Functions Question 267
If $ f(x)= \begin{cases} & ,\frac{\sin x}{x}
[x]}{[x]+1},for,x>0 \\ & \frac{\cos \frac{\pi }{2}[x]}{[x]},for,x<0 \\ & k,at,x=0 \\ \end{cases} . $ ; 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 [Kurukshetra CEE 1998]
Options:
A) Equal to 0
B) Equal to 1
C) Equal to 1
D) Indeterminate
Show Answer
Answer:
Correct Answer: A
Solution:
If f is continuous at $ \underset{x\to {0^{+}}}{\mathop{\lim }},{f}’(x)=\underset{h\to 0}{\mathop{\lim }},{f}’(0+h)=\underset{h\to 0}{\mathop{\lim }},2(0+h)=0 $ , then $ \underset{x\to {0^{-}}}{\mathop{\lim }},f(x)=\underset{x\to {0^{+}}}{\mathop{\lim }},f(x)=f(0) $
$ \Rightarrow ,f(0)=,\underset{x\to {0^{-}}}{\mathop{\lim }},f(x) $ $ k=\underset{h\to 0}{\mathop{\lim }},f(0-h)=\underset{h\to 0}{\mathop{\lim }}\frac{\cos \frac{\pi }{2},[0-h]}{[0-h]} $ $ k=\underset{h\to 0}{\mathop{\lim }}\frac{\cos \frac{\pi }{2},[-h]}{[-h]}=\underset{h\to 0}{\mathop{\lim }}\frac{\cos \frac{\pi }{2},[-h]}{[-h]} $ $ k=\underset{h\to 0}{\mathop{\lim }}\frac{\cos ,( -\frac{\pi }{2} )}{-1} $ ; $ k=0 $ .
BETA
