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If 5, 5r, 5r^2 are the lengths of the sides of a triangle, then
r cannot be equal to

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5, 5r, 5r^2 sides of triangle,
5 + 5r > 5r^2....(1)
5 + 5r^2> 5r....(2)
5r + 5r^2> 5....(3)
from (1) r^2 - r - 1 < 0, (1)
[r−()][r−()]<0\left\lbrack r - \left( \right) \right\rbrack\left\lbrack r - \left( \right) \right\rbrack < 0
r∈(1−52,1+52)r \in \left( \frac{1 - \sqrt{5}}{2},\frac{1 + \sqrt{5}}{2} \right)...(4)
from (2),
r^2 - r + 1 > 0 ⇒\Rightarrow r ϵ\epsilon R....(5)
from (3),
r^2 + r - 1 > 0
So,
(r+1+52)(r+1−52)>0\left( r + \frac{1 + \sqrt{5}}{2} \right)\left( r + \frac{1 - \sqrt{5}}{2} \right) > 0
r∈(−∞,−1+52)∪(−1−52,∞)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∈(−1+52,1+52)r \in \left( \frac{- 1 + \sqrt{5}}{2},\frac{1 + \sqrt{5}}{2} \right)
now check options

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx

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