If f(x) = l , , , , , , , [x] [x] + 1 , , , for , ,x > 0 2 [x] [x… - Vidyadip

MCQ

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

✓
Equal to $ 0$
B
Equal to $1$
C
Equal to $-1$
D
Indeterminate

✓

Answer

Correct option: A.

Equal to $ 0$

a
(a) 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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