A) \[\frac{3}{4}\]
B) \[\frac{5}{4}\]
C) \[\frac{7}{4}\]
D) \[\frac{3}{2}\]
Correct Answer: C
Solution :
5, 5r, \[5{{r}^{2}}\]sides of triangle, | |
\[5+5r>5{{r}^{2}}\] | ? (1) |
\[5+5{{r}^{2}}>5r\] | ? (2) |
From \[{{r}^{2}}-r-1<0\] | |
\[\left[ r-\left( \frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2} \right) \right]\] | ? (4) |
From (2), \[{{r}^{2}}-r+1>0\] \[\Rightarrow \]\[r\in \,R\] | ? (5) |
From (3), \[{{r}^{2}}+r-1>0\] | |
So, \[\left( r+\frac{\sqrt{1+\sqrt{5}}}{2} \right)\left( r+\frac{1-\sqrt{5}}{2} \right)>0\] | |
\[r\in \left( -\infty ,\frac{1+\sqrt{5}}{2} \right)\cup \left( -\frac{1-\sqrt{5}}{2},\infty \right)\] | ? (6) |
From (4), (5), (6), \[r\in \left( \frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2} \right)\] | |
Now check options. |
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