Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
Let f: R arrow R be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)-1, ∀ x, y ∈ R. If f prime(0)=2, then |f(-2)| is equal to
Q. Let
f
:
R
→
R
be a differentiable function that satisfies the relation
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
−
1
,
∀
x
,
y
∈
R
. If
f
′
(
0
)
=
2
, then
∣
f
(
−
2
)
∣
is equal to _____
3111
172
JEE Main
JEE Main 2023
Limits and Derivatives
Report Error
Answer:
3
Solution:
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
−
1
f
′
(
x
)
=
h
→
0
lim
h
f
(
x
+
h
)
−
f
(
x
)
f
′
(
x
)
=
h
→
0
lim
h
f
(
h
)
−
f
(
0
)
=
f
′
(
0
)
=
2
f
′
(
x
)
=
2
⇒
d
y
=
2
d
x
y
=
2
x
+
C
x
=
0
,
y
=
1
,
c
=
1
y
=
2
x
+
1
∣
f
(
−
2
)
∣
=
∣
−
4
+
1∣
=
∣
−
3∣
=
3