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UPSC 2013 CDS 2 Math Paper
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© examsnet.com
Question : 19
Total: 100
x
(
y
2
−
z
2
)
+
y
(
z
2
−
x
2
)
+
z
(
x
2
−
y
2
)
is divisible by
[2013 CDS-II]
(y – z) only
(z – x) only
both (y – z) and (z – x)
neither (y – z) nor (z – x)
Validate
Solution:
Putting
y
=
z
in
x
(
y
2
–
z
2
)
+
y
(
z
2
–
x
2
)
+
z
(
x
2
–
y
2
)
,
we get
x
(
z
2
–
z
2
)
+
z
(
z
2
–
x
2
)
+
z
(
x
2
–
z
2
)
=
0
So,
x
(
y
2
–
z
2
)
+
y
(
z
2
–
x
2
)
+
z
(
x
2
–
y
2
)
is divisible by
(
y
−
z
)
Similarly,
x
(
y
2
–
z
2
)
+
y
(
z
2
–
x
2
)
+
z
(
x
2
–
y
2
)
is divisible by
(
z
–
x
)
.
© examsnet.com
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