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Question
Mathematics
Let α be the minimum value of f(x)=5 x2-2 x+(26/5) and the graph of f(x) is symmetric about x =β. Also, S n =α+(α+β)+(α+2 β)+(α+3 β)+ ldots ldots ldots upto n terms. The value of α is equal to
Q. Let
α
be the minimum value of
f
(
x
)
=
5
x
2
−
2
x
+
5
26
and the graph of
f
(
x
)
is symmetric about
x
=
β
. Also,
S
n
=
α
+
(
α
+
β
)
+
(
α
+
2
β
)
+
(
α
+
3
β
)
+
………
upto
n
terms.
The value of
α
is equal to
69
121
Sequences and Series
Report Error
A
-5
B
5
C
2
−
5
D
2
5
Solution:
f
(
x
)
=
5
x
2
−
2
x
+
5
26
=
5
(
x
−
5
1
)
2
+
5
∴
f
m
i
n
(
x
=
5
1
)
=
5
≡
α