A) Continuous at\[x=1\]
B) Discontinuous at \[x=1\]
C) Not defined at \[x=1\]
D) None of these
Correct Answer: A
Solution :
| \[f(x)=x|x-{{x}^{2}}|\] |
| \[\left[ \begin{matrix} x-x+{{x}^{2}}={{x}^{2}} & ;\,\,\,\,\,\,\,\,x-{{x}^{2}}\ge 0\Rightarrow x(x-1)\,\le 0\Rightarrow 0\le x\le 1 \\ x+x-{{x}^{2}}=2x-{{x}^{2}} & ;\,\,\,\,\,\,\,\,\,x-{{x}^{2}}<0\Rightarrow x(x-1)>0\Rightarrow x<0\,\,or\,\,x>1 \\ \end{matrix} \right.\] |
| Check continuity at \[x=1\] |
| When \[x>1,\]let \[x=1+h,\]\[h\to 0\] |
| \[f({{1}^{+}})=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(1+h)\]\[=\underset{h\to 0}{\mathop{\lim }}\,2(1+h)-{{(1+h)}^{2}}=1\] |
| When \[x<1,\]let \[x=1-h,\]\[h\to 0\] |
| \[f\left( {{1}^{-}} \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,\,f(1-h)\]\[=\underset{h\to 0}{\mathop{\lim }}\,\,{{(1-h)}^{2}}=1\] |
| and \[f(1)={{1}^{2}}=1\] \[[\because \,\,f(x)={{x}^{2}}]\] |
| \[\therefore \,\,f({{1}^{+}})=f({{1}^{-}})=f(1)\] |
| \[\therefore \] \[f(x)\] is continuous at\[x=1\]. |
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