A) \[\frac{a-b}{a+b}\]
B) \[c\]
C) \[\frac{1}{c}\]
D) \[\frac{a+b}{a-b}\]
E) \[\frac{1}{ab}\]
Correct Answer: A
Solution :
The given equation can be rewritten as \[{{x}^{2}}(\lambda +1)-(b\lambda +b+a\lambda -a)x+x(\lambda -1)=0\] \[\because \] \[\alpha +\beta =\frac{(b\lambda +b+a\lambda +a)}{\lambda +1}\] Also \[\alpha +\beta =0\] \[\therefore \] \[b\lambda +b+a\lambda -a=0\] \[\Rightarrow \] \[\lambda =\frac{a+b}{a+b}\]You need to login to perform this action.
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