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# NCERT Solutions class 12 Maths Exercise 5.1 Part-1

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## NCERT Solutions class 12 Continuity & Differentiability

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)

6.

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)

7.

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).

8.

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.

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).

9.

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

## NCERT Solutions class 12 Maths Exercise 5.1 Part-1

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 20 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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