A) Both roots in \[[a,\,b]\]
B) Both roots in \[(-\infty ,\,a)\]
C) Both roots in \[(b,\,+\infty )\]
D) One root in \[(-\infty ,\,a)\] and the other in \[(b,\,+\infty )\]
Correct Answer: D
Solution :
The equation is \[{{x}^{2}}-(a+b)\,x+ab-1=0\] \[\therefore \] discriminant \[={{(a+b)}^{2}}-4(ab-1)={{(b-a)}^{2}}+4>0\] \[\therefore \] both roots are real. Let them be \[\alpha ,\beta \] where \[\alpha =\frac{(a+b)-\sqrt{{{(b-a)}^{2}}+4}}{2}\], \[\beta =\frac{(a+b)+\sqrt{{{(b-a)}^{2}}+4}}{2}\] Clearly, \[\alpha <\frac{(a+b)-\sqrt{{{(b-a)}^{2}}}}{2}=\frac{(a+b)-(b-a)}{2}=a\,\,\] \[(\because b>a)\] and \[\beta >\frac{(a+b)+\sqrt{{{(b-a)}^{2}}}}{2}=\frac{a+b+b-a}{2}=b\] Hence, one root \[\alpha \] is less than a and the other root b is greater than b.
You need to login to perform this action.
You will be redirected in
3 sec
