A) \[c\]
B) \[\frac{1}{c}\]
C) \[\frac{a+b}{a-b}\]
D) \[1\]
E) \[\frac{a-b}{a+b}\]
Correct Answer: E
Solution :
Given, \[\frac{{{x}^{2}}-bx}{ax-c}=\frac{m-1}{m+1}\] \[\Rightarrow \]\[(m+1){{x}^{2}}-b(m+1)x\] \[=a(m-1)x-c(m-1)\] \[\Rightarrow \]\[(m+1){{x}^{2}}-(bm+b+am-a)x\] \[+c(m-1)=0\] \[\Rightarrow \]\[(m+1){{x}^{2}}-\{(a+b)m+(b-a)\}x\] \[+c(m-1)=0\] Let the roots of the equation is\[(a,-a)\]. Then sum of the roots \[=\frac{(a+b)m+(b-a)}{(m+1)}\] \[\alpha -\alpha =\frac{(a+b)m+(b-a)}{(m+1)}=0\] \[\Rightarrow \] \[(a+b)m-(a-b)=0\] \[\Rightarrow \] \[m=\frac{a-b}{a+b}\]You need to login to perform this action.
You will be redirected in
3 sec