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Mathematics
If f: R arrow R be defined by f(x)= begincases2 x &: x>3 x2 &: 1 < x ≤ 3 3 x &: x ≤ 1 endcases then f(-1)+f(2)+f(4) is
Q. If
f
:
R
→
R
be defined by
f
(
x
)
=
⎩
⎨
⎧
2
x
x
2
3
x
:
:
:
x
>
3
1
<
x
≤
3
x
≤
1
then
f
(
−
1
)
+
f
(
2
)
+
f
(
4
)
is
526
102
KCET
KCET 2022
Relations and Functions
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A
5
42%
B
9
37%
C
10
12%
D
14
9%
Solution:
f
(
−
1
)
=
3
(
−
1
)
=
−
3
f
(
2
)
=
2
2
=
4
f
(
4
)
=
2
(
4
)
=
8
f
(
−
1
)
+
f
(
2
)
+
f
(
4
)
=
−
3
+
4
+
8
=
9