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Mathematics
Mathematics
Problem
If
x
−
5
=
m
x
+
2
\sqrt{x-5} = mx + 2
x
−
5
=
m
x
+
2
has two distinct real solutions, then range of
m
m
m
is equal to (a)
(
−
1
2
,
1
10
)
(-\frac{1}{2}, \frac{1}{10})
(
−
2
1
,
10
1
)
(b)
(
0
,
1
10
)
(0, \frac{1}{10})
(
0
,
10
1
)
(c)
(
−
1
10
,
1
2
)
(-\frac{1}{10}, \frac{1}{2})
(
−
10
1
,
2
1
)
(d)
(
0
,
1
2
)
(0, \frac{1}{2})
(
0
,
2
1
)
(a)
(
−
1
2
,
1
10
)
(-\frac{1}{2}, \frac{1}{10})
(
−
2
1
,
10
1
)
(b)
(
0
,
1
10
)
(0, \frac{1}{10})
(
0
,
10
1
)
(c)
(
−
1
10
,
1
2
)
(-\frac{1}{10}, \frac{1}{2})
(
−
10
1
,
2
1
)
(d)
(
0
,
1
2
)
(0, \frac{1}{2})
(
0
,
2
1
)
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