A) domain = [4,\[\infty \]), range = (1,\[\infty \])
B) domain = [4,\[\infty \]), range = (2,\[\infty \])
C) domain = (2,\[\infty \]), range = (2,\[\infty \])
D) none of these
Correct Answer: A
Solution :
Here log \[\frac{{{y}^{2}}}{x\,(y-1)=0}\] \[\therefore \] \[\frac{{{y}^{2}}}{x\,(y-1)}=1\] or \[{{y}^{2}}=xy-x\] or \[{{y}^{2}}-xy+x=0\] \[\therefore \] \[y=\frac{x\pm \sqrt{{{x}^{2}}-4x}}{2}\] So y is real if\[{{x}^{2}}-4x\ge 0\]or\[x(x-4)\ge 0\] \[\therefore \] \[x\le 0\]or\[x\ge 4\]. But \[x>0\]for \[\log x\]to be defined. So \[x\ge 4\]. Now, \[x=\frac{{{y}^{2}}}{y-1}\ge 4\]or\[\frac{{{y}^{2}}-4y+4}{y-1}\ge 0\] or \[{{(y-2)}^{2}}\cdot (y-1)\ge 0\] \[\Rightarrow \] \[y\ge 1\]. But \[y>1\]for log\[(y-1)\]to be real.
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