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JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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© examsnet.com
Question : 59
Total: 92
Let
f
:
R
→
R
satisfy the equation
f
(
x
+
y
)
=
f
(
x
)
⋅
f
(
y
)
for all
x
,
y
∈
R
and
f
(
x
)
≠
0
for any
x
∈
R
. If the function
f
is differentiable at
x
=
0
and
f
′
(
0
)
=
3
, then
lim
h
→
0
1
h
(
f
(
h
)
−
1
)
is equal to............
[2021, 18 March Shift-II]
Your Answer:
Validate
Solution:
Method 1
Given,
f
(
x
+
y
)
=
f
(
x
)
⋅
f
(
y
)
∀
x
,
y
∈
R
∴
f
(
x
)
=
a
x
⇒
f
′
(
x
)
=
a
x
⋅
log
(
a
)
Now,
f
′
(
0
)
=
log
(
a
)
⇒
3
=
log
(
a
)
⇒
a
=
e
3
∴
f
(
x
)
=
(
e
3
)
x
=
e
3
x
f
(
h
)
=
e
3
h
Now,
lim
h
→
0
(
f
(
h
)
−
1
h
)
=
lim
h
→
0
(
e
3
h
−
1
h
)
=
lim
h
→
0
(
e
3
h
−
1
3
h
×
3
)
=
3
×
1
=
3
Method(2)
Let
L
=
lim
h
→
0
1
h
(
f
(
h
)
−
1
)
(
0
0
form
)
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
Put
x
=
y
=
0
∴
f
(
0
)
=
f
(
0
)
⋅
f
(
0
)
⇒
[
f
(
0
)
]
2
=
f
(
0
)
⇒
[
f
(
0
)
]
2
−
f
(
0
)
=
0
⇒
f
(
0
)
[
f
(
0
)
−
1
]
=
0
⇒
f
(
0
)
=
0
,
f
(
0
)
=
1
Rejected because
f
(
x
)
≠
0
,
∀
x
∈
R
∴
f
(
0
)
=
1
Using L-Hospital Rule,
L
=
lim
h
→
0
f
′
(
h
)
−
0
1
=
f
′
(
0
)
=
3
© examsnet.com
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