- Since 1=4, f′′(x) is discontinuous at x=1.
Final Conclusion
- Summary of Results:
- 1. f(x) is continuous on [0,2].
- 2. f′(x) is continuous on [0,2].
- 3. f′′(x) is discontinuous at x=1.
- Key Takeaway: Continuity of lower-order derivatives does not guarantee continuity of higher-order ones.
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