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Mathematics
Let g(x) be on integrable function and ∫ g(x) dx = g(x) . Statement-1: ∫ g(x) (f (x) - f ''(x)) dx =g(x)(f (x) - f '(x)) +C Statement-2: ∫ g(x) (f (x) + f '(x)) dx = g(x)f (x) + C
Q. Let
g
(
x
)
be on integrable function and
∫
g
(
x
)
d
x
=
g
(
x
)
.
Statement-1 :
∫
g
(
x
)
(
f
(
x
)
−
f
′′
(
x
)
)
d
x
=
g
(
x
)
(
f
(
x
)
−
f
′
(
x
)
)
+
C
Statement-2 :
∫
g
(
x
)
(
f
(
x
)
+
f
′
(
x
)
)
d
x
=
g
(
x
)
f
(
x
)
+
C
1438
206
Integrals
Report Error
A
Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement -1
29%
B
Statement -1 is true, Statement -2 is true ; Statement-2 is NOT a correct explanation for Statement - 1
49%
C
Statement -1 is false, Statement -2 is true
20%
D
Statement - 1 is true, Statement- 2 is false
2%
Solution:
∫
g
(
x
)
(
f
(
x
)
+
f
′
(
x
))
d
x
∫
g
(
x
)
f
(
x
)
d
x
+
∫
g
(
x
)
f
′
(
x
)
d
x
=
f
(
x
)
∫
g
(
x
)
d
x
−
∫
(
f
′
(
x
)
∫
g
(
x
)
d
x
)
d
x
+
∫
g
(
x
)
f
′
(
x
)
d
x
=
f
(
x
)
g
(
x
)
−
∫
f
′
(
x
)
g
(
x
)
d
x
+
∫
g
(
x
)
f
′
(
x
)
d
x
(
∵
∫
g
(
x
)
d
x
=
g
(
x
)
)
=
f
(
x
)
g
(
x
)
+
C
Now,
∫
g
(
x
)
(
f
(
x
)
−
f
′′
(
x
)
)
d
x
=
∫
g
(
x
)
{
f
(
x
)
+
f
′
(
x
)
−
(
f
′
(
x
)
+
f
′′
(
x
)
)
}
d
x
=
g
(
x
)
f
(
x
)
−
g
(
x
)
f
′
(
x
)
+
C