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JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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© examsnet.com
Question : 13
Total: 92
Let
f
:
ℝ
−
{
0
}
→
ℝ
be a function satisfying
f
(
x
y
)
=
f
(
x
)
f
(
y
)
for all
x
,
y
,
f
(
y
)
≠
0
. If
f
′
(
1
)
=
2024
, then
[30-Jan-2024 Shift 2]
xf
′
(
x
)
−
2024
f
(
x
)
=
0
xf
′
(
x
)
+
2024
f
(
x
)
=
0
xf
′
(
x
)
+
f
(
x
)
=
2024
xf
′
(
x
)
−
2023
f
(
x
)
=
0
Validate
Solution:
👈: Video Solution
f
(
x
y
)
=
f
(
x
)
f
(
y
)
f
′
(
1
)
=
2024
f
(
1
)
=
1
Partially differentiating w. r. t. x
f
′
(
x
y
)
⋅
1
y
=
1
f
(
y
)
f
′
(
x
)
y
⟶
x
f
′
(
1
)
⋅
1
x
=
f
′
(
x
)
f
(
x
)
2024
f
(
x
)
=
x
f
′
(
x
)
⇒
x
f
′
(
x
)
−
2024
f
(
x
)
=
0
© examsnet.com
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