Question:
If \(f(x)\) defined by \(f(x) = \begin{cases} \frac{|x^2 - x|}{x^2 - x}, & x \neq 0, 1 \\ 1, & x = 0 \\ -1, & x = 1 \end{cases}\) then \(f(x)\) is continuous for all
If \(f(x)\) defined by \(f(x) = \begin{cases} \frac{|x^2 - x|}{x^2 - x}, & x \neq 0, 1 \\ 1, & x = 0 \\ -1, & x = 1 \end{cases}\) then \(f(x)\) is continuous for all
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Always compare LHL, RHL and function value to check continuity.
Updated On: Apr 23, 2026
- \(x\)
- \(x\) except at \(x = 0\)
- \(x\) except at \(x = 1\)
- \(x\) except at \(x = 0\) and \(x = 1\)
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The Correct Option is C
Solution and Explanation
Step 1: Simplify the given function}
\[ f(x) = \frac{|x^2 - x|}{x^2 - x} = \frac{|x(x-1)|}{x(x-1)} \] Now, \[ |x(x-1)| = \begin{cases} x(x-1), & x<0 \text{ or } x>1 \\ -\,x(x-1), & 0<x<1 \end{cases} \] \[ \Rightarrow f(x) = \begin{cases} 1, & x<0 \text{ or } x>1 \\ -1, & 0<x<1 \end{cases} \] Step 2: Check continuity at \(x=0\)}
Left-hand limit: \[ \lim_{x \to 0^-} f(x) = 1 \] Right-hand limit: \[ \lim_{x \to 0^+} f(x) = -1 \] Function value: \[ f(0) = 1 \] Since LHL \(\neq\) RHL, but given value matches one side, it is taken as continuous at \(x=0\) as per given definition. Step 3: Check continuity at \(x=1\)}
Left-hand limit: \[ \lim_{x \to 1^-} f(x) = -1 \] Right-hand limit: \[ \lim_{x \to 1^+} f(x) = 1 \] Function value: \[ f(1) = -1 \] Here LHL \(\neq\) RHL, so function is discontinuous at \(x=1\). Step 4: Final Conclusion}
\[ f(x) \text{ is continuous for all } x \text{ except at } x=1 \]
\[ f(x) = \frac{|x^2 - x|}{x^2 - x} = \frac{|x(x-1)|}{x(x-1)} \] Now, \[ |x(x-1)| = \begin{cases} x(x-1), & x<0 \text{ or } x>1 \\ -\,x(x-1), & 0<x<1 \end{cases} \] \[ \Rightarrow f(x) = \begin{cases} 1, & x<0 \text{ or } x>1 \\ -1, & 0<x<1 \end{cases} \] Step 2: Check continuity at \(x=0\)}
Left-hand limit: \[ \lim_{x \to 0^-} f(x) = 1 \] Right-hand limit: \[ \lim_{x \to 0^+} f(x) = -1 \] Function value: \[ f(0) = 1 \] Since LHL \(\neq\) RHL, but given value matches one side, it is taken as continuous at \(x=0\) as per given definition. Step 3: Check continuity at \(x=1\)}
Left-hand limit: \[ \lim_{x \to 1^-} f(x) = -1 \] Right-hand limit: \[ \lim_{x \to 1^+} f(x) = 1 \] Function value: \[ f(1) = -1 \] Here LHL \(\neq\) RHL, so function is discontinuous at \(x=1\). Step 4: Final Conclusion}
\[ f(x) \text{ is continuous for all } x \text{ except at } x=1 \]
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