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JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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© examsnet.com
Question : 52
Total: 92
Consider the function
f
(
x
)
=
{
P
(
x
)
sin
(
x
−
2
)
x
≠
2
7
x
=
2
.
where,
P
(
x
)
is a polynomial such that
P
′
′
′
(
x
)
is always a constant and
P
(
3
)
=
9
. If
f
(
x
)
is continuous at
x
=
2
, then
P
(
5
)
is equal to ...........
[2021, 25 July Shift-11]
Your Answer:
Validate
Solution:
f
(
x
)
=
{
P
(
x
)
∕
sin
(
x
−
2
)
x
≠
2
7
x
=
2
.
Given, that
P
′
′
(
x
)
is always a constant.
⇒
P
(
x
)
is a 2 degree polynomial.
f
(
x
)
is continuous at
x
=
2
lim
x
→
2
+
P
(
x
)
∕
sin
(
x
−
2
)
=
7
⇒
lim
x
→
2
+
(
x
−
2
)
(
a
x
+
b
)
∕
sin
(
x
−
2
)
=
7
⇒
2
a
+
b
=
7
. . . (i)
Now,
P
(
x
)
=
(
x
−
2
)
(
a
x
+
b
)
P
(
3
)
=
9
(
given)
⇒
3
a
+
b
=
9
Subtracting Eq. (ii) from Eq. (i),
a
=
2
From Eq. (i),
b
=
3
Hence,
P
(
x
)
=
(
x
−
2
)
(
2
x
+
3
)
So,
P
(
5
)
=
(
5
−
2
)
(
2
×
5
+
3
)
=
3
×
13
=
39
© examsnet.com
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