Class 12 ncert solution math exercise 5.1

The  Class 12 NCERT solution math exercise 5.1 prepared by expert Mathematics teacher at gmath.in as per CBSE  guidelines. See our Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks in your Board and School exams.

Exercise 5.1 (Continuity and Differentiability)

Question 1: Prove that the function f(x)=5 x-3 is continuous at x=0, x=-3 and at x=5. (Class 12 ncert solution math exercise 5.1)

Solution: The given function is f(x)=5 x-3

At x=0, f(0)=5(0)-3=-3

\displaystyle \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(5 x-3)=5(0)-3=-3

\therefore \lim _{x \rightarrow 0} f(x)=f(0)

Therefore, f is continous at x=0.

\text { At } x=-3, f(-3)=5(-3)-3=-18

\displaystyle \lim _{x \rightarrow-3} f(x)=\lim _{x \rightarrow-3}(5 x-3)=5(-3)-3=-18

\therefore \lim _{x \rightarrow-3} f(x)=f(-3)

Therefore, f is continous at x=-3.

\text { At } x=5, f(5)=5(5)-3=22

\lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5}(5 x-3)=5(5)-3=22

\therefore \lim _{x \rightarrow 5} f(x)=f(5)

Therefore, f is continous at x=5.

Question 2: Examine the continuity of the function f(x)=2 x^2-1 at x=3.

Solution:The given function is f(x)=2 x^2-1

At x=3, f(3)=2(3)^2-1=17

\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}\left(2 x^2-1\right)=2\left(3^2\right)-1=17

\therefore \lim _{x \rightarrow 3} f(x)=f(3)

Therefore, f is continous at x=3.

Question 3: Examine the following functions for continuity.

(i) f(x)=x-5

(ii) f(x)=\frac{1}{x-5}, x \neq 5

(iii) f(x)=\frac{x^2-25}{x+5}, x \neq-5

(iv) f(x)=|x-5|, x \neq 5

Solution:(i) The given function is f(x)=x-5

It is evident that f is defined at every real number k and its value at k is k-5.

It is also observed that

\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x-5)=k-5=f(k)

\therefore \lim _{x \rightarrow k} f(x)=f(k)

Hence, f is continuous at every real number and therefore, it is a continuous function.

(ii) The given function is f(x)=\frac{1}{x-5}, x \neq 5 For any real number k \neq 5, we obtain

\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} \frac{1}{x-5}=\frac{1}{k-5}

Also,

f(k)=\frac{1}{k-5} \quad(\text { As } k \neq 5)

\therefore \lim _{x \rightarrow k} f(x)=f(k)

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.

(iii) The given function is f(x)=\frac{x^2-25}{x+5}, x \neq-5

For any real number c \neq-5, we obtain

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c} \frac{x^2-25}{x+5}

\Rightarrow\lim _{x \rightarrow c} \frac{(x+5)(x-5)}{x+5}=\lim _{x \rightarrow c}(x-5)=(c-5)

Also,

f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5)

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.

(iv) The given function is f(x)=|x-5|=\begin{cases}5-x, \text { if } x<5 \\ x-5, \text { if } x \geq 5\end{cases}

This function f is defined at all points of the real line. Let \mathrm{c} be a point on a real line. Then, c<5,  c=5 or c>5

Case I: c<5

Then, f(c)=5-c

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(5-x)=5-c

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all real numbers less than 5 .

Case II: c=5

Then, f(c)=f(5)=(5-5)=0

\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5}(5-x)=(5-5)=0

\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(x-5)=0

\therefore \lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c)

Therefore, { }^f is continuous at x=5

Case III: c>5

Then, f(c)=f(5)=c-5

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x-5)=c-5

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all real numbers greater than 5 .

Hence, f is continuous at every real number and therefore, it is a continuous function.

Question 4: Prove that the function f(x)=x^n is continuous at x=n, where n is a positive integer.

Solution: The given function is f(x)=x^n

It is observed that f is defined at all positive integers, n, and its value at n is n^n.

Then,

\lim _{x \rightarrow n} f(n)=\lim _{x \rightarrow n}\left(x^n\right)=x^n

\therefore \lim _{x \rightarrow n} f(x)=f(n)

Therefore, f is continuous at n, where n is a positive integer.

Question 5: Is the function f defined by f(x)=\begin{cases}x, \text { if } x \leq 1 \\ 5, \text { if } x>1\end{cases}. continuous at x=0 ? At x=1 ? At x=2 ?

Solution: The given function is f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5 \text {, if } x>1\end{array}\right.

At x=0,

It is evident that f is defined at 0 and its value at 0 is 0 .

Then,

\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(x)=0

\therefore \lim _{x \rightarrow 0} f(x)=f(0)

Therefore, f is continuous at x=0.

At x=1,

It is evident that f is defined at 1 and its value at 1 is 1 .

The left hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x)=1

The right hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(5)=5

\therefore \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)

Therefore, f is not continuous at x=1.

At x=2,

It is evident that f is defined at 2 and its value at 2 is 5 .

\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2}(5)=5

\therefore \lim _{x \rightarrow 1} f(x)=f(2)

Therefore, f is continuous at x=2.

Question 6: Find all points of discontinuity of f, where f is defined by f(x)=\left\{\begin{array}{l}2 x+3, \text{ if } x \leq 2 \\ 2 x-3, \text { if } x>2\end{array}\right.

Solution: The given function is f(x)= \begin{cases}2 x+3, & \text { if } x \leq 2 \\ 2 x-3, & \text { if } x>2\end{cases}

It is evident that the given function f is defined at all the points of the real line.

Let c be a point on the real line. Then three case arise

Case I: c<2

Case II: c>2

Case III: c=2

Case I: c<2

f(c)=2 c+3

Then,

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x+3)=2 c+3

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x<2.

Case II: c>2

Then,

f(c)=2 c-3

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x-3)=2 c-3

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>2

Case III: c=2

Then, the left hand limit of f at x=2 is,

\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(2 x+3)=2(2)+3=7

The right hand limit of f at x=2 is,

\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x-3)=2(2)-3=1

It is observed that the left and right hand limit of f at x=2 do not coincide.

Therefore, f is not continuous at x=2.

Hence, x=2 is the only point of discontinuity of f.

Question 7:Find all points of discontinuity of f, where f is defined by

f(x)=\left\{\begin{array}{l} |x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3<x<3 \\ 6 x+2, \text { if } x \geq 3 \end{array}\right.

Solution:The given function is

f(x)=\left\{\begin{array}{l} |x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3<x<3 \\ 6 x+2, \text { if } x \geq 3 \end{array}\right.

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:If c<-3, then f(c)=-c+3

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-x+3)=-c+3

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x<-3.

Case II:

If c=-3, then f(-3)=-(-3)+3=6

LHL=\lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{-}}(-x+3)=-(-3)+3=6

RHL=\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}(-2 x)=-2(-3)=6

\therefore \lim _{x \rightarrow-3} f(x)=f(-3)

Therefore, f is continuous at x=-3.

Case III:If -3<c<3, then f(c)=-2 c

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-2 x)=-2 c

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous in (-3,3).

Case IV:If c=3, then the left hand limit of f at x=3 is,

LHL=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(-2 x)=-2(3)=-6

The right hand limit of f at x=3 is,

RHL=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(6 x+2)=6(3)+2=20

It is observed that the left and right hand limit of f at x=3 do not coincide. Therefore, f is not continuous at

x=3.

Case V:If c>3, then f(c)=6 c+2

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(6 x+2)=6 c+2

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>3.

Hence, x=3 is the only point of discontinuity of f.

Question 8:Find all points of discontinuity of f, where f is defined by

f(x)=\left\{\begin{array}{l} \frac{|x|}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0 \end{array}\right.

Solution: The given function is
f(x)=\left\{\begin{array}{l} \frac{|x|}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0 \end{array}\right.

It is known that, x<0 \Rightarrow|x|=-x and x>0 \Rightarrow|x|=x

Therefore, the given function can be rewritten a

f(x)=\left\{\begin{array}{l}  \frac{|x|}{x}=\frac{-x}{x}=-1, \text { if } x<0 \\ 0, \text { if } x=0 \\ \frac{|x|}{x}=\frac{x}{x}=1, \text { if } x>0 \end{array}\right.

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:If c<0, then f(c)=-1

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-1)=-1

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x<0.

Case II:If c=0, then the left hand limit of f at x=0 is,

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1

The right hand limit of f at x=0 is,

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(1)=1

It is observed that the left and right hand limit of f at x=0 do not coincide.

Therefore, f is not continuous at x=0.

Case III:If c>0, then f(c)=1

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(1)=1

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>0.

Hence, x=0 is the only point of discontinuity of f.

Question 9:Find all points of discontinuity of f, where f is defined by

f(x)= \begin{cases}\frac{x}{|x|}, & \text { if } x<0 \\ -1, & \text { if } x \geq 0 .\end{cases}

Solution: f(x)=\left\{\begin{array}{l} \frac{x}{|x|}, \text { if } x<0 \\ -1, \text { if } x \geq 0 \end{array}\right.

The given function is

It is known that x<0 \Rightarrow|x|=-x

Therefore, the given function can be rewritten as

f(x)=\left\{\begin{array}{l}  \frac{x}{|x|}=\frac{x}{-x}=-1, \text { if } x<0 \\  -1, \text { if } x \geq 0  \end{array}\right.

\Rightarrow f(x)=-1 \forall x \in R

Let c be any real number.

Then, \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-1)=-1

Also, f(c)=-1=\lim _{x \rightarrow c} f(x)

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Question 10:Find all points of discontinuity of f, where f is defined by f(x)=\left\{\begin{array}{l}x+1, \text { if } x \geq 1 \\ x^2+1, \text { if } x<1\end{array}\right.

Solution: The given function is f(x)=\left\{\begin{array}{l}x+1, \text { if } x \geq 1 \\ x^2+1, \text { if } x<1\end{array}\right.

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I: If c<1, then f(c)=c^2+1

\displaystyle \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2+1\right)=c^2+1

\therefore \displaystyle\lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x<1.

Case II: If c=1, then f(c)=f(1)=1+1=2
The left hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^2+1\right)=1^2+1=2

The right hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x+1)=1+1=2

\therefore \lim _{x \rightarrow 1} f(x)=f(1)

Therefore, f is continuous at x=1.

Case III: If c>1, then f(c)=c+1

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+1)=c+1

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>1.

Hence, the given function f has no point of discontinuity.

Question 11: Find all points of discontinuity of f, where f is defined by f(x)=\left\{\begin{array}{l}x^3-3, \text { if } x \leq 2 \\ x^2+1, \text { if } x>2\end{array}\right.

Solution: The given function is f(x)=\left\{\begin{array}{l}x^3-3, \text { if } x \leq 2 \\ x^2+1, \text { if } x>2\end{array}\right.
The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I: If c<2, then f(c)=c^3-3
\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^3-3\right)=c^3-3

\therefore \lim _{x \rightarrow c} f(x)=f(c)
Therefore, f is continuous at all points x, such that x<2.

Case II: If c=2, then f(c)=f(2)=2^3-3=5

\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^3-3\right)=2^3-3=5

\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(x^2+1\right)=2^2+1=5

\therefore \lim _{x \rightarrow 2} f(x)=f(2)

Therefore, f is continuous at x=2.

Case III: If c>2, then f(c)=c^2+1

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2+1\right)=c^2+1

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>2.

Thus, the given function f is continuous at every point on the real line.

Hence, f has no point of discontinuity.

Question 12: Find all points of discontinuity of f, where f is defined by

f(x)=\left\{\begin{array}{l} x^{10}-1, \text { if } x \leq 1 \\ x^2, \text { if } x>1 \end{array}\right.

Solution: The given function is

f(x)=\left\{\begin{array}{l} x^{10}-1, \text { if } x \leq 1 \\ x^2, \text { if } x>1 \end{array}\right.

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:If c<1, then f(c)=c^{10}-1

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{10}-1\right)=c^{10}-1

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x<1.

Case II: If c=1, then the left hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{10}-1\right)=1^{10}-1=1-1=0

The right hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^2\right)=1^2=1

It is observed that the left and right hand limit of f at x=1 do not coincide.

Therefore, f is not continuous at x=1.

Case III: If c>1, then f(c)=c^2

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2\right)=c^2

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>1.

Thus from the above observation, it can be concluded that x=1 is the only point of discontinuity of f.

Question 13: Is the function defined by f(x)=\left\{\begin{array}{ll}x+5, & \text { if } x \leq 1 \\ x-5, & \text { if } x>1\end{array}\right. a continous function?

Solution: The given function is f(x)= \begin{cases}x+5, & \text { if } x \leq 1 \\ x-5, & \text { if } x>1\end{cases}
The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I: If c<1, then f(c)=c+5

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+5)=c+5

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x<1.

Case II: If c=1, then f(1)=1+5=6

The left hand limit of f at x=1 is, \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+5)=1+5=6

The right hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-5)=1-5=-4

It is observed that the left and right hand limit of f at x=1 do not coincide.

Therefore, f is not continuous at x=1.

Case III: If c>1, then f(c)=c-5

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x-5)=c-5

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>1.

From the above observation it can be concluded that, x=1 is the only point of discontinuity of f.

Question 14: f(x)=\left\{\begin{array}{l} 3, \text { if } 0 \leq x \leq 1 \\ 4, \text { if } 1<x<3 \\ 5, \text { if } 3 \leq x \leq 10 \end{array}\right.

Discuss the continuity of the function f, where f is defined by

Solution: The given function is f(x)=\left\{\begin{array}{l} 3, \text { if } 0 \leq x \leq 1 \\ 4, \text { if } 1<x<3 \\ 5, \text { if } 3 \leq x \leq 10 \end{array}\right.

The given function f is defined at all the points of the interval [0,10].

Let c be a point in the interval [0,10].

Case I: If 0 \leq c<1, then f(c)=3

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(3)=3

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous in the interval [0,1).

Case II: If c=1, then f(3)=3

The left hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(3)=3

The right hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4)=4

It is observed that the left and right hand limit of f at x=1 do not coincide.
Therefore, f is not continuous at x=1.

Case III: If 1<c<3, then f(c)=4
\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(4)=4
\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at in the interval (1,3).

Case IV: If c=3, then f(c)=5

The left hand limit of f at x=3 is,

\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(4)=4

The right hand limit of f at x=3 is,

\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(5)=5

It is observed that the left and right hand limit of f at x=3 do not coincide. Therefore, f is discontinuous at x=3.

Case V:If 3<c \leq 10, then f(c)=5

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(5)=5

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points of the interval (3,10].

Hence, f is discontinuous at x=1 and x=3.

Question 15: f(x)=\left\{\begin{array}{l} 2 x, \text { if } x<0 \\ 0, \text { if } 0 \leq x \leq 1 \\ 4 x, \text { if } x>1 \end{array}\right.

Solution:  The given function is f(x)=\left\{\begin{array}{l} 2 x, \text { if } x<0 \\ 0, \text { if } 0 \leq x \leq 1 \\ 4 x, \text { if } x>1 \end{array}\right.

The given function f is defined at all the points of the real line.

Case I: If x=0, then f(0)=f(0)=0

The left hand limit of f at x=0 is,

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x)=2(0)=0

The right hand limit of f at x=0 is,

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(0)=0

\therefore \lim _{x \rightarrow 0} f(x)=f(0)

Therefore, f is continuous at x=0

Case II: If x=1, then f(x)=f(1)=0

The left hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(0)=0

The right hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x)=4(1)=4

It is observed that the left and right hand limit of f at
x=1 do not coincide.

Therefore, f is not continuous at x=1.

Therefore, f is continuous at all points x, except x=1.

Question 16: Discuss the continuity of the function f, where f is defined by

f(x)=\left\{\begin{array}{l} -2, \text { if } x \leq-1 \\ 2 x, \text { if }-1<x \leq 1 \\ 2, \text { if } x>1 \end{array}\right.

Solution: The given function is

f(x)=\left\{\begin{array}{l} -2, \text { if } x \leq-1 \\ 2 x, \text { if }-1<x \leq 1 \\ 2, \text { if } x>1 \end{array}\right.

The given function f is defined at all the points.

Case I: If c<-1, then f(c)=-2

herefore, f is continuous at all points x, such that x<-1.

Case II: If x=-1, then f(x)=f(-1)=-2
The left hand limit of f at x=-1 is,

\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}}(-2)=-2

The right hand limit of f at x=-1 is,

\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}}(2 x)=2(-1)=-2

\therefore \lim _{x \rightarrow-1} f(x)=f(-1)

Therefore, f is continuous at x=-1

Case III: If x=1, then f(x)=f(1)=2(1)=2

The left hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(2 x)=2(1)=2

The right hand limit of f at x=1 is,

\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(2)=2

\therefore \lim _{x \rightarrow 1} f(x)=f(c)

Therefore, f is continuous at x=1.

Therefore, f is continuous at all points x, such that x>1.

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.

Question 17: Find the relationship between a and b so that the function f defined by f(x)=\left\{\begin{array}{l}a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x>3\end{array}\right. is continous at x=3.

Solution: The given function is f(x)= \begin{cases}a x+1, & \text { if } x \leq 3 \\ b x+3, & \text { if } x>3\end{cases}

For f to be continuous at x=3, then

\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=f(3) \quad \ldots(1)

Also,

\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(a x+1)=3 a+1

\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(b x+3)=3 b+3

f(3)=3 a+1

Therefore, from (1), we obtain

3 a+1=3 b+3=3 a+1

\Rightarrow 3 a+1=3 b+3

\Rightarrow 3 a=3 b+2

\Rightarrow a=b+\frac{2}{3}

Therefore, the required relationship is given by, a=b+\frac{2}{3}.

Question 18: For what value of \lambda is the function defined by f(x)=\left\{\begin{array}{l}\lambda\left(x^2-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0\end{array}\right. is continous at x=0 ? What about continuity at x=1 ?

Solution: The given function is
f(x)=\left\{\begin{array}{l} \lambda\left(x^2-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0 \end{array}\right.

If f is continuous at x=0, then

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)

\Rightarrow \lim _{x \rightarrow 0^{-}} \lambda\left(x^2-2 x\right)=\lim _{x \rightarrow 0^{+}}(4 x+1)=\lambda\left(0^2-2 \times 0\right)

\Rightarrow \lambda\left(0^2-2 \times 0\right)=4(0)+1=0

\Rightarrow 0=1=0 \quad \text { [which is not possible] }

Therefore, there is no value of \lambda for which f is

continuous at x=0.

At x=1

f(1)=4 x+1=4(1)+1=5

\lim _{x \rightarrow 1}(4 x+1)=4(1)+1=5

\therefore \lim _{x \rightarrow 1} f(x)=f(1)
Therefore, for any values of \lambda, f is continuous at x=1.
Question 19: Show that the function defined by g(x)=x-[x] is discontinuous at all integral point. Here [x] denotes the greatest integer less than or equal to x.

Solution: The given function is g(x)=x-[x]
It is evident that g is defined at all integral points.
Let n be an integer.cuemath

Then,

g(n)=n-[n]=n-n=0

The left hand limit of g at x=n is,

\lim _{x \rightarrow n^{-}} g(x)=\lim _{x \rightarrow n^{-}}(x-[x])=\lim _{x \rightarrow n^{-}}(x)-\lim _{x \rightarrow n^{-}}[x]=n-(n-1)=1

The right hand limit of g at x=n is,

\lim _{x \rightarrow n^{+}} g(x)=\lim _{x \rightarrow n^{+}}(x-[x])=\lim _{x \rightarrow n^{+}}(x)-\lim _{x \rightarrow n^{+}}[x]=n-n=0

It is observed that the left and right hand limit of g at x=n do not coincide.

Therefore, g is not continuous at x=n.

Hence, g is discontinuous at all integral points.

Question 20: Is the function defined by f(x)=x^2-\sin x+5 continuous at x=\pi ?

Solution: The given function is f(x)=x^2-\sin x+5

It is evident that f is defined at x=\pi.

At x=\pi, f(x)=f(\pi)=\pi^2-\sin \pi+5=\pi^2-0+5=\pi^2+5

Consider \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi}\left(x^2-\sin x+5\right)

Put x=\pi+h, it is evident that if x \rightarrow \pi, then h \rightarrow 0

\therefore \lim _{x \rightarrow \pi} f(x) =\lim _{x \rightarrow \pi}\left(x^2-\sin x\right)+5

=\lim _{h \rightarrow 0}\left[(\pi+h)^2-\sin (\pi+h)+5\right]

=\lim _{h \rightarrow 0}(\pi+h)^2-\lim _{h \rightarrow 0} \sin (\pi+h)+\lim _{h \rightarrow 0} 5

=(\pi+0)^2-\lim _{h \rightarrow 0}[\sin \pi \cos h+\cos \pi \sin h]+5

=\pi^2-\lim _{h \rightarrow 0} \sin \pi \cos h-\lim _{h \rightarrow 0} \cos \pi \sin h+5

=\pi^2-\sin \pi \cos 0-\cos \pi \sin 0+5

=\pi^2-0(1)-(-1) 0+5

=\pi^2+5

=f(\pi)
Therefore, the given function f is continuous at x=\pi.

Question 21: Discuss the continuity of the following functions.
(i) f(x)=\sin x+\cos x
(ii) f(x)=\sin x-\cos x
(iii) f(x)=\sin x \times \cos x

Solution: It is known that if g and h are two continuous functions, then g+h, g-h and g, h are also continuous.
Let g(x)=\sin x and h(x)=\cos x are continuous functions.

It is evident that g(x)=\sin x is defined for every real number.

Let c be a real number. Put x=c+h
If x \rightarrow c, then h \rightarrow 0

g(c)=\sin c

\lim _{x \rightarrow c} g(x) =\lim _{x \rightarrow c} \sin x

=\lim _{h \rightarrow 0} \sin (c+h)

=\lim _{h \rightarrow 0}[\sin c \cos h+\cos c \sin h]

=\lim _{h \rightarrow 0}(\sin c \cos h)+\lim _{h \rightarrow 0}(\cos c \sin h)

=\sin c \cos 0+\cos c \sin 0

=\sin c(1)+\cos c(0)

=\sin c

\therefore \lim _{x \rightarrow c} g(x) =g(c)

Therefore, g(x)=\sin x is a continuous function.
Let h(x)=\cos x

It is evident that h(x)=\cos x is defined for every real number.

Let c be a real number. Put x=c+h

If x \rightarrow c, then h \rightarrow 0

h(c)=\cos c

\lim _{x \rightarrow c} h(x) =\lim _{x \rightarrow c} \cos x

=\lim _{h \rightarrow 0} \cos (c+h)

=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h]

=\lim _{h \rightarrow 0}(\cos c \cos h)-\lim _{h \rightarrow 0}(\sin c \sin h)

=\cos c \cos 0-\sin c \sin 0

=\cos c(1)-\sin c(0)

=\cos c

\therefore \lim _{x \rightarrow c} h(x) & =h(c)

Therefore, h(x)=\cos x is a continuous function.

Therefore, it can be concluded that,

(i) f(x)=g(x)+h(x)=\sin x+\cos x is a continuous function.

(ii) f(x)=g(x)-h(x)=\sin x-\cos x is a continuous function.

(iii) f(x)=g(x) \times h(x)=\sin x \times \cos x is a

continuous function.

Question 22: Discuss the continuity of the cosine, cosecant, secant, and cotangent functions.

Solution: It is known that if g and h are two continuous functions, then

\frac{h(x)}{g(x)}, g(x) \neq 0

is continuous.

\frac{1}{g(x)}, g(x) \neq 0

is continuous.

\frac{1}{h(x)}, h(x) \neq 0

is continuous.

Let g(x)=\sin x and h(x)=\cos x are continuous functions.

It is evident that g(x)=\sin x is defined for every real number.

Let c be a real number. Put x=c+h
If x \rightarrow c, then h \rightarrow 0

g(c) =\sin c

\lim _{x \rightarrow c} g(x) =\lim _{x \rightarrow c} \sin x

=\lim _{h \rightarrow 0} \sin (c+h)

=\lim _{h \rightarrow 0}[\sin c \cos h+\cos c \sin h]

=\lim _{h \rightarrow 0}(\sin c \cos h)+\lim _{h \rightarrow 0}(\cos c \sin h)

=\sin c \cos 0+\cos c \sin 0

=\sin c(1)+\cos c(0)

=\sin c

\therefore \lim _{x \rightarrow c} g(x) =g(c)

Therefore, g(x)=\sin x is a continuous function.

Let h(x)=\cos x

It is evident that h(x)=\cos x is defined for every real number.

Let c be a real number. Put x=c+h

If x \rightarrow c, then h \rightarrow 0

h(c) =\cos c

\lim _{x \rightarrow c} h(x) =\lim _{x \rightarrow c} \cos x

=\lim _{h \rightarrow 0} \cos (c+h)

=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h]

=\lim _{h \rightarrow 0}(\cos c \cos h)-\lim _{h \rightarrow 0}(\sin c \sin h)

=\cos c \cos 0-\sin c \sin 0

=\cos c(1)-\sin c(0)

=\cos c

\therefore \lim _{x \rightarrow c} h(x) & =h(c)

Therefore, h(x)=\cos x is a continuous function.

Therefore, it can be concluded that, \operatorname{cosec} x=\frac{1}{\sin x}, \sin x \neq 0 is continuous.

\Rightarrow \operatorname{cosec} x, x \neq n \pi(n \in Z) is continuous.

Therefore, cosecant is continuous except at x=n \pi(n \in Z)

\sec x=\frac{1}{\cos x}, \cos x \neq 0 is continuous.

\Rightarrow \sec x, x \neq(2 n+1) \frac{\pi}{2}(n \in Z) is continuous.

Therefore, secant is continuous except at x=(2 n+1) \frac{\pi}{2}(n \in Z) \cot x=\frac{\cos x}{\sin x}, \sin x \neq 0 is continuous. \Rightarrow \cot x, x \neq n \pi(n \in Z) is continuous.

Therefore, cotangent is continuous except at x=n \pi(n \in Z).
Question 23: f(x)=\left\{\begin{array}{l} \frac{\sin x}{x}, \text { if } x<0 \\ x+1, \text { if } x \geq 0 \end{array}\right.

Find the points of discontinuity of f, where

Solution: The given function is f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \\ x+1, \text { if } x \geq 0\end{cases}
The given function f is defined at all the points of the real line. Let c be a point on the real line.

Case I: If c<0, then f(c)=\frac{\sin c}{c}

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(\frac{\sin x}{x}\right)=\frac{\sin c}{c}

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x<0.

Case II: If c>0, then f(c)=c+1

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+1)=c+1

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x>0.

Case III: If c=0, then f(c)=f(0)=0+1=1

The left hand limit of f at x=0 is,

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\frac{\sin x}{x}\right)=1

The right hand limit of f at x=0 is,

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(x+1)=1

\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)

Therefore, f is continuous at x=0

From the above observations, it can be concluded that f is
continuous at all points of the real line.

Thus, f has no point of discontinuity.

Question 24: Determine if f defined by \quad\{0, if x=0 \quad is a continuous function?

Solution: f(x)=\left\{\begin{array}{l} x^2 \sin \frac{1}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0 \end{array}\right.

The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case I: If c \neq 0, then f(c)=c^2 \sin \frac{1}{c}

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2 \sin \frac{1}{x}\right)=\left(\lim _{x \rightarrow c} x^2\right)\left(\lim _{x \rightarrow c} \sin \frac{1}{x}\right)=c^2 \sin \frac{1}{c}

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x \neq 0.

Case II: If c=0, then f(0)=0

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^2 \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)

It is known that, -1 \leq \sin \frac{1}{x} \leq 1, x \neq 0

\Rightarrow-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2

\Rightarrow \lim _{x \rightarrow 0}\left(-x^2\right) \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^2

\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq 0

\Rightarrow \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)=0

\therefore \lim _{x \rightarrow 0^{-}} f(x)=0

Similarly,

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^2 \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)=0

\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x)

Therefore, f is continuous at x=0.

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

Question 25: Examine the continuity of f, where f is defined by f(x)=\left\{\begin{array}{l}\sin x-\cos x, \text { if } x \neq 0 \\ -1, \text { if } x=0\end{array}\right.

Solution: The given function is

f(x)=\left\{\begin{array}{l} \sin x-\cos x, \text { if } x \neq 0 \\ -1, \text { if } x=0 \end{array}\right.

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:If c \neq 0, then f(c)=\sin c-\cos c

\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(\sin x-\cos x)=\sin c-\cos c

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x, such that x \neq 0.

Case II:If c=0, then f(0)=-1

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)

Therefore, f is continuous at x=0.

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

Question 26: Find the values of k so that the function f is continuous at the indicated point f(x)=\left\{\begin{array}{l}\frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\ 3, \text { if } x=\frac{\pi}{2}\end{array}\right. at x=\frac{\pi}{2}

Solution: The given function is

f(x)=\left\{\begin{array}{l} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right.

The given function f is continuous at x=\frac{\pi}{2}, if f is defined at x=\frac{\pi}{2} and if the value of the f at x=\frac{\pi}{2} equals the limit of f at x=\frac{\pi}{2}.

It is evident that f is defined at x=\frac{\pi}{2} and f\left(\frac{\pi}{2}\right)=3

\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}

Put x=\frac{\pi}{2}+h

Then x \rightarrow \frac{\pi}{2} \Rightarrow h \rightarrow 0

\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}

\quad=k \lim _{h \rightarrow 0} \frac{-\sin h}{-2 h}=\frac{k}{2} \lim _{h \rightarrow 0} \frac{\sin h}{h}=\frac{k}{2} \cdot 1=\frac{k}{2}

\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)

\Rightarrow \frac{k}{2}=3
\Rightarrow k=6

Therefore, the value of k=6.

Question 27: Find the values of k so that the function f is continuous at the indicated point. f(x)=\left\{\begin{array}{l}k x^2, \text { if } x \leq 2 \\ 3, \text { if } x>2\end{array}\right. at x=2

Solution: The given function is f(x)=\left\{\begin{array}{l}k x^2, \text { if } x \leq 2 \\ 3, \text { if } x>2\end{array}\right.
The given function f is continuous at x=2, if f is defined at x=2 and if the value of the f at x=2 equals the limit of f at x=2.
It is evident that f is defined at x=2 and f(2)=k(2)^2=4 k

\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)

\Rightarrow \lim _{x \rightarrow 2^{-}}\left(k x^2\right)=\lim _{x \rightarrow 2^{+}}(3)=4 k

\Rightarrow k \times 2^2=3=4 k

\Rightarrow 4 k=3

\Rightarrow k=\frac{3}{4}

Therefore, the value of k=\frac{3}{4}.

Question 28: Find the values of k so that the function f is continuous at the indicated point f(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq \pi \\ \cos x, \text { if } x>\pi\end{array}\right. at x=\pi

Solution: The given function is f(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq \pi \\ \cos x, \text { if } x>\pi\end{array}\right.

The given function f is continuous at x=\pi, if f is defined at x=\pi and if the value of the f at x=\pi equals the limit of f at x=\pi.
It is evident that f is defined at x=\pi and f(\pi)=k \pi+1

\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)=f(\pi)

\Rightarrow \lim _{x \rightarrow \pi^{-}}(k x+1)=\lim _{x \rightarrow \pi^{+}}(\cos x)=k \pi+1.

\Rightarrow k \pi+1=\cos \pi=k \pi+1

\Rightarrow k \pi+1=-1=k \pi+1

\Rightarrow k=-\frac{2}{\pi}

Therefore, the value of k=-\frac{2}{\pi}.

Question 29: Find the values of k so that the function f is continuous at the indicated point f(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq 5 \\ 3 x-5, \text { if } x>5\end{array}\right. at x=5.

Solution: The given function is f(x)= \begin{cases}k x+1, & \text { if } x \leq 5 \\ 3 x-5, & \text { if } x>5\end{cases}

The given function f is continuous at x=5, if f is defined at x=5 and if the value of the f at x=5 equals the limit of f at x=5.

It is evident that f is defined at x=5 and f(5)=k x+1=5 k+1

\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=f(5)

\Rightarrow \lim _{x \rightarrow 5^{-}}(k x+1)=\lim _{x \rightarrow 5^{+}}(3 x-5)=5 k+1

\Rightarrow 5 k+1=3(5)-5=5 k+1

\Rightarrow 5 k+1=15-5=5 k+1

\Rightarrow 5 k+1=10=5 k+1

\Rightarrow 5 k+1=10

\Rightarrow 5 k=9

\Rightarrow k=\frac{9}{5}

Therefore, the value of k=\frac{9}{5}.

Question 30:Find the values of a \& b such that the function defined by
f(x)=\left\{\begin{array}{l} 5, \text { if } x \leq 2 \\ a x+b, \text { if } 2<x<10 \\ 21, \text { if } x \geq 10, \end{array}\right.
is a continuous function.

Solution: f(x)=\left\{\begin{array}{l} 5, \text { if } x \leq 2 \\ a x+b, \text { if } 2<x<10 \\ 21, \text { if } x \geq 10 \end{array}\right.

The given function is It is evident that f is defined at all points of the real line.

If f is a continuous function, then f is continuous at all real numbers.

In particular, f is continuous at x=2 and x=10
Since f is continuous at x=2, we obtain

\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)

\Rightarrow \lim _{x \rightarrow 2^{-}}(5)=\lim _{x \rightarrow 2^{+}}(a x+b)=5

\Rightarrow 5=2 a+b=5

\Rightarrow 2a+b=5

Since f is continuous at x=10, we obtain

\lim _{x \rightarrow 10^{-}} f(x)=\lim _{x \rightarrow 10^{+}} f(x)=f(10)

\Rightarrow \lim _{x \rightarrow 10^{-}}(a x+b)=\lim _{x \rightarrow 10^{+}}(21)=21

\Rightarrow 10 a+b=21=21

\Rightarrow 10 a+b=21

On subtracting equation (1) from equation (2), we obtain

8a=16

\Rightarrow a=2

By putting a=2 in equation (1), we obtain

2(2)+b=5

\Rightarrow 4+b=5

\Rightarrow b=1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.

Question 31: Show that the function defined by f(x)=\cos \left(x^2\right) is a continuous function.

Solution: The given function is f(x)=\cos \left(x^2\right).
This function f is defined for every real number and f can be written as the composition of two functions as,
f=g o h, where g(x)=\cos x and h(x)=x^2

\left[\because(g o h)(x)=g(h(x))=g\left(x^2\right)=\cos \left(x^2\right)=f(x)\right]

It has to be proved first that g(x)=\cos x and h(x)=x^2 are continuous functions.

It is evident that g is defined for every real number.
Let c be a real number.

Let g(c)=\cos c. Put x=c+h
If x \rightarrow c, then h \rightarrow 0

\lim _{x \rightarrow c} g(x) =\lim _{x \rightarrow c} \cos x

=\lim _{h \rightarrow 0} \cos (c+h)

=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h]

=\lim _{h \rightarrow 0}(\cos c \cos h)-\lim _{h \rightarrow 0}(\sin c \sin h)

=\cos c \cos 0-\sin c \sin 0

=\cos c(1)-\sin c(0)

=\cos c

\therefore \lim _{x \rightarrow c} g(x) & =g(c)

Therefore, g(x)=\cos x is a continuous function.

Let h(x)=x^2

It is evident that h is defined for every real number.

Let k be a real number, then h(k)=k^2

\lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k} x^2=k^2

\therefore \lim _{x \rightarrow k} h(x)=h(k)

Therefore, h is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g(c), then (f o g) is continuous at c.

Therefore, f(x)=(g o h)(x)=\cos \left(x^2\right) is a continuous function.

Question 32: Show that the function defined by f(x)=|\cos x| is a continuous function.

Solution: The given function is f(x)=|\cos x|.
This function f is defined for every real number and f can be written as the composition of two functions as,
f=g o h, where g(x)=|x| and h(x)=\cos x

[\because(g \circ)(x)=g(h(x))=g(\cos x)=|\cos x|=f(x)]

It has to be proved first that g(x)=|x| and h(x)=\cos x are continuous functions.

g(x)=|x| \text { can be written as } g(x)=\left\{\begin{array}{l} -x, \text { if } x<0 \\ x, \text { if } x \geq 0 \end{array}\right.

It is evident that g is defined for every real number.

Let c be a real number.

Case I: If c<0, then g(c)=-c

\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c

\therefore \lim _{x \rightarrow c} g(x)=g(c)

Therefore, g is continuous at all points x, such that x<0.

Case II:If c>0, then g(c)=c

\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(x)=c

\therefore \lim _{x \rightarrow c} g(x)=g(c)

Therefore, g is continuous at all points x, such that x>0.

Case III: If c=0, then g(c)=g(0)=0

\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0

\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0

\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0).

Therefore, g is continuous at all x=0.

From the above three observations, it can be concluded that g is continuous at all points.

It is evident that h(x)=\cos x is defined for every real number.
Let c be a real number. Put x=c+h

If x \rightarrow c, then h \rightarrow 0

h(c) =\cos c

\lim _{x \rightarrow c} h(x) =\lim _{x \rightarrow c} \cos x

=\lim _{h \rightarrow 0} \cos (c+h)

=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h]

=\lim _{h \rightarrow 0}(\cos c \cos h)-\lim _{h \rightarrow 0}(\sin c \sin h)

=\cos c \cos 0-\sin c \sin 0

=\cos c(1)-\sin c(0)

=\cos c

\therefore \lim _{x \rightarrow c} h(x) =h(c)

Therefore, h(x)=\cos x is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at

c and if f is continuous at g(c), then (f o g) is continuous at c.

Therefore, f(x)=(g o h)(x)=g(h(x))=g(\cos x)=|\cos x| is a continuous function.

Question 33: Show that the function defined by f(x)=|\sin x| is a continuous function.

Solution: The given function is f(x)=|\sin x|.
This function f is defined for every real number and f can be written as the composition of two functions as,
f=g o h, where g(x)=|x| and h(x)=\sin x

\lceil\because(g \circ h)(x)=g(h(x))=g(\sin x)=|\sin x|=f(x)]

It has to be proved first that g(x)=|x| and h(x)=\sin x are continuous functions.

g(x)=|x| can be written as g(x)=\left\{\begin{array}{l}-x, \text { if } x<0 \\ x, \text { if } x \geq 0\end{array}\right.

It is evident that g is defined for every real number.
Let c be a real number.

Case I: If c<0, then g(c)=-c

\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c

\therefore \lim _{x \rightarrow c} g(x)=g(c)

Therefore, g is continuous at all points x, such that x<0.

Case II: If c>0, then g(c)=c

\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(x)=c

\therefore \lim _{x \rightarrow c} g(x)=g(c)

Therefore, g is continuous at all points x, such that x>0.

Case III: If c=0, then g(c)=g(0)=0

\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0

\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0

\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0)

Therefore, g is continuous at all x=0.

From the above three observations, it can be concluded that g is continuous at all points.

Let h(x)=\sin x

It is evident that h(x)=\sin x is defined for every real number.

Let c be a real number. Put x=c+k
If x \rightarrow c, then k \rightarrow 0

h(c) =\sin c

\lim _{x \rightarrow c} h(x) =\lim _{x \rightarrow c} \sin x

=\lim _{k \rightarrow 0} \sin (c+k)

=\lim _{k \rightarrow 0}[\sin c \cos k+\cos c \sin k]

=\lim _{k \rightarrow 0}(\sin c \cos k)+\lim _{k \rightarrow 0}(\cos c \sin k)

=\sin c \cos 0+\cos c \sin 0

=\sin c(1)+\cos c(0)

=\sin c

\therefore \lim _{x \rightarrow c} h(x) =h(c)

Therefore, h(x)=\sin x is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g(c), then (f o g) is continuous at c.

Therefore, f(x)=(g o h)(x)=g(h(x))=g(\sin x)=|\sin x| is a continuous function.

Question 34: Find all the points of discontinuity of f defined by f(x)=|x|-|x+1|.

Solution: The given function is f(x)=|x|-|x+1|.

The two functions, g and h are defined as g(x)=|x| and h(x)=|x+1|.

Then, f=g-h

The continuity of g and h are examined first.

g(x)=|x| can be written as g(x)=\left\{\begin{array}{l}-x, \text { if } x<0 \\ x, \text { if } x \geq 0\end{array}\right.

It is evident that g is defined for every real number.

Let c be a real number.

Case I: If c<0, then g(c)=-c

\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c

\therefore \lim _{x \rightarrow c} g(x)=g(c)

Therefore, g is continuous at all points x, such that x<0.

Case II: If c>0, then g(c)=c

\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(x)=c

\therefore \lim _{x \rightarrow c} g(x)=g(c)

Therefore, g is continuous at all points x, such that x>0.

Case III: If c=0, then g(c)=g(0)=0

\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0

\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0

\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0)

Therefore, g is continuous at all x=0.

From the above three observations, it can be concluded that g is continuous at all points.

h(x)=|x+1| \text { can be written as } h(x)=\left\{\begin{array}{l} -(x+1), \text { if } x<-1 \\ x+1, \text { if } x \geq-1 \end{array}\right.

It is evident that h is defined for every real number.
Let c be a real number.

Case I:If c<-1, then h(c)=-(c+1)

\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c}[-(x+1)]=-(c+1)

\therefore \lim _{x \rightarrow c} h(x)=h(c)

Therefore, h is continuous at all points x, such that x<-1.

Case II:If c>-1, then h(c)=c+1

\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c}(x+1)=c+1

\therefore \lim _{x \rightarrow c} h(x)=h(c)

Therefore, h is continuous at all points x, such that x>-1.

Case III: If c=-1, then h(c)=h(-1)=-1+1=0

\lim _{x \rightarrow-1^{-}} h(x)=\lim _{x \rightarrow-1^{-}}[-(x+1)]=-(-1+1)=0

\lim _{x \rightarrow-1^{+}} h(x)=\lim _{x \rightarrow-1^{+}}(x+1)=(-1+1)=0

\therefore \lim _{x \rightarrow-1^{-}} h(x)=\lim _{x \rightarrow-1^{+}} h(x)=h(-1)

Therefore, h is continuous at x=-1.

From the above three observations, it can be concluded that h is continuous at all points. It concludes that g

and h are continuous functions. Therefore, f=g-h is also a continuous function.

Therefore, f has no point of discontinuity.

NCERT solution chapter 5 continuity and differentiability

1. Ncert solution  Exercise 5.2 continuity and differentiability

2. Ncert solution  Exercise 5.3 continuity and differentiability

3. Ncert solution  Exercise 5.4 continuity and differentiability

4. Ncert solution  Exercise 5.5 continuity and differentiability

5. Ncert solution  Exercise 5.6 continuity and differentiability

6. Ncert solution  Exercise 5.7 continuity and differentiability

7. Ncert solution  Chapter 5 Miscellaneous continuity and differentiability

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