A) \[-\frac{9}{8}\]
B) \[\frac{9}{8}\]
C) \[-\frac{8}{9}\]
D) \[\frac{8}{9}\]
Correct Answer: B
Solution :
Given equation is \[(\ell -m){{x}^{2}}+\ell x+1=0\] Roots are \[\alpha ,\,\,\beta .\] \[\because \] One root is double the other. \[\beta =2\alpha \] Sum of roots \[=\alpha +\beta \] \[3\alpha =\frac{-\ell }{\ell -m}\] \[\alpha (2\alpha )=\frac{1}{(\ell -m)}\] \[\Rightarrow \,\,\,{{\alpha }^{2}}=\frac{{{\ell }^{2}}}{9{{(\ell -m)}^{2}}}\] \[2\,{{\alpha }^{2}}=\frac{1}{\ell -m}\] \[\Rightarrow \,\,\,2\frac{{{\ell }^{2}}}{9{{(\ell -m)}^{2}}}=\frac{1}{(\ell -m)}\] \[\Rightarrow \,\,\frac{2{{\ell }^{2}}}{9(\ell -m)}=1\] \[\Rightarrow \,\,\,2{{\ell }^{2}}=9(\ell -m)\Rightarrow 2{{\ell }^{2}}-9\ell +9m=0\] For \[\ell \] to be real discriminant should be \[{{b}^{2}}-4ac\ge 0\] \[81-4\times 2\times 9m\ge 0\] \[m\le \frac{9}{8}.\]
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