If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,\frac{{\sin [x]}}{{[x] + 1}},\,\,{\rm{for}}\,\,x > 0\\\frac{{\cos \frac{\pi }{2}[x]}}{{[x]}},\,\,{\rm{for}}\,\,x < 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k,\,\,{\rm{at}}\,x = 0\end{array} \right.;$ where $[x]$ denotes the greatest integer less than or equal to $x$, then in order that $f$ be continuous at $x = 0$, the value of $k$ is
- A
Equal to $ 0$
- B
Equal to $1$
- C
Equal to $-1$
- D
✓
Answer
If $f$ is continuous at $x = 0$, then
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$
$ \Rightarrow f(0) = \mathop {\lim }\limits_{x \to {0^ - }} f(x)$
$k = \mathop {\lim }\limits_{h \to 0} f(0 - h) = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \frac{\pi }{2}[0 - h]}}{{[0 - h]}}$
$k = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \frac{\pi }{2}[ - h]}}{{[ - h]}} = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \frac{\pi }{2}[ - h - 1]}}{{[ - h - 1]}}$
$k = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( { - \frac{\pi }{2}} \right)}}{{ - 1}};$
$k = 0$.
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