NCERT Solutions Class 12

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NCERT Solutions Class 12

Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)

Exercise 5.1

1. Prove that the function is continuous at at and at

Ans. Given:

Continuity at = 5 (0) – 3 = 0 – 3 = – 3

And 5 (0) – 3 = 0 – 3 = – 3

Since , therefore, is continuous at .

Continuity at = 5 (– 3) – 3 = – 15 – 3 = – 18

And 5 (– 3) – 3 = – 15 – 3 = – 18

Since , therefore, is continuous at

Continuity at = 5 (5) – 3 = 25 – 3 = 2

And 5 (5) – 3 = 25 – 3 = 22

Since , therefore, is continuous at .

2. Examine the continuity of the function at

Ans. Given:

Continuity at , =

And

Since , therefore, is continuous at .

3. Examine the following functions for continuity:

(a)

(b)

(c)

(d)

Ans. (a) Given:

It is evident that is defined at every real number and its value at is

It is also observed that

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

(b) Given:

For any real number , we get

And

Since , therefore, is continuous at every point of domain of and it is a continuous function.

(c) Given:

For any real number , we get

And

Since , therefore, is continuous at every point of domain of and it is a continuous function.

(d) Given:

Domain of is real and infinite for all real

Here is a modulus function.

Since, every modulus function is continuous, therefore, is continuous in its domain R.

4. Prove that the function is continuous at where is a positive integer.

Ans. Given: where is a positive integer.

Continuity at ,

And

Since , therefore, is continuous at .

5. Is the function defined by continuous at at at

Ans. Given:

At , It is evident that is defined at 0 and its value at 0 is 0.

Then and

Therefore, is continuous at .

At , Left Hand limit of

Right Hand limit of

Here

Therefore, is not continuous at .

At , is defined at 2 and its value at 2 is 5.

, therefore,

Therefore, is not continuous at .

Find all points of discontinuity of where is defined by: (Exercise 6 to 12)

Ans. Given:

Here is defined for i.e., on and also for i.e., on

Domain of is = R

For all is a polynomial and hence continuous and for all is a continuous and hence it is also continuous on R – {2}.

Now Left Hand limit = = 2 x 2 + 3 = 4 + 3 = 7

Right Hand limit = = 2 x 2 – 3 = 4 – 3 = 1

Since

Therefore, does not exist and hence is discontinuous at (only)

Ans. Given:

Here is defined for i.e., on and for and also for i.e., on

Domain of is = R

For all is a polynomial and hence continuous and for all is a continuous and hence it is also continuous and also for all . Therefore, is continuous on R –

It is observed that and are partitioning points of domain R.

Now Left Hand limit =

Right Hand limit =

And

Therefore, is continuous at .

Again Left Hand limit =

Right Hand limit =

Since

Therefore, does not exist and hence is discontinuous at (only).

Ans. Given: i.e., if and if

if , if and if

It is clear that domain of is R as is defined for , and .

For all , is a constant function and hence continuous.

Therefore is continuous on R – {0}.

Now Left Hand limit =

Right Hand limit =

Since

Therefore, does not exist and hence is discontinuous at (only).

Ans. Given:

At L.H.L. = And

R.H.L. =

Since L.H.L. = R.H.L. =

Therefore, is a continuous function.

Now, for

Therefore, is a continuous at

Now, for

Therefore, is a continuous at

Hence, the function is continuous at all points of its domain.

10.

Ans. Given:

It is observed that being polynomial is continuous for and for all R.

Continuity at R.H.L. =

L.H.L. =

And

Since L.H.L. = R.H.L. =

Therefore, is a continuous at for all R.

Hence, has no point of discontinuity.

11.

Ans. Given:

At L.H.L. =

R.H.L. =

Since L.H.L. = R.H.L. =

Therefore, is a continuous at

Now, for and

Therefore, is a continuous for all R.

Hence the function has no point of discontinuity.

12.

Ans. Given:

At L.H.L. =

R.H.L. =

Since L.H.L. R.H.L.

Therefore, is discontinuous at

Now, for and for

Therefore, is a continuous for all R – {1}

Hence for all given function is a point of discontinuity.

13. Is the function defined by a continuous function?

Ans. Given:

At L.H.L. =

R.H.L. =

Since L.H.L. R.H.L.

Therefore, is discontinuous at

Now, for and for

Therefore, is a continuous for all R – {1}

Hence is not a continuous function.

Discuss the continuity of the function where is defined by:

14.

Ans. Given:

In the interval

is continuous in this interval.

At L.H.L. = and R.H.L. =

Since L.H.L. R.H.L.

Therefore, is discontinuous at

At L.H.L. = and R.H.L. =

Since L.H.L. R.H.L.

Therefore, is discontinuous at

Hence, is discontinuous at and

15.

Ans. Given:

At L.H.L. = and R.H.L. =

Since L.H.L. = R.H.L.

Therefore, is continuous at

At L.H.L. = and R.H.L. =

Since L.H.L. R.H.L.

Therefore, is discontinuous at

When being a polynomial function is continuous for all

When . It is being a polynomial function is continuous for all

Hence is a point of discontinuity.

16.

Ans. Given:

At L.H.L. = and R.H.L. =

Since L.H.L. = R.H.L.

Therefore, is continuous at

At L.H.L. = and R.H.L. =

Since L.H.L. = R.H.L.

Therefore, is continuous at

17. Find the relationship between and so that the function defined by is continuous at

Ans. Given:

Continuity at

Also

18. For what value of is the function defined by continuous at What about continuity at

Ans. Since is continuous at

And

Here, therefore should be L.H.L. = R.H.L.

0 = 1, which is not possible.

Again Since is continuous at

And

Here, therefore should be L.H.L. = R.H.L.

19. Show that the function defined by is discontinuous at all integral points. Here denotes the greatest integer less than or equal to

Ans. For any real number we use the symbol to denote the fractional part or decimal part of For example,

[3.45] = 0.45

[–7.25] = 0.25

[3] = 0

[–7] = 0

The function : R R defined by is called the fractional part function. It is observed that the domain of the fractional part function is the set R of all real numbers and the range of the set [0, 1).

Hence given function is discontinuous function.

20. Is the function continuous at ?

Ans. Given:

L.H.L. =

R.H.L. =

And

Since L.H.L. = R.H.L. =

Therefore, is continuous at

21. Discuss the continuity of the following functions:

(a)

(b)

(c)

Ans. (a) Let be an arbitrary real number then

Similarly, we have

Therefore, is continuous at

Since, is an arbitrary real number, therefore, is continuous.

(b) Let be an arbitrary real number then

Similarly, we have

Therefore, is continuous at

Since, is an arbitrary real number, therefore, is continuous.

(c) Let be an arbitrary real number then

Similarly, we have

Therefore, is continuous at

Since, is an arbitrary real number, therefore, is continuous.

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

Ans. (a) Let be an arbitrary real number then

= =

for all R

Therefore, is continuous at

Since, is an arbitrary real number, therefore, is continuous.

(b) and domain I

Therefore, is continuous at

Since, is an arbitrary real number, therefore, is continuous.

(c) and domain I

= =

Therefore, is continuous at

Since, is an arbitrary real number, therefore, is continuous.

(d) and domain I

= =

Therefore, is continuous at

Since, is an arbitrary real number, therefore, is continuous.

23. Find all points of discontinuity of where .

Ans. Given:

At L.H.L. =

R.H.L. =

is continuous at

When and are continuous, then is also continuous.

When is a polynomial, then is continuous.

Therefore, is continuous at any point.

24. Determine if defined by is a continuous function.

Ans. Here, = 0 x a finite quantity = 0

Also

Since, therefore, the function is continuous at

25. Examine the continuity of where is defined by .

Ans. At L.H.L. =

R.H.L. =

And

Therefore, is discontinuous at

Find the values of so that the function is continuous at the indicated point in Exercise 26 to 29.

26. at

Ans. Here,

Putting where

= =

= ……….(i)

And ……….(ii)

when [Given]

Because is continuous at

From eq. (i) and (ii),

27. at

Ans. Here,

and

Now, when , we have

Therefore, is continuous at when .

28. at

Ans. Here,

And

Also

Since the given function is continuous at

29. at

Ans. When we have which being a polynomial is continuous at each point

And, when we have which being a polynomial is continuous at each point

Now

……….(i)

…….(i)

Since function is continuous, therefore, eq. (i) = eq. (ii)

30. Find the values of and such that the function defined by is a continuous function.

Ans. For function is constant, therefore it is continuous.

For function polynomial, therefore, it is continuous.

For function is constant, therefore it is continuous.

For continuity at

……….(i)

For continuity at

……….(ii)

Solving eq. (i) and eq. (ii), we get

and

31. Show that the function defined by is a continuous function.

Ans. Let and , then

Now and being continuous it follows that their composite is continuous.

Hence is continuous function.

32. Show that the function defined by is a continuous function.

Ans. Given: ….(i)

has a real and finite value for all R.

Domain of is R.

Let and

Since and being cosine function and modulus function are continuous for all real

Now, being the composite function of two continuous functions is continuous, but not equal to

Again, [Using eq. (i)]

Therefore, being the composite function of two continuous functions is continuous.

33. Examine that is a continuous function.

Ans. Let and , then

Now, and being continuous, it follows that their composite, is continuous.

Therefore, is continuous.

34. Find all points of discontinuity of defined by

Ans. Given:

When =

When

At L.H.L. =

R.H.L. =

And

Therefore, at is continuous.

At L.H.L. =

R.H.L. =

And

Therefore, at is continuous.

Hence, there is no point of discontinuity.

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