Class 12 ncert solution math exercise 5.1

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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.

 

 

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