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KVPY SA Exam 2018 Question Paper
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© examsnet.com
Question : 10
Total: 80
Let
x
0
,
y
0
be fixed real numbers such that
x
0
2
+
y
0
2
>
1
. If x,y are arbitrary real numbers such that
x
2
+
y
2
≤
1
,
then the minimum value of
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
is
(
√
x
0
2
+
y
0
2
−
1
)
2
x
0
2
+
y
0
2
−
1
(
|
x
0
|
+
|
y
0
|
−
1
)
2
(
|
x
0
|
+
|
y
0
|
)
2
−
1
Validate
Solution:
x
0
2
+
y
0
2
>
1
x
0
−
y
0
fixed
x
,
y
arbitrary
x
0
2
+
y
0
2
≤
1
,
Let
x
=
cos
θ
,
y
=
sin
θ
for men
z
=
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
z
=
x
2
+
x
0
2
+
y
2
+
y
0
2
−
2
(
x
x
0
+
y
y
0
)
put x - cosθ , y = sinθ
z
=
x
0
2
+
y
0
2
−
2
(
x
0
cos
θ
+
y
0
sin
θ
)
d
z
d
θ
⇒
0
−
2
(
−
x
0
sin
θ
+
y
0
cos
θ
)
d
z
d
θ
=
0
−
x
0
sin
θ
=
−
y
0
cos
θ
tan
θ
=
y
0
x
0
sin
θ
=
y
0
√
x
0
2
+
y
0
2
cos
θ
=
x
0
√
x
0
2
+
y
0
2
x
=
x
0
√
x
0
2
+
y
0
2
,
y
=
y
0
√
x
0
2
+
y
0
2
z
=
(
x
0
√
x
0
2
+
y
0
2
−
x
0
)
2
+
(
y
0
√
x
0
2
+
y
0
2
−
y
0
)
2
=
x
0
2
(
x
0
√
x
0
2
+
y
0
2
−
1
)
2
+
y
0
2
(
y
0
√
x
0
2
+
y
0
2
−
1
)
2
=
(
x
0
2
+
y
0
2
)
(
1
−
√
x
0
2
+
y
0
2
)
(
√
x
0
2
+
y
0
2
)
2
(
1
−
√
x
0
2
+
y
0
2
)
⇒
(
√
x
0
2
+
y
0
2
−
1
)
2
© examsnet.com
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