A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
We have,\[\log \,(-2x)=2\log \,(x+1)\] |
\[-2x>0x<0\]and \[x+1>0x>-\,1\] |
\[\therefore \]\[x\in \,(-1,0)\] |
\[-2\,x={{(x+1)}^{2}}\]\[\Rightarrow \]\[{{x}^{2}}+4x+1=0\]\[\Rightarrow \]\[x=\frac{-\,4\pm 2\sqrt{3}}{2}=-2\pm \sqrt{3}\] |
So, \[x=-\,2+\sqrt{3}\]only one solution lies in \[(-1,0).\] |
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