A) \[\left] -\frac{4}{3},\frac{4}{3} \right[\]
B) \[\left] \frac{4}{3},\frac{3}{8} \right[\]
C) \[\left. \left] -\infty ,\frac{-4}{3}, \right[\cup \right]\frac{4}{3},\infty \left[ \begin{align} & \\ & \\ \end{align} \right.\]
D) \[\left] \frac{4}{3},\infty \right[\]
Correct Answer: C
Solution :
Give lines are \[y=mx,y=2\]and \[y=6\] \[\therefore \]Coordinates of points A and B are \[\left( \frac{2}{m},2 \right)\]and \[\left( \frac{6}{m},6 \right)\], respectively. \[\therefore \]\[AB=\sqrt{{{\left( \frac{2}{m}-\frac{6}{m} \right)}^{2}}+{{(2-6)}^{2}}<5}\] \[\Rightarrow \] \[{{\left( \frac{2}{m}-\frac{6}{m} \right)}^{2}}+{{4}^{2}}<25\] \[\Rightarrow \] \[{{\left( \frac{2}{m}-\frac{6}{m} \right)}^{2}}<9\] \[\Rightarrow \] \[-3<\frac{2}{m}-\frac{6}{m}<3\] \[\Rightarrow \] \[-3<-\frac{4}{m}<3\Rightarrow -\frac{4}{3}>m\frac{4}{3}\] \[\therefore \] \[m\in \left. \left] -\infty ,\frac{-4}{3} \right[\cup \right]\frac{4}{3},\infty \left[ _{_{_{{}}^{{}}}^{{}}}^{{}} \right.\]You need to login to perform this action.
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