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Question 7 - Geometry & Mensuration Practice | Prasnya
Q7
00:00
26 Nov 2017
Q7.
The shortest distance of the point
(
1
2
,
1
)
\left(\frac12,1\right)
(
2
1
,
1
)
from the curve
y
=
∣
x
−
1
∣
+
∣
x
+
1
∣
y = |x -1| + |x + 1|
y
=
∣
x
−
1∣
+
∣
x
+
1∣
is
A
1
B
0
C
2
\sqrt2
2
D
3
/
2
\sqrt{3/2}
3/2
Save
Check
Details
Q7
00:00
26 Nov 2017
Q7.
The shortest distance of the point
(
1
2
,
1
)
\left(\frac12,1\right)
(
2
1
,
1
)
from the curve
y
=
∣
x
−
1
∣
+
∣
x
+
1
∣
y = |x -1| + |x + 1|
y
=
∣
x
−
1∣
+
∣
x
+
1∣
is
A
1
B
0
C
2
\sqrt2
2
D
3
/
2
\sqrt{3/2}
3/2
Save
Check
Details
