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प्रश्न
Show that the function
\[f\left( x \right) = \begin{cases}\left| 2x - 3 \right| \left[ x \right], & x \geq 1 \\ \sin \left( \frac{\pi x}{2} \right), & x < 1\end{cases}\] is continuous but not differentiable at x = 1.
उत्तर
Given:
(LHL at x = 1) =
Hence, (LHL at x = 1) = (RHL at x = 1)
Differentiability at x = 1:
\[\left(\text { LHD at x } = 1 \right) = \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[\left( \text { LHD at x } = 1 \right) = \lim_{h \to 0} \frac{f\left( 1 - h \right) - f\left( 1 \right)}{1 - h - 1}\]
\[\left( \text { LHD at x = 1 } \right) = \lim_{h \to 0} \frac{f\left( 1 - h \right) - f\left( 1 \right)}{- h}\]
\[\left( \text { LHD at x } = 1 \right) = \lim_{h \to 0} \frac{\sin\left( \frac{\pi\left( 1 - h \right)}{2} \right) - 1}{- h}\]
\[\left( \text { LHD at x = 1 } \right) = \lim_{h \to 0} \frac{\cos\frac{\ pih}{2} - 1}{- h}\]
\[\left( \text { LHD at x = 1 } \right) = - \frac{\pi}{2} \lim_{h \to 0} \frac{\cos\frac{\ pih}{2} - 1}{\frac{\pi}{2}h} = 0\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{h \to 0} \frac{f\left( 1 + h \right) - f\left( 1 \right)}{1 + h - 1}\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{h \to 0} \frac{f\left( 1 + h \right) - f\left( 1 \right)}{h}\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{h \to 0} \frac{- \left( 2\left( 1 + h \right) - 3 \right) - 1}{h}\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{h \to 0} \frac{- 2h}{h} = - 2\]
LHD ≠ RHD
Hence, the function is continuous but not differentiable at x = 1.
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