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AP EAPCET 24-Aug-2021 Shift 1 Paper
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© examsnet.com
Question : 67
Total: 160
A function
f
:
R
→
R
satisfies the relation
f
(
x
+
y
)
=
f
(
x
)
.
f
(
y
)
,
∀
x
,
y
∈
R
and
f
(
x
)
≠
0
,
∀
x
∈
R
. If
f
' is differentiable at
x
=
0
,
f
′
(
0
)
=
4
and
f
(
6
)
=
3
, then
f
′
(
6
)
=
______
0
12
3
6
Validate
Solution:
f
=
R
→
R
f
(
x
+
y
)
=
f
(
x
)
.
f
(
y
)
.
∀
x
,
y
∈
R
f
(
x
)
≠
0
,
∀
x
∈
R
and
f
′
(
0
)
=
4
and
f
(
6
)
=
3
∵
f
(
x
+
y
)
=
f
(
x
)
.
f
(
y
)
,...(i)
Put
x
=
0
,
y
=
0
⇒
f
(
0
)
[
f
(
0
)
−
1
]
=
0
not possible and
f
(
0
)
=
1
∵
f
′
(
0
)
=
4
⇒
lim
h
→
0
f
(
0
+
h
)
−
f
(
0
)
h
=
4
⇒
lim
h
→
0
(
f
(
h
)
−
1
)
h
=
4
...(ii)
Now,
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
f
(
x
)
.
f
(
h
)
−
f
(
x
)
h
=
lim
h
→
0
f
(
x
)
.
[
f
(
h
)
−
1
h
]
=
f
(
x
)
lim
h
→
0
f
(
h
)
−
1
h
=
f
(
x
)
×
4
[by Eq. (ii)]
f
′
(
x
)
=
4
f
(
x
)
∴
f
;
(
6
)
=
4
f
(
6
)
=
4
×
3
=
12
© examsnet.com
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