Chapter 5: Continuity and Differentiability

Ans: It is known that if two functions $\text{g}$ and $\text{h}$ are continuous, then $\text{g+h, g-h}$and $\text{g,h}$ are also continuous.

So, let us assume that, $\text{g(x)=sinx}$ and $\text{h(x)=cosx}$ are two continuous functions.

Now, as $\text{g(x)=sinx}$ is defined for every real number, so let $\text{c}$ be a real number. Substitute $\text{x=c+h}$ into the function $\text{g}$.

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

So, \[\text{g(c)=sinc}\].

Also,

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sinx}\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{c+h} \right)\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{sinc}\cdot \text{cosh+cosc}\cdot \text{sinh} \right]\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{cosh} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{sinh} \right)\]

\[\text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0}\]

\[\text{=sinc+0}\]

 \[\text{=sinc}\]

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

Hence, the function $\text{g}$ is continuous.

Again, let us assume that $\text{h(x)=cosx}$.

Note that, the function $\text{h(x)=cosx}$ is defined for every real number.

Now, let $\text{c}$ be a real number. 

Substitute $\text{x=c+h}$ into the function.

When $\text{x}\to \text{c}$, then \[\text{h}\to 0\].

So, $\text{h(c)=cosc}$ and

$\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos(c+h)} \\ & \text{=}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}$

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

Thus, the function $\text{h}$ is continuous.

Hence, we conclude that all the following functions are continuous.

(a) \[\text{f(x)=g(x)+h(x)=sinx+cosx}\].

(b) \[\text{f(x)=g(x)-h(x)=sinx-cosx}\].

(c) \[\text{f(x)=g(x) }\!\!\times\!\!\text{ h(x)=sinx }\!\!\times\!\!\text{ cosx}\].

Ans: We know that if two functions say $\text{g}$ and $\text{h}$ are continuous, then

        i.  $\frac{\text{h(x)}}{\text{g(x)}}\text{,g(x)}\ne \text{0}$ is continuous.

      ii.  \[\frac{\text{1}}{\text{g(x)}}\text{,g(x)}\ne \text{0}\] is continuous.

    iii.  \[\frac{\text{1}}{\text{h(x)}}\text{,h(x)}\ne \text{0}\] is continuous.

It can be observed that the function $\text{g(x)=sinx}$ is defined for all real numbers.

Now, let's consider $\text{c}$ to be a real number and substitute $\text{x=c+h}$ into the function $\text{g}$.

When, $\text{x}\to \text{c}$, then \[\text{h}\to 0\].

So, \[\text{g(c)=sinc}\] and

\[\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sinx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{sinc}\cdot \text{cosh+cosc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{cosh} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{sinh} \right) \\ & \text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0} \\ & \text{=sinc+0} \\ & \text{=sinc} \\ \end{align}\]

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

Thus, the function $\text{g}\left( \text{x} \right)\text{=sinx}$ is continuous.

Again, let $\text{h(x)=cosx}$.

It can be noted that $\text{h(x)=cosx}$ is defined for all real numbers.

Now, let's consider $\text{c}$ to be a real number and substitute $\text{x=c+h}$ into the function $\text{h}$.

Then, \[\text{h(c)=cosc}\] and

\[\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}\]

Therefore, \[\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{h(x)=h(c)}\].

Thus, the function $\text{h(x)=cosx}$ is continuous.

Now note that,

$\text{cosec x=}\frac{\text{1}}{\text{sinx}}\text{,}$ and $\text{sinx}\ne \text{0}$ is a continuous function.

Thus, the cosecant function is continuous except at $\text{x=n }\!\!\pi\!\!\text{ ,}\,\,\text{n}\in \mathbb{Z}$.

Again, $\text{secx=}\frac{\text{1}}{\text{cosx}}\text{,}\,\,\text{cosx}\ne \text{0}$ is continuous.

$\Rightarrow \text{secx,}\,\,\text{x}\ne \text{(2n+1)}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{, }\,\text{n}\in \mathbb{Z}$  is a continuous function.

Thus, secant function is also continuous except at $\text{x=(2n+1)}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{, }\,\text{n}\in \mathbb{Z}$.

And the cotangent function is

$\text{cotx=}\frac{\text{cosx}}{\text{sinx}}$ and where $\text{sinx}\ne 0$ is a continuous function.

$\Rightarrow \text{cotx,}\,\,\text{x}\ne \text{n }\!\!\pi\!\!\text{ ,}\,\,\text{n}\in \mathbb{Z}$ is a continuous function.

Hence, the cotangent function is continuous except at $\text{x=n }\!\!\pi\!\!\text{ ,}\,\,\text{n}\in \mathbb{Z}$.

Ans: The given function is  $\text{f(x)=}\left\{ \begin{align} & \frac{\text{sinx}}{\text{x}}\text{, if x}\,<\,\text{0} \\ & \text{x+1, }\,\text{if x}\ge \text{0} \\ \end{align} \right.$

Note that, the function $\text{f}$ is defined at every point on the real number line.

Now, let's consider $\text{c}$ be a real number.

Then there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c}=\text{0}$.

Let us discuss one after another.

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

Then, $\text{f(c)=}\frac{\text{sinc}}{\text{c}}$.

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

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 \[\text{x0}\].

Case II: When $\text{c}>\text{0}$.

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

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

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

Hence, the function $\text{f}$ is continuous at every point, where \[\text{x0}\].

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

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

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

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

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

Therefore, \[\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(0)}\].

So, the function $\text{f}$ is continuous at \[\text{x = 0}\].

Thus, the function $\text{f}$ is continuous at every real point.

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

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x=0} \\ \end{align} \right\}$

We can observe that the function $\text{f}$ is defined at every point on the real number line.

Now, let's consider $\text{c}$ to be a real number.

Then, there may arise two cases, either $\text{c}\ne \text{0}$ or $\text{c=0}$.

Let us discuss the cases one after another.

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

Then $\text{f(c)=}{{\text{c}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{c}}$.

Also,

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)=\left( \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,{{\text{x}}^{\text{2}}} \right)\left( \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sin}\frac{\text{1}}{\text{x}} \right)={{\text{c}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{c}}\].

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

Hence, the function $\text{f}$ is continuous at every point \[\text{x}\ne 0\].

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

Then $\text{f(0)=0}$ and also

$\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\text{=}\underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)$

Now, we know that,

 $\text{-1}\le \text{sin}\frac{\text{1}}{\text{x}}\le 1,\text{ x}\ne 0$.

$\begin{align} & \Rightarrow \text{-}{{\text{x}}^{\text{2}}}\le \text{sin}\frac{\text{1}}{\text{x}}\le {{\text{x}}^{\text{2}}} \\ & \Rightarrow \underset{\text{x}\to 0}{\mathop{\lim }}\,\text{(-}{{\text{x}}^{\text{2}}}\text{)}\le \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\le 0 \\ & \Rightarrow 0\le \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\le 0 \\ & \Rightarrow \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)=0 \\ \end{align}$

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

Similarly, we have, 

$\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)=\underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\text{=0}$

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

Thus, the function $\text{f}$ is continuous at the point \[\text{x = 0}\].

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

Hence, the function $\text{f}$ is continuous.

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{sinx-cosx, if x}\ne \text{0} \\ & \text{ 1 if x=0} \\ \end{align} \right.$

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

Now, let's consider $\text{c}$ to be a real number.

Then, there may arise two cases, either $\text{c}\ne \text{0}$ or $\text{c=0}$.

Let us discuss the cases one after another.

Case I: When $\text{c}\ne 0$.

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

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

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 $\text{x}\ne 0$.

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

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

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

\[\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{(sinx-cosx)=sin0-cos0=0-1=-1}\] and the right-hand-limit is 

\[\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{(sinx-cosx)=sin0-cos0=0-1=-1}\].

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

So, the function $\text{f}$ is continuous at \[\text{x = 0}\].

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

Hence, the function $\text{f}$ is continuous.

Ans: The given function is $\text{f(x)}\mathbf{=}\left\{ \begin{align} & \frac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}\text{, if x}\ne \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ & \text{ 3, if x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ \end{align} \right.$

Observe that, $\text{f}$ is defined and continuous at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ , since the value of the $\text{f}$ at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ is equal with the limiting value of $\text{f}$ at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Since, $\text{f}$ is defined at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ and $\text{f}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)\text{=3}$, so

$\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\frac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}$.

Substitute $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h}$ into the function $\text{f}\left( \text{x} \right)$.

So, we have, \[\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\Rightarrow \text{h}\to 0\].

Then,

$\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\frac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}=\underset{\text{h}\to 0}{\mathop{\lim }}\,\frac{\text{kcos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h} \right)}{\text{ }\!\!\pi\!\!\text{ -2}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h} \right)}$.

 $\Rightarrow \text{k}\underset{\text{h}\to 0}{\mathop{\lim }}\,\frac{\text{-sinh}}{\text{-2h}}\text{=}\frac{\text{k}}{\text{2}}\underset{\text{h}\to 0}{\mathop{\lim }}\,\frac{\text{sinh}}{\text{h}}\text{=}\frac{\text{k}}{\text{2}}\text{.1}=\frac{\text{k}}{\text{2}}$

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

 $\begin{align} & \Rightarrow \frac{\text{k}}{\text{2}}\text{=3} \\ & \Rightarrow \text{k=6} \\ \end{align}$

Hence, the value of $\text{k}$ is $\text{6}$ for which the function $\text{f}$ is continuous.

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

It is known that, $\text{f}$ is continuous at \[\text{x = 2}\] only if $\text{f}$ is defined at $\text{x=2}$ and if the value of $\text{f}$ at \[\text{x = 2}\] is equal with the limiting value of $\text{f}$ at \[\text{x = 2}\].

So, at $x=2$,

Now, the left-hand-limit and right-hand-limit of the function $\text{f}\left( \text{x} \right)$ at $\text{x}=\text{2}$respectively are,

$\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( \text{k}{{\text{x}}^{\text{2}}} \right)=\text{k}{{\left( \text{2} \right)}^{\text{2}}}=\text{4k}$, 

and $\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\left( 3 \right)=3$.

Since, the function is continuous at $\text{x}=\text{2}$, so

\[\begin{align} & \underset{\text{x}\to {{2}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(2)} \\ & \Rightarrow \text{4k=3} \\ & \Rightarrow \text{k=}\frac{\text{3}}{\text{4}} \\ \end{align}\]

Hence, the value of $\text{k}$ is $\frac{\text{3}}{\text{4}}$ for which the function $\text{f}$ is continuous.

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

It is known that, $\text{f}$ is continuous at $\text{x= }\!\!\pi\!\!\text{ }$ only if the value of $\text{f}$ at $\text{x= }\!\!\pi\!\!\text{ }$ is equal with the limiting value of $\text{f}$ at $\text{x= }\!\!\pi\!\!\text{ }$.

It is provided that the function $\text{f}$ is defined at $\text{x= }\!\!\pi\!\!\text{ }$.

Also, $\text{f( }\!\!\pi\!\!\text{ )=k }\!\!\pi\!\!\text{ +1}$.

Now, the left-hand-limit,

\[\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{(kx+1)=k }\!\!\pi\!\!\text{ +1}\].

Also, the right-hand-limit,

\[\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{cosx=cos }\!\!\pi\!\!\text{ =-1}\].

Since, the function $\text{f}$ is continuous, so

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

\[\begin{align} & \Rightarrow \text{k }\!\!\pi\!\!\text{ +1=-1} \\ & \Rightarrow \text{k }\!\!\pi\!\!\text{ =-2} \\ & \Rightarrow \text{k=-}\frac{\text{2}}{\text{ }\!\!\pi\!\!\text{ }} \\ \end{align}\]

Hence, the value of $\text{k}$ is $\text{-}\frac{\text{2}}{\text{ }\!\!\pi\!\!\text{ }}$ for which the function $\text{f}$ is continuous at $\text{x= }\!\!\pi\!\!\text{ }$. 

Ans: The given function is  $\text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \text{5} \\ & \text{3x-5, }\,\,\text{if x}>\text{5} \\ \end{align} \right.$

Recall that, the function $\text{f}$ is continuous at \[\text{x = 5}\] only if the value of $\text{f}$ at \[\text{x = 5}\] is equal to the limiting value of $\text{f}$ at \[\text{x = 5}\].

Note that, the function $\text{f}$ is defined at \[\text{x = 5}\].

Also, $\text{f(5)=kx+1=5k+1}$.

Then, the left-hand-limit of the function,

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

The right-hand-limit of the function,

$\underset{\text{x}\to {{\text{5}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{\text{5}}^{+}}}{\mathop{\lim }}\,\left( \text{3x-5} \right)\text{=3}\left( 5 \right)-5=15-5=10$.

Since, the function $\text{f}$ is continuous, so

$\begin{align} & \underset{\text{x}\to {{\text{5}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{5}^{+}}}{\mathop{\lim }}\,\text{(3x-5)=5k+1} \\ & \Rightarrow \text{5k+1=10} \\ & \Rightarrow \text{5k=9} \\ & \Rightarrow \text{k=}\frac{\text{9}}{\text{5}} \\ \end{align}$

Hence, the value of $\text{k}$ is $\frac{9}{5}$ for which the function $\text{f}$ is continuous at $\text{x=5}$.

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{5, }\,\,\,\text{if x}\le \text{2} \\ & \text{ax+b,}\,\text{ if }\,\,\text{2}<\text{x}<\text{10} \\ & \text{ 21, if x}\ge \text{10} \\ \end{align} \right.$

Note that, $\text{f}$ is defined at every point on the real number line.

Now, realise that if the function $\text{f}$ is continuous then $\text{f}$ is continuous at every real number.

So, let $\text{f}$ satisfies continuity at \[\text{x=2}\] and \[\text{x=10}\].

Then, since $\text{f}$ is continuous at $\text{x=2}$, so

\[\begin{align} & \underset{\text{x}\to {{\text{2}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(2)} \\ & \Rightarrow \underset{\text{x}\to {{\text{2}}^{\text{-}}}}{\mathop{\lim }}\,\text{(5)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{(ax+b)=5} \\ & \Rightarrow \text{5=2a+b} \\ \end{align}\]

$\Rightarrow \text{2a+b=5}$                                                   …… (1)

Again, since $\text{f}$ attains continuity at \[\text{x=10}\], so

\[\begin{align} & \underset{\text{x}\to 1{{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{10}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(10)} \\ & \Rightarrow \underset{\text{x}\to 1{{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{(ax+b)=}\underset{\text{x}\to {{10}^{+}}}{\mathop{\lim }}\,\text{(21)=21} \\ \end{align}\]

\[\Rightarrow 1\text{0a+b=21}\]                                   …… (2)

Subtracting the equation (1) from the equation (2), gives

$\text{8a=16}\Rightarrow \text{a=2}$

Substituting \[\text{a=2}\] in the equation (1), gives

$\begin{align} & \text{2 }\!\!\times\!\!\text{ 2+b=5} \\ & \Rightarrow \text{4+b=5}\Rightarrow \text{b=1} \\ \end{align}$

Hence, the values of $\text{a}$ and $\text{b}$ are $2$ and $1$ respectively for which $\text{f}$ is a continuous function.