A) All integ ers
B) All integers except 0 and 1
C) All integers except 0
D) All integers except 1
Correct Answer: C
Solution :
| \[f(x)={{[x]}^{2}}-[{{x}^{2}}]\] |
| Check continuity at \[\operatorname{x} = 0\] |
| \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\, f(x)=\,\,\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,{{\left[ x \right]}^{2}}-\,\,[{{x}^{2}}]=\,\,0\] |
| \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\, f(x)=\,\,\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,{{\left[ x \right]}^{2}}-\,\,[{{x}^{2}}]\] |
| \[=\,\,\,{{(-1)}^{2}}-0=1\] |
| Thus, discontinuous at \[\operatorname{x} = 0\] |
| Check continuity at \[\operatorname{x} = 1\] |
| \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\, f(x)=\,\,1-1=0\] |
| \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\, f(x)=\,\,0-0=0\] |
| Also \[\operatorname{f}(1)=0\] |
| Hence continuous at \[\operatorname{x} = 1.\] |
You need to login to perform this action.
You will be redirected in
3 sec
