Class-XII-Maths
C
ontinuity and Differentiability
1
Practice more on Continuity and Differentiability
CBSE NCERT Solutions for Class 12 Maths Chapter 05
Back of Chapter Questions
Exercise
?. ?
1.
Prove that the function
?(?) = 5? − 3
is continuous at
? = 0,
at
? = −3
and at
? = 5.
Solution:
Given function is
?(?) = 5? − 3
At
? = 0, ?(0) = 5(0) − 3 = −3
LHL = lim
?→0
−
?(?) = lim
?→0
−
(5? − 3) = −3
RHL = lim
?→0
+
?(?) = lim
?→0
+
(5? − 3) = −3
Here, at
? = 0, LHL = RHL = ?(0) = −3
Hence, the function
?
is continuous at
? = 0.
Now at
? = −3, ?(−3) = 5(−3) − 3 = −18
LHL =
lim
?→−3
−
?(?) =
lim
?→−3
−
(5? − 3) = −18
RHL =
lim
?→−3
+
?(?) =
lim
?→−3
+
(5? − 3) = −18
Here, at
? = −3, LHL = RHL = ?(−3) = −18
Hence, the function
?
is continuous at
? = −3.
At
? = 5, ?(5) = 5(5) − 3 = 22
𝐿𝐻𝐿 = lim
?→5
−
?(?) = lim
?→5
−
(5? − 3) = 22
RHL = lim
?→5
+
?(?) = lim
?→5
+
(5? − 3) = 22
Here, at
? = 5, LHL = RHL = ?(5) = 22
Hence, the function
?
is continuous at
? = 5.
2.
Examine the continuity of the function
?(?) = 2?
2
− 1
at
? = 3.

Class-XII-Maths
C
ontinuity and Differentiability
2
Practice more on Continuity and Differentiability
Solution:
Given function is
?(?) = 2?
2
− 1.
At
? = 3, ?(3) = 2(3)
2
− 1 = 17
LHL = lim
?→3
−
?(?) = lim
?→3
−
(2?
2
− 1) = 17
RHL = lim
?→3
+
?(?) = lim
?→3
+
(2?
2
− 1) = 17
Here, at
? = 3, LHL = RHL = ?(3) = 17
Therefore, the function
?
is continuous at
? = 3.
3.
Examine the following functions for continuity:
(a)
?(?) = ? − 5
(b)
?(?) =
1
?−5
, ? ≠ 5
(c)
?(?) =
?
2
−25
?+5
, ? ≠ −5
(d)
?(?) = |? − 5|
Solution:
(a) Given function
?(?) = ? − 5
Let
?
be any real number. At
? = ?, ?(?) = ? − 5
LHL = lim
?→?
−
?(?) = lim
?→?
−
(? − 5) = ? − 5
RHL = lim
?→?
+
?(?) = lim
?→?
+
(? − 5) = ? − 5
At
? = ?, LHL = RHL = ?(?) = ? − 5
Hence, the function
?
is continuous for all real numbers.
(b) Given function
?(?) =
1
?−5
, ? ≠ 5
Let
?(? ≠ 5)
be any real number. At
? = ?, ?(?) =
1
?−5
LHL = lim
?→?
−
?(?) = lim
?→?
−
(
1
?−5
) =
1
?−5
RHL = lim
?→?
+
?(?) = lim
?→?
+
(
1
?−5
) =
1
?−5
At
? = ?, LHL = RHL = ?(?) =
1
?−5
Hence, the function
?
is continuous for all real numbers (except
5
).

Class-XII-Maths
C
ontinuity and Differentiability
3
Practice more on Continuity and Differentiability
(c) Given function
?(?) =
?
2
−25
?+5
, ? ≠ −5
Let
?(? ≠ −5)
be any real number.
At
? = ?, ?(?) =
?
2
−25
?+5
=
(?+5)(?−5)
(?+5)
= (? + 5)
LHL = lim
?→?
−
?(?) = lim
?→?
−
(
?
2
−25
?+5
) = lim
?→?
−
(
(?+5)(?−5)
(?+5)
) = ? + 5
RHL = lim
?→?
+
?(?) = lim
?→?
+
(
?
2
−25
?+5
) = lim
?→?
+
(
(?+5)(?−5)
(?+5)
) = ? + 5
At
? = ?, LHL = RHL = ?(?) = ? + 5
Hence, the function
?
is continuous for all real numbers (except
−5
).
(d) Given function is
?(?) = |? − 5| = {
5 − ?, ? < 5
? − 5, ? ≥ 5
Let
?
be any real number. According to question,
?
can be
? < 5
or
? = 5
or
? > 5.
First case:
If
? < 5,
?(?) = 5 − ?
and
lim
?→?
?(?) = lim
?→?
(5 − ?) = 5 − ?.
Here,
lim
?→?
?(?) = ?(?)
Hence, the function
?
is continuous for all real numbers less than
5.
Second case: If
? = 5,
?(?) = ? − 5
and
lim
?→?
?(?) = lim
?→?
(? − 5) = ? − 5.
Here,
lim
?→?
?(?) = ?(?)
Hence, the function
?
is continuous at
? = 5.
Third case: If
? > 5,
?(?) = ? − 5
and
lim
?→?
?(?) = lim
?→?
(? − 5) = ? − 5.
Here,
lim
?→?
?(?) = ?(?)
Hence, the function
?
is continuous for all real numbers greater than
5.
Hence, the function
?
is continuous for all real numbers.
4.
Prove that the function
?(?) = ?
?
,
is continuous at
? = ?,
where
?
is a positive integer.
Solution:
Given function is
?(?) = ?
?
.
At
? = ?, ?(?) = ?
?
lim
?→?
?(?) = lim
?→?
(?
?
) = ?
?
Here, at
? = ?, lim
?→?
?(?) = ?(?) = ?
?
Since
lim
?→?
?(?) = ?(?) = ?
?

Class-XII-Maths
C
ontinuity and Differentiability
4
Practice more on Continuity and Differentiability
Hence, the function
?
is continuous at
? = ?,
where
?
is positive integer.
5.
Is the function
?
defined by
?(?) = {
?, ? ≤ 1
5, ? > 1
continuous at
? = 0?
At
? = 1?
At
? = 2?

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