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UPSC CDS 1 2021 Math Paper
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© examsnet.com
Question : 41
Total: 100
For how many real values of k is
6
k
x
2
+
12
k
x
−
24
x
+
16
a perfect square for every integer
x
?
0
1
2
4
Validate
Solution:
👈: Video Solution
for a quadratic equation
a
x
2
+
b
x
+
c
=
0
to be perfect square,
the discriminant,
D
=
0
⇒
b
2
=
4
a
c
(
12
k
−
24
)
2
=
4
(
6
k
)
(
16
)
12
×
12
(
k
2
−
4
k
+
4
)
=
4
×
6
×
16
k
3
k
2
−
12
k
+
12
=
8
k
3
k
2
−
20
k
+
12
=
0
(
k
−
6
)
(
3
k
−
2
)
=
0
k
=
6
,
2
3
© examsnet.com
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