# NCERT Solutions For Class 12 Maths Chapter 5

## NCERT Solutions Class 12 Maths Continuity and Differentiability

### Ncert Solutions For Class 12 Maths Chapter 5 PDF Free Download

Class 12 is an important phase of a student’s life because after class 12th most of the students go for higher education and few opt for a job. To crack these competitive examinations and to score good marks in class 12 one must solve the NCERT class 12th maths questions. Solving these questions will help you to understand the chapter in a better way. This chapter is one of the most important chapters given in the syllabus of mathematics and questions are expected in various examinations from this chapter. This chapter will prove to be helpful and will act as the fundamental pillar in building greater concepts on differentiability and continuity in higher grades such as engineering etc. There are various questions given in this chapter for getting clarity in this subject.

### NCERT Solutions Class 12 Maths Chapter 5 Exercises

We will be seeing questions on  NCERT solutions for class 12 maths chapter 5 which contains questions on proving an equation is continuous if given with different values of ‘x’. We will be examining certain functions of continuity and we will be generating conclusions such is one number is a real number etc. You will come across various explanation type of questions where you will providing the reasons if one quantity is replaced with different values.

You will be finding the points of discontinuity for a certain variable when it is defined by a function. You will be discussing the continuity of various trigonometric functions and you will be differentiating functions with respect to x. You will solve the question such as find,

dy/dx , if 2x+3y=siny. These type of questions will ensure you get a good grasp of this subject and you will be scoring good marks in the exam if you are practicing these questions well.

Exercise 5.1

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

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

Sol:

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:

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 $x=3$

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:

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

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

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

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

Sol:

Given function  $f(x)=x^{n}$

It is obvious that 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.

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:

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:

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:

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:

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

It is known 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 rewritten 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,

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

It is observed that 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 $f$.

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:

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

It is known that, $x<0\Rightarrow \begin{vmatrix} x \end{vmatrix}=-x$

Therefore, the given function can be rewritten 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: The 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: The 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:

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 coincide.

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, it can be concluded 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:

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:

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,

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

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

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:

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,

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

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 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:

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:

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:

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:

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:

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:

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:

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 $g(x)=\sin x$

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

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

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

$g(x)=\sin x$

$\lim_{x \to c}g(x) = \lim_{x \to c}\sin x\\ =\lim_{h \to 0}\sin (c+h)\\ =\lim_{h \to 0}[\sin c\cos h+\cos c\sin h]\\ = \sin c\cos 0+\cos c\sin 0\\ =\sin c+0 \\ =\sin c$

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

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,

$\csc x-\frac{1}{sinx}, \sin x\neq 0 is continuous$

$\Rightarrow \csc x, x\neq n\pi (n\in \mathbb{Z})$

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

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

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 $x=0$

From the above observation, 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, $\lim_{x \to 0^{+}}f(x)= \lim_{x \to 0^{+}}x^{2}\sin \frac{1}{x} =0$

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

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

From the above observation, 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 $x=0$

From the above observation, 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:

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.

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:

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:

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:

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:

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:

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:

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 $x=0$

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

$h(x)=\cos x$

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 $x\: \tilde{A}\: c\: \hat{a}\: \in \; ‘c$, 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.

It is known 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:

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.

It is known 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:

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

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

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 $x=-1$

From the above three observation, 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,

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

And 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:

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:

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:

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:

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 written in the form of two composite function $p and q$.

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

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

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:

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:

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:

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

It is 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:

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:

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:

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:

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:

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:

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:

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, we obtain

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

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

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:

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:

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:

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:

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:

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:

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:

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:

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:

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

Using chain rule, we have

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[\sin (\tan^{-1} e^{-x})]$

$=\cos (\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1} e^{-x})$

$=\cos (\tan^{-1} e^{-x}).\frac{1}{1+(e^{-x})^{2}}(\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(e^-{x})$

$=\frac{\cos (\tan^{-1} e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{\mathrm{d} }{\mathrm{d} x}(-x)$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)$

$=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}$

$=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}$

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

Class 12 maths chapter 5 NCERT solutions is provided here so that students can have a look at these solutions whenever they are facing any difficulties. Along with NCERT solutions one must solve the previous year questions and sample papers, these question will help you to know about the types of question asked in the examination.