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KEAM 2014 Math Paper
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© examsnet.com
Question : 24
Total: 120
Let
f
(
x
)
=
{
a
x
+
3
,
x
≤
2
a
2
x
−
1
,
x
>
2
Then the values of
a
for which
f
is continuous for all x are
1 and - 2
1 and 2
-1 and 2
-1 and - 2
0 and 3
Validate
Solution:
Given,
f
(
x
)
=
{
a
x
+
3
x
≤
2
a
2
x
−
1
x
>
2.
Continuity at
x
=
2
.
L
H
L
=
lim
x
→
2
−
f
(
x
)
=
lim
x
→
2
(
a
x
+
3
)
=
2
a
+
3
R
H
L
=
lim
x
→
2
+
f
(
x
)
=
lim
x
→
2
(
a
2
x
−
1
)
=
2
a
2
−
1
since,
f
(
x
)
is continuous for all values of
x
.
∴
L
H
L
=
R
H
L
⇒
2
a
+
3
=
2
a
2
−
1
⇒
2
a
2
−
2
a
−
4
=
0
⇒
a
2
−
a
−
2
=
0
⇒
a
2
−
2
a
+
a
−
2
=
0
⇒
a
(
a
−
2
)
+
1
(
a
−
2
)
=
0
⇒
(
a
+
1
)
(
a
−
2
)
=
0
∴
a
=
−
1
,
2
© examsnet.com
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