Question:
The function f(x)=x-|x-x²|,-1≤ x\le1 is
The function f(x)=x-|x-x²|,-1≤ x\le1 is
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Absolute value functions are often continuous but fail differentiability at sign-change points.
Updated On: Mar 19, 2026
- [-1,1]
- (-1,1)
- [-1,1] but not differentiable at 0
- (-1,1)setminus0
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The Correct Option is C
Solution and Explanation
x-x² = x(1-x) f(x)= begincases 2x-x², & -1≤ x<0
x², & 0≤ x\le1 endcases Continuity: limₓtₒ₀⁻f(x)=limₓtₒ₀⁺f(x)=0=f(0) So f is continuous on [-1,1].
Differentiability at 0: f'_-(0)=2, f'_+(0)=0 Since derivatives are unequal, f is not differentiable at x=0.
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