**NCERT solutions of chapter 5-Continuity and Differentiability**

Class 12 maths NCERT solutions of chapter 5- Continuity and Differentiability are published here for helping class 12 students in their preparation of forthcoming examinations and CBSE board exams of 2020-21. All the questions of chapter 5 – Continuity and Differentiability are solved by an expert teacher of CBSE maths as per the norms of the CBSE board. The NCERT solutions of chapter 5-Continuity and Differentiability of class 12 maths will help you a complete understanding of the value of the function and limits at a particular point the function is defined for. You will learn in chapter 5-Continuity and Differentiability of class 12 maths textbook that a particular function is differentiable or not means the derivatives of the function exist or not at the given point.

**NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability**

**Exercise 5.1-Continuity and Differentiability**

**Exercise 5.2-Continuity and Differentiability**

**Exercise 5.3-Continuity and Differentiability**

**Exercise 5.4 – Continuity and Differentiability**

**NCERT Solutions Class 10 Science from chapter 1 to 16**

**Class 12 maths NCERT solutions of Exercise 5.1 chapter 5-Continuity and Differentiability**

**Q1. Prove that the function f(x) = 5x – 3 is continuous at x= o, x = -3 and x = 5.**

Ans. We are given the function f(x) = 5x – 3

The value of f(x) at x = 0

f(0) = 5 × 0 – 3 = -3

The continuety at x = 0

So, f(x) is continuous at x = 0

at x = -3, f(-3) = 5 × -3 – 3 = -15 -3 = -18

Therefore f(x) is continuous at x = -3

The value of f(x) at x = 5

f(0) = 5 × 5 – 3 = 22

The continuety at x = 5

Therefore f(x) is continuous at x = 5

**Q2. Examine the continuety of the function f(x) = 2x² – 1, at x = 3.**

Ans. The value of f(x) at x = 3

f(3) = 2×3² – 1 = 2 × 9 – 1 = 18 – 1 = 17

Therefore the given function is continuous at x = 3

**Q3.Examine the following function for continuty.**

Ans. (a) Given function f(x) = x – 5, we know f is defined at every real number k

The value of the function at k

f(k) = k – 5

The continuity of f(x) at k

Therefore f(x) is continuous at every real number, so f(x) is continuous

The value of f(x) for any real number k ≠ 5

The continuity of the function for a real number k ≠ 5

Therefore f(x) is continuous for any real number x ≠ 5 or of the domain of the function

The value of f(x) for any real number k ≠ -5

The continuity of the function for any real number k ≠ -5

The given function is

The given function f(x) is a modulus function

The given function is defined for all real numbers

Let k be a real number such that k< 5

Then f(x) = 5 – x

The value of f(x) at x = k

f(k) = 5 – k

The continuity of f(x) at x = k

Therefore f(x) is continuous at all real numbers less than 5.

In the case when k = 5

f(x) = x – 5

The value of f(x) at x =k

f(k) = k -5

The continuity of f(x) at x = k

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

Ans. The given function is f(x) =x^{n }

The value of f(x) at x = n

f(n) = n^{n}

The continuty of f(x) at x = n

Therefore the given function is continuous at x = n

**Q5. Is the given function f defined by**

continuous at x = 0 at x = 1 at x = 2.

Ans. The given function is

The function is defined for x< 1

f (x) = x

So, its value at x = 0 is f(0) = 0

The continuty of f(x) at x= 0

Therefore f(x) is continuous at x = 0

The function is defined for x = 1

f (x) = x

The value of f(x) at x = 1

f(1) = 1

The continuty of the function f(x) at x = 1 since f(x) can be approached from LHS and RHS of 1, so calculating LHS limit and RHS limit of the f(x) at x = 1

LHS limit of f(x) at x = 1

RHS limit of f(x) at x = 1

LHS limit ≠ RHS limit

Therefore f(x) is not continuous at x = 1

For x >1 the function is defined f(x) = 5

The value of f(x) at x = 2 is f(2) = 5

Therefore f(x) is continuous at x = 2

**Q6. Find all the points of discontinuty of f , where f is defined by**

Ans. The given function f(x) is defined as follows

It is clear that the function is valid for x ≤ 2 and x > 2, so is defined for all values of real numbers

Let there is a real number k, so there arises 3 cases as follows

First case, when k < 2

The value of f(x) at x = k

f(k) = 2k + 3

The continuty of f(x) at x = k

So, f(x) is continuous at x = k

Second case, when x = k

The value of f(x) at x = k

f(k) = 2k + 3

Checking the LHS and RHS limit of the f(x) at k = 2

LHS limit

RHS limit

LHS limit ≠ RHS

Therefore the function f(x) is discontinuous at x = 2

The third case , when k > 2

The value of f(x) at x = k

f(k) = 2k + 3

The continuty of f(x) at x = k

So, f(x) is continuous at x = k

Therefore f(x) is discontinuous at x = 2

**Q7. Find all the points of discontinuity of f, where f is defined by**

Ans. The given function f is defined as follows

As it is seen that the function f(x) is defined for all values of real numbers

Let k is a point on the real number line, then, there arise following cases

**First case when k < 3**

The value of f(x) at x = k

For k < 3

So, f(x) is continuous at all points when x < -3.

**Second case when k = -3**

The value of f(x) at k = -3

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

LHS limit of f(x) at x = -3

RHS limit of f(x) at x = -3

f(x) = LHS limit = RHS limit

So, the f(x) is continuous at x = -3

**Third case when -3 < x < 3**

Let – 3 < k < 3

The value of f(x) at x = k

f(k) = -2k

Thus

So, f(x) is continuous at -3 < x < 3

**Fourth case when k > 3**

f(x) = 6x + 2

The value of f(x) at x = k

f(k) = 6 × k + 2 = 6k + 2

The limit of the f(x) at x = k

So, f(x) is continuous at k > 3

Fifth case when k = 3

LHS limit at k = 3

RHS limit at k = 3

RHS limit ≠ LHS limit

So, f(x) is not continuous at k = 3

**Q8. Find all the points of discontinuity of f, where f is defined by**

Ans. The given function is as follows

There arises the following cases

For x ≠ 0

When x < 0, then

When x > 0

First case, let there is a point on real number line k <0

The value of f(x) at x = k

f(k) = -1

Limit of f(x) at x = k

So, f(x) is continuous for all values x < 0

Second case, when k > o

Value of f(x) at x = k

f(k) = 1

The limit of f(x) at x = k

So, f(x) is continuous for all values x > 0

Third case when x = 0

LHS limit at x = 0

RHS limit at x = 0

LHS limit ≠ RHS limit

So, f(x) is not continuous at x = 0

**Q9.Find all the points of discontinuity of f, where f is defined by**

Ans. The given function is f(x) is defined as follows

If x < 0

Let there is a point k on the real number line such that k < 0

The value of f(x) at x = k

f(k) = -1

The limit of f(x) at x = k

So, f(x) is continuous at x = k

If x ≥ 0, then f(x) = -1

Let there is a point k on real number line such that k > 0

f(k) = -1

The limit at x = k, will be = -1

So, the f(x) is continuous for all values x > 0

If x = 0, then f(x) = -1

Let there is a point k on real number line such that k = 0

f(k) =f(0) =-1

LHS limit

RHS limit

Since LHS limit = RHS limit = f(k) = -1

So, the function f(x) is continuous at x = 0

Hence the f(x) is continuous at all points ,thus f(x) is not discontinuous at any point.

**Q10.Find all the points of discontinuity of f, where f is defined by**

Ans. The given function f(x) is defined as follows

First case when x > 1

f(x) = x + 1

Let there is point on real number line such that k >1

The value of f(x) at x = k

f(k) = k + 1

The limit of f(x) at x = k

So, f(x) is continuous at all values of x > 1

When x = 1, f(x) = x + 1

The value of f(x) at x = 1

f(1) = 1 + 1 = 2

RHS limit of f(x)

LHS limit of f(x) at x = 1

So, f(x) = LHS limit = RHS limit

So, f(x) is continuous at x = 1

If x <1, then f(x) = x² + 1

Let there is a point k on real number line such that k < 1

The value of f(x) at x = k

f(k) = k² + 1

The limit of f(x) at x = k

So, the f(x) is continuous at all values of x> 1

Hence f(x) is continuous at all real values of x, so f(x) is discontinuous at any point.

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