A) \[(-\infty ,\infty )\]
B) \[(0,\infty )\]
C) \[(-\infty ,0)\]
D) \[(2,\infty )\]
Correct Answer: A
Solution :
[a] Given, \[f(xy)=f(x).f(y)\] for all x, y, ?.(1) |
\[f(x)\] is continuous at x = 2, |
i.e.,\[\underset{x\to 2}{\mathop{Lt}}\,\,\,f(x)=f(2)...(2)\] |
Let \[a\ne 0\] |
Now, \[\underset{x\to a}{\mathop{Lt}}\,\,\,f(x)=\underset{h\to 2}{\mathop{Lt}}\,\,f\left( \frac{ah}{2} \right)\] |
\[\left[ putting\,\,x=\frac{ah}{2}so\,\,that\,\,h=\frac{2x}{a} \right]\] |
\[=f\left( \frac{a}{2} \right)\underset{h\to 2}{\mathop{Lt}}\,f(h)=f\left( \frac{a}{2} \right).f(2)=f\left( \frac{a}{2}.2 \right)=f(a)\] |
Hence, \[f(x)\] is necessarily continuous at x = a for all \[a\ne 0\]. |
At x = 0, f(x) may or may not be continuous |
Hence f(x) is not necessarily continuous in\[(-\infty ,+\infty )\]. |
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