In this page we have *NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability* for
EXERCISE 5.1 . Hope you like them and do not forget to like , social share
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Prove that the function

$\lim_{x \rightarrow 0} f(x)=\lim_{x \rightarrow 0} (5x-3)=5(0) -3=-3$

Therefore, limit and f (0) are same. So, it is continuous

At x=-3

$\lim_{x \rightarrow -3} f(x)=\lim_{x \rightarrow -3} (5x-3)=5(-3) -3=-18$

Therefore, limit and f (-3) are same. So, it is continuous

At x=5

$\lim_{x \rightarrow 5} f(x)=\lim_{x \rightarrow 5} (5x-3)=5(5) -3=22$

Therefore limit and f(5) are same. So, it is continuous

Examine the continuity of the function

At x=3

$\lim_{x \rightarrow 3} f(x)=\lim_{x \rightarrow 3} (2x^2 -1)=2(9) -1=17$ Therefore limit and f(3) are same. So, it is continuous at x=3

Examine the following functions for continuity.

(a)

(b)

(c)

(d)

The function is clearly defined at every point and

Also,

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

Since

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

So, function is continuous at every real number

b)

The function is clearly defined at every point except x=5 and

$\lim_{x \rightarrow c} f(x)=\lim_{x \rightarrow c} \frac {1}{x-5}=\frac {1}{c-5}$

Since

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

So, function is continuous at every real number in its domain

c)

The function is clearly defined at every point except x=-5 and

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

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

So, function is continuous at every real number in its domain

This function can re-written as

f(x)=\left\{\begin{matrix} 5-x, & x <5 \\ x- 5, & x \geq 5 \end{matrix}\right. This function f is defined at all points of the real line.

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

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

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

Since

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

Function is continuous at c< 5

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

Since

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

Function is continuous at c > 5

$\lim_{x \rightarrow 5^-} f(x)=\lim_{x \rightarrow 5^-} (5 -x)=5-5=0$

$\lim_{x \rightarrow 5^+} f(x)=\lim_{x \rightarrow 5^+} (x -5)=5-5=0$

Since

Function is continuous at c =5

Prove that the function

The given function is f (x) =

It is evident that f is defined at all positive integers, n, and its value at n is n

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

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

Is the function

continuous at

At x = 0,

f(0) = 0.

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

Therefore, f is continuous at x = 0

f(1) = 1.

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

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

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

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

Therefore, function is not continuous at x=1

f(2) = 5.

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

Therefore, f is continuous at x = 2

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 2 or c = 2 or c > 2

It can be easily proved for c < 2 and c> 2. We have seen that in lot of previous example

For c=2

f(2) =2(2)+3=7

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

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

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

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

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

Therefore, f is not continuous at x = 2

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

For x≤ -3

F(x) = |x| +3

Since x is negative in that, we can re-write this as

F(x) = 3-x

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < -3 or c = -3 or -3 < c < 3 or c=3 or c> 3

It can be easily proved for c < -3 , -3 < c < 3 And c> 3 We have seen that in lot of previous example, we will look at common points

For c=-3

f(-3) =3-(-3) =6

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

$\lim_{x \rightarrow -3^-} f(x)=\lim_{x \rightarrow -3^-} (3-x)=6$

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

$\lim_{x \rightarrow -3^+} f(x)=\lim_{x \rightarrow -3^+} (-2x)=6$

Therefore, f is continuous at x = -3

For c=3

f(3) =6(3)+2 =20

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

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

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

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

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

Therefore, f is not continuous at x = 3

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

We know for x < 0 , |x|=-x and x> 0, |x|=x, So this function can be written as

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

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0

It can be easily proved for c < 0 And c> 0 We have seen that in lot of previous example, we will look at common points

For c=0

F(0) =0

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

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

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

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

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

Therefore, f is not continuous at x = 0

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

We know for x < 0 , |x|=-x So this function can be written as

F(x) =-1 if x < 0

And

F(x) =-1 if x≥0

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0

It can be easily proved for c < 0 And c> 0 . We have seen that in lot of previous example, we will look at common points

For c=0

F(0) =0

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

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

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

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

Therefore, f is continuous at x = 0

Hence, there is no point of discontinuity of f.

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 1 or c = 1 or c> 1

It can be easily proved for c < 1 And c> 1 .We have seen that in lot of previous example, we will look at common points

For c=1

f(1) =1+1=2

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

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

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

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

Therefore, f is continuous at x = 1

Hence, there is no point of discontinuity of f.

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 2 or c = 2 or c> 2

It can be easily proved for c < 2 And c> 2 .We have seen that in lot of previous example, we will look at common points

For c=2

f(1) =x

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

$\lim_{x \rightarrow 2^-} f(x)=\lim_{x \rightarrow 2^-} (x^3 -5)=5$

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

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

Therefore, f is continuous at x = 2

Hence, there is no point of discontinuity of f.

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 1 or c = 1 or c> 1

It can be easily proved for c < 1 And c> 1 . We have seen that in lot of previous example, we will look at common points

For c=1

F(1) =x

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

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

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

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

Therefore, f is discontinuous at x = 1

Hence, x = 1 is the only point of discontinuity of f.

a continuous function?

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 1 or c = 1 or c> 1

It can be easily proved for c < 1 And c> 1. We have seen that in lot of previous example, we will look at common points

For c=1

f(1) =x+5 =6

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

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

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

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

Therefore, f is discontinuous at x = 1

Hence, x = 1 is the only point of discontinuity of f.

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, 0 ≤ c < 1 or c = 1 or 1<c < 3 or c=3 or 3 <c < 10

It can be easily proved for 0 ≤ c < 1 , 1< c < 3 , 3 <c < 10 We have seen that in lot of previous example, we will look at common points

For c=1

f(1) = 3

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

$\lim_{x \rightarrow 1^-} f(x)=\lim_{x \rightarrow 1^-} (3)=4$

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

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

Therefore, f is discontinuous at x = 1

For c=3

f(3)= 5

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

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

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

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

Therefore, f is discontinuous at x = 3

Hence, f is not continuous at x = 1 and x = 3

This can be solved similarly above

Answer is f is not continuous only at x = 1

This can be solved similarly above

Answer is f is continuous at all the points

Find the relationship between

is continuous at

When the function is continuous at any point, it Left-hand limit, right-hand limit and function value are equal

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

F(3) =3a+1

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

$\lim_{x \rightarrow 3^+} f(x)=\lim_{x \rightarrow 3^+} (bx+3) = 3b+3$

So

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

Or a =b + (2/3)

For what value of λ is the function defined by

continuous at

If f is continuous at x=0, then

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

f(0) =0

$\lim_{x \rightarrow 0^-} f(x) =\lim_{x \rightarrow 0^-} \lambda(x^2 -2x) = 9$

$\lim_{x \rightarrow 0^+} f(x) =\lim_{x \rightarrow 0^+} \lambda(4x+1) = 1$

Since Left hand limit is not equal to right hand limit, Therefore, there is no value of λ for which the function f is continuous at x=0.

At x=1,

f(1)=4x+1=4×1+1=5

$\lim_{x \rightarrow 1} f(x) =\lim_{x \rightarrow 1}(4x+1) = 5$

Therefore, for any values of λ, f is continuous at x=1.

Show that the function defined by

This function 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 \rightarrow n^-} f(x) =\lim_{x \rightarrow n^-}(x -[x]) =\lim_{x \rightarrow n^-} (x) - \lim_{x \rightarrow n^-} [x]=n -(n-1)=1 $

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

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

So, Left hand limit and Right hand limit are not equal

Therefore, f is not continuous at x=n

Hence, g is discontinuous at all integral points.

Is the function defined by

At x=π, f(x)=f(π)= π

Consider

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

Put x=π+h

If x→π, then it is evident that h→0

$\lim_{x \rightarrow \pi} f(x)= \lim_{x \rightarrow \pi} (x^2 - sin x + 5)=\lim_{h \rightarrow 0}[(\pi + h)^2 -sin (\pi + h) + 5] $

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

$=\lim_{h \rightarrow 0}(\pi + h)^2 - \lim_{h \rightarrow 0} (sin \pi cos h + cos \pi sin h) + \lim_{h \rightarrow 0} 5$

$=\pi^2 + 5$

So, function is continuous at

Discuss the continuity of the following functions:

(a)

(b)

(c)

It is known that if g and h are two continuous functions, then g+ h , g − h ,g.h are also continuous functions

.So, It has to prove 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.

g(c) =sin(c)

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

If x → c, then h → 0

$\lim_{x \rightarrow c} g(x) =\lim_{x \rightarrow c} sinx$

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

$=\lim_{h \rightarrow 0} sin c cos h + \lim_{h \rightarrow 0} cos c sin h= sin c + 0 =sin c$

So, g(x) is continuous function

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

h(c) =cos(c)

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

If x → c, then h → 0

$\lim_{x \rightarrow c} h(x) =\lim_{x \rightarrow c} cos x$

$=\lim_{h \rightarrow 0} cos(c+h)=\lim_{h \rightarrow 0} [cos c cos h - sin c sin h]$

$=\lim_{h \rightarrow 0} cos c cos h + \lim_{h \rightarrow 0} sin c sin h= cos c + 0 =cos c$

So, h(x) is a continuous function

Therefore, it can be concluded that

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

(b) f (x) = g (x) − h (x) = sin x − cos x is a continuous function

(c) f (x) = g (x) × h (x) = sin x × cos x is a continuous function

Now these functions can be rewritten as

Cosec (x) = 1/ sin (x) sin(x) ≠0

Sec (x) = 1/cos (x) cos (x) ≠0

Cot(x) = cos(x)/sin(x) sin(x) ≠0

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

(i)$\frac {h(x)}{g(x)}$ , $g(x) \ne 0$ is continous

(ii)$\frac {1}{g(x)}$ , $g(x) \ne 0$ is continous

(iii)$\frac {1}{h(x)}$ , $h(x) \ne 0$ is continous

Comparing to this, we have h(x) = cos (x) and g(x) =sin(x)

We have already proved the previous question about continuity of these function

So, we can say that

Cosec (x) is continuous at all the point except where sin(x) ≠0

i.e x ≠nπ where n ∈ Z

sec (x) is continuous at all the point except where cos(x) ≠0

i.e x ≠(2n+1)π/2 where n ∈ Z

Cot(x) is continuous at all the point except where sin(x) ≠0

i.e x ≠nπ where n ∈ Z

Find all points of discontinuity of

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

For continuity in these type of question, we should look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0

It can be easily proved for c < 0 And c> 0. We have seen that in lot of previous example, we will look at common points

f(0) = 0+1=1

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

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

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

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

Therefore, f is continuous at x = 1

Hence, There is no point of discontinuity of f.

Determine if

is a continuous function?

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0

It can be easily proved for c < 0 And c> 0. We have seen that in lot of previous example, we will look at common points

f(0) = 0

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

$\lim_{x \rightarrow 0^-} f(x) =\lim_{x \rightarrow 0^-} \frac x^2 $ Now we know that, for x≠0

$-1 \leq sin \frac {1}{x} \leq 1$

Multiplying by x

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

$\lim_{x \rightarrow 0^-} -x^2 \leq \lim_{x \rightarrow 0^-} x^2sin \frac {1}{x} \leq \lim_{x \rightarrow 0^-} x^2$

Or

$0 \leq \lim_{x \rightarrow 0^-} x^2sin \frac {1}{x} \leq 0$

So

$\lim_{x \rightarrow 0^-} x^2sin \frac {1}{x}=0$

Similarly The right-hand limit of f at x = 0 can be proved

$\lim_{x \rightarrow 0^+} x^2sin \frac {1}{x}=0$

Therefore, f is continuous at x = 0

Hence, There is no point of discontinuity of f.

Examine the continuity of

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

For continuity in these type of question, we must look for the continuity at the common point and point outside the common point

Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0

It can be easily proved for c < 0 And c> 0. We have seen that in lot of previous example, we will look at common points

At c=0

f(0)=-1

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

$\lim_{x \rightarrow 0^-} f(x) =\lim_{x \rightarrow 0^-} sin x -cos x = -1$

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

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

Therefore, f is continuous at x = 0

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

Thus, f is a continuous function.

For the function to be continuous at x=π/2, we have

$f(\pi/2) = \lim_{x \rightarrow /pi/2} f(x)$

F(π/2) =3

$\lim_{x \rightarrow /pi/2} f(x)=\lim_{x \rightarrow /pi/2} \frac {kcosx}{\pi -2x}$

Let x = π/2 + h

When x → π/2, then h → 0

$\lim_{x \rightarrow /pi/2} f(x)= \lim_{h \rightarrow 0}\frac {kcos(\pi/2 + h}{\pi -2(\pi/2 +h)}$

$=\lim_{h \rightarrow 0} \frac{-k sinh}{-2h}= \frac {k}{2} \lim_{h \rightarrow 0} \frac {sin h}{h}= \frac {k}{2}$

Now k/2=3

K=6

For the function to be continuous at x=2, we have

$\lim_{x \rightarrow 2} f(x) = f(2)$ F(2) =4k

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

$\lim_{x \rightarrow 2^-} f(x)=4k$

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

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

Now

4k=4k=3

Or

K=3/4

For the function to be continuous at x=π, we have

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

F(π) =k π+1

The left-hand limit of f at x = π is,

$\lim_{x \rightarrow \pi^-} f(x)=\lim_{x \rightarrow \pi^-} (kx+1)=\pi k + 1$

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

$\lim_{x \rightarrow \pi^+} f(x)=\lim_{x \rightarrow \pi^+} (cos x)=-1$

Now

k π+1=-1

k=-2/ π

For the function to be continuous at x=5, we have

$\lim_{x \rightarrow 5} f(x)= f(5)$ F(5) =5k+1

The left-hand limit of f at x = 5 is,

$\lim_{x \rightarrow 5^-} f(x)=\lim_{x \rightarrow 5^-} (kx+1)=5 k + 1$

The right-hand limit of f at x = 5 is,

$\lim_{x \rightarrow 5^+} f(x)=\lim_{x \rightarrow 5^+} (3x-5)=10$

Now

5k+1=10

k=9/5

Find the values of

is a continuous function.

The function is defined for all the real values and it is given as continuous

It is also continuous at x=2 and x=10

Let’s check the continuity at those values

At x=2

f(2)=5

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

$\lim_{x \rightarrow 2^-} f(x)=\lim_{x \rightarrow 2^-} 5=5$

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

$\lim_{x \rightarrow 2^+} f(x)=\lim_{x \rightarrow 2^+} (ax+b)=2a + b$

Now

2a+b=5 -(P)

At x=10

F(10)=21

The left-hand limit of f at x = 10 is,

$\lim_{x \rightarrow 10^-} f(x)=\lim_{x \rightarrow 10^-} (ax+b)=10a + b$

The right-hand limit of f at x = 10 is,

$\lim_{x \rightarrow 10^+} f(x)=\lim_{x \rightarrow 10^+} 21=21$

Now

10a+b=21 --(Q)

From (p) and (q)

a=2 and b=1

Show that the function defined by

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

f = g o h, where g (x) = cos x and h (x) = x

Because g o h =g[h(x)] =cos (x

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

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

So, we just need to prove that g(x) =cos(x) and h(x) =x

First let us prove for g(x)=cos (x)

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

g(c) =cos(c)

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

If x → c, then h → 0

$\lim_{x \rightarrow c} g(x) =\lim_{x \rightarrow c} cos x$

$=\lim_{h \rightarrow 0} cos(c+h)=\lim_{h \rightarrow 0} [cos c cos h - sin c sin h]$

$=\lim_{h \rightarrow 0} cos c cos h + \lim_{h \rightarrow 0} sin c sin h= cos c + 0 =cos c$

So, g(x) is continuous

Now h (x) = x

It is evident that h (x) = x

h(c) =c

$\lim_{x \rightarrow c} h(x) =\lim_{x \rightarrow c} x^2=c^2$

So, h(x) is continuous

Since both g(x) and h(x) are continuous. F(x) is continuous function

Show that the function defined by

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

of two functions, as,

f = g o h, where g (x) = |x| and h (x) = cos x

Because g o h =g[h(x)] =| cos

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

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

So, we just need to prove that g(x) =|x| and h(x) =cos(x) are continuous function

First let us prove for g(x)=|x|

It can be re-written as

g(x) = -x for x < 0

=+x for x ≥0

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

For c < 0

g(c) =-c

$\lim_{x \rightarrow c} g(x) =\lim_{x \rightarrow c} -x=-c$

So, it is continuous

For c = 0

g(c) =0

$\lim_{x \rightarrow 0^-} g(x) =\lim_{x \rightarrow 0^-} -x=0$

$\lim_{x \rightarrow 0^+} g(x) =\lim_{x \rightarrow 0^+} x=0$

So, it is continuous

For c > 0

g(c) =c

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

So, it is continuous

let us prove for h(x)=cos (x)

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

h(c) =cos(c)

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

If x → c, then h → 0

$\lim_{x \rightarrow c} h(x) =\lim_{x \rightarrow c} cos x$

$=\lim_{h \rightarrow 0} cos(c+h)=\lim_{h \rightarrow 0} [cos c cos h - sin c sin h]$

$=\lim_{h \rightarrow 0} cos c cos h + \lim_{h \rightarrow 0} sin c sin h= cos c + 0 =cos c$

So, h(x) is continuous

Since both g(x) and h(x) are continuous. F(x) is continuous function

Examine that sin |

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

of two functions, as,

f = g o h, where g (x) = sin(x) and h (x) = |x|

Because g o h =g[h(x)] = sin |

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

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

So, we just need to prove that g(x) =sin (x) and h(x) =|x| are continuous function

Let g (x) = sin x

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

g(c) =sin(c)

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

If x → c, then h → 0

$\lim_{x \rightarrow c} g(x) =\lim_{x \rightarrow c} sinx$

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

$=\lim_{h \rightarrow 0} sin c cos h + \lim_{h \rightarrow 0} cos c sin h= sin c + 0 =sin c$

So, g(x) is continuous function

Now let us prove for h(x)=|x|

It can be re-written as

h(x) = -x for x < 0

=+x for x ≥0

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

For c < 0

h(c) =-c

$\lim_{x \rightarrow c} h(x) =\lim_{x \rightarrow c} -x=-c$

So, it is continuous

For c = 0

h(c) =0

$\lim_{x \rightarrow 0^-} h(x) =\lim_{x \rightarrow 0^-} -x=0$

$\lim_{x \rightarrow 0^+} h(x) =\lim_{x \rightarrow 0^+} x=0$

So, it is continuous

For c > 0

h(c) =c

$\lim_{x \rightarrow c} h(x) =\lim_{x \rightarrow c} x=-c$

So, it is continuous

So h(x) is continous

Since both g(x) and h(x) are continuous. F(x) is continuous function

Find all the points of discontinuity of

The two functions, g and h, are defined as

g(x) =|x|

h(x) =|x+1|

Then, f = g – h

It is known that if g and h are two continuous functions, then g + h,g − h,g.h are also continuous functions.

So, we just need to prove that g(x) and h(x) is continuous

The continuity of g and h is examined first.

First let us prove for g(x)=|x|

It can be re-written as

g(x) = -x for x < 0

=+x for x ≥0

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

For c < 0

g(c) =-c

$\lim_{x \rightarrow c} g(x) =\lim_{x \rightarrow c} -x=-c$

So, it is continuous

For c = 0

g(c) =0

$\lim_{x \rightarrow 0^-} g(x) =\lim_{x \rightarrow 0^-} -x=0$

$\lim_{x \rightarrow 0^+} g(x) =\lim_{x \rightarrow 0^+} x=0$

So, it is continuous

For c > 0

g(c) =c

$\lim_{x \rightarrow c} g(x) =\lim_{x \rightarrow c} x=-c$

So, it is continuous

Now let us look

h(x)= |x+1|

It can be re-written as

h(x) = -(x+1) for x < -1

=x+1 for x ≥-1

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

For c < -1

h(c) =-(c+1)

$\lim_{x \rightarrow c} h(x) =\lim_{x \rightarrow c} -(x+1)=-(c+1)$

So, it is continuous

For c = -1

h(-1) =0

$\lim_{x \rightarrow -1^-} h(x) =\lim_{x \rightarrow -1^-} -(x+1)=0$

$\lim_{x \rightarrow -1^+} h(x) =\lim_{x \rightarrow -1^+} (x+1)=0$

So it is continuous

For c > -1

h(c) =c+1

$\lim_{x \rightarrow c} h(x) =\lim_{x \rightarrow c} (x+1)=(c+1)$

So, it is continuous

So, both g(x) and h(x) are continuous

Therefore, f = g − h is also a continuous function.

Therefore, f has no point of discontinuity.

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**Notes**-
**NCERT Solutions & Assignments**- NCERT Solution for Continuity And differentiability Exercise 5.1
- NCERT Solution for Continuity And differentiability Exercise 5.2
- NCERT Solution for Continuity And differentiability Exercise 5.3
- NCERT Solution for Continuity And differentiability Exercise 5.4
- Continuity and Differentiability class 12 Important questions

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