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# 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

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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

$\boldsymbol{\lim_{x\rightarrow 0}f\left ( x \right )=\lim_{x\rightarrow 0}\left ( 5x-3 \right )=5\times 0-3=-3}$

$\boldsymbol{\therefore \lim_{x\rightarrow 0}f\left ( x \right )=f\left ( 0 \right )}$

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

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

$\boldsymbol{\lim_{x\rightarrow -3}f\left ( x \right )=\lim_{x\rightarrow -3}\left ( 5x -3 \right )= 5\times-3-3=-15-3=-18 }$

$\boldsymbol{\therefore \lim_{x\rightarrow -3}f\left ( x \right ) =f\left ( -3 \right )}$

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

$\boldsymbol{ \lim_{x\rightarrow 5}f\left ( x \right ) = \lim_{x\rightarrow 5}\left ( 5x-3 \right )=5\times 5-3=25-3=22}$

$\boldsymbol{\therefore \lim_{x\rightarrow 5}f\left ( x \right ) = x\left ( 5 \right )}$

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

$\boldsymbol{ \lim_{x\rightarrow 3}f\left ( x \right ) = \lim_{x\rightarrow 3}\left ( 2x^{2}-1 \right )=2\times 3^{2}-1=18-1=17}$

$\boldsymbol{ \therefore \lim_{x\rightarrow 3}f\left ( x \right ) = f\left ( 3 \right )}$

Therefore the given function is continuous at x = 3

Q3.Examine the following function for continuty.

$\boldsymbol{\left ( a \right )\: f\left ( x \right )=\left ( x-5 \right )\: \: \: \: \left ( b \right )f\left ( x \right )=\frac{1}{x-5},x\neq 5}$

$\boldsymbol{\left ( c \right )f\left ( x \right )=\frac{x^{2}-25}{x+5}, x\neq -5\: \: \left ( d \right )f\left ( x \right )=\left | x -5 \right |}$

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

$\boldsymbol{\lim_{x\rightarrow 5}f\left ( x \right )=\lim_{x\rightarrow 5}\left ( x-5 \right )=k-5=f\left ( k \right )}$

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

$\boldsymbol{The\: given \: function \: is\: f\left ( x \right )=\frac{1}{x-5}, x\neq 5}$

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

$\boldsymbol{f\left ( k \right )=\frac{1}{k-5}}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( \frac{1}{x-5} \right )=\frac{1}{k-5}}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

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

$\boldsymbol{\left ( c \right )The \: given \: function\: is\: f\left ( x \right )=\frac{x^{2}-25}{x+5},x\neq -5}$

The value of f(x) for any real number k ≠ -5
$\boldsymbol{ f\left ( k \right )=\frac{k^{2}-25}{k+5}=\frac{k^{2}-5^{2}}{k+5}=\frac{\left ( k+5 \right )\left ( k-5 \right )}{k+5}=k-5}$
The continuity of the function  for any real number k ≠ -5
$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( \frac{x^{2}-25}{x+5} \right )=\lim_{x\rightarrow k}\frac{x^{2}-5^{2}}{x+5}=\lim_{x\rightarrow k}\frac{\left ( x+5 \right )\left ( x-5 \right )}{x+5}=\lim_{x\rightarrow k}\left ( x-5 \right )}$
$\boldsymbol{=k-5=f\left ( k \right )}$

$\boldsymbol{\left ( d \right )\: f\left ( x \right )=\left | x-5 \right |}$

The given function is    $\boldsymbol{f\left ( x \right )=\left | x-5 \right |}$

The given function f(x) is a modulus function

$\boldsymbol{f\left ( x \right )=\left | x-5 \right |=\left\{\begin{matrix}5-x \: if\:x < 5 & \\ x-5 \: if \: x\geq 5& \end{matrix}\right.}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( 5-x \right )=5-k}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( x-5 \right )=k-5}$

Q4.Prove that the f(x) =xn is continuous at x = n where n is a positive integer.

Ans. The given function is f(x) =x

The value of f(x) at x = n

f(n) = nn

The continuty of f(x) at x = n

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

$\boldsymbol{\lim_{x\rightarrow n}f\left ( x \right )=f\left ( x \right )}$

Therefore the given function is continuous at x = n

Q5. Is the given function f defined by

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x, if\: x\leq 1 & \\ 5,if\: x>1& \end{matrix}\right.}$

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

Ans. The given function is

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x, if\: x\leq 1 & \\ 5,if\: x>1& \end{matrix}\right.}$

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

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

$\boldsymbol{\therefore \lim_{x\rightarrow 0}f\left ( x \right )=f\left ( x \right )}$

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

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

RHS limit of f(x) at x = 1

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

$\boldsymbol{ \lim_{x\rightarrow 1^{-}}f\left ( x \right )\neq \lim_{x\rightarrow 1^{+}}f\left ( x \right )}$

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

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

$\boldsymbol{\lim_{x\rightarrow 2}f\left ( x \right )=f\left ( x \right )}$

Therefore f(x) is  continuous at x = 2

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}2x+3 \: if\: x\leq 2 & & \\ 2x-3 \: if \: x> 2 & & \end{matrix}\right.}$

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}2x+3 \: if\: x\leq 2 & & \\ 2x-3 \: if \: x> 2 & & \end{matrix}\right.}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( 2x+3 \right )=2k+3}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

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

$\boldsymbol{ \: \lim_{x\rightarrow 2^{-}}f\left ( x \right )=\lim_{x\rightarrow 2^{-}}\left ( 2x+3 \right )=2\times 2+3=7}$

RHS limit

$\boldsymbol{ \: \lim_{x\rightarrow 2^{+}}f\left ( x \right )=\lim_{x\rightarrow 2^{+}}\left ( 2x-3 \right )=2\times 2-3=1}$

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

$\boldsymbol{ \: \lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( 2x-3 \right )=2\times k-3=2k-3}$

$\boldsymbol{ \therefore \: \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\left | x \right |+3,\: if\: x\leq -3 & \\-2x,if -3< x< 3 \\6x+2,\: if\: x\geq 3 & \end{matrix}\right.}$

Ans. The given function f is defined as follows

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\left | x \right |+3,\: if\: x\leq -3 & \\-2x,if -3< x< 3 \\6x+2,\: if\: x\geq 3 & \end{matrix}\right.}$

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

$\boldsymbol{f\left ( k \right )= \left | k \right |+1}$

For k < 3

$\boldsymbol{f\left ( k \right )=\left | k \right |+3=-k+3}$

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( \left | x \right |+3 \right )=\lim_{x\rightarrow k}-x+3=-k+3}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

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

Second case when k  = -3

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

$\boldsymbol{f\left ( k \right )=\left | k \right |+3=-k+3}$

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

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

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

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

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

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}-2x=-2k}$

Thus

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

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

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

So, f(x) is continuous at  k > 3

Fifth case when  k = 3

LHS limit at k = 3

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

RHS limit at k = 3

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

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{\left | x \right |}{x}\: if \: x\neq 0 & \\ & \\ 0\: if\: x=0& \end{matrix}\right.}$

Ans. The given function is as follows

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{\left | x \right |}{x}\: if \: x\neq 0 & \\ & \\ 0\: if\: x=0& \end{matrix}\right.}$

There arises the following cases

For x ≠ 0

$\boldsymbol{f\left ( x \right )=\frac{\left | x \right |}{x}}$

When x < 0, then    $\boldsymbol{\left | x \right |=-x}$

$\boldsymbol{f\left ( x \right )=\frac{\left | x \right |}{x}=\frac{-x}{x}=-1}$

When x > 0

$\boldsymbol{f\left ( x \right )=\frac{\left | x \right |}{x}=\frac{x}{x}=1}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}-1=-1}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}1=1}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

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

Third case when x = 0

LHS limit at x = 0

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

RHS limit at x = 0

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

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{x}{\left | x \right |},\: if\: x<0 & & \\ -1,if\: x\geq 0& & \end{matrix}\right.}$

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{x}{\left | x \right |},\: if\: x<0 & & \\ -1,if\: x\geq 0& & \end{matrix}\right.}$

If x < 0

$\boldsymbol{f\left ( x \right )=\frac{x}{\left | x \right |}=\frac{x}{-x}=-1}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}-1=-1}$

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

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

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

RHS limit

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

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x+1\: if\: x\geq 1 & \\ x^{2}+1,if\: x<1& \end{matrix}\right.}$

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

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x+1\: if\: x\geq 1 & \\ x^{2}+1,if\: x<1& \end{matrix}\right.}$

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

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}x+1=k+1}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

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)

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

LHS limit of f(x) at x = 1

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

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

$\boldsymbol{ \lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}x^{2}+1=k^{2}+1}$

$\boldsymbol{ \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

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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