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Test Index
JEE Main Math Class 12 Application of Derivatives Part 3 Questions
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© examsnet.com
Question : 36
Total: 40
If the equation of the normal to the curve
y
=
x
−
a
(
x
+
b
)
(
x
−
2
)
at the point
(
1
,
−
3
)
is
x
−
4
y
=
13
then the value of
a
+
b
is equal to _______.
[29-Jan-2023 Shift 2]
Your Answer:
Validate
Solution:
y
=
x
−
a
(
x
+
b
)
(
x
−
2
)
At point
(
1
,
−
3
)
,
−
3
=
1
−
9
(
1
+
b
)
(
1
−
2
)
⇒
1
−
a
=
3
(
1
+
b
)
. . . (1)
Now,
y
=
x
−
a
(
x
+
b
)
(
x
−
2
)
⇒
d
y
d
x
=
(
x
+
b
)
(
x
−
2
)
×
(
1
)
−
(
x
−
a
)
(
2
x
+
b
−
2
)
(
x
+
b
)
2
(
x
−
2
)
2
At
(
1
,
−
3
)
slope of normal is
1
4
hence
d
y
d
x
=
−
4
, So,
−
4
=
(
1
+
b
)
(
−
1
)
−
(
1
−
a
)
b
(
1
+
b
)
2
(
−
1
)
2
Using equation (1)
⇒
−
4
=
(
1
+
b
)
(
−
1
)
−
3
(
b
+
1
)
b
(
1
+
b
)
2
⇒
−
4
=
(
−
1
)
−
3
b
(
1
+
b
)
(
b
≠
−
1
)
⇒
b
=
−
3
So,
a
=
7
Hence,
a
+
b
=
7
−
3
=
4
© examsnet.com
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