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BITSAT 2023 Shift 2 Solved Paper
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© examsnet.com
Question : 99
Total: 130
If
f
(
x
)
=
{
x
2
−
1
0
<
x
<
2
2
x
+
3
2
≤
x
<
3
.
, then the quadratic equation whose roots are
lim
x
→
2
−
f
(
x
)
and
lim
x
→
2
+
f
(
x
)
is
x
2
−
6
x
+
9
=
0
x
2
−
7
x
+
8
=
0
x
2
−
14
x
+
49
=
0
x
2
−
10
x
+
21
=
0
Validate
Solution:
👈: Video Solution
Since,
f
(
x
)
=
{
x
2
−
1
0
<
x
<
2
2
x
+
3
2
≤
x
<
3
.
Now,
lim
x
→
2
−
f
(
x
)
=
lim
x
→
2
−
(
x
2
−
1
)
=
lim
h
→
0
[
(
2
−
h
)
2
−
1
]
=
lim
h
→
0
[
4
+
h
2
−
4
h
−
1
]
=
lim
h
→
0
[
h
2
−
4
h
+
3
]
=
0
−
0
+
3
=
3
and
lim
x
→
2
+
f
(
x
)
=
lim
x
→
2
+
(
2
x
+
3
)
=
lim
h
→
0
[
2
(
2
+
h
)
+
3
]
=
lim
h
→
0
[
4
+
2
h
+
3
]
=
7
Hence the quadratic equation whose roots are 3 and 7 is given by
x
2
−
(
3
+
7
)
x
+
(
3
+
7
)
=
0
⇒
x
2
−
10
x
+
21
=
0
© examsnet.com
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