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KVPY SX SB Exam 2015 Question Paper
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© examsnet.com
Question : 14
Total: 120
Let x > 0 be a fixed real number. Then the integral
∞
∫
0
e
–
1
|
x
−
t
|
d
t
is equal to -
x
+
2
e
−
x
−
1
x
−
2
e
−
x
+
1
x
+
2
e
−
x
+
1
−
x
−
2
e
−
x
+
1
Validate
Solution:
f
(
x
)
=
∞
∫
0
e
–
t
|
x
−
t
|
d
t
f
(
x
)
=
x
∫
0
e
–
t
(
x
−
t
)
d
t
+
∞
∫
x
e
–
t
(
t
−
x
)
d
t
f
′
(
x
)
=
e
–
x
(
x
−
x
)
−
e
−
0
(
x
−
0
)
.
0
+
x
∫
0
e
–
t
(
1
)
d
t
+
0
−
e
−
x
(
x
−
x
)
.1
+
∞
∫
x
−
e
–
t
d
t
=
[
−
e
−
t
]
0
x
+
[
−
e
−
t
]
x
∞
=
–
e
–
x
+
1
+
0
−
e
–
x
f
′
(
x
)
=
1
−
2
e
–
x
d
y
=
1
−
2
e
–
x
d
x
y
=
x
+
2
e
–
x
+
c
f
(
0
)
=
∞
∫
0
e
−
t
t
d
t
=
[
(
−
e
−
t
)
t
]
0
∞
+
∞
∫
0
e
−
t
d
t
=
[
0
−
e
−
t
]
0
∞
= 0 + 1
f(0) = 1
f(0) = 1 = 0 + 2e
–
0
+ c
c = –1
f(x) = x + 2e
–
x
– 1
© examsnet.com
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