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Test Index
TS EAMCET 4 May 2018 Shift 1 Solved Paper
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© examsnet.com
Question : 1
Total: 160
Let f : R → R , g : R → R be differentiable functions such that (fog) (x) = x . If f (x) = 2x + cos x +
s
i
n
2
x , then the value of
99
Σ
n
=
1
g (1 + (2n - )π) is
1250 π
(
99
)
2
π
2
(
99
)
2
π
2500 π
Validate
Solution:
It is given that,
fog
(
x
)
=
x
And,
f
(
x
)
=
2
x
+
cos
x
+
sin
2
x
Therefore,
g
(
x
)
is inverse function of
f
(
x
)
So,
1
+
(
2
n
−
1
)
π
=
2
x
+
cos
x
+
sin
2
x
1
+
(
2
n
−
1
)
π
=
2
x
+
cos
x
+
1
−
cos
2
x
2
x
+
cos
x
−
cos
3
x
=
(
2
−
1
)
π
For
n
=
1
x
=
π
2
For
n
=
2
x
=
3
π
2
For
n
=
3
,
x
=
7
π
2
For
n
=
n
x
=
2
n
−
1
2
π
Hence,
99
∑
n
=
1
g
(
1
+
(
2
n
−
1
)
π
)
=
π
2
+
3
π
2
+
.
.
.
.
.
+
up to 99 terms
=
π
2
[
1
+
3
+
7
+
.
.
.
.
.
+
up to 99 terms
]
=
(
99
)
2
π
2
© examsnet.com
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