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

Continuity and Differentiability are important foundational steps for advanced calculus. Continuity of a function means that the function's graph can be drawn without any break or the graph can be drawn without lifting the pen. Differentiability means the function derivative exists at every point of the given interval, or we can also define differentiability if there is only a tangent to the given point in an interval. In this article, we will see many continuous and differentiable functions, and we will learn many counter-examples.

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1. NCERT Class 12 Maths Chapter 5- PDF Free Download
2. Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae
3. NCERT Continuity and Differentiability Class 12 Questions And Answers (Exercise)
4. NCERT solutions for class 12 maths - Chapter-wise
5. Importance of solving NCERT Class 12 Maths Chapter 5

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These NCERT solutions are created by a team of experts at Careers360. These experts also prepared the Class 12 Maths Chapter 5 continuity and differentiability Notes to clarify the concepts. These solutions for class 12 are designed according to the latest curriculum designed by the CBSE, ensuring that students grasp the concepts efficiently and easily. For a better understanding, you can practice NCERT Exemplar Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability to enhance your knowledge.

NCERT Class 12 Maths Chapter 5- PDF Free Download

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Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae

Continuity: A function $f(x)$ is continuous at a point $x=a$ if:

• $f(a)$ exists (finite, definite, and real).
• $\underset{x\to a}{lim}f(x)$ exists.
• $\underset{x\to a}{lim}f(x)=f(a)$.
• $\underset{x\to 0}{lim}f(x)=f(0)$
• $\underset{x\to a}{lim}f(x)$ exists

Discontinuity: $f(x)$ is discontinuous in an interval if it is discontinuous at any point in that interval.

Algebra of Continuous Functions:

The sum, difference, product, and quotient of continuous functions are continuous.

Differentiation:

The derivative of $f(x)$ at $x=a$, denoted as $f^{\prime }(a)$, represents the slope of the tangent line to the graph.

Chain Rule:

If $f=vou$, where $t=u(x)$, and if both $\frac{dt}{dx}$ and $\frac{dv}{dx}$ exist, then: $\frac{df}{dx}=\frac{dv}{dt}\cdot \frac{dt}{dx}$.

Derivatives of Some Standard Functions:

• $\frac{d}{dt}(x^n)=n{x}^{n-1}$
• $\frac{d}{dt}(\sin x)=\cos x$
• $\frac{d}{dt}(\cos x)=-\sin x$
• $\frac{d}{dt}(\tan x)={\sec }^{2}x$
• $\frac{d}{dt}(\cot x)=-{\csc }^{2}x$
• $\frac{d}{dt}(\sec x)=\sec x\cdot \tan x$
• $\frac{d}{dt}(\csc x)=-\csc x\cdot \cot x$
• $\frac{d}{dt}(a^x)=a^x\cdot ln(a)$
• $\frac{d}{dt}(e^x)=e^x$
• $\frac{d}{dt}(lnx)=\frac{1}{x}$

Mean Value Theorem:

The mean Value Theorem states that if $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists some $c$ in $(a,b)$ such that: $f^{\prime }(c)=\frac{(f(b)-f(a))}{(b-a)}$.

Rolle's Theorem:

Rolle's Theorem states that if $f(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists some $c$ in $(a,b)$ such that $f^{\prime }(c)=0$.

Lagrange's Mean Value Theorem:

Lagrange's Mean Value Theorem states that if $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists some c in $(a,b)$ such that:

$f^{\prime }(c)=\frac{(f(b)-f(a))}{(b-a)}$.

NCERT Continuity and Differentiability Class 12 Questions And Answers (Exercise)

Question 1. Prove that the function $f(x)=5x-3$ is continuous at $x=0,atx=-3$ and at $x=5$

Answer:

Given function is
$f(x)=5x-3$
$f(0)=5(0)-3=-3$
$\underset{x\to 0}{lim}f(x)=5(0)-3=-3$
$\underset{x\to 0}{lim}f(x)=f(0)$
Hence, the function is continuous at x = 0
$f(-3)=5(-3)-3=-15-3=-18\Rightarrow \underset{x\to -3}{lim}f(x)=5(-3)-3=-15-3=-18$

$\Rightarrow \underset{x\to -3}{lim}f(x)=f(-3)$
Hence, the function is continuous at $x=-3$
$f(5)=5(5)-3=25-3=22\Rightarrow \underset{x\to 5}{lim}f(x)=5(5)-3=25-3=-22$

$\Rightarrow \underset{x\to 5}{lim}f(x)=f(5)$
Hence, the function is continuous at $x=5$

Question 2. Examine the continuity of the function $f(x)=2x^2-1atx=3.$

Answer:

Given function is
$f(x)=2x^2-1$
at $x=3$
$f(3)=2(3{)}^2-1=2\times 9-1=18-1=17$

$\underset{x\to 3}{lim}f(x)=2(3{)}^2-1=2\times 9-1=18-1=17$
$\underset{x\to 3}{lim}f(x)=f(3)$
Hence, the function is continuous at $x=3$

Question 3Examine the following functions for continuity.
$(a)f(x)=x-5$

Answer:

Given function is
$f(x)=x-5$
Our function is defined for every real number, say k
and value at $x=k$ , $f(k)=k-5$
and also,
$\underset{x\to k}{lim}f(x)=k-5$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the function $f(x)=x-5$ is continuous at every real number

Question 3 b) Examine the following functions for continuity.

$f(x)=\frac{1}{x-5},x\ne 5$

Answer:

Given function is
$f(x)=\frac{1}{x-5}$
For every real number k, $k\ne 5$
We get,
$f(k)=\frac{1}{k-5}$

$\underset{x\to k}{lim}f(x)=\frac{1}{k-5}$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, function $f(x)=\frac{1}{x-5}$ continuous for every real value of $x$, $x\ne 5$

Question 3 c) Examine the following functions for continuity.

$f(x)=\frac{x^2-25}{x+5},x\ne -5$

Answer:

Given function is
$f(x)=\frac{x^2-25}{x+5}$
For every real number k, $k\ne -5$
We get,
$f(k)=\frac{k^2-5^2}{k+5}=\frac{(k+5)(k-5)}{k+5}=k-5$

$\underset{x\to k}{lim}f(x)=\frac{k^2-5^2}{k+5}$

$=\frac{(k+5)(k-5)}{k+5}=k-5$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, function $f(x)=\frac{x^2-25}{x+5}$ continuous for every real value of x , $x\ne -5$

Question 3 d) Examine the following functions for continuity. $f(x)=|x-5|$

Answer:

Given function is
$f(x)=|x-5|$
for $x>5,f(x)=x–5$
for $x<5,f(x)=5–x$
So, there are different cases.
case(i) $x>5$
for every real number $k>5$ , $f(x)=x–5$ is defined
$f(k)=k-5$

$\underset{x\to k}{lim}f(x)=k-5$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, function $f(x)=x–5$ is continuous for $x>5$
case (ii) $x<5$
for every real number $k<5$ , $f(x)=5–x$ is defined
$$\begin{gathered} f(k)=5-k \\ \underset{x\to k}{lim}f(x)=5-k \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, function $f(x)=5–x$ is continuous for $x<5$
case(iii) $x=5$
for $x=5$ , $f(x)=x–5$ is defined
$$\begin{gathered} f(5)=5-5=0 \\ \underset{x\to 5}{lim}f(x)=5-5=0 \\ \underset{x\to 5}{lim}f(x)=f(5) \end{gathered}$$
Hence, function $f(x)=x–5$ is continous for $x=5$

Hence, the function $f(x)=|x-5|$ is continuous for every real number.

Question 4. Prove that the function $f(x)=x^n$ is continuous at x = n, where n is a positive integer

Answer:

Given function is
$f(x)=x^n$
The function $f(x)=x^n$ is defined for all positive integer, n
$f(n)=n^n$

$\underset{x\to n}{lim}f(x)=n^n$

$\underset{x\to n}{lim}f(x)=f(n)$
Hence, the function $f(x)=x^n$ is continuous at x = n, where n is a positive integer

Question 5.Is the function f defined by
$f(x)=\{\begin{array}{cc}x,& ifx\le 1\\ 5& ifx\ge 1\end{array}$
continuous at x = 0? At x = 1? At x = 2?

Answer:

Given function is
$f(x)=\{\begin{array}{cc}x,& ifx\le 1\\ 5& ifx\ge 1\end{array}$
The function is defined at $x=0$ and its value is $0$
$f(0)=0$

$\underset{x\to 0}{lim}f(x)=f(x)=0$

$\underset{x\to 0}{lim}f(x)=f(0)$
Hence, the given function is continuous at $x=0$
Given function is defined for $x=1$
Now, for $x=1$ Right-hand limit and left-hand limit are not equal.
$$\begin{gathered} f(1)=1 \\ \underset{x\to 1^{-}}{lim}f(x)=f(x)=1 \end{gathered}$$

$\underset{x\to 1^{+}}{lim}f(x)=f(5)=5$
R.H.L $\ne$ L.H.L.
Therefore, the given function is not continuously $x=1$
Given function is defined for $x=2$ and its value at $x=2$ is $5$
$$\begin{gathered} f(2)=2 \\ \underset{x\to 2}{lim}f(x)=f(5)=5 \\ \underset{x\to 2}{lim}f(x)=f(2) \end{gathered}$$
Hence, the given function is continuous at $x=2$

Question:6.Find all points of discontinuity of f, where f is defined by

$f(x)=\{\begin{array}{cc}2x+3& ifx\le 2\\ 2x-3& ifx\ge 2\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}2x+3& ifx\le 2\\ 2x-3& ifx\ge 2\end{array}$
Given function is defined for every real number k
There are different cases for the given function.
case(i) $k>2$
$$\begin{gathered} f(k)=2k-3 \\ \underset{x\to k}{lim}f(x)=2k-3 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of $k>2$

case(ii) $k<2$
$$\begin{gathered} f(k)=2k+3 \\ \underset{x\to k}{lim}f(x)=2k+3 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of $k<2$

case(iii) $x=2$
$\underset{x\to 2^{-}}{lim}f(x)=2x+3=2\times 2+3=4+3=7$

$\underset{x\to 2^{+}}{lim}f(x)=2x-3=2\times 2-3=4-3=1$
Right hand limit at x= 2 $\ne$ Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity

Question:7.Find all points of discontinuity of f, where f is defined by

$f(x)=\{\begin{array}{ccc}|x|+3& ifx\le -3& \\ -2x& if-3<x<3& \\ 6x+2& ifx\ge 3& \end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{ccc}|x|+3& ifx\le -3& \\ -2x& if-3<x<3& \\ 6x+2& ifx\ge 3& \end{array}$
Given function is defined for every real number k
There are different cases.
case (i) $k<-3$
$$\begin{gathered} f(k)=-k+3 \\ \underset{x\to k}{lim}f(x)=-k+3 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for every value of k < -3

case(ii) $k=-3$
$$\begin{gathered} f(-3)=-(-3)+3=6 \\ \underset{x\to -3^{-}}{lim}f(x)=-k+3=-(-3)+3=6 \end{gathered}$$

$\underset{x\to -3^{+}}{lim}f(x)=-2x=-2(-3)=6$

$R.H.L.=L.H.L.=f(-3)$
Hence, the given function is continuous for x = -3

case(iii) $-3<k<3$
$$\begin{gathered} f(k)=-2k \\ \underset{x\to k}{lim}f(x)=-2k \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, for every value of k in -3 < k < 3 given function is continuous.

case(iv) $k=3$
$f(3)=6x+2=6\times 3+2=18+2=20$

$\underset{x\to 3^{-}}{lim}f(x)=-2x=-2(3)=-6$

$\underset{x\to 3^{+}}{lim}f(x)=6x+2=6\times 3+2=20$

$R.H.L.=f(3)\ne L.H.L.$
Hence, $x=3$ is the point of discontinuity

case(v) $k>3$
$f(k)=6k+2$

$\underset{x\to k}{lim}f(x)=6k+2$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for every value of k > 3

Question:8.Find all points of discontinuity of f, where f is defined by

$f(x)=\{\begin{array}{cc}\frac{|x|}{x}& ifx\ne 0\\ 0& ifx=0\end{array}$

Answer:

Given function is
$f(x)\{\begin{array}{cc}\frac{|x|}{x}& ifx\ne 0\\ 0& ifx=0\end{array}$
if x > 0 , $f(x)=\frac{x}{x}=1$
if x < 0 , $f(x)=\frac{-(x)}{x}=-1$
Given function is defined for every real number k
Now,
case(i) k < 0
$$\begin{gathered} f(k)=-1 \\ \underset{x\to k}{lim}f(x)=-1 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for every value of k < 0
case(ii) k > 0
$$\begin{gathered} f(k)=1 \\ \underset{x\to k}{lim}f(x)=1 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for every value of k > 0
case(iii) x = 0
$$\begin{gathered} f(0)=0 \\ \underset{x\to 0^{-}}{lim}f(x)=-1 \end{gathered}$$

$$\begin{gathered} \underset{x\to 0^{+}}{lim}f(x)=1 \\ f(0)\ne R.H.L.\ne L.H.L. \end{gathered}$$
Hence, 0 is the only point of discontinuity

Question:9.Find all points of discontinuity of f, where f is defined by

$f(x)=\{\begin{array}{cc}\frac{x}{|x|}& ifx<0\\ -1& ifx\ge 0\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}\frac{x}{|x|}& ifx<0\\ -1& ifx\ge 0\end{array}$
if x < 0 , $f(x)=\frac{x}{|x|}=\frac{x}{-(x)}=-1$
Now, for any value of x, the value of our function is -1
Therefore, the given function is continuous for every value of x
Hence, no point in discontinuity

Question:10.Find all points of discontinuity of f, where f is defined by

$f(x)=\{\begin{array}{cc}x+1& ifx\ge 1\\ x^2+1& ifx<1\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}x+1& ifx\ge 1\\ x^2+1& ifx<1\end{array}$
Given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k>1$
$$\begin{gathered} f(k)=k+1 \\ \underset{x\to k}{lim}f(x)=k+1 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of $k>1$

case(ii) $k<1$
$$\begin{gathered} f(k)=k^2+1 \\ \underset{x\to k}{lim}f(x)=k^2+1 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of $k<1$

case(iii) $x=1$

$\underset{x\to 1^{-}}{lim}f(x)=x^2+1=1^2+1=1+1=2$

$$\begin{gathered} \underset{x\to 1^{+}}{lim}f(x)=x+1=1+1=2 \\ f(1)=1^2+1=2 \end{gathered}$$

$R.H.L.=L.H.L.=f(1)$

Hence, at x = 2 given function is continuous.
Therefore, no point of discontinuity

Question 11.Find all points of discontinuity of f, where f is defined by

$f(x)=\{\begin{array}{cc}{x}^3-3& ifx\le 2\\ x^2+1& ifx>2\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}{x}^3-3& ifx\le 2\\ x^2+1& ifx>2\end{array}$
Given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k>2$
$$\begin{gathered} f(k)=k^2+1 \\ \underset{x\to k}{lim}f(x)=k^2+1 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of $k>2$

case(ii) $k<2$
$$\begin{gathered} f(k)=k^3-3 \\ \underset{x\to k}{lim}f(x)=k^3-3 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of $k<2$

case(iii) $x=2$
$\underset{x\to 2^{-}}{lim}f(x)=x^3-3=2^3-3=8-3=5$

$\underset{x\to 2^{+}}{lim}f(x)=x^2+1=2^2+1=4+1=5$

$$\begin{gathered} f(2)=2^3-3=8-3=5 \\ f(2)=R.H.L.=L.H.L. \end{gathered}$$
Hence, the given function is continuous at $x=2$
There is no point of discontinuity

Question 12: Find all points of discontinuity of f, where f is defined by

$f(x)=\{\begin{array}{cc}{x}^{10}-1& ifx\le 1\\ x^2& x>1\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}{x}^{10}-1& ifx\le 1\\ x^2& x>1\end{array}$
Given function is defined for every real number k
There are different cases for the given function.
case(i) $k>1$
$$\begin{gathered} f(k)=k^2 \\ \underset{x\to k}{lim}f(x)=k^2 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of k > 1

case(ii) $k<1$
$f(k)=k^{10}-1$

$\underset{x\to k}{lim}f(x)=k^{10}-1$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of k < 1

case(iii) x = 1
$\underset{x\to 1^{-}}{lim}f(x)=x^{10}-1=1^{10}-1=1-1=0$

$\underset{x\to 1^{+}}{lim}f(x)=x^2=1^2=1$

$f(1)=x^{10}-1=0f(1)=L.H.L.\ne R.H.L.$
Hence, x = 1 is the point of discontinuity.

Question 13. Is the function defined by

$f(x)=\{\begin{array}{cc}x+5& ifx\le 1\\ x-5& ifx>1\end{array}$

A continuous function?

Answer:

Given function is
$f(x)=\{\begin{array}{cc}x+5& ifx\le 1\\ x-5& ifx>1\end{array}$
Given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k>1$
$$\begin{gathered} f(k)=k-5 \\ \underset{x\to k}{lim}f(x)=k-5 \end{gathered}$$

$\underset{x\to k}{lim}f(x)=f(k)$
Hence, the given function is continuous for each value of $k>1$

case(ii) $k<1$
$$\begin{gathered} f(k)=k+5 \\ \underset{x\to k}{lim}f(x)=k+5 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, the given function is continuous for each value of $k<1$

case(iii) $x=1$
$$\begin{gathered} \underset{x\to 1^{-}}{lim}f(x)=x+5=1+5=1+5=6 \\ \underset{x\to 1^{+}}{lim}f(x)=x-5=1-5=-4 \\ f(1)=x+5=1+5=6 \\ L.H.L.=f(1)\ne R.H.S. \end{gathered}$$

Hence, $x=1$ is the point of discontinuity.

Question:14.Discuss the continuity of the function f, where f is defined by

$f(x)\{\begin{array}{cc}3& if0\le x\le 1\\ 4& if1<x<3\\ 5& if3\le x\le 10\end{array}$

Answer:

Given function is
$f(x)\{\begin{array}{cc}3& if0\le x\le 1\\ 4& if1<x<3\\ 5& if3\le x\le 10\end{array}$
Given function is defined for every real number k
Different cases are their
case (i) $k<1$
$$\begin{gathered} f(k)=3 \\ \underset{x\to k}{lim}f(x)=3 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, the given function is continuous for every value of $k<1$

case(ii) $k=1$
$$\begin{gathered} f(1)=3 \\ \underset{x\to 1^{-}}{lim}f(x)=3 \\ \underset{x\to 1^{+}}{lim}f(x)=4 \\ R.H.L.\ne L.H.L.=f(1) \end{gathered}$$
Hence, the given function is discontinuous at $x=1$
Therefore, $x=1$ is the point of discontinuity.

case(iii) $1<k<3$
$$\begin{gathered} f(k)=4 \\ \underset{x\to k}{lim}f(x)=4 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, for every value of $k$ in $1<k<3$ given function is continuous.

case(iv) $k=3$
$$\begin{gathered} f(3)=5 \\ \underset{x\to 3^{-}}{lim}f(x)=4 \\ \underset{x\to 3^{+}}{lim}f(x)=5 \\ R.H.L.=f(3)\ne L.H.L. \end{gathered}$$
Hence, $x=3$ is the point of discontinuity

case(v) $k>3$
$$\begin{gathered} f(k)=5 \\ \underset{x\to k}{lim}f(x)=5 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, the given function is continuous for every value of $k>3$
case(vi) when $k<3$

$$\begin{gathered} f(k)=4 \\ \underset{x\to k}{lim}f(x)=4 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, for every value of $k$ in $k<3$ given function is continuous

Question:15 Discuss the continuity of the function f, where f is defined by $f(x)\{\begin{array}{ccc}2x& if& x<0\\ 0& if& 0\le x\le 1\\ 4x& if& x>1\end{array}$

Answer:
Given function is satisfied for all real values of $x$
case (i) $k<0$
Hence, the function is continuous for all values of $x<0$

case (ii) $x=0$
L.H.L at $x=0$

R.H.L. at $x=0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous at $x=0$

case (iii) $k>0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous for all values of x > 0

case (iv) k < 1

Hence, the function is continuous for all values of x < 1

case (v) k > 1

Hence, the function is continuous for all values of x > 1

case (vi) x = 1
Hence, the function is not continuous at x = 1

Question:16.Discuss the continuity of the function f, where f is defined by

$f(x)=\{\begin{array}{cc}-2& ifx\le -1\\ 2x& if-1<x\le 1\\ 2& ifx>1\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}-2& ifx\le -1\\ 2x& if-1<x\le 1\\ 2& ifx>1\end{array}$
Given function is defined for every real number $k$
Different cases are their
case (i) $k<-1$
$$\begin{gathered} f(k)=-2 \\ \underset{x\to k}{lim}f(x)=-2 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, the given function is continuous for every value of $k<-1$
case(ii) k = -1
$$\begin{gathered} f(-1)=-2 \\ \underset{x\to -1^{-}}{lim}f(x)=-2 \\ \underset{x\to -1^{+}}{lim}f(x)=2x=2(-1)=-2 \\ R.H.L.=L.H.L.=f(-1) \end{gathered}$$
Hence, the given function is continuous at $x=-1$
case(iii) $k>-1$
$$\begin{gathered} f(k)=2k \\ \underset{x\to k}{lim}f(x)=2k \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, the given function is continuous for all values of $x>-1$

case(vi) $-1<k<1$
$$\begin{gathered} f(k)=2k \\ \underset{x\to k}{lim}f(x)=2k \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, for every value of $k$ in $-1<k<1$ given function is continuous.

case(v) $k=1$
$$\begin{gathered} f(1)=2x=2(1)=2 \\ \underset{x\to 1^{-}}{lim}f(x)=2x=2(1)=2 \\ \underset{x\to 1^{+}}{lim}f(x)=2 \\ R.H.L.=f(1)=L.H.L. \end{gathered}$$
Hence, at x =1 function is continuous

case(vi) $k>1$
$$\begin{gathered} f(k)=2 \\ \underset{x\to k}{lim}f(x)=2 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, the given function is continuous for every value of k > 1
case(vii) when $k<1$
$$\begin{gathered} f(k)=2k \\ \underset{x\to k}{lim}f(x)=2k \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, for every value of k in $k<1$ given function is continuous.

Therefore, continuous at all points

Question:17.Find the relationship between a and b so that the function f is defined by
$f(x)=\{\begin{array}{cc}ax+1,& ifx<3\\ bx+3& ifx>3\end{array}$
is continuous at x = 3.

Answer:

Given function is
$f(x)=\{\begin{array}{cc}ax+1,& ifx<3\\ bx+3& ifx>3\end{array}$
For the function to be continuous at $x=3$, R.H.L. must be equal to L.H.L.
$$\begin{gathered} \underset{x\to 3^{-}}{lim}f(x)=ax+1=3a+1 \\ \underset{x\to 3^{+}}{lim}f(x)=bx+3=3b+3 \end{gathered}$$
For the function to be continuous
$$\begin{gathered} \underset{x\to 3^{-}}{lim}f(x)=\underset{x\to 3^{+}}{lim}f(x) \\ 3a+1=3b+3 \\ 3(a-b)=2 \\ a-b=\frac{2}{3} \\ a=b+\frac{2}{3} \end{gathered}$$

Question:18.For what value of l is the function defined by
$f(x)=\{\begin{array}{cc}\lambda (x^2-2x)& ifx\le 0\\ 4x+1& ifx>0\end{array}$
continuous at x = 0? What about continuity at x = 1?

Answer:

Given function is
$f(x)=\{\begin{array}{cc}\lambda (x^2-2x)& ifx\le 0\\ 4x+1& ifx>0\end{array}$
For the function to be continuous at $x=0$, R.H.L. must be equal to L.H.L.
$$\begin{gathered} \underset{x\to 0^{-}}{lim}f(x)=\lambda (x^2-2x)=0 \\ \underset{x\to 0^{+}}{lim}f(x)=4x+1=1 \end{gathered}$$
For the function to be continuous
$$\begin{gathered} \underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{+}}{lim}f(x) \\ 0\ne 1 \end{gathered}$$
Hence, if no value of the function is continuous at $x=0$

For $x=1$
$$\begin{gathered} f(1)=4x+1=4(1)+1=5 \\ \underset{x\to 1}{lim}f(x)=4+1=5 \\ \underset{x\to 1}{lim}f(x)=f(x) \end{gathered}$$
Hence, the given function is continuous at $x=1$

Question:19.Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here, [x] denotes the greatest integer less than or equal to x.

Answer:

Given function is
$g(x)=x-[x]$
Given is defined for all real numbers k
$$\begin{gathered} \underset{x\to k^{-}}{lim}f(x)=k-(k-1)=k-k+1=1 \\ \underset{x\to k^{+}}{lim}f(x)=k-k=0 \\ \underset{x\to k^{-}}{lim}f(x)\ne \underset{x\to k^{+}}{lim}f(x) \end{gathered}$$
Hence, by this, we can say that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points

Question 20.Is the function defined by $f(x)=x^2-sinx+5$ continuous at x = $\pi$ ?

Answer:

Given function is
$f(x)=x^2-sinx+5$
Clearly, the Given function is defined at x = $\pi$
$$\begin{gathered} f(\pi )={\pi }^2-\sin \pi +5={\pi }^2-0+5={\pi }^2+5 \\ \underset{x\to \pi }{lim}f(x)={\pi }^2-\sin \pi +5={\pi }^2-0+5={\pi }^2+5 \\ \underset{x\to \pi }{lim}f(x)=f(\pi ) \end{gathered}$$
Hence, the function defined by $f(x)=x^2-sinx+5$ continuous at x = $\pi$