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JEE Main Math Class 11 Relations and Functions Part 2 Questions
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© examsnet.com
Question : 7
Total: 95
Let the sum of the maximum and the minimum values of the function
f
(
x
)
=
2
x
2
−
3
x
+
8
2
x
2
+
3
x
+
8
be
m
n
, where
gcd
(
m
,
n
)
=
1
. Then
m
+
n
is equal to :
[4 Apr 2024 Shift 1]
182
217
195
201
Validate
Solution:
👈: Video Solution
y
=
2
x
2
−
3
x
+
8
∕
2
x
2
+
3
x
+
8
x
2
(
2
y
−
2
)
+
x
(
3
y
+
3
)
+
8
y
−
8
=
0
use
D
≥
0
(
3
y
+
3
)
2
−
4
(
2
y
−
2
)
(
8
y
−
8
)
≥
0
(
11
y
−
5
)
(
5
y
−
11
)
≤
0
⇒
y
∈
[
5
∕
11
,
11
∕
5
]
y
=
1
is also included
© examsnet.com
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