Chapter 5: Continuity and Differentiability

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & {{\text{x}}^{\text{3}}}\text{-3, if x}\le \text{2} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}>\text{2} \\ \end{align} \right.$

Observe that, the function $\text{f}$ is defined at all points in the real line.

Now, let assume $\text{c}$ as a point on the real line.

Case I: When $\text{c}<\text{2}$.

Then, we have  $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{3}}}\text{-3}$ and also \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{3}}}\text{-3} \right)\text{=}{{\text{c}}^{\text{3}}}\text{-3}\].

Therefore, the function $\text{f}$ attains continuity at all the points $\text{x}$, where $\text{x}<\text{2}$.

Case II: When \[\text{c=2}\].

Then, we have $\text{f}\left( \text{c} \right)\text{=f}\left( \text{2} \right)\text{=}{{\text{2}}^{\text{3}}}\text{-3=5}$.

Now the left-hand limit of the function is

$\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( {{\text{x}}^{3}}-3 \right)\text{=}{{\text{2}}^{3}}\text{-3=5}$ and the right-hand limit is

$\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{2}}^{\text{2}}}\text{+1=5}$.

Therefore, $\underset{\text{x}\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{2} \right)$.

Hence, the function $\text{f}$ is continuous at \[\text{x=2}\].

Case III: When $\text{c}>\text{2}$.

Then, $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{2}}+1$ and

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, $\text{f}$ attains continuity at all the points $\text{x}$, where $\text{x}>\text{2}$.

Thus, the function $\text{f}$ is continuous at all the points on the real line.

Hence, $\text{f}$ does not have any point of discontinuity.

Ans: The given function is $\text{f}\left( \text{x} \right)=\left\{ \begin{align} & {{\text{x}}^{\text{10}}}\text{-1, if x}\le \text{1} \\ & {{\text{x}}^{\text{2}}}\text{, if x}\,>\text{1} \\ \end{align} \right.$

Observe that, the function $\text{f}$ is defined at every point of the real line.

Now, let assume $\text{c}$ as a point on the real number line.

Case I: When $\text{c}<\text{1}$.

Then $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{10}}}\text{-1}$.

Also,  $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{10}}}\text{-1} \right)\text{=}{{\text{c}}^{\text{10}}}\text{-1}$

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, the function $\text{f}$ attains continuity at every point $\text{x}$, for $\text{x}<\text{1}$.

Case II: When \[\text{c=1}\].

Then the left-hand limit of the function $\text{f}\left( \text{x} \right)$ at \[\text{x=1}\] is

\[\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\left( {{\text{x}}^{10}}-1 \right)={{1}^{10}}-1=1-1=0\] and

the right-hand limit of the function $\text{f}$ at $\text{x=1}$ is

$\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}} \right)\text{=}{{\text{1}}^{\text{2}}}\text{=1}$.

So, we can notice that, $\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ does not satisfy continuity at \[\text{x=1}\].

Case III: When $\text{c}>\text{1}$.

Then, $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{2}}}$.

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}} \right)\text{=}{{\text{c}}^{\text{2}}}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Thus, the function $\text{f}$ attains continuity at every point $\text{x}$, for $\text{x}>\text{1}$.

Hence, we can conclude that \[\text{x=1}\] is the only point of discontinuity for the function $\text{f}$.

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{x+5, if x}\le \text{1} \\ & \text{x-5, if x}\,>\text{1} \\ \end{align} \right.$

It can be noted that the function $\text{f}$ is defined at every point on the real line. 

Now, let assume $\text{c}$ as a point on the real line.

Case I: When $\text{c}<\text{1}$.

Then, $\text{f}\left( \text{c} \right)\text{=c+5}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x+5} \right)\text{=c+5}$.

Hence, $\text{f}$ satisfies continuity at every point $\text{x}$, for $\text{x}<\text{1}$.

Case II: When \[\text{c=1}\].

Then, $\text{f}\left( \text{1} \right)\text{=1+5=6}$.

Now, the left-hand limit of the function $\text{f}$ at \[\text{x=1}\] is

$\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\left( \text{x+5} \right)\text{=1+5=6}$ and the right-hand limit of the function at \[\text{x=1}\] is $\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(x-5)=1-5=4}$.

Thus, it is seen that, $\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, $\text{f}$ does not attain continuity at \[\text{x=1}\].

Case III: When $\text{c}>\text{1}$.

Then $\text{f}\left( \text{c} \right)\text{=c-5}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=c-5}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Thus, the function $\text{f}$ is continuous at every point \[\text{x}\], for $\text{x}>\text{1}$.

Hence, we can conclude that \[\text{x=1}\] is the only point of discontinuity for the function $\text{f}$.

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{3, if 0}\le \text{x}\le \text{1} \\ & \text{4, if 1}\,<\,\text{x}\,<\,\text{3} \\ & \text{5, if 3}\le \text{x}\le \text{10} \\ \end{align} \right.$

Therefore, $\text{f}$ is defined in the interval $\text{ }\!\![\!\!\text{ 0,10 }\!\!]\!\!\text{ }$.

Now let's assume $\text{c}$ as a point in the interval $\text{ }\!\![\!\!\text{ 0,10 }\!\!]\!\!\text{ }$.

Then there may arise five cases.

Case I: When \[\text{0}\le \text{c}<\text{1}\].

Then $\text{f}\left( \text{c} \right)\text{=3}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{3} \right)\text{=3}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, the function $\text{f}$ attains continuity at the interval $\text{ }\!\![\!\!\text{ 0,1 }\!\!]\!\!\text{ }$.

Case II: When \[\text{c=1}\].

Then $\text{f}\left( \text{3} \right)\text{=3}$.

Also, the left-hand-limit of the function at \[\text{x=1}\] is

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{3} \right)\text{=3}$ and the right-hand-limit of the function at \[\text{x=1}\] is

$\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4}$.

Thus, it is noticed that $\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ does not satisfy continuity at \[\text{x=1}\].

Case III: When \[\text{1}<\text{c}<\text{3}\].

Then $\text{f}\left( \text{c} \right)\text{=4}$.

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4}\].

Thus, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, the function $\text{f}$ attains continuity at every point in the interval $\text{ }\!\![\!\!\text{ 1,3 }\!\!]\!\!\text{ }$.

Case IV: When \[\text{c=3}\].

Then $\text{f}\left( \text{c} \right)\text{=5}$.

Now, the left-hand-limit of the function $\text{f}$ at \[\text{x=3}\] is

$\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4}$ and the right-hand-limit of the function $\text{f}$ at \[\text{x=3}\] is

\[\underset{\text{x}\to {{\text{3}}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\left( \text{5} \right)\text{=5}\].

Therefore, it is noted that $\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{3}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$ .

Hence, the function $\text{f}$ is not continuous at \[\text{x=3}\].

Case V: When $\text{3}<\text{c}\le \text{10}$.

Then $\text{f}\left( \text{c} \right)\text{=5}$.

Also,  $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{5} \right)\text{=5}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, the function $\text{f}$ attains continuity at every point in the interval $\text{ }\!\![\!\!\text{ 3,10 }\!\!]\!\!\text{ }$.

Hence, the function $\text{f}$ is not continuous at \[\text{x=1}\] and \[\text{x=3}\].

Ans: The given function is 

$\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x, if x}<\text{0} \\ & \text{0, if 0}\le \text{x}\le \text{1} \\ & \text{4x, if x}>\text{1} \\ \end{align} \right.$

Now, let consider $\text{c}$ be a point on the real number line. 

Then, five cases may arrive.

Case I: When \[\text{c}<\text{0}\].

Then, $\text{f}\left( \text{c} \right)\text{=2c}$.

Also,  $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{2x} \right)\text{=2c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, the function $\text{f}$ attains continuity at every point $\text{x}$ whenever \[\text{x}<\text{0}\].

Case II: When \[\text{c = 0}\].

Then, \[\text{f}\left( \text{c} \right)\text{=f}\left( \text{0} \right)\text{=0}\].

Now, the left-hand-limit of the function \[\text{f}\]at \[\text{x = 0}\] is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{2x} \right)=0$ and the right-hand limit of the function \[\text{f}\]at \[\text{x = 0}\] is, 

$\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{=}\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{0} \right)\text{=0}$.

Therefore, $\underset{\text{x}\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right)$.

Thus, the function \[\text{f}\] attains continuity at \[\text{x = 0}\].

Case III: When $\text{0}<\text{c}<\text{1}$ 

Then, $\text{f(x)=0}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(0)=0}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, $\text{f}$ attains continuity at every point in the interval $\text{(0,1)}$.

Case IV: When \[\text{c =1}\].

Then, $\text{f(c)=f(1)=0}$.

Now, the left-hand-limit at \[\text{x = 1}\] is

\[\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(0)=0}\] and the right-hand-limit at \[\text{x = 1}\] is

\[\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(4x)=4 }\!\!\times\!\!\text{ 1=4}\].

Thus, it is noticed that, \[\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)}\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f(x)}\].

Hence, the function $\text{f}$ is not continuous at \[\text{x = 1}\].

Case V: When \[\text{c}>\text{1}\].

Then, \[\text{f(c)=f(1)=0}\].

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(4x)=4c}\]

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

So, the function $\text{f}$ attains continuity at every point $\text{x}$, for \[\text{c}>\text{1}\].

Hence, the function $\text{f}$ is discontinuous only at \[\text{x = 1}\].

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{-2, if x}\le \text{-1} \\ & \text{2x, if -1}<\text{x}\le \text{1} \\ & \text{2, }\,\,\,\,\text{if x}>\text{1} \\ \end{align} \right.$

Note that, $\text{f}$ is defined at every point in the interval $\left[ -1,\infty  \right)$.

Now, let assume $\text{c}$ is a point on the real number line.

Case I: When \[\text{c}<-\text{1}\].

Then, $\text{f(c)=-2}$.

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(-2)=-2}\].

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ attains continuity at every point $\text{x}$ , for \[\text{x}<-\text{1}\].

Case II: When \[\text{c=-1}\].

Then, $\text{f(c)=f(-1)=-2}$.

Now, the left-hand-limit of the function at \[\text{x=-1}\] is

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(-2)=-2}$ and the right-hand-limit at \[\text{x=-1}\] is

$\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{=2 }\!\!\times\!\!\text{ (-1)=-2}$.

Therefore, $\underset{\text{x}\to -1}{\mathop{\lim }}\,\text{f(x)=f(-1)}$.

Hence, the function \[\text{f}\] satisfies continuity at \[\text{x=-1}\].

Case III: When $\text{-1}<\text{c}<\text{1}$.

Then, $\text{f(c)=2c}$ and \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(2x)=2c}\].

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ attains continuity at every point in the interval $\text{(-1,1)}$.

Case IV: When \[\text{c=1}\].

Then, $\text{f(c)=f(1)=2 }\!\!\times\!\!\text{ 1=2}$

Now, the left-hand-limit of the function at \[\text{x = 1}\] is 

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(2x)=2 }\!\!\times\!\!\text{ 1=2}$ and the right-hand-limit at \[\text{x = 1}\] is

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

Therefore, $\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Thus, the function $\text{f}$ attains continuity at \[\text{x=2}\].

Case V: When \[\text{c}>\text{1}\].

Then \[\text{f(c)=2}\].

Also, $\underset{\text{x}\to 2}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to 2}{\mathop{\lim }}\,\text{(2)=2}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ is continuous at every point $\text{x}$, for \[\text{x}>\text{1}\].

Thus, it can be concluded that the function $\text{f}$ is continuous for all the points.

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{ax+1, if x}\le \text{3} \\ & \text{bx+3, if x}\,>\text{3} \\ \end{align} \right.$

The function $\text{f}$ will be continuous at \[\text{x = 3}\] if

$\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(3)}$,                               ..….. (1)

$\underset{x\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(ax+1)=3a+1}$,

$\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f(bx+1)=3b+3}$,                    …… (2)

 and

$\text{f(3)=3a+1}$.                                                     …… (3)

Therefore, from the equation (1), (2), and (3) gives

$\Rightarrow \text{3a+1=3b+3}$

$\Rightarrow \text{3a=3b+2}$

$\Rightarrow \text{a=b+}\frac{\text{2}}{\text{3}}$

Hence, the required relationship between $\text{a}$ and $\text{b}$ is given by $\text{a=b+}\frac{\text{2}}{\text{3}}$.

Ans: The given function is\[\text{f(x)=}\left\{ \begin{align} & \text{ }\!\!\lambda\!\!\text{ (}{{\text{x}}^{\text{2}}}\text{-2x), if x}\le \text{0} \\ & \text{ 4x+1, if x}\,>\,\text{0} \\ \end{align} \right.\]

Now the function will be continuous at \[\text{x = 0}\] if

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(0)}$.

Also, the R.H.L and L.H.L are given by,

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( 4x+1 \right)=4\left( 0 \right)+1=1$,

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{ }\!\!\lambda\!\!\text{ (}{{\text{x}}^{\text{2}}}\text{-2x)}\,\text{=}\,\text{ }\!\!\lambda\!\!\text{ (}{{\text{0}}^{\text{2}}}\text{-2 }\!\!\times\!\!\text{ 0)=0}$.

So, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$.

Thus, there does not exist any value of $\text{ }\!\!\lambda\!\!\text{ }$ for which $\text{f}$ is continuous at \[\text{x = 0}\].

Now, at \[\text{x = 1}\],

$\text{f(1)=4x+1=4 }\!\!\times\!\!\text{ 1+1=5}$ and

$\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{(4x+1)=4 }\!\!\times\!\!\text{ 1+1=5}$.

Therefore, $\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f(x)=f(1)}$.

Hence, the function $\text{f}$ is continuous at \[\text{x = 1}\], for all values of $\text{ }\!\!\lambda\!\!\text{ }$.

Ans: The given function is $\text{g(x)=x-}\left[ \text{x} \right]$.

Note that the function is defined at every integral point.

Now, let assume that $\text{n}$ is an integer.

Then, $\text{g(n)=n-}\left[ \text{n} \right]\text{=n-n=0}$.

Now taking left-hand-limit as $\text{x}\to \text{n}$ to the function $\text{g}$ gives 

$\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\left[ \text{x-}\left[ \text{x} \right] \right]\text{=}\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{-}\underset{\text{x}\to \text{n-}}{\mathop{\lim }}\,\left[ \text{x} \right]\text{=n-}\left( \text{n-1} \right)\text{=1}$.

Again, the right-hand-limit on the function at $\text{x=n}$ is

\[\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\left[ \text{x-}\left[ \text{x} \right] \right]\text{=}\underset{\text{x}\to {{\text{n}}^{\text{+}}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{-}\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\left[ \text{x} \right]\text{=n-n=0}\].

Note that, $\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\text{g(x)}\ne \underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\text{g(x)}$.

Thus, the function $\text{f}$ is cannot be continuous at $\text{x=n,}$ 

Hence, the function $\text{g}$ is not continuous at any integral point. 

Ans: The given function is $\text{f(x)=}{{\text{x}}^{\text{2}}}\text{-sinx+5}$.

Now, at \[\text{x= }\!\!\pi\!\!\text{ }\], 

\[\text{f(x)}\,\text{=}\,\text{f( }\!\!\pi\!\!\text{ )}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-sin }\!\!\pi\!\!\text{ }\,\text{+}\,\text{5}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-0+5}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{+5}\]

Taking limit as $\text{x}\to \text{ }\!\!\pi\!\!\text{ }$ on the function $\text{f}\left( \text{x} \right)$ gives

\[\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{(}{{\text{x}}^{\text{2}}}\text{-sinx+5)}\].

Now substitute $\text{x= }\!\!\pi\!\!\text{ +h}$ into the function $\text{f}\left( \text{x} \right)$.

When $\text{x}\to \text{ }\!\!\pi\!\!\text{ }$, then $\text{h}\to 0$.

Therefore, 

\[\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{-sinx} \right)\text{+5}\].

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \left( \text{ }\!\!\pi\!\!\text{ +}{{\text{h}}^{\text{2}}} \right)\text{-sin}\left( \text{ }\!\!\pi\!\!\text{ +h} \right)\text{+5} \right]\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,{{\left( \text{ }\!\!\pi\!\!\text{ +h} \right)}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{ }\!\!\pi\!\!\text{ +h} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,5\]

\[\text{=}{{\left( \text{ }\!\!\pi\!\!\text{ +0} \right)}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{ }\!\![\!\!\text{ sin }\!\!\pi\!\!\text{ }\cdot \text{cosh+cos }\!\!\pi\!\!\text{ }\cdot \text{sinh }\!\!]\!\!\text{ +5}\]

\[\text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sin }\!\!\pi\!\!\text{ }\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cos }\!\!\pi\!\!\text{ }\cdot \text{sinh} \right)\text{+5}\]

\[\text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\text{sin }\!\!\pi\!\!\text{ }\cdot \text{cos0}-\text{cos }\!\!\pi\!\!\text{ }\cdot \text{sin0+5}\]

\[\text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\text{0 }\!\!\times\!\!\text{ 1}-\text{(}-\text{1) }\!\!\times\!\!\text{ 0+5=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{+5}\].

So, \[\underset{\text{x}\to \text{x}}{\mathop{\lim }}\,\text{f(x)=f( }\!\!\pi\!\!\text{ )}\].

Hence, it is concluded that the function $\text{f}$ is continuous at $\text{x=n}$.