Suppose f is a real function on a subset of the real numbers and let c be a point in the domain of f. Then f is continuous at c if

$$\lim_{x \rightarrow c} f(x) = f(c)$$

1. More elaborately, if the left hand limit, right hand limit and the value of the function at x = c exist and equal to each other, then f is said to be continuous at x = c.

2. if the right hand and left hand limits at x = c coincide, then we say that the common value is the limit of the function at x = c.

3. continuity as follows: a function is continuous at x = c if the function is defined at x = c and if the value of the function at x = c equals the limit of the function at x = c. If f is not continuous at c, we say f is discontinuous at c and c is called a point of discontinuity of f.

Q. Check the continuity of the function f given by f (x) = 2x + 3 at x = 1.

1. First note that the function is defined at the given point x = 1 and its value is 5. Then find the limit of the function at x = 1. Clearly

2. $$\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} (2x+3)$$

3. = 2(1) + 3 = 5

4. Thus, $$\lim_{x \rightarrow 1} f(x) = 5 = f(1)$$

Q. Examine whether the function f given by f (x) = x2 is continuous at x = 0.

1. First note that the function is defined at the given point x = 0 and its value is 0. Then find the limit of the function at x = 0. Clearly

2. $$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} x^2 = 0^2 = 0$$

3. $$\lim_{x \rightarrow 0} f(x) = 0 = f(0)$$

4. Hence, f is continuous at x = 0.

Q. Discuss the continuity of the function f given by f(x) = | x | at x = 0.

1. f(x) = $$\begin{cases}-x & if x < 0\\ x & if x \ge 0\end{cases}$$

2. Clearly the function is defined at 0 and f (0) = 0. Left hand limit of f at 0 is

3. $$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (-x) = 0$$

4. Similarly, the right hand limit of f at 0 is : $$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} (+x) = 0$$

5. Thus, the left hand limit, right hand limit and the value of the function coincide at x = 0. Hence, f is continuous at x = 0.

Q. Check the points where the constant function f (x) = k is continuous.

1. The function is defined at all real numbers and by definition, its value at any real number equals k. Let c be any real number. Then

2. $$\lim_{x \rightarrow c} f(x) = \lim_{x \rightarrow c} k = k$$

3. Since f (c) = k = $$\lim_{x \rightarrow c} f(x)$$ for any real number c, the function f is continuous at every real number.

#### Class 11 - Continuity and Differentiability

Exercise 1

Q1: Prove that the function $$f(x)=5x-3$$ is continuous

at $$x=0$$, at $$x=-3$$ and at $$x=5$$.

Sol: Based on formulae given in Continuity and Differentiability

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

At $$x=0$$, $$f(0)=5\times 0-3=-3$$

$$\lim_{x \to 0 } f(x)=\lim_{x \to 0 }(5x-3)=5\times 0-3=-3$$ $$therefore,\; \lim_{x \to 0 } f(x)=f(0)$$

So, f is continuous at x=0

At $$x=-3, f(-3)=5\times (-3)-3=-18$$

$$\lim_{x \to -3} f(x)=\lim_{x \to -3 }(5x-3)-3=-18$$ $$therefore,\; \lim_{x \to -3} f(x)=f(-3)$$

So , f is continuous at x=-3

At x=5,f(x)=f(5)= (5×5) – 3=25-3=22

$$\lim_{x \to 5} f(x)=\lim_{x \to 5}(5x-3)=5\times 5-3=22$$ $$therefore,\; \lim_{x \to 0}f(x)=f(5)$$

So , f is continuous at x=5

Q2 : Examine the continuity of the function $$f(x)=2x_{2}- 1$$ at $$x=3$$.

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=2x_{2}-1$$

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

$$\lim_{x \to 3}f(x)=\lim_{x \to 3}(2x^{2}-1) =2\times 3^{^{2}}-1=17$$ $$therefore,\; \lim_{x \to 3}f(x)=f(3)$$

Thus, f is continuous at $$x=3$$ hence solved

Q3: Examine the following function for continuity.

1. $$f(x)=x-5$$
2. $$f(x)=\frac{1}{x-5}, x\neq 5$$
3. $$f(x)=\frac{x^{2}-25}{x+5}, x\neq -5$$
4. $$f(x)=\left | x-5 \right |$$

Sol:Based on formulae given in Continuity and Differentiability

1. The given function is $$f(x)=x-5$$

It is evident that $$f$$ is defined at every real number

$$k$$ is k-5.

It is also observed that $$\lim_{x \to k} f(x)=\lim_{x \to k}(x-5)=k-5=f(k)$$.

$$therefore,\; \lim_{x \to k} f(x)=f(k)$$

Hence, $$f$$ is continuous at every real number and

therefore, it is continuous function.

(b) The given function is $$f(x)=\frac{1}{x-5}, x\neq 5$$

For any real number k $$\neq 5$$, we obtain

$$\lim_{x \to k} f(x)=\lim_{x \to k}\frac{1}{x-5}=\frac{1}{k-5}$$

Also, $$f(k)=\frac{1}{k-5} (as k\neq 5)$$

$$therefore,\;\lim_{x \to k} f(x)=f(k)$$

Hence, $$f$$ is continuous at every point in the domain of

$$f$$ and therefore, it is a continuous function.

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

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

$$\lim_{x \to c} f(x)=\lim_{x \to c}\frac{x^{2}-25}{x+5}=\lim_{x \to c}\frac{(x+5)(x-5)}{x+5}=\lim_{x \to c}(x-5)=(c-5)$$

Also,$$f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) (as c\neq -5)$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

(d) The function is $$f(x)=\left | x-5 \right |$$

$$= \left\{\begin{matrix} 5-x, & if x< 5\\ x-5 ,& if x\geq 5 \end{matrix}\right.$$

This function $$f$$ is determined at all points of the real line.

Let c be a point on a real line. Then, $$c<5$$ or $$c= 5$$

or $$c> 5$$

Case 1: $$c<5$$

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

$$\lim_{x \to c}f(x)=\lim_{x \to c}(5-x)=5-c$$ $$therefore,\; \lim_{x \to c} f(x)=f(c)$$

Therefore f is continuous at all real number less than 5.

Case 2: $$c=5$$

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

$$\lim_{x \to 5}f(x)=\lim_{x \to 5}(5-x)=(5-5)=0$$ $$\lim_{x \to 5}f(x)=\lim_{x \to 5}(x-5)=(5-5)=0$$ $$therefore,\; \lim_{x \to c^{-}}f(x)=\lim_{x \to c^{+}}=f(c)$$

So $$f$$ is continuous at $$x=5$$

Case 3: $$c>5$$

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

$$\lim_{x \to c}f(x)=\lim_{x \to c}(x-5)=c-5$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$Based on formulae given in Continuity and Differentiability

Q4: Prove that the function $$f(x)=x^{n}$$ is continuous at $$x=n$$, where n is a positive integer.

Sol: Based on formulae given in Continuity and Differentiability

Given function $$f(x)=x^{n}$$

f is to be defined at all positive integers, n and its value at n is $$n^{n}$$.

Then $$\lim_{x \to n}f(n)=\lim_{x \to n}(x^{n})=n^{n}$$

$$therefore,\; \lim_{x \to n}f(x)=f(n)$$

So, f is continuous at n for all positive values of n. Based on formulae given in Continuity and Differentiability

Q5: Is the function $$f$$ given by $$f(x)=\begin{cases} x & \text{ if } x\leq 1 \\ 5 & \text{ if } x> 1 \end{cases}$$

continuous at $$x=0,1,2?$$

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} x & \text{ if } x\leq 1 \\ 5 & \text{ if } x> 1 \end{cases}$$

Case 1: At $$x=0$$

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

Then,$$\lim_{x \to 0}f(x)=\lim_{x \to 0}x=0$$

$$therefore,\; \lim_{x \to 0}f(x)=f(0)$$

Therefore, f is continuous at x=0

Case 2: At $$x=1$$,

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

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

$$\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(5)=5$$

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

$$\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(5)=5$$ $$therefore,\; \lim_{x \to 1^{-}}f(x)\neq \lim_{x \to 1^{+}}f(x)$$

So, $$f$$ is not continuous at $$x=1$$.

Case 3: At $$x=2$$

$$f$$ is defined at 2 and its value at 2 is 5.

Then $$\lim_{x \to 2}f(x)=\lim_{x \to 2}(5)=5$$

$$therefore,\; \lim_{x \to 2}f(x)=f(2)$$

Therefore, f is continuous at $$x=2$$

Q6: Find all the points of discontinuity of $$f$$, where $$f$$

Is defined by

$$f(x)=\begin{cases} 2x+3 & \text{ if } x\leq 2 \\ 2x-3& \text{ if } x> 2 \end{cases}$$

Sol: Based on formulae given in Continuity and Differentiability

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

It is necessary that the given function $$f(x)$$ is defined at all the points of the real line.

(i) $$c<2$$

(ii) $$c>2$$

(iii) $$c=2$$

Case (i) $$c<2$$

Then,$$f(c)=2c+3$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(2x+3)=2c+3$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at all points $$x$$, such that $$x<2$$

Case (ii) $$c>2$$

Then, $$f(c)=2c+3$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(2c+3)=2c+3$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ 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,

$$\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{-}}(2x+3))=2\times 2+3=7$$

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

$$\lim_{x \to 2^{+}}f(x)=\lim_{x \to 2^{+}}(2x-3))=2\times 2-3=1$$

It is observed that the left 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 $$f$$.

Q7: Find all points of discontinuity of $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} \begin{vmatrix} x \end{vmatrix}+3, & \text{ if } x\leq -3 \\ -2x,& \text{ if } -3< x< 3 \\ 6x+2,& \text{ if } x\geq 3 \end{cases}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} \begin{vmatrix} x \end{vmatrix}+3, & \text{ if } x\leq -3 \\ -2x,& \text{ if } -3< x< 3 \\ 6x+2,& \text{ if } x\geq 3 \end{cases}$$

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

Let c be a point on the real line.

Case 1:

If $$c< -3,$$ then $$f(c)=-c+3$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(-x+3)=-c+3$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at all points $$x$$, such that $$x<-3$$

Case 2:

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

$$\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(-x+3)=-(-3)+3=6$$ $$\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(-2x)=-2\times(-3)=6$$ $$therefore,\; \lim_{x \to 3^{+}}f(x)=f(-3)$$

So, $$f$$ is continuous at $$x=-3$$.

Case 3:

If $$-3<x<3,$$ then $$f(c)=-2c$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(-2x)=-2c$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous in $$(-3,3)$$.

Case 4:

If $$c=3$$, then the left hand limit of $$f$$ at $$x=3$$ is,

$$\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(-2x)=-2\times 3=-6$$

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

$$\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(6x+2)=6\times +2=20$$

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

Therefore, $$f$$ is not continuous at $$x=3$$.

Case 5:

If $$c> 3$$, then $$f(c)=5c+2$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(6x+2)=6c+2$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at all points $$x$$, such that $$x>3$$

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

Q8: Find all points of discontinuity of $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} \frac{\begin{vmatrix} x \end{vmatrix}}{x} & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} \frac{\begin{vmatrix} x \end{vmatrix}}{x} & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}$$

We know that, $$x<0\Rightarrow \begin{vmatrix} x \end{vmatrix}=-x$$ and $$x>0\Rightarrow \begin{vmatrix} x \end{vmatrix}=x$$

Therefore, the given function can be also written as

$$f(x)=\begin{cases} \frac{\begin{vmatrix} x \end{vmatrix}}{x}=\frac{-x}{x}=-1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \frac{\begin{vmatrix} x \end{vmatrix}}{x}=\frac{x}{x}=1, & \text{ if } x>0 \end{cases}$$

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

Let $$c$$ be a point on a real line.

Case 1:

If $$c<0$$, then $$f(c)=-1$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(-1)=-1$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 2:

If $$c=0$$ then the left hand limit of $$f$$ at $$x=0$$ is,

$$\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}(-1)=-1$$

the right hand limit of $$f$$ at $$x=0$$ is,

$$\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(1)=1$$

The left hand limit and right hand limit of $$f$$ at $$x=0$$ do not coincide.

Therefore, $$f$$ is not continuous at $$x=0$$.

Case 3:

If $$c>0,$$ then $$f(c)=1$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(1)=1$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at $$x>0$$

Hence, $$x=0$$ is the only point of discontinuity of $$f$$. Based on formulae given in Continuity and Differentiability

Q9: Find all points of discontinuity of $$f$$, where $$f$$ is defined by

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

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f(x)=\begin{cases} \frac{x}{\begin{vmatrix} x \end{vmatrix}}, & \text{ if } x<0 \\ -1,& \text{ if } x\geq 0 \end{cases}$$

We know that $$x<0\Rightarrow \begin{vmatrix} x \end{vmatrix}=-x$$

Therefore, the given function can be also written as

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

$$\Rightarrow f(x)=-1$$ for all $$x\in \mathbb{R}$$

Let c be any real number. Then, $$\lim_{x \to c}f(x)=\lim_{x \to c}(-1)=-1$$

Also, $$f(c)=-1=\lim_{x \to c}f(x)$$

Therefore, the given function is a continuous function.

Hence the given function has no point of discontinuity .

Q10: Find all the points of discontinuity of f, where $$f$$ is defined by

$$f(x)=\begin{cases} x+1 & \text{ if } x\geq 1 \\ x^{2}+1 & \text{ if } x<1 \end{cases}$$.

Sol: Based on formulae given in Continuity and DifferentiabilityThe given function $$f$$ is $$f(x)=\begin{cases} x+1 & \text{ if } x\geq 1 \\ x^{2}+1 & \text{ if } x<1 \end{cases}$$.

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

Let c be a point on the real line.

Case 1:

If $$c<1,$$ then $$f(c)=c^{2}+1$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(x^{2}+1)=(c^{2}+1)$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 2:

If $$c=1,$$ then $$f(c)=f(1)=1+1=2$$

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

$$\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(x^{2}+1)=1^{2}+1=2$$

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

$$\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(x^{2}+1)=1^{2}+1=2$$ $$therefore,\; \lim_{x \to 1}f(x)=f(c)$$

So, $$f$$ is continuous at $$x=1$$.

Case 3:

If $$c>1$$, then $$f(c)=c+1$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(x+1)=c+1$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at all points of $$x$$ such that $$x>1$$.

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

Q11: Find all the points of discontinuity of $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} x^{3}-3 & \text{ if } x\leq 2 \\ x^{2}+1 & \text{ if } x>2 \end{cases}$$ .

Sol: Based on formulae given in Continuity and DifferentiabilityThe given function $$f$$ is $$f(x)=\begin{cases} x^{3}-3 & \text{ if } x\leq 2 \\ x^{2}+1 & \text{ if } x>2 \end{cases}$$

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

Let c be a point on the real line.

Case1:

If $$c<2$$, then $$f(c)=c^{3}-3$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(x^{3}-3)=c^{3}-3$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at all points of $$x$$, such that $$x<2$$.

Case 2:

If $$c=2$$, then $$f(c)=f(2)=2^{3}-3=5$$

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

$$\lim_{x \to 2^{+}}f(x)=\lim_{x \to 2^{+}}(x^{2}+1)=2^{2}+1=5$$

The left hand limit of $$f$$ at $$x=2$$ is,

$$\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{-}}(x^{3}-3)=2^{3}-3=5$$ $$therefore,\; \lim_{x \to 2}f(x)=f(2)$$

So, $$f$$ is continuous at $$x=2$$.

Case 3:

If $$c>2$$, then $$f(c)=c^{2}+1$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(x^{2}+1)=c^{2}+1$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at all points of $$x$$, such that $$x>2$$.

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

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

Q12: Find all points of discontinuity of $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} x^{10}-1, & \text{ if } x\leq 1 \\ x^{2}, & \text{ if } x>1 \end{cases}$$.

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} x^{10}-1, & \text{ if } x\leq 1 \\ x^{2}, & \text{ if } x>1 \end{cases}$$

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

Let c be a point on the real line.

Case1:

If $$c<1$$, then $$f(c)=c^{10}-1$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(x^{10}-1)=(c^{10}-1)$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at $$x$$ such that $$x<1$$

Case 2:

If $$c=1$$, then the left hand limit of $$f$$ at $$x=1$$ is,

$$\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(x^{10}-1)=1^{10}-1=1-1=0$$

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

$$\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(x^{2})=1^{2}=1$$

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

Therefore, $$f$$ is not continuous at $$x=1$$.

Case 3:

If $$c>1$$, then $$f(c)=c^{2}$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(x^{2})=c^{2}$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at $$x$$ such that $$x>1$$

Thus, we say that $$x=1$$ is the only point of discontinuity of $$f$$.

Q13: Is the function defined by

$$f(x)=\begin{cases} x+5, & \text{ if } x\leq 1 \\ x-5, & \text{ if } x> 1 \end{cases}$$

a continuous function?

Sol: Based on formulae given in Continuity and Differentiability

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

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

Let $$c$$ be a points on the real line.

Case 1:

If $$c<1$$, then $$f(c)=c+5$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(x+5)=c+5$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

So, $$f$$ is continuous at all the points $$x$$, such that $$x<1$$

Case 2:

If $$c=1$$, then $$f(1)=1+5=6$$

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

$$\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(x+5)=1+5=6$$

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

$$\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(x-5)=1-5=-4$$

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

Therefore, $$f$$ is not continuous at $$x=1$$.

Case 3:

If $$c>1$$, then $$f(c)=c-5$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(x-5)=c-5$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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 $$f$$.

Q14: Discuss the continuity of the function $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} 3 & \text{ if } 0\leq x\leq 1 \\ 4 & \text{ if } 1< x< 3 \\ 5 & \text{ if } 3\leq x\leq 10 \end{cases}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=\begin{cases} 3 & \text{ if } 0\leq x\leq 1 \\ 4 & \text{ if } 1< x< 3 \\ 5 & \text{ if } 3\leq x\leq 10 \end{cases}$$

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

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

Case 1:

If $$0\leq c\leq 1$$, then $$f(c)=3$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(3)=3$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 2:

If $$c=1$$, then $$f(3)=3$$

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

$$\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(3)=3$$

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

$$\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(4)=4$$

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

Therefore, $$f$$ is not continuous at $$x=1$$.

Case 3:

If $$1<c<3,$$, then $$f(c)=4$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(4)=4$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

Case 4:

If $$c=3$$, then $$f(c)=5$$

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

$$\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(4)=4$$

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

$$\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(5)=5$$

The left hand and right hand limits of $$f$$ at $$x=3$$ do not overlap.

Therefore, $$f$$ is not continuous at $$x=3$$.

Case 5:

If $$3<c\leq10$$ then $$f(c)=5$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}f(5)=5$$

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

Hence $$f$$ is not continuous at $$x=1$$ and $$x=3$$.

Q15: Discuss the continuity of the function $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} 2x, & \text{ if } x<0 \\ 0,& \text{ if } 0\leq x\leq 1\\ 4, & \text{ if } x>1 \end{cases}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=\begin{cases} 2x, & \text{ if } x<0 \\ 0,& \text{ if } 0\leq x\leq 1\\ 4, & \text{ if } x>1 \end{cases}$$

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

Let $$c$$ be a point on the real line.

Case 1:

If c<0, then $$f(c)=2c$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(2x)=2c$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 2:

If $$c=0$$, then $$f(c)f(0)=0$$

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

$$\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}(2x)=2\times0=0$$

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

$$\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(0)=0$$ $$therefore,\; \lim_{x \to 0}f(x)=f(0)$$

Therefore, $$f$$ is continuous at $$x=0$$

Case 3:

If $$0<c<1$$, then $$f(x)=0$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(0)=0$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

Therefore, $$f$$ is continuous at all points of the interval $$(0,1)$$.

Case 4:

If $$c=1$$, then $$f(c)=f(1)=0$$

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

$$\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(0)=0$$

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

$$\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(4x)=4\times 1=4$$

The left hand and right hand limits of $$f$$ at $$x=1$$ do not overlap.

Therefore, $$f$$ is not continuous at $$x=1$$.

Case 5:

If c<1, then $$f(c)=4c$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(4x)=4c$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Hence, $$f$$ is discontinuous only at $$x=1$$.

Q16: Discuss the continuity of the function $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} -2, & \text{ if } x\leq -1\\ 2x, & \text{ if } -1<x\leq 1 \\ 2, & \text{ if } x>1 \end{cases}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} -2, & \text{ if } x\leq -1\\ 2x, & \text{ if } -1<x\leq 1 \\ 2, & \text{ if } x>1 \end{cases}$$

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

Let $$c$$ be a point on the real line.

Case 1:

If $$c<-1$$, then $$f(c)=-2$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(-2)=-2$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 2:

If $$c=-1$$, then $$f(c)=f(-1)=-2$$

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

$$\lim_{x \to -1^{-}}f(x)=\lim_{x \to -1^{-}}(-2)=-2$$

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

$$\lim_{x \to -1^{+}}f(x)=\lim_{x \to -1^{+}}(2x)=2\times (-1)=-2$$ $$therefore,\;\lim_{x \to -1}f(x)=f(-1)$$

Therefore, $$f$$ is continuous at $$x=-1$$.

Case 3:

If $$-1<c<1$$, then $$f(c)=2c$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(2x)=2c$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

Therefore, $$f$$ is continuous at all points of the interval $$(-1,1)$$.

Case 4:

If $$c=1$$, then $$f(c)=f(1)=2\times 1=2$$

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

$$\lim_{x \to 1^{-}}f(x)=\lim_{x \to 1^{-}}(2x)=2\times 1=2$$

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

$$\lim_{x \to 1^{+}}f(x)=\lim_{x \to 1^{+}}(2)=2$$ $$therefore,\; \lim_{x \to 1}f(x=f(c))$$

Therefore, f is continuous at $$x=2$$

Case 5:

If $$c>1$$, then $$f(c)=2$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(2)=2$$ $$\lim_{x \to c}f(x)=f(c)$$

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

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

Q17: Find the relationship between a and b so that the function $$f$$ is defined by

$$f(x)=\begin{cases} ax-1, & \text{ if } x\leq 3 \\ bx+3, & \text{ if } x>3 \end{cases}$$

is continuous at $$x=3$$.

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} ax-1, & \text{ if } x\leq 3 \\ bx+3, & \text{ if } x>3 \end{cases}$$

If $$f$$ is continuous at $$x-3$$, then

$$\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}f(x)=f(3)\; \; \; \; \; \; …..(1)$$

Also,

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

$$\lim_{x \to 3^{-}}f(x)=\lim_{x \to 3^{-}}(ax+1)=3a+1$$

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

$$\lim_{x \to 3^{+}}f(x)=\lim_{x \to 3^{+}}(bx+3)=3b+3$$ $$f(3)=3a+1$$

Therefore, from (1) we obtain

3a+1-3b+3-3a+1

$$\Rightarrow 3a+1-3b-3$$ $$\Rightarrow 3a=3b+2$$ $$\Rightarrow a-b=\frac{2}{3}$$

Therefore, the required relationship is given by, $$\Rightarrow a-b=\frac{2}{3}$$

Q18: For what value of $$\lambda$$, is the function defined by

$$f(x)=\begin{cases} \lambda (x^{2}-2x), & \text{ if } x\leq 0 \\ 4x+1, & \text{ if } x> 0 \end{cases}$$

continuous at x=0? What about continuity at $$x=1$$?

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} \lambda (x^{2}-2x), & \text{ if } x\leq 0 \\ 4x+1, & \text{ if } x> 0 \end{cases}$$

If $$f$$ is continuous at $$x=0$$, then

$$\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{+}}f(x)=f(0)$$ $$\Rightarrow \lim_{x \to 0^{-}}\lambda(x^{2}-2x)=\lim_{x \to 0^{+}}(4x+1)$$ $$\Rightarrow \lambda (0^{2}-2\times 0)=(4\times 0-1)$$

$$\Rightarrow 0=1$$ which is not possible.

Therefore, there is no value of $$\lambda$$ for which the function $$f$$ is continuous at $$x=0$$.

At $$x=1$$,

$$f(1)=4x+1=4\times 1+1=5$$ $$\lim_{x \to 1}f(1)=4x+1=4\times 1+1=5$$ $$therefore,\; \lim_{x \to 1}f(x)=f(1)$$

Therefore, for any values of $$\lambda$$, $$f$$ is continuous at $$x=1$$.

Q19: Show that the function defined by $$g(x)=x-[x]$$ is discontinuous at all integral points. Hence $$[x]$$ denotes the greatest integer less than or equal to $$x$$.

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$g(x)=x-[x]$$

It is evident that $$g$$ is defined at all integral points.

Let $$n$$ be an integer.

Then,

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

Then left hand limit of $$f$$ at $$x=n$$ is,

$$\lim_{x \to n^{-}}g(x)=\lim_{x \to n^{-}}(x-[x])=\lim_{x \to n^{-}}(x)-\lim_{x \to n^{-}}[x]=n-(n-1)=1$$

Then right hand limit of $$f$$ at $$x=n$$ is,

$$\lim_{x \to n^{+}}g(x)=\lim_{x \to n^{+}}(x-[x])=\lim_{x \to n^{+}}(x)-\lim_{x \to n^{+}}[x]=n-n=0$$

It is observed that the left and right hand limits of $$f$$ at $$x=n$$ do not coincide.

Therefore, $$f$$ is not continuous at $$x=n$$

Hence, $$g$$ is discontinuous at all integral points.

Q20: Is the function defined by $$f(x)=x^{2}-\sin x+5$$ is continuous at x?

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=x^{2}-\sin x+5$$

It is evident that $$f$$ is defined at x

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

Consider $$\lim_{x \to \pi }f(x)=\lim_{x \to \pi }(x^{2}-sin x +5)$$

Put $$x=\pi +h$$

If $$x\rightarrow \pi$$, then it is evident that $$h\rightarrow 0$$

$$therefore,\; \lim_{x \to \pi }f(x)=\lim_{x \to \pi }(x^{2}-\sin x+5)$$ $$=\lim_{x \to 0 }[(\pi +h)^{2}-\sin (\pi +h)+5]$$ $$=\lim_{x \to 0 }(\pi +h)^{2}-\lim_{x \to 0 }\sin (\pi +h)+\lim_{x \to 0 }5$$ $$=(\pi +0)^{2}-\lim_{h \to 0}[\sin\pi\cosh+\cos\pi\sinh]+5$$ $$=\pi ^{2}-\sin\pi\cos0-\cos\pi\sin0+5$$ $$=\pi ^{2}-0\times 1-(-1)\times 0+5$$ $$=\pi ^{2}+5$$ $$=therefore,\; \lim_{x \to \pi }f(x)=f(\pi )$$

Therefore, the given function is continuous at $$x=\pi$$

Q21: Discuss the continuity of the following functions.

1. $$f(x)=\sin x+\cos x$$
2. $$f(x)=\sin x-\cos x$$
3. $$f(x)=\sin x\times \cos x$$

Sol: Based on formulae given in Continuity and Differentiability

It is known that if $$g$$ and $$g$$ are two continuous functions, then

$$g+h, g-h, g.h$$ are also continuous functions.

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

Let $$g(x)=\sin x$$

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 $$X\;\tilde{A}\;C\;\hat{a}\;\epsilon \;c,\;then\;h\;\tilde{A}\;C\;\hat{a}\;\epsilon \;0$$

$$h(c)-\cos c$$ $$\lim_{x \to c}h(x)=\lim_{x \to c}\cos x$$ $$=\lim_{h \to 0}\cos (c+h)$$ $$=\lim_{h \to 0}\cos [\cos c\cos h-\sin c\sin h]$$ $$=\lim_{h \to 0}\cos \cos c\cos h-\lim_{h \to 0}\sin c\sin h$$ $$=\cos \cos c\cos 0-\sin c\sin 0$$ $$=\cos c\times 1-\sin c\times 0$$ $$=\cos c$$ $$therefore,\; \lim_{x \to c}h(x)=h(c)$$

Therefore, $$h$$ is continuous function.

Therefore, it can be said that

1. $$f(x)=\sin x+\cos x$$ is a continuous function.
2. $$f(x)=\sin x-\cos x$$ is a continuous function.
3. $$f(x)=\sin x\times \cos x$$ is a continuous function.

Q22: Discuss the continuity of the cosine, cosecant, secant and cotangent function.

Sol: Based on formulae given in Continuity and Differentiability

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

(i) $$\frac{h(x)}{g(x)},g(x)\neq 0$$ is continuous

(ii) $$\frac{1}{g(x)},g(x)\neq 0$$ is continuous

(iii) $$\frac{1}{h(x)},h(x)\neq 0$$ is continuous

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

Let $$g(x)=\sin x$$

Clearly $$g(x)=\sin x$$ is defined for every real number.

Let $$c$$ be a real number. Put $$x\rightarrow c+h$$

If $$x\rightarrow c$$, then $$h\rightarrow 0$$

$$g(x)=\sin x$$ $$\lim_{x \to c}g(x) = \lim_{x \to c}\sin x\\ =\lim_{h \to 0}\sin (c+h)\\ =\lim_{h \to 0}[\sin c\cos h+\cos c\sin h]\\ = \sin c\cos 0+\cos c\sin 0\\ =\sin c+0 \\ =\sin c$$ $$therefore,\; \lim_{x \to c}g(x)=g(c)$$ Based on formulae given in Continuity and Differentiability

Therefore, $$g$$ 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 $$x\rightarrow c+h$$

$$x\;\hat{A}\;c,\;then\;h\;\hat{A}$$ $$h(c)=\cos x$$ $$\lim_{x \to c}h(x) = \lim_{x \to c}\cos x\\ =\lim_{h \to 0}\cos (c+h)\\ =\lim_{h \to 0}[\cos c\cos h-\sin c\sin h]\\ = \cos c\cos 0+\sin c\sin 0\\ =\cos c+0 \\ =\cos c$$ $$therefore,\; \lim_{x \to c}h(x)=h(c)$$

Therefore, $$h$$ is a continuous function.

It can be concluded that,

$$\csc x-\frac{1}{sinx}, \sin x\neq 0 is continuous$$ $$\Rightarrow \csc x, x\neq n\pi (n\in \mathbb{Z})$$Based on formulae given in Continuity and Differentiability

Q23: Find the points of discontinuity of $$f$$, where

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

Sol:

The given function $$f$$ is $$f(x)=\begin{cases} \frac{\sin x}{x}, & \text{ if } x<0 \\ x-1, & \text{ if } x\geq 0 \end{cases}$$

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

Let $$c$$ be a real number.

Case 1:

If $$c<0, then f(c)=\frac{\sin c }{c}$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(\frac{\sin x}{x})=(\frac{\sin c }{c})$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$Based on formulae given in Continuity and Differentiability

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

Case 2:

If $$c>0, then f(c)=c+1$$ and $$\lim_{x \to c}f(x)=\lim_{x \to c}(x+1)=(c+1)$$

$$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 3:

If $$c=0, then f(c)=f(0)=0+1=1$$

The left hand limit of $$f$$ at $$x$$ is,

$$\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}\frac{\sin x}{x}=1$$

The right hand limit of $$f$$ at $$x$$ is,

$$\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(x+1)=1$$ $$therefore,\; \lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{+}}f(x)=f(0)$$

Therefore, $$f$$ is continuous at $$x=0$$

Based on formulae given in Continuity and Differentiability

It can be concluded that $$f$$ is continuous at all points of the real line.

Thus, $$f$$ has no point of discontinuity.

Q24: Determine if $$f$$ defined by

$$f(x)=\begin{cases} x^{2}\sin \frac{1}{x}, & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}$$

Sol:

The given function $$f$$ is $$f(x)=\begin{cases} x^{2}\sin \frac{1}{x}, & \text{ if } x\neq 0 \\ 0, & \text{ if } x= 0 \end{cases}$$

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

Let $$c$$ be a real number.

Case 1:

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

$$\lim_{x \to c}f(x)=\lim_{x \to c}\left ( x^{2}\sin \frac{1}{x}\right )=\left ( \lim_{x \to c}x^{2} \right )\left ( \lim_{x \to c}\sin \frac{1}{x} \right )=c^{2}\sin \frac{1}{c}$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 2:

If $$c= 0, then f(0)=0$$

$$\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}\left ( x^{2}\sin \frac{1}{x} \right )=\lim_{x \to 0^{-}}\left ( x^{2} \sin \frac{1}{x}\right )$$

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

$$\Rightarrow -x^{2}\leq \sin \frac{1}{x}\leq x^{2}$$ $$\Rightarrow \lim_{x \to 0}(-x^{2})\lim_{x \to 0}\leq \lim_{x \to 0}(\sin \frac{1}{x})\leq \lim_{x \to 0}(x^{2})$$ $$\Rightarrow 0\lim_{x \to 0}\leq \lim_{x \to 0}(\sin \frac{1}{x})\leq 0$$ $$\Rightarrow \lim_{x \to 0}(\sin \frac{1}{x})= 0$$ $$therefore,\; \lim_{x \to 0^{-}}f(x)= 0$$

Similarly, $$\lim_{x \to 0^{+}}f(x)= \lim_{x \to 0^{+}}x^{2}\sin \frac{1}{x} =0$$

Based on formulae given in Continuity and Differentiability $$therefore,\; \lim_{x \to 0^{-}}f(x)=f(0)=\lim_{x \to 0^{+}}f(0)$$

Therefore, $$f$$ is continuous at $$x=0$$

It can be concluded that $$f$$ is continuous at all points of the real line.

Thus, $$f$$ has no point of discontinuity.

Q25: Determine the continuity of $$f$$, where $$f$$ is defined by

$$f(x)=\begin{cases} \sin x- \cos x,& \text{ if } x\neq 0 \\ -1,& \text{ if } x= 0 \end{cases}$$

Sol:

The given function $$f$$ is $$f(x)=\begin{cases} \sin x- \cos x,& \text{ if } x\neq 0 \\ -1,& \text{ if } x= 0 \end{cases}$$

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

Let $$c$$ be a real number.

Case 1:

If $$c\neq 0, then f(c)= \sin c- \cos c$$

$$\lim_{x \to c}f(x)=\lim_{x \to c}(\sin x- \cos x)=\sin c- \cos c$$ $$therefore,\; \lim_{x \to c}f(x)=f(c)$$

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

Case 2:

If $$c= 0, then f(0)=-1$$

$$\lim_{x \to 0^{-}}f(x)=\lim_{x \to 0^{-}}(\sin x- \cos x)=\sin 0- \cos 0=0-1=-1$$ $$\lim_{x \to 0^{+}}f(x)=\lim_{x \to 0^{+}}(\sin x- \cos x)=\sin 0- \cos 0=0-1=-1$$ $$therefore,\; \lim_{x \to 0^{-}}f(x)=f(0)=\lim_{x \to 0^{+}}f(0)$$

Therefore, $$f$$ is continuous at $$x=0$$

It can be concluded that $$f$$ is continuous at all points of the real line.

Thus, $$f$$ is a continuous function.

Q26: Find the values of $$k$$ so that the function $$f$$ is continuous at the indicated points

$$f(x)=\begin{cases} \frac{k\cos x}{\pi -2x} ,& \text{ if } x\neq \frac{\pi }{2} \\ 3,& \text{ if} x= \frac{\pi }{2} \end{cases}$$

At $$x=\frac{\pi }{2}$$.

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f$$ is $$f(x)=\begin{cases} \frac{k\cos x}{\pi -2x} ,& \text{ if } x\neq \frac{\pi }{2} \\ 3,& \text{ if} x= \frac{\pi }{2} \end{cases}$$

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

It is evident that the limit of $$f$$ is defined at $$x=\frac{\pi }{2}$$ and $$f(\frac{\pi }{2})=3$$.

$$\lim_{x \to \frac{\pi }{2}}f(x)=\lim_{x \to \frac{\pi }{2}}\frac{k\cos x}{\pi -2x}$$ $$\lim_{x \to \frac{\pi }{2}}f(x)=\lim_{x \to \frac{\pi }{2}}\frac{k\cos x}{\pi -2x}$$

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

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

$$therefore,\; \lim_{x \to \frac{\pi }{2}}f(x)=\lim_{x \to \frac{\pi }{2}}\frac{k \cos x}{\pi -2x}=\lim_{h \to 0}\frac{k \cos (\frac{\pi }{2}+h)}{\pi -2(\frac{\pi }{2}+h)}$$ $$=k\lim_{h \to 0}\frac{-\sin h}{-2h}=\frac{k}{2}\lim_{h \to 0}\frac{\sin h}{h}=\frac{k}{2}$$ $$therefore,\; \lim_{x \to \frac{\pi }{2}}f(x)=f(\frac{\pi }{2})$$ $$\Rightarrow \frac{k}{2}=3$$ $$\Rightarrow k=6$$

therefore , the required value of $$k$$ is 6 Based on formulae given in Continuity and Differentiability.

Q27: Find the values of $$k$$ so that the function $$f$$ is continuous at the indicated points

$$f(x)=\begin{cases} kx^{2,} & \text{ if } x=2 \\ 3, & \text{ if } x>2 \end{cases}$$

At $$x=2$$

Sol: Based on formulae given in Continuity and Differentiability

The given function is$$f(x)=\begin{cases} kx^{2,} & \text{ if } x=2 \\ 3, & \text{ if } x>2 \end{cases}$$

The given function $$f$$ is continuous at $$x=2$$. If $$f$$ is defined at $$x=2$$ and if the value of $$f$$ at $$x=2$$ equals the limit of $$f at x=2$$

It is evident that $$f$$ is defined at $$x=2$$ and $$f(2)=k\left ( 2 \right )^{2}=4k$$

$$\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{+}}f(x)=f(2)$$ $$\Rightarrow \lim_{x \to 2^{-}}(kx^{2})=\lim_{x \to 2^{+}}3=4k$$ $$\Rightarrow (k\times 2^{2})=3=4k$$ $$\Rightarrow 4k=3$$ $$\Rightarrow k=\frac{3}{4}$$

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

Q28: Find the values of $$k$$ so that the function $$f$$ is continuous at the indicated points

$$f(x)=\begin{cases} kx+1, & \text{ if } x\leq \pi \\ \cos x,& \text{ if } x>\pi \end{cases}$$

At $$x=5$$

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=\begin{cases} kx+1, & \text{ if } x\leq \pi \\ \cos x,& \text{ if } x>\pi \end{cases}$$

The given function $$f$$ is continuous at $$x=c$$, If $$f$$ is defined at $$x=c$$ and if the value of $$f$$ at $$x=c$$ equals the limit of $$f at x=c$$

It is evident that $$f$$ is defined at $$x=c$$ and $$f(\pi)=k\pi +1$$

$$\lim_{x \to \pi^{-}}f(x)=\lim_{x \to \pi^{+}}f(x)=f(\pi )$$ $$\Rightarrow \lim_{x \to \pi^{-}}(kx+1)=\lim_{x \to \pi^{+}}\cos x=k\pi +1$$ $$\Rightarrow (k\pi +1)=\cos \pi =k\pi +1$$ $$\Rightarrow (k\pi +1)=-1 =k\pi +1$$ $$\Rightarrow k=-\frac{2}{\pi }$$

Therefore, the required value of $$k is -\frac{2}{\pi }$$

Q29: Find the values of $$k$$ so that the function $$f$$ is continuous at the indicated points

$$f(x)=\begin{cases} kx+1, & \text{ if } x\leq 5\\ 3x-5, & \text{ if } x>5 \end{cases}$$

At $$x=5$$

Sol: Based on formulae given in Continuity and Differentiability

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

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

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

$$\lim_{x \to 5^{-}}f(x)=\lim_{x \to 5^{+}}f(x)=f(5 )$$ $$\Rightarrow \lim_{x \to 5^{-}}(kx+1)=\lim_{x \to 5^{+}}(3x-5)=5k+1$$ $$\Rightarrow (5k +1)=15-5 =5k+1$$ $$\Rightarrow 5k+1=10$$ $$\Rightarrow 5k=9$$ $$\Rightarrow k=\frac{9}{5 }$$

Therefore, the required value of $$k is \frac{9}{5 }$$

Q30: Find the values a and b such that the function defined by

$$f(x)=\begin{cases} 5, & \text{ if } x\leq 2 \\ ax+b, & \text{ if } 2<x<10 \\ 21, & \text{ if } x\geq 10 \end{cases}$$

Is a continuous function.

Sol: Based on formulae given in Continuity and Differentiability

The given function $$k$$ is $$f(x)=\begin{cases} 5, & \text{ if } x\leq 2 \\ ax+b, & \text{ if } 2<x<10 \\ 21, & \text{ if } x\geq 10 \end{cases}$$

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

If $$f$$ is 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

$$\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{+}}f(x)=f(2)$$ $$\Rightarrow \lim_{x \to 2^{-}}(5)=\lim_{x \to 2^{+}}(ax+b)=5$$ $$\Rightarrow 5=2a+b=5$$ $$\Rightarrow 2a+b=5 \; \; \; \; \; \; \; \; \; \; ……(1))$$

Since $$f$$ is continuous at $$x=10$$, we obtain

$$\lim_{x \to 10^{-}}f(x)=\lim_{x \to 10^{+}}f(x)=f(10)$$ $$\Rightarrow \lim_{x \to 10^{-}}(ax+b)=\lim_{x \to 10^{+}}(21)=21$$ $$\Rightarrow 10a+b=21=21$$ $$\Rightarrow 10a+b=21 \; \; \; \; \; \; \; \; \; \; ……(2))$$

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

$$8a=16$$ $$a=2$$

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

$$2\times 2+b=5$$ $$\Rightarrow 4+b=5$$ $$\Rightarrow b=1$$

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

Q31: Show that the function defined by $$f(x)=\cos (x^{2})$$ is a continuous function.

Sol: Based on formulae given in Continuity and Differentiability

The given function is$$f(x)=\cos (x^{2})$$.

This function $$f$$ is defined for every real number and $$f$$ be written as the composition of two function as,

$$f-g\, o\, h \; where \; g(x)=\cos x \; and\; h(x)=x^{2}$$ $$[because \; (goh)(x)= g(h(x))=g(x^{2})=\cos (x^{2})=f(x)]$$

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

Let $$c$$ be a real number.

Then, $$g(x)=\cos c$$

Put $$x=c+h$$

If $$x\rightarrow c, then h\rightarrow 0$$

$$\lim_{x \to c}g(x)=\lim_{x \to c}\cos x$$ $$\lim_{x \to c}g(x)=\lim_{x \to c}\cos x\\ =\lim_{h \to 0}\cos(c+h)\\ =\lim_{h \to 0}[\cos c\cos h- \sin c\sin h]\\ =\lim_{h \to 0}\cos c\cos h-\lim_{h \to 0} \sin c\sin h\\ =\cos c\cos 0-\sin c\sin 0\\ =\cos c\times 1-\sin c\times 0\\ =\cos c$$ $$therefore,\; \lim_{x \to c}=g(c)$$

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

Clearly, $$h$$ is defined for every real number

Let $$k$$ be a real number, then $$h(k)=k^{2}[$$

$$\lim_{x \to k}h(x)=x^{2}=k^{2}$$ $$therefore,\; \lim_{x \to k}h(x)=h(k)$$

Therefore, $$h$$ is continuous function.

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

Therefore, $$f(x)=(g\; o\; h)(x)=cos(x^{2})$$ is a continuous function.

Q32: Show that the function defined by $$f(x)=\left | \cos x \right |$$ is a continuous function.

Sol: Based on formulae given in Continuity and Differentiability

The given function$$f(x)=\left | \cos x \right |$$

This function $$f$$ is defined for every real number and $$f$$ can be written as the composition of two function as,

$$f=g\, o\, h \; where \; g(x)=\left | x \right |; and\; h(x)=\cos x$$ $$[because \; (goh)(x)= g(h(x))=g(\cos x)= \left |\cos x \right |=f(x)]$$

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

$$g(x)=\left |x \right |$$ can be written as

$$g(x)=\begin{cases} -x, & \text{ if } x<0 \\ x, & \text{ if } x\geq 0 \end{cases}$$

Clearly $$g$$ is for all real numbers.

Let $$c$$ be a real number.

Case 1:

If $$c< 0, then g(c)=-c$$ and $$\lim_{x \to c}g(x)=\lim_{x \to c}(-x)=-c$$

$$therefore,\; \lim_{x \to c}g(x)=g(c)$$

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

Case 2:

If $$c> 0, then g(c)=c$$ and $$\lim_{x \to c}g(x)=\lim_{x \to c}(x)=c$$

$$therefore,\; \lim_{x \to c}g(x)=g(c)$$

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

Case 3:

If $$c= 0, then g(c)=g(0)=0$$

$$\lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{-}}(-x)=0$$ $$\lim_{x \to 0^{+}}g(x)=\lim_{x \to 0^{+}}(x)=0$$ $$therefore,\; \lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{+}}g(x)=g(0)$$

Therefore, $$g$$ is continuous at $$x=0$$

It can be concluded that $$g$$ is continuous at all points.

$$h(x)=\cos x$$

Clearly $$h(x)=\cos x$$ is defined for every real number.

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

If $$x\: \tilde{A}\: c\: \hat{a}\: \in \; ‘c$$, then $$h\: \tilde{A}\: c\: \hat{a}\: \in \; ‘0$$

$$h(c)=\cos c$$ $$\lim_{x \to c}h(x)=\lim_{x \to c}\cos x$$ $$\lim_{x \to c}h(x)=\lim_{x \to c}\cos x \\ =\lim_{h \to 0} \cos (c+h)\\ =\lim_{h \to 0}[ \cos c \cos h- \sin c \sin h]\\ =\lim_{h \to 0} \cos c \cos h- \lim_{h \to 0} \sin c \sin h\\ =\cos c \cos 0-\sin c \sin 0\\ =\cos c \times 1-\sin c\times 0\\ =\cos c$$ $$therefore,\; \lim_{x \to c}h(x)=h(c)$$

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

We know that for every real valued function $$g$$ 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,

$$f(x)= (goh)(x)= g(h(x))=g(\cos x)= \left |\cos x \right |$$ is a continuous function.

Q33: Examine that $$\ sin\left | x \right |$$ is a continuous function.

Sol: Based on formulae given in Continuity and Differentiability

Let $$f(x)=\ sin\left | x \right |$$

The function $$f$$ is defined for every real number and $$f$$ can be written as the composition of two function as,

$$f=g\, o\, h \; where \; g(x)=\left | x \right |; and\; h(x)=\sin \left | x \right |$$ $$[because \; (goh)(x)= g(h(x))=g(\sin x)= \left |\sin x \right |=f(x)]$$

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

$$g(x)=\left |x \right |$$ can be written as

$$g(x)=\begin{cases} -x, & \text{ if } x<0 \\ x, & \text{ if } x\geq 0 \end{cases}$$

Clearly $$g$$ is defined for all real numbers.

Let $$c$$ be a real number.

Case 1:

If $$c< 0, then g(c)=-c$$ and $$\lim_{x \to c}g(x)=\lim_{x \to c}(-x)=-c$$

$$therefore,\; \lim_{x \to c}g(x)=g(c)$$

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

Case 2:

If $$c> 0, then g(c)=c$$ and $$\lim_{x \to c}g(x)=\lim_{x \to c}(x)=c$$

$$therefore,\; \lim_{x \to c}g(x)=g(c)$$

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

Case 3:

If $$c= 0, then g(c)=g(0)=0$$

$$\lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{-}}(-x)=0$$ $$\lim_{x \to 0^{+}}g(x)=\lim_{x \to 0^{+}}(x)=0$$ $$therefore,\; \lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{+}}g(x)=g(0)$$

Therefore, $$g$$ is continuous at $$x=0$$

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

$$h(x)=\sin x$$

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

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

If $$x\: \tilde{A}\: c\: \hat{a}\: \in \; ‘c$$, then $$h\: \tilde{A}\: c\: \hat{a}\: \in \; ‘0$$

$$h(c)=\sin c$$ $$\lim_{x \to c}h(x)=\lim_{x \to c}\sin x$$ $$\lim_{x \to c}h(x)=\lim_{x \to c}\sin x \\ =\lim_{h \to 0} \sin (c+h)\\ =\lim_{h \to 0}[ \sin c \cos h- \cos c \sin h]\\ =\lim_{h \to 0} \sin c \cos h- \lim_{h \to 0} \cos c \sin h\\ =\sin c \cos 0-\cos c \sin 0\\ =\sin c \times 1-\cos c\times 0\\ =\sin c$$ $$therefore,\; \lim_{x \to c}h(x)=h(c)$$

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

We know that for every real valued function $$g$$ 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,$$f(x)= (goh)(x)= g(h(x))=g(\sin x)= \left |\sin x \right |$$ is a continuous function.

Q34: Find all the points of discontinuity of $$f$$ defined by $$f(x)=\left | x \right |-\left | x+1 \right |$$

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=\left | x \right |-\left | x+1 \right |$$

The two functions, $$g and h$$, are defined as

$$g(x)=\left | x \right | and h(x)=\left | x+1 \right |$$

Then, $$f=g\, o\, h \;$$

The continuity of $$g \;and\; h$$ is examined first.

$$g(x)=\left |x \right |$$ can be written as

$$g(x)=\begin{cases} -x, & \text{ if } x<0 \\ x, & \text{ if } x\geq 0 \end{cases}$$

Clearly $$g$$ is defined for all real numbers.

Let $$c$$ be a real number.

Case 1:

If $$c< 0, then g(c)=-c$$ and $$\lim_{x \to c}g(x)=\lim_{x \to c}(-x)=-c$$

$$therefore,\; \lim_{x \to c}g(x)=g(c)$$

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

Case 2:

If $$c> 0, then g(c)=c$$ and $$\lim_{x \to c}g(x)=\lim_{x \to c}(x)=c$$

$$therefore,\; \lim_{x \to c}g(x)=g(c)$$

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

Case 3:

If $$c= 0, then g(c)=g(0)=0$$

$$\lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{-}}(-x)=0$$ $$\lim_{x \to 0^{+}}g(x)=\lim_{x \to 0^{+}}(x)=0$$ $$therefore,\; \lim_{x \to 0^{-}}g(x)=\lim_{x \to 0^{+}}g(x)=g(0)$$

Therefore, $$g$$ is continuous at $$x=0$$

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

$$h(x)=\left | x+1 \right |$$

Which can be written as also second form:

$$h(x)=\begin{cases} -(x+1), & \text{ if } x<-1 \\ x+1, & \text{ if } x\geq -1 \end{cases}$$

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

Let $$c$$ be a real number.

Case 1:

If $$c< -1, then h(c)=-(c+1)$$ and $$\lim_{x \to c}h(x)=\lim_{x \to c}[-(x+1)]=-(c+1)$$

$$therefore,\; \lim_{x \to c}h(x)=h(c)$$

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

Case 2:

If $$c> -1, then h(c)=-c+1$$ and $$\lim_{x \to c}h(x)=\lim_{x \to c}(x+1)=c+1$$

$$therefore,\; \lim_{x \to c}h(x)=h(c)$$

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

Case 3:

If $$c= -1, then h(c)=h(-1)=-1+1=0$$

$$\lim_{x \to -1^{-}}h(x)=\lim_{x \to -1^{-}}[-(x+1)]=(-1+1)=0$$ $$\lim_{x \to -1^{+}}h(x)=\lim_{x \to -1^{+}}(x+1)=(-1+1)=0$$ $$therefore,\; \lim_{x \to -1^{-}}h(x)=\lim_{x \to -1^{+}}h(x)=h(-1)$$

Therefore, $$h$$ is continuous at $$x=-1$$

It can be concluded that $$h$$ is continuous at all the real points.

Exercise 5.2 :

Q1: Differentiate the function with respect to $$x$$

$$f(x)=\ sin (x^{2}+5)$$

Sol:The given function is $$f(x)=\ sin (x^{2}+5)=y$$

Let $$t=(x^{2}+5)$$

$$So f(t)=\sin t$$

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$ ……..(i)

$$\frac{\mathrm{d} y}{\mathrm{d} t}=\cos t$$ ……..(ii)

And $$\frac{\mathrm{d} t}{\mathrm{d} x}=2x$$ …….(iii)

Substituting equation (ii) and (iii) in (i) we have,

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\cos t \times (2x)$$

We know $$t=x^{2}+5$$

Thus $$\frac{\mathrm{d} y}{\mathrm{d} x}=(2x)\times \cos x$$

Q2: Differentiate the function with respect to $$x$$

$$f(x)=\cos (\sin x)$$

Sol: Based on formulae given in Continuity and Differentiability

Let $$f(x)=\cos (\sin x)$$

Here $$f$$ is a composite function which can be written in the form of two composite function $$u and v$$.

$$u(x)=\sin x\; \; and\; \; v(t)=\cos t$$ $$(vou)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$$

Put $$t=u(x)=\sin x$$

$$therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\cos t)}{\mathrm{d} x}=-\sin t = -\sin(\sin x)$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (\sin x)}{\mathrm{d} x}=\cos x$$

By chain rule ,

$$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$ $$\frac{\mathrm{d} f}{\mathrm{d} x}=-\sin (\sin x)\times \ cos x=-\cos x\sin (\sin x)$$

Q3: Differentiate the function with respect to $$x$$

$$f(x)=\sin (ax+b)$$

Sol: Based on formulae given in Continuity and Differentiability

Let $$f(x)=\sin (ax+b)$$

Here $$f$$ is a composite function which can be written in the form of two composite function $$u and v$$.

$$u(x)=ax+b\; \; and\; \; v(t)=\sin t$$ $$(vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)=f(x)$$

Put $$t=u(x)=ax+b$$

$$therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} x}=\cos t = \cos(ax+b)$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a$$

By chain rule ,

$$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$ $$\frac{\mathrm{d} f}{\mathrm{d} x}=a\times \ cos (ax+b)$$

Q4: Differentiate the function with respect to $$x$$

$$\sec (\tan (\sqrt{x}))$$.

Sol: Based on formulae given in Continuity and Differentiability

Let $$\sec (\tan (\sqrt{x}))$$

Here $$f$$ is a composite function which can be written in the form of three composite function $$u,v and w$$.

$$u(x)=\sqrt{x}\; \; v(t)=\tan t\; \; and w(s)=\sec s$$ $$(wovou)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w[\tan( \sqrt{x})]=\sec (\tan (\sqrt{x}))=f(x)$$

Put $$s=v(t)=\tan t and t=u(x)=\sqrt{x}$$

then ,$$\frac{\mathrm{d} w}{\mathrm{d} s}=\frac{\mathrm{d} (\sec s)}{\mathrm{d} s}=\sec s\tan s =\sec(\tan t). \tan (\tan t)$$ ( as $$s=\tan t$$

$$=\sec(\tan \sqrt{x}). \tan (\tan \sqrt{x})$$ $$\frac{\mathrm{d} s}{\mathrm{d} t}=\frac{\mathrm{d} (\tan t)}{\mathrm{d} t}=\sec ^{2}t=\sec ^{2}\sqrt{t}$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} \sqrt{x}}{\mathrm{d} x}=\frac{\mathrm{d} (x^{\frac{1}{2}})}{\mathrm{d} x}=\frac{1}{2}.x^{\frac{1}{2}-1}=\frac{1}{2\sqrt{x}}$$

By chain rule ,

$$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} w}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$

=$$\sec (\tan\sqrt{x}).\tan (\tan\sqrt{x})\times \sec ^{2} \sqrt{x}\times \frac{1}{2\sqrt{x}}$$

=$$\frac{\sec ^{2}\sqrt{x}.\sec (\tan \sqrt{x}.\tan (\tan \sqrt{x}))}{2\sqrt{x}}$$

Q5: Differentiate the function with respect to $$x$$

$$f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}= \frac{g(x)}{h(x)}$$,

where $$g(x)= \sin (ax+b)$$ and $$h(x)=\cos (cx+d)$$

Consider $$g(x)=\sin (ax+b)$$

Here $$g$$ is a composite function which can be written in the form of two composite function $$u and v$$.

$$u(x)=ax+b\; \; v(t)=\sin t$$ $$(vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)]=g(x)$$

Put $$t=u(x)=ax+b$$

$$\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} t}=\cos t=\cos (ax+b)$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a$$

By chain rule ,

$$g^{‘}=\frac{\mathrm{d} g}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}=\cos (ax+b).a=a \cos (ax+b)$$

Consider $$h(x)=\cos (cx+d)$$

Here $$h$$ is a composite function which can be also written in the form of two composite function $$p and q$$ of type.

$$p(x)=cx+d\; \; q(y)=\cos y$$ $$(qop)(x)=q(p(x))=q(cx+d)=\cos (cx+d)]=h(x)$$

Put $$y=p(x)=cx+d$$

$$\frac{\mathrm{d} q}{\mathrm{d} y}=\frac{\mathrm{d} (\cos y)}{\mathrm{d} y}=-\sin y=-\sin (cx+d)$$ $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} (cx+d)}{\mathrm{d} x}=c$$

By chain rule ,

$$h^{‘}=\frac{\mathrm{d} h}{\mathrm{d} x}=\frac{\mathrm{d} q}{\mathrm{d} y}\times \frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (cx+d).c=-c \sin (cx+d)$$

Therefore by chain rule , we obtain

$$f^{‘}=\frac{a\cos (ax+b).\ cos (cx+d)-\sin (ax+b)(-\sin(cx+d))}{[\cos(cx+d)]^{2}}$$

=$$=\frac{a\cos (ax+b)}{\ cos (cx+d)}+c\sin (ax+b).\frac{\sin (cx+d)}{\cos (cx+d)}\times \frac{1}{\cos (cx+d)}$$

=$$=a\cos (ax+b)\ sec (cx+d)+c\sin (ax+b).\tan (cx+d).\sec (cx+d)$$

Q6: Differentiate the function with respect to $$x$$

$$f(x)=\cos x^{3}.\sin ^{2}(x^{5})$$

Sol:

The given function is $$f(x)=\cos x^{3}.\sin ^{2}(x^{5})$$

$$\frac{\mathrm{d} }{\mathrm{d} x}\left [ \cos x^{3}.\sin ^{2}(x^{5}) \right ]=\sin ^{2}(x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\cos x^{3}+\cos x^{3}\times \frac{\mathrm{d} }{\mathrm{d} x}\sin ^{2}(x^{5})$$

=$$\sin ^{2}(x^{5})\times (-\sin x^{3})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{3})+\cos x^{3}\times 2\sin (x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \sin (x^{5}) \right ]$$

$$=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\frac{\mathrm{d} }{\mathrm{d} x} (x^{5})$$ $$=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\times 5x^{4}$$ $$=10x^{4}\sin x^{5}.\cos x^{3}.\cos x^{5}\times -3x^{2}\sin x^{3}.\sin ^{2}(x^{5} )$$

Q7: Differentiate the functions with respect to x.

$$2\sqrt{\cot (x^{2})}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$2\sqrt{\cot (x^{2})}$$

$$\frac{\mathrm{d} }{\mathrm{d} x}2\sqrt{\cot (x^{2})}$$ $$=2.\frac{1}{2\sqrt{\cot (x^{2})}}\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \cot (x^{2}) \right ]$$ $$=\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times -cosec^{2} (x^{2})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{2})$$ $$=-\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times \frac{1}{\sin ^{2(x^{2})}}\times(2x)$$ $$=\frac{-2x}{\sqrt{\cos (x^{2})}\sqrt{\sin (x^{2})}\sin (x^{2})}$$ $$=\frac{-2\sqrt{2}x}{\sqrt{2\cos (x^{2})\sin (x^{2})}\sin (x^{2})}$$ $$=\frac{-2\sqrt{2}x}{\sin (x^{2})\sqrt{\sin 2(x^{2})}}$$

Q8: Differentiate the functions with respect to x.

$$\cos\sqrt{x}$$

Sol: Based on formulae given in Continuity and Differentiability

The given function $$f(x)$$ is $$\cos\sqrt{x}$$.

Let $$u(x)=\sqrt{x}$$

And $$v(t)=\cos t$$

$$(vou)(x)=v(u(x))$$ $$=v(\sqrt{x})$$ $$=\cos (\sqrt{x})$$

=$$f(x)$$

Clearly, $$f$$ is a composite function of two functions, $$u\; and \; v$$, such that

$$t=u(x)=\sqrt{x}$$

Then, $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}$$

$$=\frac{1}{2\sqrt{x}}$$

And, $$\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} }{\mathrm{d} t}(\cos t)=-\sin t$$

$$=-\sin (\sqrt{x})$$

By chain rule we have,

$$\frac{\mathrm{d} v}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}.\frac{\mathrm{d} t}{\mathrm{d} x}$$.

$$=-\sin (\sqrt{x}).\frac{1}{2\sqrt{x}}$$ $$=-\frac{1}{2\sqrt{x}}\sin (\sqrt{x})$$ $$=-\frac{\sin (\sqrt{x})}{2\sqrt{x}}$$

Q9: Prove that the function f given by $$f(x)=\left | x-1 \right |,x\in \mathbb{R}$$, is not differentiable at $$x=1$$.

Sol: Based on formulae given in Continuity and Differentiability

The given function is $$f(x)=\left | x-1 \right |,x\in \mathbb{R}$$.

We have known that a function $$f$$ is differentiable at a point $$x=c$$ in its domain if the right hand limit and the left hand limit are finite and equal.

To check the differentiability of the given function at x=1,

The right hand and the left hand limits where x=c are

$$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}$$ and $$\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}$$

Considering the right hand limit of the given function at x=1

$$\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{+}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}$$ $$=\lim_{h \to 0^{+}}\frac{\left | h \right |-\left | 0 \right |}{h}$$ $$=\lim_{h \to 0^{+}}\frac{h}{h}$$ $$=1$$

Considering the left hand limit of the given function at x=1

$$\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{-}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}$$ $$=\lim_{h \to 0^{-}}\frac{\left | h \right |-\left | 0 \right |}{h}$$ $$=\lim_{h \to 0^{-}}\frac{-h}{h}$$ $$=-1$$

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

Q10: Prove that the greatest integer function defined by $$f(x)=\left [ x \right ], 0<x<3$$ is not differentiable at x = 1 and x = 2.

Sol:

The function f is $$f(x)=\left [ x \right ], 0<x<3$$

It is known that a function f is differentiable at a point $$x=c$$ in its domain if both the left hand and the left hand limit are equal

$$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}$$ and $$\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}$$ are finite and equal.

To check the differentiability of the given function at $$x=1$$, consider the right hand limit of f at $$x=1$$

$$\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{+}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}$$ $$=\lim_{h \to 0^{+}}\frac{1-1}{h}$$ $$=\lim_{h \to 0^{+}}\frac{1-1}{h}=\lim_{h \to 0^{+}}(0)=0$$

Now consider the left hand limit of f at $$x=1$$

$$\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{-}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}$$ $$=\lim_{h \to 0^{-}}\frac{0-1}{h}$$ $$=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty$$

Since left hand and the right hand limit of f at x=1 are not equal, f is not differentiable at x=1.

Now to check the differentiability of the given function at x=2,

consider the left hand limit at x=2.

$$\lim_{h \to 0^{-}}\frac{f(2+h)-f(2)}{h}$$ $$\lim_{h \to 0^{-}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}$$ $$=\lim_{h \to 0^{-}}\frac{1-2}{h}$$ $$=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty$$

Now consider the right hand limit of f at $$x=2$$

$$\lim_{h \to 0^{+}}\frac{f(2+h)-f(2)}{h}$$ $$\lim_{h \to 0^{+}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}$$ $$=\lim_{h \to 0^{+}}\frac{2-2}{h}$$ $$=\lim_{h \to 0^{+}}\frac{0}{h}=\lim_{h \to 0^{+}}(0)=0$$

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2

Exercise 5.3

Q1: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$2x+3y=\sin x$$

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$2x+3y=\sin x$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)$$ $$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)$$ $$\Rightarrow 2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos x$$ $$\Rightarrow 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos x-2$$

$$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\cos x-2}{3}$$.

Q2: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$2x+3y=\sin y$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$2x+3y=\sin y$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)$$ $$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)$$ $$\Rightarrow 2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos y \frac{\mathrm{d} y}{\mathrm{d} x}$$ $$\Rightarrow 2=(\cos y-3) \frac{\mathrm{d} y}{\mathrm{d} x}$$

$$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\cos y -3}$$.

Q3: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$ax+by^{2}=\cos y$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$ax+by^{2}=\cos y$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(ax+by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)$$ $$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (ax) + \frac{\mathrm{d} }{\mathrm{d} x}(by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)$$ $$\Rightarrow a + 2by\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin y \frac{\mathrm{d} y}{\mathrm{d} x}$$ $$\Rightarrow a=-(\sin y+2by) \frac{\mathrm{d} y}{\mathrm{d} x}$$

$$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-a}{(\sin y+2by )}$$.

Q4: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$xy+y^{2}=\tan x+ y$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$xy+y^{2}=\tan x+ y$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x + y)$$ $$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x)+ \frac{\mathrm{d} y}{\mathrm{d} x}$$ $$\Rightarrow y.\frac{\mathrm{d} }{\mathrm{d} x} (x)+x .\frac{\mathrm{d} y}{\mathrm{d} x} +2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x}$$ $$\Rightarrow y.1 + x.\frac{\mathrm{d} y}{\mathrm{d} x}+2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x}$$ $$\Rightarrow (x+2y-1)\frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x-y$$

$$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sec ^{2}x-y}{(x+2y-1 )}$$.

Q5: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$x^{2}+xy+y^{2}=100$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$x^{2}+xy+y^{2}=100$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}+xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(100)$$

$$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (x^{2})+\frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=0$$ (derivatives of constant function is 0)

$$\Rightarrow 2x+ y.1+x.\frac{\mathrm{d} y}{\mathrm{d} x} +2y.\frac{\mathrm{d} y}{\mathrm{d} x}=0$$ $$\Rightarrow 2x+ y+(x+2y)\frac{\mathrm{d} y}{\mathrm{d} x} =0$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2x+y}{x+2y}$$

Q6: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$x^{3}+x^{2}y+xy^{2}+y^{3}$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$x^{3}+x^{2}y+xy^{2}+y^{3}=81$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(x^{3}+x^{2}y+xy^{2}+y^{3})=\frac{\mathrm{d} }{\mathrm{d} x}(81)$$

$$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (x^{3})+\frac{\mathrm{d} }{\mathrm{d} x} (x^{2}y) + \frac{\mathrm{d} }{\mathrm{d} x}(xy^{2})+\frac{\mathrm{d} }{\mathrm{d} x}(y^{3})=0$$ (derivatives of constant function is 0)

$$\Rightarrow 3x^{2}+ y.\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}\frac{\mathrm{d} }{\mathrm{d} x}(x)+x\frac{\mathrm{d} }{\mathrm{d} x}(y^{2}+3y^{2}.\frac{\mathrm{d} y}{\mathrm{d} x} =0$$ $$\Rightarrow 3x^{2}+ y.2x+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}.1+x.2y.\frac{\mathrm{d} y}{\mathrm{d} x}+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x} =0$$ $$\Rightarrow (x^{2}+2xy+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+(3x^{2}+2xy+y^{2}=0$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-(3x^{2}+2xy+y^{2})}{x^{2}+2xy+3y^{2}}$$

Q7: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$\sin ^{2}y +\cos xy=\Pi$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$\sin ^{2}y+\cos xy=\Pi$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y+\cos xy)=\frac{\mathrm{d} }{\mathrm{d} x}(\Pi)$$

$$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos xy) =0$$ ……(i) (derivatives of constant function is 0)

Using chain rule,we get

$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y=2 \sin y \frac{\mathrm{d} }{\mathrm{d} x}(\sin y) =2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}$$ …….(ii)

$$\frac{\mathrm{d} }{\mathrm{d} x}(\cos xy)=-\sin xy \frac{\mathrm{d} }{\mathrm{d} x}(xy)= -\sin xy (y. \frac{\mathrm{d} }{\mathrm{d} x}(x)+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -\sin xy (y. 1+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -y \sin xy -x\sin xy \frac{\mathrm{d} y}{\mathrm{d} x}$$ …….(iii)

From (i), (ii) and (iii) we have

$$2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}- y\sin xy- x \sin xy \frac{\mathrm{d} y}{\mathrm{d} x}=0$$ $$\Rightarrow (2\sin y \cos y- x \sin xy)\frac{\mathrm{d} y}{\mathrm{d} x}= y \sin xy$$ $$\Rightarrow (\sin 2y-x\sin xy )\frac{\mathrm{d} y}{\mathrm{d} x}=y \sin xy$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y \sin xy}{\sin 2y -x \sin xy}$$

Q8: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$\sin ^{2}x +\cos ^{y}=1$$.

Sol:

The given relationship is $$\sin ^{2}x +\cos ^{y}=1$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}x+\cos ^{y})=\frac{\mathrm{d} }{\mathrm{d} x}(1)$$

$$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos ^{y}) =0$$ (derivatives of constant function is 0)

$$\Rightarrow 2\sin x \frac{\mathrm{d} }{\mathrm{d} x} (\sin x)+2\cos y.\frac{\mathrm{d} }{\mathrm{d} x} (\cos y) =0$$ $$\Rightarrow 2\sin x \cos x+2\cos y(-\sin y).\frac{\mathrm{d} y}{\mathrm{d} x} =0$$ $$\Rightarrow \sin 2x-\sin 2y.\frac{\mathrm{d} y}{\mathrm{d} x} =0$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin 2x}{\sin 2y}$$

Q9: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$y=\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right )$$.

Sol:

The given relationship is $$y=\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right )$$

$$\sin y=\left ( \frac{2x}{1+x^{2}} \right )$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )$$

$$\Rightarrow \cos y \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{2x}{1+x^{2}} \right )$$ ………..(i)

The right side function $$\left ( \frac{2x}{1+x^{2}} \right )$$ is of the form $$\frac{u}{v}$$

So, by quotient rule, Based on formulae given in Continuity and Differentiability

$$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )=\frac{(1+x^{2}).\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}$$

$$=\frac{(1+x^{2}).2-2x.(0+2x)}{(1+x^{2})^{2}}=\frac{2+2x^{2}-4x^{2}}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}$$ ………(ii)

Also $$\sin y=\frac{2x}{1+x^{2}}$$

$$\Rightarrow \cos y=\sqrt{1-\sin^{2}y}=\sqrt{1-\left ( \frac{2x}{1+x^{2}} \right )^{2}}=\sqrt{\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}}}$$

$$=\sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\frac{1-x^{2}}{1+x^{2}}$$ ……….(iii)

From (i) , (ii) and (iii) we obtain

$$=\frac{1-x^{2}}{1+x^{2}}\times \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x^{2}}$$

Q10: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right ),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )$$

$$\tan y=\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )$$ ………(i)

Differentiating the equation with respect to x, we have

We know that, $$\tan y=\frac{3\tan \frac{y}{3}-\tan ^{3}\frac{y}{3}}{1-3\tan ^{2}\frac{y}{3}}$$ …………(ii)

Comparing equation (i) and (ii), we have

$$x=\tan \frac{y}{3}$$

Differentiating this relationship w.r.t. x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \tan \frac{y}{3} \right )$$ $$\Rightarrow 1=\sec ^{2}\frac{y}{3}.\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{y}{3} \right )$$ $$\Rightarrow 1=\sec ^{2}\frac{y}{3}.\frac{1}{3}\frac{\mathrm{d} y}{\mathrm{d} x}$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{\sec ^{2}\frac{y}{3}}=\frac{3}{1+\tan ^{2}\frac{y}{3}}$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{1+x^{2}}$$

Q11: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$$

$$\cos y= \left ( \frac{1-x^{2}}{1+x^{2}} \right )$$ $$\frac{1-\tan ^{2}\frac{y}{2}}{1+\tan ^{2}\frac{y}{2}}=\frac{1-x^{2}}{1+x^{2}}$$

Comparing both sides equation

$$\tan \frac{y}{2}=x$$

Differentiating the equation with respect to x, we have

$$\sec ^{2} \frac{y}{2}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )=\frac{\mathrm{d} }{\mathrm{d} x}(x)$$ $$\sec ^{2} \frac{y}{2}\times \frac{1}{2}.\frac{\mathrm{d} y}{\mathrm{d} x}=1$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sec ^{2}\frac{y}{2}}$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+\tan ^{2}\frac{y}{2}}$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x ^{2}}$$

Q12: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$$

$$\sin y= \left ( \frac{1-x^{2}}{1+x^{2}} \right )$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )$$ ……..(i)

Using chain rule

$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\cos y \frac{\mathrm{d}y }{\mathrm{d} x}$$ $$\cos y=\sqrt{1-\sin ^{2}y}=\sqrt{1-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}$$ $$=\sqrt{({1+x^{2}})^{2}-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}$$ $$=\sqrt{ \frac{({1+x^{2}})^{2}-(1-x^{2})}{(1+x^{2})^{2}} ^{2}}=\sqrt{\frac{4x^{2}}{(1+x^{2})^{2}}}=\frac{2x}{1+x^{2}}$$

$$therefore,\; \frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{2x}{1+x^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}$$ …..(ii)

$$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )=\frac{(1+x^{2}).(1-x^{2})-(1-x^{2}).(1+x^{2})}{(1+x^{2})^{2}}$$ [using quotient rule]

$$=\frac{(1+x^{2}).(-2x)-(1-x^{2}).(2x)}{(1+x^{2})^{2}}$$ $$=\frac{-2x-2x^{3}-2x+2x^{3}}{(1+x^{2})^{2}}$$

$$=\frac{-4x}{(1+x^{2})^{2}}$$ ……..(iii)

From (i),(ii) and (iii) we have

$$\frac{2x}{(1+x^{2})}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-4x}{(1+x^{2})^{2}}$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{(1+x^{2})}$$

Q13: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$y=\cos ^{-1}\left ( \frac{2x}{1+x^{2}} \right ), -1<x<1$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$y=\cos ^{-1}\left ( \frac{2x}{1+x^{2}} \right )$$

$$\cos y= \left ( \frac{2x}{1+x^{2}} \right )$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )$$ $$-\sin y \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{(1+x^{2}.\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}$$ $$\Rightarrow -\sqrt{1-\cos^{2}y}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{(1+x^{2}.2-2x.2x)}{(1+x^{2})^{2}}$$ $$\Rightarrow \sqrt{1-\left ( \frac{2x}{1+x^{2}} \right )^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]$$ $$\Rightarrow \sqrt{\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]$$ $$\Rightarrow \sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]$$ $$\Rightarrow \frac{(1-x^{2})}{(1+x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}}$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{1+x^{2}}$$

Q14: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$y=\sin^{-1}(2x\sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$y=\sin^{-1}(2x\sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$

$$\sin y= (2x\sqrt{1-x^{2}})$$

Differentiating the equation with respect to x, we have

$$\cos y{\mathrm{d} y}{\mathrm{d} x}=2\left [ x.\frac{\mathrm{d} }{\mathrm{d} x}\sqrt{1-x^{2}} +\sqrt{1-x^{2}}\frac{\mathrm{d} x}{\mathrm{d} x}\right ]$$ $$\sqrt{1-\sin^{2}y} \frac{\mathrm{d}y }{\mathrm{d} x}=2\left [ \frac{x}{2}.\frac{-2x}{\sqrt{1-x^{2}}} +\sqrt{1-x^{2}}\right ]$$ $$\Rightarrow \sqrt{1-(2x\sqrt{1-x^{2}})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}} \right ]$$ $$\Rightarrow \sqrt{1-4x^{2}(1-x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]$$ $$\Rightarrow \sqrt{(1-2x^{2})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sqrt{1-x^{2}}}$$

Q15: Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

$$y=\sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}$$.

Sol: Based on formulae given in Continuity and Differentiability

The given relationship is $$y=\sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}$$

$$\sec y= \left ( \frac{1}{2x^{2}-1} \right )$$ $$\Rightarrow \cos y=2x^{2}-1$$ $$\Rightarrow 2x^{2}=1+\cos y$$ $$\Rightarrow 2x^{2}=2\cos {2}\frac{y}{2}$$ $$\Rightarrow x=cos \frac{y}{2}$$

Differentiating the equation with respect to x, we have

$$\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos \frac{y}{2} \right )$$ $$\Rightarrow 1=-\sin \frac{y}{2}.\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )$$ $$\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2}\frac{\mathrm{d} y}{\mathrm{d} x}$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos^{2}\frac{y}{2}}}$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sqrt{1-x^{2}}}$$

Exercise 5.4

Q1: Differentiate the following w.r.t. x

$$\frac{e^{x}}{\sin x}$$

Sol: Based on formulae given in Continuity and Differentiability

Let y=$$\frac{e^{x}}{\sin x}$$

Using quotient rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin x.\frac{\mathrm{d} }{\mathrm{d} x}(e^{x})-e^{x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)}{\sin^{2} x}$$ $$=\frac{\sin x.(e^{x})-e^{x}.(\cos x)}{\sin^{2} x}$$ $$=\frac{e^{x}(\sin x-\cos x)}{\sin^{2} x},x\neq n\pi ,n\in \mathbb{Z}$$

Q2: Differentiate the following w.r.t. x

$$e^{\sin ^{-1}x}$$

Sol: Based on formulae given in Continuity and Differentiability

Let y=$$y=e^{\sin ^{-1}x}$$

Using chain rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{\sin ^{-1}x})$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)$$ $$=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}$$ $$=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)$$

Q3: Differentiate the following w.r.t. x

$$e^{x^{3}}$$

Sol: Based on formulae given in Continuity and Differentiability

Let y=$$e^{x^{3}}$$

Using chain rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{x^{3}})=e^{x^{3}}.3x^{2}=3x^{2}.e^{x^{3}}$$

Q4: Differentiate the following w.r.t. x

$$\sin (\tan^{-1} e^{-x})$$

Sol: Based on formulae given in Continuity and Differentiability

Let y=$$\sin (\tan^{-1} e^{-x})$$

Using chain rule, we have

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[\sin (\tan^{-1} e^{-x})]$$ $$=\cos (\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1} e^{-x})$$ $$=\cos (\tan^{-1} e^{-x}).\frac{1}{1+(e^{-x})^{2}}(\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(e^-{x})$$ $$=\frac{\cos (\tan^{-1} e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{\mathrm{d} }{\mathrm{d} x}(-x)$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)$$ $$=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}$$ $$=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}$$ $$therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)$$