MCQ
If $f(x)=\left\{\begin{array}{c}\frac{\sin [x]}{[x]+1}, \text { for } x>0 \\ \frac{\cos \frac{\pi}{2}[x]}{[x]}, \text { for } x<0 ; \\ k, \text { for } x=0\end{array}\right.$
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)
For $f (x)$ to be continuous at $x=0$,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0^{-}} f (x)$
$\Rightarrow k =\lim _{ h \rightarrow 0} f (0- h )=\lim _{ h \rightarrow 0} \frac{\cos \frac{\pi}{2}[0- h ]}{[0- h ]}$
$\Rightarrow k =\lim _{ h \rightarrow 0} \frac{\cos _2^\pi[ h ]}{[- h ]}$
$\Rightarrow k =\lim _{ h \rightarrow 0} \frac{\cos \left(-\frac{\pi}{2}\right)}{-1}$
$\Rightarrow k =0$

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