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JEE Main 7 Jan 2020 Shift 1 Solved Paper
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Section:
Mathematics
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© examsnet.com
Question : 65
Total: 75
If
f
(
a
+
b
+
1
−
x
)
=
f
(
x
)
, for all x, where a and b are fixed positive real numbers, then
1
a
+
b
‌
b
∫
a
x
(
f
(
x
)
+
f
(
x
+
1
)
)
d
x
is equal to :
[7-Jan-2020 Shift 1]
b
−
1
∫
a
−
1
f
(
x
)
d
x
b
+
1
∫
a
+
1
f
(
x
+
1
)
d
x
b
−
1
∫
a
−
1
f
(
x
+
1
)
d
x
b
+
1
∫
a
+
1
f
(
x
)
d
x
Validate
Solution:
f
(
a
+
b
+
1
−
x
)
=
f
(
x
)
.
.
.
(
i
)
f
(
a
+
b
−
x
)
=
f
(
x
+
1
)
.
.
.
(
i
i
)
I
=
‌
1
(
a
+
b
)
‌
b
∫
a
x
(
f
(
x
)
+
f
(
x
+
1
)
)
d
x
.
.
.
.(iii)
From (i) and (ii)
I
=
‌
1
(
a
+
b
)
‌
b
∫
a
(
a
+
b
−
x
)
(
f
(
x
+
1
)
+
f
(
x
)
)
d
x
.
.
.
‌
(iv)
‌
Adding (iii) and (iv)
2
I
=
b
∫
a
(
f
(
x
)
+
f
(
x
+
1
)
)
d
x
2
I
=
b
∫
a
f
(
a
+
b
−
x
+
1
)
d
x
+
b
∫
a
f
(
x
)
d
x
2
I
=
2
‌
b
∫
a
f
(
x
)
d
x
I
=
b
∫
a
f
(
x
)
d
x
‌
‌
;
‌
‌
x
=
t
+
1
,
d
x
=
d
t
I
=
b
−
1
∫
a
−
1
f
(
t
+
1
)
d
t
I
=
b
−
1
∫
a
−
1
f
(
x
+
1
)
d
x
© examsnet.com
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