Question:
Let \(\alpha\in\mathbb{R}\) be such that the function
\(f(x) = \begin{cases}\dfrac{\cos^{-1}(1 - \{x\}^2)\,\sin^{-1}(1 - \{x\})}{\{x\} - \{x\}^3}, & x \ne 0\\[2mm] \alpha, & x = 0\end{cases}\)
is continuous at \(x = 0\), where \(\{x\} = x - [\,x\,]\) and \([\,x\,]\) is the greatest integer less than or equal to \(x\). Then :
Let \(\alpha\in\mathbb{R}\) be such that the function
\(f(x) = \begin{cases}\dfrac{\cos^{-1}(1 - \{x\}^2)\,\sin^{-1}(1 - \{x\})}{\{x\} - \{x\}^3}, & x \ne 0\\[2mm] \alpha, & x = 0\end{cases}\)
is continuous at \(x = 0\), where \(\{x\} = x - [\,x\,]\) and \([\,x\,]\) is the greatest integer less than or equal to \(x\). Then :
\(f(x) = \begin{cases}\dfrac{\cos^{-1}(1 - \{x\}^2)\,\sin^{-1}(1 - \{x\})}{\{x\} - \{x\}^3}, & x \ne 0\\[2mm] \alpha, & x = 0\end{cases}\)
is continuous at \(x = 0\), where \(\{x\} = x - [\,x\,]\) and \([\,x\,]\) is the greatest integer less than or equal to \(x\). Then :
Show Hint
Compare the left-hand and right-hand limits at 0, taking the fractional part \(\{x\}\) carefully on each side.
Updated On: Nov 4, 2026
- no such \(\alpha\) exists
- \(\alpha = 0\)
- \(\alpha = \frac{\pi}{4}\)
- \(\alpha = \frac{\pi}{\sqrt{2}}\)
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The Correct Option is A
Solution and Explanation
Right-hand limit (\(x\to0^+\), \(\{x\} = x\)): \(\cos^{-1}(1 - x^2)\sim\sqrt{2}\,x\) (since \(\cos^{-1}(1-u)\sim\sqrt{2u}\)), \(\sin^{-1}(1-x)\to\frac{\pi}{2}\), denominator \(x - x^3\sim x\). So the limit is \(\sqrt{2}\cdot\frac{\pi}{2} = \frac{\pi}{\sqrt{2}}\).
Left-hand limit (\(x\to0^-\), \(\{x\} = 1 + x = 1 - h\), \(h\to0^+\)): \(\cos^{-1}(1 - \{x\}^2)\to\cos^{-1}(0) = \frac{\pi}{2}\); \(\sin^{-1}(1 - \{x\}) = \sin^{-1}h\sim h\); denominator \(\{x\}(1 - \{x\})(1 + \{x\}) \sim 1\cdot h\cdot 2 = 2h\). So the limit is \(\frac{\pi}{2}\cdot\frac{h}{2h} = \frac{\pi}{4}\).
The two one-sided limits differ (\(\frac{\pi}{\sqrt{2}}\ne\frac{\pi}{4}\)), so the limit at 0 does not exist and no \(\alpha\) makes \(f\) continuous.
Answer: (1) no such \(\alpha\) exists
Left-hand limit (\(x\to0^-\), \(\{x\} = 1 + x = 1 - h\), \(h\to0^+\)): \(\cos^{-1}(1 - \{x\}^2)\to\cos^{-1}(0) = \frac{\pi}{2}\); \(\sin^{-1}(1 - \{x\}) = \sin^{-1}h\sim h\); denominator \(\{x\}(1 - \{x\})(1 + \{x\}) \sim 1\cdot h\cdot 2 = 2h\). So the limit is \(\frac{\pi}{2}\cdot\frac{h}{2h} = \frac{\pi}{4}\).
The two one-sided limits differ (\(\frac{\pi}{\sqrt{2}}\ne\frac{\pi}{4}\)), so the limit at 0 does not exist and no \(\alpha\) makes \(f\) continuous.
Answer: (1) no such \(\alpha\) exists
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