A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 0
D) 1
Correct Answer: B
Solution :
We have, \[\frac{x(x-1)-(m+1)}{(x-1)(m-1)}=\frac{x}{m}\] \[\Rightarrow \] \[({{x}^{2}}-x)m-({{m}^{2}}+m)=x(x-1)(m-1)\] \[\Rightarrow \] \[{{x}^{2}}m-mx-{{m}^{2}}-m={{x}^{2}}m-{{x}^{2}}-xm+x\] \[\Rightarrow \] \[({{x}^{2}}-{{m}^{2}})-(m+x)=0\] \[\Rightarrow \] \[(x+m)\,(x-m-1)=0\] Now, since roots are equal \[\Rightarrow \] \[-m=m+1\] \[\Rightarrow \] \[m=\frac{-1}{2}\]
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