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**Download NCERT solutions for Continuity and Differentiability as PDF.**

## 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 .

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

**2. Examine the continuity of the function **** at **

**Ans. **Given:

Continuity at , =

And

Since , therefore, is continuous at .

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

**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.

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

**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 .

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

**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.

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

**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

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