Question 15 Marks
If the functions f(x), defined below is continuous at x = 0, find the value of k.
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\\frac{\text{x}}{|\text{x}|},&\text{x}>0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\\frac{\text{x}}{|\text{x}|},&\text{x}>0\end{cases}$
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\1,&\text{x}>0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{1-\cos2(-\text{h})}{2(-\text{h})^2}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{1-\cos2\text{h}}{2\text{h}^2}\Big)$
$=\frac{1}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{2\sin^2\text{h}}{\text{h}^2}\Big)$
$=\frac{2}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{\sin^2\text{h}}{\text{h}^2}\Big)$
$=\frac{2}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{\sin^2\text{h}}{\text{h}}\Big)^2$
$=1\times1$
$=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow 0}\text{f(h)}=\lim_\limits{\text{h}\rightarrow 0}(1)=1$
Also, $\text{f}(0)=\text{k}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow 0^+}=\text{f}(0)$
$\Rightarrow1=1=\text{k}$
Hence, the required value of k is 1
View full question & answer→Question 25 Marks
Find the value of 'a' for which the function f defined by
$\text{f}\text{(x)}=\begin{cases}\text{a}\sin\frac{\pi}{2}(\text{x}+1),& \text{x}\leq0 \\\frac{\tan\text{x-sin}\text{x}}{\text{x}^3} &\text{x} > 0\end{cases}$ is discontinuous at x = 0.
AnswerSince f(x) is continuous at x = 0, L.H.L = R.H.L.
Thus, we have
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{a}\sin\frac{\pi}{2}(\text{x}+1)=\lim\limits_{\text{x} \rightarrow 0^+}\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}\times1=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\cos\text{x}}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\text{x}}\Big(\frac{1}{\cos\text{x}}-1\Big)}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\text{x}}\Big(\frac{1-\cos\text{x}}{\cos\text{x}}\Big)}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{\cos\text{x}}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{a}=1\times1\times\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos^2\text{x}}{\text{x}^2(1+\cos\text{x)}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^2\text{x}}{\text{x}^2(1+\cos\text{x)}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^2\text{x}}{\text{x}^2}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{1+\cos\text{x}}$
$\Rightarrow\text{a}=1\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{1+\cos\text{x}}$
$\Rightarrow\text{a}=1\times\frac{1}{1+1}$
$\Rightarrow\text{a}=\frac{1}{2}$
View full question & answer→Question 35 Marks
Determine the values of a, b, c for which the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}},&\text{for x}>0\end{cases}$ is continuous at x = 0.
AnswerThe given function can be rewritten as,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}},&\text{for x}>0\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{1+\text{bx}}-1}{\text{bx}},&\text{for x}>0\end{cases}$
We observe
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}-\sin(-\text{h})}{\text{h}}\Big]=\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}}{\text{h}}-\frac{\sin\text{h}}{\text{h}}\Big]$
$=-(\text{a}-1)\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}}{(\text{a}+1)\text{h}}\Big]-\lim\limits_{\text{h} \rightarrow 0}\frac{\sin\text{h}}{\text{h}}=-\text{a}-1$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sqrt{1+\text{bh}}-1}{\text{bh}}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\text{bh}}{\text{bh}(\sqrt{1+\text{bh}}+1)}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{1}{\sqrt{1+\text{bh}}+1}\Big)=\frac{1}{2}$
And, f(0) = c
If f(x) is continuous at x = 0, then
$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x )}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x )}=\text{f}(0)$
$\Rightarrow-\text{a}-1=\frac{1}{2}=\text{c}$
$\Rightarrow-\text{a}-1=\frac{1}{2}$ and $\text{c}=\frac{1}{2}$
$\Rightarrow\text{a}=\frac{-3}{2},\text{c}=\frac{1}{2}$
Now, $\frac{\sqrt{1+\text{bx}}-1}{\text{bx}}$ exists only if $\text{bx}\neq0\Rightarrow\text{b}\neq0.$
$\therefore\text{b}\in\text{R}-\{0\}$
View full question & answer→Question 45 Marks
Find the value of a and b so that the function f(x) defind by $\text{f(x)}=\begin{cases}\text{x}+\text{a}\sqrt{2}\sin\text{x},&\text{if }0\leq\text{x}<\frac{\pi}{4}\\2\text{x}\cot\text{ x}+\text{b},&\text{if }\frac{\pi}{4}\leq\text{x}<\frac{\pi}{2}\\\text{a}\cos2\text{x}-\text{b}\sin\text{x},&\text{if }\frac{\pi}{2}\leq\text{x}\leq\pi\end{cases}$ becomes continuous on $[0,\pi]$
AnswerGiven, f is continuous on $[0,\pi]$
$\therefore$ f is continuous at $\text{x }\frac{\pi}{4}$ and $\frac{\pi}{2}$
At $\text{x}=\frac{\pi}{4},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\Big(\frac{\pi}{4}-\text{h}\Big)+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}-\text{h}\Big)\Big]\\=\Big[\frac{\pi}{4}+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}\Big)\Big]=\Big[\frac{\pi}{4}+\text{a}\Big]$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{4}+\text{h}\Big)\cot\Big(\frac{\pi}{4}+\text{h}\Big)+\text{b}\Big]\\=\Big[\frac{\pi}{2}\cot\Big(\frac{\pi}{4}\Big)+\text{b}\Big]=\Big[\frac{\pi}{2}+\text{b}\Big]$
At $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{2}-\text{h}\Big)\cot\Big(\frac{\pi}{2}-\text{h}\Big)+\text{b}\Big]=\text{b}$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\text{a}\cos2\Big(\frac{\pi}{2}+\text{h}\Big)-\text{b}\sin\Big(\frac{\pi}{2}+\text{h}\Big)\Big]=-\text{a}-\text{b}$
Since f is continuous at $\text{x}=\frac{\pi}{4}$ and $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}$
$\Rightarrow-\text{b}-\text{a}=\text{b}$ and $\frac{\pi}{4}+\text{a}=\frac{\pi}{2}+\text{b}$
$\Rightarrow\text{b}=\frac{-\text{a}}{2}\ ...(\text{i})$ and $\frac{-\pi}{4}=\text{b}-\text{a}\ ...(\text{ii})$
$\Rightarrow\frac{-\pi}{4}=\frac{-3\text{a}}{2}$ [Substituting the value of b in eq. (ii)]
$\Rightarrow\text{a}=\frac{\pi}{6}$
$\Rightarrow\text{b}=\frac{-\pi}{12}$ [From eq. (i)]
View full question & answer→Question 55 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}|\text{x}-\text{a|}\sin(\frac{1}{\text{x}-\text{a}}), &\text{for} \text{ x} \neq\text{a}\\0,&\text{for} \text{ x} = \text{a}\end{cases}\text{ at x}=0$
Answer$\text{f}\text{(x)}=\begin{cases}|\text{x}-\text{a|}\sin\Big(\frac{1}{\text{x}-\text{a}}\Big),&\text{for }\text{x }\neq \text{a}\\0, &\text{for x } = \text{a}\end{cases}$ $(\text{LHL at x}=\text{a})\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow \text{0}}\text{f}\text{(a}-\text{h)}$ $=\lim\limits_{\text{h} \rightarrow0}\text{|a}-\text{h}-\text{a|}\sin\Big(\frac{1}{\text{a}-\text{h}-\text{a}}\Big)=\lim\limits_{\text{h} \rightarrow0}\text{h}\sin\Big(\frac{1}{-\text{h}}\Big)$= 0 × (an oscillating number between -1 and 1)
= 0 $(\text{RHL at x}=\text{a})\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow \text{0}}\text{f}\text{(a}+\text{h)}$ $=\lim\limits_{\text{h} \rightarrow0}\text{|a}+\text{h}-\text{a|}\sin\Big(\frac{1}{\text{a}+\text{h}-\text{a}}\Big)=\lim\limits_{\text{h} \rightarrow0}\text{h}\sin\Big(\frac{1}{\text{h}}\Big)$= 0 × (an oscillating number between -1 and 1)
Thus, we obtian $\lim\limits_{\text{x} \rightarrow\text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow\text{a}^+}\text{f}\text{(x)}=\text{f}\text{(a)}$ $\therefore$ f(x) is continuous at x = a.
View full question & answer→Question 65 Marks
Extend the definition of the following by continuity $\text{f(x)}=\frac{1-\cos7(\text{x}-\pi)}{5(\text{x}-\pi)^2}$ at the point $\text{x}=\pi.$
AnswerGiven,
$\text{f(x)}=\frac{1-\cos7(\text{x}-\pi)}{5(\text{x}-\pi)^2},\text{ x}=\pi$
If f(x) is continuous at $\text{x}=\pi,$ then
$\lim_\limits{\text{x}\rightarrow\pi}\text{f(x)}=\text{f}(\pi)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow\pi}\frac{1-\cos7(\text{x}-\pi)}{5(\text{x}-\pi)^2}=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin^2\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{(\text{x}-\pi)^2}=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\times\frac{49}{4}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin^2\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{\frac{49}{4}(\text{x}-\pi)^2}=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\times\frac{49}{4}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin^2\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{\Big(\frac{7}{2}(\text{x}-\pi)\Big)^2}=\text{f}(\pi)$
$\Rightarrow\begin{bmatrix}\frac{2}{5}\times\frac{49}{4}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{\Big(\frac{7}{2}(\text{x}-\pi)\Big)}\end{bmatrix}^2=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\times\frac{49}{4}\times1=\text{f}(\pi)$
$\Rightarrow\frac{1}{5}\times\frac{49}{2}\times1=\text{f}(\pi)$
$\Rightarrow\frac{49}{10}=\text{f}(\pi)$
Hence, the given function will be continuous at $\text{x}=\pi,$ if $\text{f}(\pi)=\frac{49}{10}$
View full question & answer→Question 75 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\text{x}^4+\text{x}^3+2\text{x}^2}{\tan^{-1}\text{x}},&\text{if }\text{ x}\neq0\\10,&\text{if }\text{ x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$ we have $\text{f(x)}=\frac{\text{x}^4+\text{x}^3+2\text{x}^2}{\tan^{-1}\text{x}}$
We know that a polynomial is continuous for $x < 0$ and $x > 0$,
Also the inverse trignometric function is continuous in its domain.
Here, $X^4 + x^3 + 2x^2$ is polnomial,
so is continuous for $x < 0$ and $x > 0$ and $\tan^{-1}\text{x}$ is also continuous for $x < 0$ and $x > 0$
So, the quotient function $\text{f(x)}=\frac{\text{x}^4+\text{x}^3+2\text{x}^2}{\tan^{-1}\text{x}}$ is continuous for each $x < 0$ and $x > 0$
Now, consider the point $x = 0$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^4+(-\text{h})^3+2(-\text{h})^2}{\tan^{-1}(-\text{h})}=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^4-\text{h}^3+2\text{h}^2}{\tan^{-1}\text{h}}=0$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^4+\text{h}^3+2\text{h}^2}{\tan^{-1}\text{h}}=0$
$\text{f}(0)=10$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence, the function is not continuous at $x = 0$
View full question & answer→Question 85 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}|\text{x}-3|,&\text{if }\text{ x}\geq1\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},&\text{if }\text{ x}<1\end{cases}$
AnswerWhen $x >,$ then
$f(x) = |x - 3|$
Since modulus function is a continuous function, $f(x)$ is continuous for each $x > 1$
When $x < 1,$ then
$\text{f(x)}=\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}$
Since, $x^2 $ & $3x$ are continuous being polynomial functions, $x^2 $ & $3x$ will also be continuous.
Also, $\frac{13}{4}$ is continuous being a polymomial function.
$\Rightarrow\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}$ is continuous for eqch $x < 1$
$\Rightarrow f(x)$ is continuous for each $x < 1$
At $x = 1$, we have
$(\text{LHL at x}=1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\Big[\frac{(1-\text{h}^2)}{4}-\frac{3(1-\text{h})}{2}+\frac{13}{4}\Big]=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|1+\text{h}-3|\big]=|-2|=2$
Also, $\text{f}(1)=|1-3|=|-2|=2$
Thus, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\text{f}=(1)$
Hence, $f(x)$ is continuous at $x = 1$
Thus, the given function is now where discontinuous.
View full question & answer→Question 95 Marks
Discuss the continuity of the f(x) at the indicated points f(x) = |x - 1| + |x + 1| at x = -1, 1.
AnswerGiven,$\text{f(x)}=|\text{x}-1|+|\text{x}+1|$
We have,
$(\text{LHL at x}=-1)=\lim_\limits{\text{x}\rightarrow-1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-1-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|-1-\text{h}-1|+|-1-\text{h}+1|\big]=2+0=2$
$(\text{RHL at x}= -1)=\lim_\limits{\text{x}\rightarrow-1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-1+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|-1+\text{h}-1|+|-1+\text{h}+1|\big]=2+0=2$
Also, $\text{f}(-1)=|-1-1|+|-1+1|=|-2|=2$
Now,
$(\text{LHL at x}=1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|1-\text{h}-1|+|1-\text{h}+1|\big]=0+2=2$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|1+\text{h}-1|+|1+\text{h}+1|\big]=0+2=2$
$\therefore\ \lim_\limits{\text{x}\rightarrow-1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow-1^+}\text{f(x)}=\text{f}(-1)$ and $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\text{f}=(1)$
Hence, f(x) is continuous at x = -1, 1
View full question & answer→Question 105 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\text{x}^4-16}{\text{x}-2},&\text{if }\text{ x}\neq2\\16,&\text{if }\text{ x}=2\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{x}^4-16}{\text{x}-2},&\text{if }\text{ x}\neq2\\16,&\text{if }\text{ x}=2\end{cases}$
When $\text{x}\neq2,$ then
$\text{f(x)}=\frac{\text{x}^4-16}{\text{x}-2}=\frac{\text{x}^4-2^4}{\text{x}-2}=\frac{(\text{x}^2-4)(\text{x}-2)(\text{x}+2)}{\text{x}-2}=(\text{x}^2+4)(\text{x}+2)$
We know that a polynomial function is everywhere continuous.
Therefore, the functions $(x^2 + 4)$ and $(x + 2)$ are everywhere continuous.
So, the product function $x^2 + 4x + 2$ is everywhere continuous.
Thus, $f(x)$ is continuous at every $\text{x}\neq2$
At x = 2, we have
$(\text{LHL at x}=2)=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[(2-\text{h})^2+4\big](2-\text{h}+2)=8(4)=32$
$(\text{RHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[(2+\text{h})^2+4\big](2+\text{h}+2)=8(4)=32$
Also, f(2) = 16
$\therefore\ \lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}\neq\text{f}(2)$
Thus, f(x) is discontinuous at x = 2
Hence, the only point of discontinuity for f(x) is x = 2
View full question & answer→Question 115 Marks
Find all the points of discontinuity of f defined by f (x) = |x| - |x + 1|.
AnswerThe given function is f(x) = |x| - |x + 1| The two functions, g and are defined as g(x) = |x| and h(x) = |x + 1| Then, f = g - h The continuity of g and h is examined first. g(x) = |x| can be written as $\text{g(x)}=\begin{cases}-\text{x},&\text{if }\text{ x}<0\\\text{x},&\text{if }\text{ x}\geq0\end{cases}$ Clearly, g is defined for all real numbers. Let c be real number.Case I:
If c < 0, then g(c) = -c and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}(-\text{x)}=-\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$ Therefore, g is continuous at all points x, such that x < 0Case II:
If c < 0, then g(c) = c and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}(\text{x)}=\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$ Therefore, g is continuous at all points x, such that x > 0Case III:
If c = 0, then g(c) = g(0) = 0 $\lim\limits_{{\text{x}}\rightarrow0^-}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow0^-}(-\text{x)}=0$ $\lim\limits_{{\text{x}}\rightarrow0^+}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow0^+}\text{g(x)}=0$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow0^-}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow0^+}\text{(x)}=\text{g}(0)$ Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is continuous at all points. h(x) = |x + 1| can be written as $\text{h(x)}=\begin{cases}-(\text{x}+1),&\text{if }\text{ x}<-1\\\text{x}+1,&\text{if }\text{ x}\geq-1\end{cases}$ Clearly, h is defined for every real number. Let c be a real number.Case I:
If c < -1, then h(c) = -(c + 1) and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\big[-(\text{x}+1)\big]=-(\text{c}+1)$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\text{h(c)}$ Therefore, h is continuous at all points x, such that x < -1Case II:
If c > -1, then h(c) = c + 1 and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}(\text{x}+1)=(\text{c}+1)$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\text{h(c)}$ Therefore, h is continuous at all points x, such that x > -1Case III:
If c = -1, then h(c) = h(-1) = -1 + 1 = 0 $\lim\limits_{{\text{x}}\rightarrow1^-}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow1^-}\text{h(x)}\big[-(\text{x}+1)\big]=-(-1+1)=0$ $\lim\limits_{{\text{x}}\rightarrow1^+}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow1^+}(\text{x}+1)=(-1+1)=0$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow1^+}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow1^+}\text{h(x)}=\text{h}(-1)$ Therefore, h is continuous at x = -1 From the above three observation, it can be concluded that h is continuous at all points on the real line. g and h are continuse functions. Therefore f = g - h is also a continuse function. Therefore, f has no point of discontinuity.
View full question & answer→Question 125 Marks
Prove that the function $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.
AnswerWhen x < 0, we have
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ as well as the identity function x are everywhere continuous.
So, the quotient function $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ is continuous at each x < 0
When x > 0, we have f(x) = x + 1, which is a polynomial function.
Therefore, f(x) is continuous at each x > 0
Now, Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}(\text{h}+1)=1$
Also,
$\text{f}(0)=0+1=1$
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0
Hence, f(x) is everywhere continuous.
View full question & answer→Question 135 Marks
Find all point of discontinuity of the function $\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
Answer$\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
Clearly $\text{t}=\frac{1}{\text{x}-1}$ is discontinuous at x = 1
For $\text{x}\neq1,$ we get
$\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2}=\frac{1}{(\text{t}-2)(\text{t}-1)}$
This is discontinuous at t = -2 and t = 1
For t = -2, $\text{t}=\frac{1}{\text{x}-1}$
$\Rightarrow\text{x}=\frac{1}{2}$
For t = 1, $\text{t}=\frac{1}{\text{x}-1}$
⇒ x = 2
Hence, f is discontinuous at $\text{x}=\frac{1}{2},$ x = 1 and x = 2
View full question & answer→Question 145 Marks
Show that $\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$ is discontinuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$
We observe
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3(-\text{h})}{\tan2(-\text{h})}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{\tan2\text{h}}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{3\sin3\text{h}}{3\text{h}}}{\frac{2\tan2\text{h}}{2\text{h}}}\Bigg)$
$=\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{3\sin3\text{h}}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{2\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{3\text{h}}\Big)}{2\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{\text{e}^{2\text{h}}-1}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{3\text{h}\frac{\log(1+3\text{h})}{3\text{h}}}{2\text{h}\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}}\Bigg)$
$\frac{3}{2}\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{\log(1+3\text{h})}{3\text{h}}}{\frac{(\text{e}^{2\text{h}}-1)}{2\text{h}}}\Bigg)$
$=\frac{3}{2}\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
And, $\text{f}(0)=\frac{3}{2}$
$\therefore\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0.
View full question & answer→Question 155 Marks
Find the relationship between 'a' and 'b' so that the function 'f' defind by $\text{f(x)}=\begin{cases}\text{ax}+1,&\text{if }\text{ x}\leq3\\\text{bx}+3,&\text{if }\text{ x}>3\end{cases}$ is continuous at x = 3.
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{ax}+1,&\text{if }\text{ x}\leq3\\\text{bx}+3,&\text{if }\text{ x}>3\end{cases}$
We have,
$(\text{LHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{a}(3-\text{h})+1=3\text{a}+1$
$(\text{RHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{b}(3+\text{h})+3=3\text{b}+3$
If f(x) is continuous at x = 3, then
$\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$
$\Rightarrow3\text{a}+1=3\text{b}+3$
$\Rightarrow3\text{a}-3\text{b}=2$
Hence, the required relationship between a & b is 3a - 2b = 2
View full question & answer→Question 165 Marks
Let $\text{f}\text{(x)}=\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}},\text{x}\neq0$ Find the value of f at x = 0. So that f becomes continuous at x = 0.
AnswerGiven, $\text{f}\text{(x)}=\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}},\text{x}\neq0$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{ax}}{\text{a}}}-\frac{\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\frac{\text{bx}}{\text{b}}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}\end{pmatrix}-\Big(-\frac{1}{\text{b}}\Big)\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\frac{-\text{x}}{\text{b}}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\frac{1}{\text{a}}\times1-\Big(-\frac{1}{\text{b}}\Big)\times1=\text{f}(0)$ $\Big[\text{Using :}\lim_{\text{x} \rightarrow 0}\frac{\text{log(1}+\text{x)}}{\text{x}}=1\Big]$
$\Rightarrow\frac{1}{\text{a}}+\frac{1}{\text{b}}=\text{f}(0)$
$\Rightarrow\frac{\text{a}+\text{b}}{\text{ab}}=\text{f}(0)$
View full question & answer→Question 175 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}-2,&\text{if }\text{ x}\leq-1\\2\text{x},&\text{if }-1<\text{x}\leq1\\2,&\text{if }\text{ x}>1\end{cases}$
AnswerThe given function f is $\text{f(x)}=\begin{cases}-2,&\text{if }\text{ x}\leq-1\\2\text{x},&\text{if }-1<\text{x}\leq1\\2,&\text{if }\text{ x}>1\end{cases}$ The given function is defined at all points of the real line. Let c be a point on the real line.Case I:
If c < -1, then f(c) = -2 and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(-2)=-2$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x < -1Case II:
If c = -1, then f(c) = f(-1) = -2 The left hand limit of f at x = -1 is $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(-2)=-2$ The right hand limit of f at x = -1 is $\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}(2\text{x})=2\times(-1)=-2$ $\therefore\ \lim_\limits{\text{x}\rightarrow-1}\text{f(x)}=\text{f}(-1)$ Therefore, f is continuous at x = -1Case III:
If -1 < c < 1, then f(c) = 2c $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(2\text{x})=2\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points of the interval (-1, 1) Case IV: If c = 1, then f(c) = f(1) = 2 × 1 = 2 The left hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(2\text{x})=2\times1=2$ The right hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}2=2$ $\therefore\ \lim_\limits{\text{x}\rightarrow1}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at x = 2Case V:
If c > 1, then f(c) = 2 and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(2)=2$ $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.
View full question & answer→Question 185 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text {x}^2}, & \text{when} \text{ x}\neq 0\\1, & \text{when}\text{ x} = 0\end{cases}$ Show that f(x) is discontinuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\frac{1-\cos \text{x}}{\text{x}^2}, \text{when x}\neq0$
$\text{f}\text{(x)}=1, \text{when x}=0$
consider
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos \text{x}}{\text{x}^2}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin^2\frac{\text{x}}{2}}{\frac{4\text{x}^2}{4}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\frac{\text{x}^2}{2}}{\frac{4\text{x}^2}{2}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2}{4}\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{1}{2}(1)$
Given f(0) = 1
$\therefore\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}\neq\text{f}(0)$
Thus, f(x) is discontinuous at x = 0.
View full question & answer→Question 195 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}},&\text{x}<\frac{{\pi}}{2}\\3,&\text{x}=\frac{\pi}{2}\\\frac{3\tan\text{x}}{2\text{x}-\pi},&\text{x}>\frac{\pi}{2}\end{cases}$
AnswerSince the f(x) function is continuous at $\text{x}=\frac{\pi}{2}$ therefore
LHL of f(x) at $\text{x}=\frac{\pi}{2}$ is
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\text{h}-\frac{\pi}{2}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{k}\cos\text{f}\Big(\text{h}-\frac{\pi}{2}\Big)}{\pi-2\text{f}\Big(\text{h}-\frac{\pi}{2}\Big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{k}\sin\text{h}}{2\pi-2\text{h}}$
$=\frac{\text{k}}{2}\lim\limits_{\text{h}\rightarrow0}\frac{\sin(\pi-\text{h})}{(\pi-\text{h})}$
$=\frac{\text{k}}{2}$
Again,
$\text{f}\Big(\frac{\pi}{2}\Big)=3$
Hence,
$\text{LHL}=\text{f}\Big(\frac{\pi}{3}\Big) $
$\frac{\text{k}}{2}=3$
$\text{k}=6$
View full question & answer→Question 205 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\text{x}^3-\text{x}^2+2\text{x}-2,&\text{if }\text{ x}\neq1\\4,&\text{if }\text{ x}=1\end{cases}$
AnswerWhen $\text{x}\neq1,$ then
$\text{f(x)}=\text{x}^3-\text{x}^2+2\text{x}-2$
We know that a polynomial function is everywhere continuous.
So, $\text{f(x)}=\text{x}^3-\text{x}^2+2\text{x}-2$ is continuous at each $\text{x}\neq1$
At x = 1, we have
$(\text{LHL at x}=1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\Big((1-\text{h})^3-(1-\text{h})^2+2(1-\text{h})-2\Big)=1-1+2-2=0$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\Big((1+\text{h})^3-(1+\text{h})^2+2(1+\text{h})-2\Big)=1-1+2-2=0$
Also, f(1) = 4
$\therefore\ \lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}\neq\text{f}(1)$
Thus, f(x) is discontinuous at x = 1
Hence, the only point of discontinuity for f(x) is x = 1
View full question & answer→Question 215 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{if }\text{ x}\neq0\\4,&\text{if }\text{ x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$ then
$\text{f(x)}=\frac{\sin3\text{x}}{\text{x}}$
We know that $\sin3\text{x}$ as well as the identity function x are everwhere continuous.
So, the quotient function $\frac{\sin3\text{x}}{\text{x}}$ is continuous at each $\text{x}\neq0$
Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{if }\text{ x}\neq0\\4,&\text{if }\text{ x}=0\end{cases}$
We have
$(\text{LHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-3\text{h})}{-\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{3\sin(3\text{h})}{3\text{h}}\Big)=3$
$(\text{RHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(3\text{h})}{\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{3\sin(3\text{h})}{3\text{h}}\Big)=3$
Also, f(0) = 4
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}\neq\text{f}(0)$
Thus, f(x) is discontinuous at x = 0
Hence, the only point of discontinuity for f(x) is x = 0
View full question & answer→Question 225 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{(\text{x}-2)^2},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{(\text{x}-2)^2},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{\text{x}^2-4\text{x}+4},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\text{x}+5,&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
If f(x) is continuous at x = 2, then
$\lim_\limits{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow2}\text{(x}+5)=\text{k}$
$\Rightarrow\text{k}=2+5=7$
View full question & answer→Question 235 Marks
Show that $\text{f}\text{ (x)}=\begin{cases}\frac{\text{|x}-\text{a}|}{|\text{x}-\text{a}|}, & \text{when} \text{ x}\neq 0\\2, & \text{when}\text{ x} = 0\end{cases}$ is discontinuous at x = a.
AnswerThe given function can be rewritten as:
$\text{f}\text{(x)}=\begin{cases}\frac{\text{z}-\text{a}}{\text{z}-\text{a}}, & \text{when} \text{ x}> 0\\\frac{\text{a}-\text{x}}{\text{z}-\text{a}}, & \text{when}\text{ x} < 0\\ 1,&\text{when}\text{ x} = \text{a}\end{cases}$
$\text{f}\text{(x)}=\begin{cases}1, & \text{when} \text{ x}> \text{a}\\-1, & \text{when}\text{ x} < \text{a}\\ 1,&\text{when x}= \text{a}\end{cases}$
$\text{f}\text{(x)}=\begin{cases}1, & \text{when} \text{ x}\geq \text{a}\\-1, & \text{when}\text{ x} < \text{a}\end{cases}$
We observe
$(\text{LHL at x}=\text{a})=\lim\limits_{\text{z} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(a}-\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}(-1)=-1$
$(\text{RHL at x}=\text{a})=\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(a}+\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}(1)=1$
$\therefore\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}$
Thus, f(x) is discontinuous at x = a
View full question & answer→Question 245 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}(\text{x}-1)\tan\frac{\pi\text{x}}{2},&\text{if}\text{ x}\neq1\\\text{k},&\text{if}\text{ x}=1\end{cases}\text{at x} = 1$
AnswerLet $\text{x}-1=\text{y}$
$\Rightarrow\text{x}=\text{y}+1$
Thus,
$\lim_\limits{\text{x}\rightarrow 1}(\text{x}-1)\tan\frac{\pi\text{x}}{2}=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\frac{\pi(\text{y}+1)}{2}$
$=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\Big(\frac{\pi\text{y}}{2}+\frac{\pi}{2}\Big)$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\cot\frac{\pi\text{y}}{2}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\sin\frac{\pi\text{y}}{2}}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{2}{\pi}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\frac{2}{\pi}\lim_\limits{\text{y}\rightarrow0}\cos\frac{\pi\text{y}}{2}$
$=-\frac{2}{\pi}$
Since the function is continuous, LHL = RHL.
Thus, $\text{k}=-\frac{2}{\pi}$
View full question & answer→Question 255 Marks
Find the value of a and b so that the function f given by $\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$ is continuous x = 3 and x = 5.
AnswerGiven,
$\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$
We have,
$(\text{LHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})=\lim_\limits{\text{h}\rightarrow0}(1)=1$
$(\text{RHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{a}(3+\text{h})+\text{b}=3\text{a}+\text{b}$
$(\text{LHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{a}(5-\text{h})+\text{b})=5\text{a}+\text{b}$
$(\text{RHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}7=7$
If f(x) is continuous at x = 3 and 5, then
$\therefore\ \lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$ and $\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}$
$\Rightarrow1=3\text{a}+\text{b}\ .... (\text{i})$ and $5\text{a}+\text{b}=7\ .... (\text{ii})$
On solving eqs. (i) and (ii) we get
$\text{a}=3$ and $\text{b}=-8$
View full question & answer→Question 265 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-4}{\text{|x}-4|}+\text{a}, &\text{if x} <4\\\text{a}+\text{b},&\text{if x}=4\\\frac{\text{x}-4}{\text{|x}-4|}+\text{b}, & \text{if x}>4\end{cases}$ is continuous at $x = 4$. Find $a, b$.
AnswerGiven,
f(x) is continuous at x = 4 & f(4) = a + b
For f(x) to be continuous at $x = 4, f(4)^- = f(4)^+= f(4)$
$\text{L.H.L}=\text{f(4)}^-=\lim\limits_{\text{x} \rightarrow 4}\frac{\text{x}-4}{|\text{x}-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h})-4}{|(4-\text{h})-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h}-4)}{|(4-\text{h}-4)|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{|(-\text{h})|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{\text{h}}+\text{a}$
$\Rightarrow\text{a}-1$
$\text{L.H.L}=\text{f(4)}^+=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}-4}{|\text{x}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4+\text{h})-4}{|(4+\text{h})-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{4+\text{h}-4}{|4+\text{h}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{|\text{h}|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{1}{|1|}$
$\Rightarrow1$
Since, f(x) is is continuous at x = 4 & f(4) = a + b
$\text{f(4)}^-=\text{f(4)}^+=\text{f(4)}$
$\therefore\ \text{a}-1=\text{a}+\text{b}=1$
$\Rightarrow\text{a}-1=1$
$\Rightarrow\text{a}=2$
$\Rightarrow\text{a}+\text{b}=1$
$\Rightarrow\text{b}=1-2$
$\Rightarrow\text{b}=-1$
View full question & answer→Question 275 Marks
Prove that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}}{|\text{x|+2}\text{x}^2}, &\text{ x}\neq0\\\text{k}, &\text{ x}=0\end{cases}$
remains discontinuous at x = 0, regardless the choice of k.
AnswerThe given function can be rewritten as,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}}{\text{x}+2\text{x}^2}, & \text{x} > 0\\\frac{-\text{x}}{\text{x}-2\text{x}^2}, &\text{x} <0\\ \text{k},&\text{x}=0\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\frac{1}{2\text{x}+1}, &\text{x} > 0\\\frac{1}{2\text{x}-1}, & \text{x} <0\\\text{k},&\text{x} =0\end{cases}$
We observe
$\text{(LHL at x}=0)\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{1}{-2\text{h}-1}=-1$
$\text{(RHL at x}=0)\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{1}{2\text{h}+1}=1$
So,$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$ Such that
$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\&\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
are independent of k.
Thus, f(x) is discontinuous at x = 0,
regardless of the choice of k.
View full question & answer→Question 285 Marks
If $\text{f(x)}=\begin{cases}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, find k.
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\sin^2\text{x}-\sin^2\text{x}-1}{{\sqrt{\text{x}^2+1}-1}}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2\sin^2\text{x}}{{\sqrt{\text{x}^2+1}-1}}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\big({\sqrt{\text{x}^2+1}-1}\big)\big({\sqrt{\text{x}^2+1}+1}\big)}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\text{x}^2}=\text{k}$
$\Rightarrow-2\lim_\limits{\text{x}\rightarrow 0}\frac{(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\text{x}^2}=\text{k}$
$\Rightarrow-2\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^2\lim_\limits{\text{x}\rightarrow 0}\Big({\sqrt{\text{x}^2+1}-1}\Big)$
$\Rightarrow-2\times-1\times(1+1)=\text{k}$
$\Rightarrow\text{k}=-4$
View full question & answer→Question 295 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\text{x}^{10}-1,&\text{if }\text{ x}\leq1\\\text{x}^2,&\text{if }\text{ x}>1\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\text{x}^{10}-1,&\text{if }\text{ x}\leq1\\\text{x}^2,&\text{if }\text{ x}>1\end{cases}$ The given function f is defind at all the points of the relline. Let c be a point on the real line.Case I:
If c < 1, then $\text{f(c)}=\text{c}^{10}-1$ and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(\text{x}^{10}-1)=\text{c}^{10}-1$ $\therefore\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x such, that x < 1Case II:
If c = 1, then the left hand limit of at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}(\text{x}^{10}-1)=1^{10}-1=1-1=0$ The right hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}(\text{x}^2)=1^1=1$ It is observed that the left and right hand limit of at x = 1 do not coincide. Therefore, f is not continuouse at x = 1Case III:
If c > 1, then $f(c) = c^2$ $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(\text{x}^2)=\text{c}^2$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observation, if can be concluded that x = 1 is the only point of discontinuity of y.
View full question & answer→Question 305 Marks
Show that $\text{f(x)}=\cos\text{x}^2$ is a continuous function.
AnswerGiven, $\text{f(x)}=\cos\big(\text{x}^2\big)$
This function f is defined for every real number and f can be written as the composition of two functions as
f = goh, where $\text{g(x)}=\cos\text{x}$ and $h(x) = x^2$
$\big[\because(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\text{x}^2)=\cos(\text{x}^2)=\text{f(x)}\big]$
It has to be first proved that $\text{g(x)}=\cos\text{x}$ and $h(x) = x^2$ are continuous functions.
It is evident that g is defined for every real number.
Let c be a real number.
Then, $\text{g(c)}=\cos\text{c}$
Put x = c + h
If x → c, then h→ 0
$=\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x})=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
So, $\text{g(x)}=\cos\text{x}$ is a continuous function.
Now,
$h(x) = x^2$
Clearly, h is defined for every real number.
Let k be a real number, then $h (k) = k^2$
$=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{x}^2=\text{k}^2$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\text{h}(\text{k})$
So, h is a continuous function.
It is known that for real valued functions g and h, such that (goh) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (fog) is continuous at x = c.
Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\cos(\text{x}^2)$ is a continuous function.
View full question & answer→Question 315 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}|\text{x}|+3,&\text{if }\text{ x}\geq-3\\-2\text{x},&\text{if }-3<\text{ x}<3\\6\text{x}+2,&\text{if }\text{ x}>3\end{cases}$
AnswerWhen x < -3,
f(x) = |x| + 3
We know that |x| is continuous for x < -3
$\therefore$ |x| + 3 is continuous for x < -3
When x > 3
f(x) = 6x + 2 which is a polynomial of degree 1, so f(x) = 6x + 2 is continuous for x > 3
When -3 < x < 3
f(x) = -2x which is again a polynomial so, it is continuous for -3 < x < 3
Now, consider the point x = -3
$\text{LHL}=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}|-3-\text{h}|+3=\lim_\limits{\text{h}\rightarrow0}|3+\text{h}|+3=6$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}-2(-3+\text{h})=6$
$\text{f}(-3)=|-3|+3=6 $
Thus, LHL = RHL = f(-3) = 6
So, the function is continuous at x = 3
Now, consider the point x = 3
$\text{LHL}=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}-2(3-\text{h})=-6$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}6(3+\text{h})+2=20$
Thus, $\text{LHL}\neq\text{RHL}$
Hence, f(x) is discontinuous at x = 3
View full question & answer→Question 325 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}4,&\text{if }\text{ x}\leq-1\\\text{ax}^2+\text{b},&\text{if }-1<\text{ x}<0\\\cos\text{x},&\text{if }\text{ x}\geq0\end{cases}$
AnswerIt is given that the function is continuous
At x = -1
f(-1) = 4
$\text{RHL}=\lim_\limits{\text{x}\rightarrow-1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{a}(-1+\text{h})^2+\text{b}=\text{a}+\text{b}$
Since, f(x) is continuous at x = -1
$\therefore$ a + b = 4 ...(i)
Now, at x = 0
$\text{f}(0)=\cos0^\circ=1$
$\text{LHL}=\lim_\limits{\text{h}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{a}(-\text{h})^2+\text{b}=\text{b}$
$\therefore$ f(0) = LHL
b = 1
$\therefore$ from (i)
a = 3
Thus, a = 3, b = 1
View full question & answer→Question 335 Marks
Determine if $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$ is a continuous function?
AnswerThe given function f is $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$
It is evident that f is defind at all points of the real line.
Let c be a real number.
Case I:
If $\text{c}\neq0,$ then $\text{f(c)}=\text{c}^2\sin\frac{1}{\text{c}}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\Big(\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\frac{1}{\text{x}}\Big)=\text{c}^2\sin\frac{1}{\text{c}}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$
So, f is continuouse at all points $\text{x}\neq0$
Case II:
If c = 0, then f(0) = 0
$\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow0^-}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)$
It is know that $-1\leq\sin\frac{1}{\text{x}}\leq1,\text{x}\neq0$
$\Rightarrow-\text{x}^2\leq\text{x}^2\sin\frac{1}{\text{x}}\leq\text{x}^2$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}(-\text{x}^2)\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq\lim\limits_{{\text{x}}\rightarrow0}\text{x}^2$
$\Rightarrow0\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=0$
Similarly, $\lim\limits_{{\text{x}}\rightarrow0^+}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}$
So, f is continuous at x = 0
From the above observations, it can be continuouse that f is continuous at every point of the real line.
Thus, f is a continuous function.
View full question & answer→Question 345 Marks
Examine the continuity of the function
$\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0$
Also sketch the graph of this function.
AnswerGiven function is, $\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0\ ...(\text{ii})$ We need to check whether f(x) is continuous at x = 0 or not. For this we need to check L.H.L, R.H.L and value of function at x = 0 Clearly, f(0) = 3*0 - 2 = -2 [from equation ii] $\text{L.H.L}=\lim\limits_{\text{h} \rightarrow 0}(0-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h})$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{3(-\text{h})-2\big\}=-2$ $\text{R.H.L}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f(h)}$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{\text{h}+1\big\}=0+1=1$ As, $\text{L.H.L}\neq\text{R.H.L}$ $\therefore\ \text{f(x)}$ is discontinuous at x = 0 This can also be proved by plotting f(x) on cartesian plane. For x >0 ,we need to plot y = x + 1 put y = 0, we get x = -1 and for second point we put x = 0 and thus get y = 1 Two points are enough to plot the straight line. Two coordinates are (-1, 0) and (0, 1) For $\text{x}\leq0,$ we need to plot y = 3x - 2 put x = 0 then y = -2 On putting y = 0 we get $\text{x}=\frac{2}{3}$ Two coordinates are (0, -2) and $\Big(\frac{2}{3},0\Big)$ It can be seen from graph that there is breakage in curve at (0, 0) Thus, it is discontinuous at x = 0
View full question & answer→Question 355 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{2\text{x}+\text{x}^2}{\text{x}}, & \text{x} \neq0\\0,&\text{ x} = 0\end{cases}\text{at x}=0$
AnswerGiven,
$\Rightarrow\text{f}\text{(x)}=\frac{2\text{x}+\text{x}^2}{\text{x}},\text{x}>0$
$\Rightarrow\text{f}\text{(x)}=\frac{-2\text{x}+\text{x}^2}{\text{x}},\text{x}>0$
$\Rightarrow\text{f}\text{(x)}=0,\ \text{x}=0$
$\Rightarrow\text{f}\text{(x)}=\text{x+2},\ \text{x}>0$
$\Rightarrow\text{f}\text{(x)}=\text{x-2},\ \text{x}<0$
$\Rightarrow\text{f}\text{(x)}=0,\ \text{x}=0$
We observe
$(\text{LHL at x }=0)$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(0-h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}=\lim\limits_{\text{h} \rightarrow 0}-\text{h-2}$
$=-2$
$(\text{RHL at x}=0)$
$\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(0+h)}$
$\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(0+h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(h)}=\lim\limits_{\text{h} \rightarrow 0}\text{h+2}$
$=2$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 365 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}\sqrt{1-\text{px}}}{\text{x}},&\text{if }-1\leq\text{ x}\leq-0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{ x}\leq1\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}\sqrt{1-\text{px}}}{\text{x}},&\text{if }-1\leq\text{ x}\leq-0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{ x}\leq1\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\bigg(\frac{\sqrt{1-\text{px}}\sqrt{1+\text{px}}}{-\text{h}}\bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(\sqrt{1-\text{ph}}\big)-(\sqrt{1+\text{ph}})\big(\sqrt{1-\text{ph}}\big)+\big(1+\text{ph}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}+\sqrt{1+\text{ph}}\big)}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(1-\text{ph}-1-\text{ph}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}+\sqrt{1+\text{ph}}\big)}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(-2\text{ph}\big)}{-\text{h}\sqrt{1-\text{ph}+\sqrt{1+\text{ph}}}}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(2\text{p}\big)}{\big(\sqrt{1-\text{ph}}+\sqrt{1+\text{ph}}\big)}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\Big(\frac{(2\text{p})}{(2)}\Big)=\Big(\frac{1}{-2}\Big)$
$\Rightarrow\text{p}=\frac{-1}{2}$
View full question & answer→Question 375 Marks
The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&\text{if }0\leq\text{ x}<1\\\text{a},&\text{if }1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\text{if }\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous on $(0,\infty),$ then find the most suitable value of a and b.
AnswerGiven, f is continuous on $(0,\infty)$
$\therefore$ f is continuous at x = 1 and $\sqrt{2}$
Ar x = 1, we have
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{(1-\text{h})^2}{\text{a}}\bigg]=\frac{1}{\text{a}}$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
Also,
At $\text{x}=\sqrt{2},$ we have
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{2\text{b}^2-4\text{b}}{(\sqrt{2}+\text{h})^2}\bigg]=\frac{2\text{b}^2-4\text{b}}{2}=\text{b}^2-2\text{b}$
f is continuous at x = 1 and $\sqrt{2}$
$\therefore\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}$
$\Rightarrow\frac{1}{\text{a}}=\text{a}$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}^2=1$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}=\pm1$ and $\text{b}^2-2\text{b}=\text{a}\ ...(\text{i})$
If a = 1, then
$\text{b}^2-2\text{b}=\text{a}$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}-1=0$
$\Rightarrow\text{b}=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm2\sqrt{2}}{2}\\=1\pm\sqrt{2}$
If a = -1, then
$\Rightarrow\text{b}^2-2\text{b}=-1$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}+1=0$
$\Rightarrow(\text{b}-1)^2=0$
$\Rightarrow\text{b}=1$
Hence, the most suitable value of a and b are
$\text{a}=-1,\text{ b}=1$ or $\text{a}=1,\text{ b}=1\pm\sqrt{2}$
View full question & answer→Question 385 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}5,&\text{if }\text{ x}\leq2\\\text{ax}+\text{b},&\text{if }2<\text{x}<10\\21,&\text{if }\text{ x}\geq10\end{cases}$
AnswerGiven,
$\text{f(x)}=\begin{cases}5,&\text{if }\text{ x}\leq2\\\text{ax}+\text{b},&\text{if }2<\text{x}<10\\21,&\text{if }\text{ x}\geq10\end{cases}$
If f(x) is continuous x = 2 and 10, then
$\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow10^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow10^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$ and $\lim_\limits{\text{h}\rightarrow0}\text{f}(10-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(10+\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}=\lim_\limits{\text{h}\rightarrow0}\big[\text{a}(2+\text{h})+\text{b}\big]$ and $\lim_\limits{\text{h}\rightarrow0}\big[\text{a}(10-\text{h})+\text{b}\big]=\lim_\limits{\text{h}\rightarrow0}(21)$
$\Rightarrow5=2\text{a}+\text{b}\ ....(\text{i})$ and $10\text{a}+\text{b}=21\ ....(\text{ii})$
On solving eqs. (i) and (ii) we get
$\text{a}=2$ and $\text{b}=1$
View full question & answer→Question 395 Marks
Discuss the continuity of the following functions:
- $\text{f(x)}=\sin\text{x}+\cos\text{x}$
- $\text{f(x)}=\sin\text{x}-\cos\text{x}$
- $\text{f(x)}=\sin\text{x}\cos\text{x}$
AnswerIt is know that if g and h are two continuous functions, then
g + h, g - h, and g, h are also continuous.
It has to proved first that $\text{g(x)}=\sin\text{x}$ and $\text{h(x)}=\cos\text{x}$ are continuous functions.
Let $\text{g(x)}=\sin\text{x}$
It is evident that $\text{g(x)}=\sin\text{x}$ is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
$\text{g(c)}=\sin\text{c}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\text{x}$
$=\lim\limits_{\text{h}\rightarrow0}\sin(\text{c}+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\big[\sin\text{c}\cos\text{h}+\cos\text{c}\sin\text{h}\big]$
$=\lim\limits_{\text{h}\rightarrow0}(\sin\text{c}\cos\text{h})+\lim\limits_{\text{h}\rightarrow0}(\cos\text{c}\sin\text{h})$
$=\sin\text{c}\cos0+\cos\text{c}\sin0$
$=\sin\text{c}+0$
$=\sin\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
Therefore, g is a continuous function.
Let $\text{h(x)}=\cos\text{x}$
It is evident that $\text{h(x)}=\cos\text{x}$ is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
$\text{h(c)}=\cos\text{c}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\text{h(c)}$
Therefore, h is a continuous function.
Therefore, it can be concluded that
- $\text{f(x)}=\text{g(x)}+\text{h(x)}=\sin\text{x}+\cos\text{x}$ is a continuous function.
- $\text{f(x)}=\text{g(x)}-\text{h(x)}=\sin\text{x}-\cos\text{x}$ is a continuous function.
- $\text{f(x)}=\text{g(x)}\times\text{h(x)}=\sin\text{x}\times\cos\text{x}$ is a continuous function.
View full question & answer→Question 405 Marks
The function f(x) is befined as follows: $\text{f(x)}=\begin{cases}\text{x}^2+\text{ax}+\text{b},&0\leq\text{x}<2\\3\text{x}+2,&2\leq\text{x}\leq4\\2\text{ax}+5\text{b},&4<\text{x}\leq8\end{cases}$ if is continuous on [0, 8] find the value of a and b.
AnswerIt is given that the f(x) is continuous on [0, 8]
f(x) is continuous at x = 2 and x = 4
Now, At x = 2
LHL = RHL = f(2) ....(A)
f(2) = 3 × 2 + 2 = 8 ....(i)
$\text{LHL}=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2-\text{h}^2)+\text{a}(2-\text{h})+\text{b}=4+2\text{a}+\text{b}$
From (A)
4 + 2a + b = 8
2a + b = 4 ....(B)
Now, At x = 4
LHL = RHL = f(4) ....(C)
f(4) = 3 × 4 + 2 = 14 ....(ii)
$\text{RHL}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(4+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}2\text{a}(4+\text{h})+5\text{b}=8\text{a}+5\text{b}$
From (c) we get
8a + 5b = 14 ....(D)
Solving (B) and (D) we get
a = 3 and b = -2
View full question & answer→Question 415 Marks
Find the values of a so that the function
$\text{f}\text{(x)}=\begin{cases}\text{ax}+5, &\text{if}\text{ x}\leq2\\\text{x}-1, &\text{if}\text{ x}>2\end{cases}$ is continuous at x = 2.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\text{ax}+5, &\text{if}\text{ x}\leq2\\\text{x}-1, &\text{if}\text{ x}>2\end{cases}$
We observe
$\text{(LHL at x}=2)=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{a}(2-\text{h)}+5=2\text{a}+5$
$\text{(RHL at x}=2)=\lim\limits_{\text{x} \rightarrow 2^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(2+\text{h}-1)=1$
And, $\text{f}(2)=\text{a}(2)+5=2\text{a}+5$
Since f(x) is continuous at x = 2, we have
$=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 2^+}\text{f}\text{(x)}=\text{f}(2)$
$\Rightarrow2\text{a}+5=1$
$\Rightarrow2\text{a}=-4$
$\Rightarrow\text{a}=-2$
View full question & answer→Question 425 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1},& \text{for }\text{ x}\neq1 \\2,&\text{for }\text{ x}=1\end{cases}$ Find whether f (x) is continuous at x = 1.
AnswerGiven,
$\text{f}\text{(x)}=\frac{\text{x}^2-1}{\text{x}-1},\text{ if}\text{ x}\neq1$
$\text{f}\text{(x)}=2,\text{ if}\text{ x}=1$
We observe
$\text{(LHL at x = 1)}$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{ (x)}=\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})$
$\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})=\lim\limits_{\text{x} \rightarrow 0}\frac{(1-\text{h})^2-1}{(1-\text{h})^2-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{1-\text{h}^2-2\text{h}-1}{1-\text{h}-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{\text{h}^2-2\text{h}}{-\text{h}}$
$\lim\limits_{\text{x} \rightarrow 0}2-\text{h}$
$= 2$
$(\text{RHL at x}=1)$
$\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}(1+\text{h)}$
$\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1+\text{h})^2-1}{(1+\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{1+\text{h}^2+2\text{h}-1}{1+\text{h}-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+2\text{h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}\text{h}+2$
$=2$
Also f(x) = 2
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\text{f}(1)$
Hence f(x) is continuous at x = 1.
View full question & answer→Question 435 Marks
Discuss the continuity of $\text{f(x)}=\sin|\text{x}|$
AnswerLet $\text{f(x)}=\sin|\text{x}|$ This function f is defined for every real number and f can be written as the composition of two functions as, f = goh, where g(x) = |x| and $\text{h(x)}=\sin\text{x}$ $\big[\because (\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\sin\text{x})=|\sin\text{x}|=\text{f(x)}]\Big]$ It has to be proved first that g(x) = |x| and $\text{h(x)}=\sin\text{x}$ are continuous functions. g(x) = |x| can be written as $\text{g(x)}=\begin{cases}-\text{x},&\text{if }\text{ x}<0\\\text{x},&\text{if }\text{ x}\geq0\end{cases}$ Clearly, g is defined for all real numbers. Let c be real number.Case I:
If c < 0, then g(c) = -c and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}(-\text{x})=-\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$ Therefore, g is continuous at all points x, such that x < 0Case II:
If c > 0, then g(c) = c and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}(\text{x})=\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$ Therefore, g is continuous at all points x, such that x > 0Case III:
If c = 0, then g(c) = g(0) = 0 $\lim\limits_{\text{x}\rightarrow0^-}\text{g(x)}=\lim\limits_{\text{x}\rightarrow0^-}\text{g}(-\text{x})=0$ $\lim\limits_{\text{x}\rightarrow0^+}\text{g(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{g(x)}=0$ $\therefore\ \lim\limits_{\text{x}\rightarrow0^-}\text{g(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{g(x)}=\text{g}(0)$ Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is continuous at all points. $\text{h(x)}=\sin\text{x}$ It is evident that $\text{h(x)}=\sin\text{x}$ is defined for every real number. Let c be a real number. Put x = c + k If x → c, then k → 0 $\text{h(c)}=\sin\text{c}$ $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\text{x}$ $=\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin(\text{c}+\text{k})$ $=\lim\limits_{{\text{x}}\rightarrow\text{c}}\big[\sin\text{c}\cos\text{k}+\cos\text{c}\sin\text{k}\big]$ $=\lim\limits_{{\text{x}}\rightarrow\text{c}}(\sin\text{c}\cos\text{k})+\lim\limits_{{\text{x}}\rightarrow\text{c}}(\cos\text{c}\sin\text{k})$ $=\sin\text{c}\cos0+\cos\text{c}\sin0$ $=\sin\text{c}+0$ $=\sin\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\text{g(c)}$ Therefore, h is a continuous function. It is know that for real valued functions g and h, such that (goh) is defind at c, if g is continuse at c and if f is continuous at g(c), then (fog) is continuous at c. Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\sin\text{x})=|\sin\text{x}|$ is a continuous function.
View full question & answer→Question 445 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^\text{x}-1}{\log(1+2\text{x})}&,\text{ if x}\neq0&\\ \ &&\text{at x}=0\\7&,\text{ if x}=0\end{cases}$
AnswerIn this problem we need to check continuity at x = 0
Given function is,
$\text{f(x)}=\begin{cases}\frac{\text{x}^\text{x}-1}{\log(1+2\text{x})}&,\text{ if x}\neq0&\\ \ &&\text{at x}=0\\7&,\text{ if x}=0\end{cases}$
$\therefore$ We need to check L.H.L, R.H.L and value of function at x = 0
Idea of logarithmic limit and exponential limit,
$\lim\limits_{\text{x} \rightarrow 0}\frac{\log(1+\text{x})}{\text{x}}=1\ ...(\text{i})$
$\lim\limits_{\text{x} \rightarrow 0}\frac{\text{e}^{\text{x}-1}}{\text{x}}=1\ ...(\text{ii})$
You must have read such limits. You can verify these by expanding log(1 + x) and $e^x$ in its taylor form.
Numerator and denominator conditions also hold for this limit like sandwich theorem.
E.g: $\lim\limits_{\text{x} \rightarrow 0}\frac{\log(1+2\text{x})}{2\text{x}}=1$
But, $\lim\limits_{\text{x} \rightarrow 0}\frac{\log(1+2\text{x})}{\text{x}}\neq1$ as denominator does not have 2x
Now we are ready to solve the problem.
Given function is,
$\text{f(x)}=\begin{cases}\frac{\text{e}^\text{x}-1}{\log(1+2\text{x})}&,\text{ if x}\neq0&\\ \ \ \ \ &&\text{at x}=0\\7&,\text{ if x}=0\end{cases}\ ...(\text{iii})$
Clearly,
f(0) = 7 [from equation 2]
$\text{L.H.L}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h})$ [putting x = -h in equation iii]
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{-\text{h}}-1}{\log1+2(-\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{-\text{h}}-1}{\log(1-2\text{h})}$
Using logarithmic and exponential limit as explained above, we have,
$\text{L.H.L}=\frac{1}{2}\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{(\text{e}^{-\text{h}}-1)}{-\text{h}}}{\frac{\log(1-2\text{h})}{-2\text{h}}}=\frac{1}{2}$
$\text{RHL}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f(h)}$ [putting x = h in equation iii]
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{\text{h}}-1}{\log1+2\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{e}^{\text{h}}-1}{\log(1+2\text{h})}$
Using logarithmic and exponential limit as explained above, we have,
$\text{R.H.L}=\frac{1}{2}\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{(\text{e}^\text{h}-1)}{\text{h}}}{\frac{\log(1+2\text{h})}{2\text{h}}}=\frac{1}{2}$
Thus, $\text{L.H.L}=\text{R.H.L}\neq\text{f(0)}$
$\therefore\ \text{f(x)}$ is discontinuous at x = 0
View full question & answer→Question 455 Marks
Show that $\text{f(x)}=|\cos\text{x}|$ is a continuous function.
AnswerIf f is a real function on a subset of the real numbers and c be a point in the domain off, then f is
continuous at c is $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$
Step I:
Let $\text{g(x)}=|\text{x}|$
$\text{h(x)}=\cos\text{x}$
$\text{f(x)}=(\text{goh})(\text{x})$
$=\text{g}(\text{h(x)})$
$=\text{g}(\cos\text{x})$
$={|\cos\text{x}|}$
$\text{g(x)}=|\text{x}|$ and $\text{h(x)}=\cos\text{x}$
Both are continuous for all values of $\text{x}\in\text{R}$
Step II:
(goh)(x) is also continuous
$\text{f(x)}=(\text{goh})(\text{x})$
$={|\cos\text{x}|}$
$\text{f(x)}={|\cos\text{x}|}$ is continuous for all values of $\text{x}\in\text{R}$
View full question & answer→Question 465 Marks
Prove that $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.
AnswerWhen x < 0, we have
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\sin\text{x}}{\text{x}}$ is continuous at each x < 0
When x > 0, we have
f(x) = x + 1, which is a polynomial function.
Therefore, f(x) is continuous at each x > 0
Now,
Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$
We have,
$(\text{LHL at x = 0})=\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)=1$
$(\text{RHL at x = 0})=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}=\lim\limits_{\text{h}\rightarrow0}(\text{h}+1)=1$
Also, $\text{f}(0)=0+1=1$
$\therefore\ \lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0
Hence, f(x) is everywherefore continuous.
View full question & answer→Question 475 Marks
Let $\text{f(x)}=\begin{cases}\frac{1-\sin^3\text{x}}{3\cos^2\text{x}},&\text{if }\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if }\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})}^2,&\text{x}>\frac{\pi}{2}\end{cases}$ if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ find a and b.
AnswerIt is given that the function is continuous at $\text{x}=\frac{\pi}{2}$
$\therefore\ \text{LHL}=\text{RHL}=\text{f}\Big(\frac{\pi}{2}\Big)\ ...(\text{i})$
Now, $\text{f}\Big(\frac{\pi}{2}\Big)=\text{a}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow\frac{\pi^-}{2}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow 0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{1-\sin^3\Big(\frac{\pi}{2}-\text{h}\Big)}{3\cos^2\Big(\frac{\pi}{2}-\text{h}\Big)}=\lim_\limits{\text{h}\rightarrow 0}\frac{1-\cos^3\text{h}}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{(1-\cos\text{h})(1+\cos^2\text{h}+\cos\text{h})}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\sin^2\frac{\text{h}}{2}(1+\cos^2\text{h}+\cos\text{h})}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\Bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg)^2\times\frac{\text{h}^2}{4}\times(1+\cos^2\text{h}+\cos\text{h})}{3\Big(\frac{\sin\text{h}}{\text{h}}\Big)^2\times\text{ h}^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\times\frac{1}{4}(1+\cos^2\text{h}+\cos\text{h})}{3}=\frac{1}{2}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow\frac{\pi^-}{2}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow 0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}\Big(1-\sin\Big(\frac{\pi}{2}+\text{h}\Big)\Big)}{\Big(\pi-2\Big(\frac{\pi}{2}+\text{h}\Big)\Big)^2}=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}(1-\cos\text{h})}{(\pi-\pi-2\text{h})^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b }\times\ 2\sin^2\frac{\text{h}}{2}}{(2\text{h})^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}}{2}\Bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg)^2\times\frac{1}{4}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}}{8}=\frac{\text{b}}{8}$
Thus, using (i) we get,
$\text{a}=\frac{1}{2}$
And $\frac{\text{b}}{8}=\frac{1}{2}$
$\Rightarrow\text{b}=4$
Thus, $\text{a}=\frac{1}{2}$ and $\text{b}=4$
View full question & answer→Question 485 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}2,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{ x}<5\\9,&\text{if }\text{ x}\geq5\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}2,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{ x}<5\\9,&\text{if }\text{ x}\geq5\end{cases}$
If f(x) is continuous at x = 3 and 5, then
$\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$ and $\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$ and $\lim_\limits{\text{h}\rightarrow0}\text{f}(5-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(5+\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}(2)=\lim_\limits{\text{h}\rightarrow0}\big(\text{a}(3+\text{h})+\text{b}\big)$ and $\lim_\limits{\text{h}\rightarrow0}\big(\text{a}(5-\text{h})+\text{b}\big)=\lim_\limits{\text{h}\rightarrow0}(9)$
$\Rightarrow2=3\text{a}+\text{b}$ and $5\text{a}+\text{b}=9$
View full question & answer→Question 495 Marks
Show that the function g(x) = x - [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.
AnswerThe given function is g(x) = x - [x]
It is evident that g is defined at all integral points.
Let n be an integer.
Then,
g(n) = n - [n] = n - n = 0
The left hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\big(\text{x}-[\text{x}]\big)=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^-}[\text{x}]\\=\text{n}-(\text{n}-1)=1$
The right hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\big(\text{x}-[\text{x}]\big)\\=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^+}[\text{x}]=\text{n}-\text{n}=0$
It is observed that the left and right hand limits of f at x = n do not coincide.
Therefore, f is not continuous at x = n
Hence, g is discontinuous at all integral points.
View full question & answer→Question 505 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}}+\cos\text{x},&\text{if }\text{ x}\neq0\\5,&\text{if }\text{ x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$ then
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}$
We know that $\sin\text{x}$ as well as the identity function x both are everwhere continuous.
So, the quotient function $\frac{\sin\text{x}}{\text{x}}$ is continuous at each $\text{x}\neq0$
Also, $\cos\text{x}$ is everwhere continuous.
Therefore, $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}$ is continuous at each $\text{x}\neq0$
Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}}+\cos\text{x},&\text{if }\text{ x}\neq0\\5,&\text{if }\text{ x}=0\end{cases}$
We have
$(\text{LHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}+\cos(-\text{h})\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)+\lim_\limits{\text{h}\rightarrow0}\cos(-\text{h})=1+1=2$
$(\text{RHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}+\cos(\text{h})\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)+\lim_\limits{\text{h}\rightarrow0}\cos(\text{h})=1+1=2$
Also, f(0)=5
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}\neq\text{f}(0)$
Thus, f(x) is discontinuous at x = 0
Hence, the only point of discontinuity for f(x) is x = 0
View full question & answer→Question 515 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{{1}-\text{x}^\text{n}}{1-\text{x}}, & \text{x} \neq1\\\text{n}-1, & \text{ x} = 1\end{cases}\text{ n }\in\ \text{N at x}=1$
AnswerWe want, to check the continuity at x = 1
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1-\text{h})^\text{n}}{1-(1-\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}-\frac{\text{n(n-1)}}{2!}\text{h}+\dots$
$=\text{n}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 1^+}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}\text{(1+h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1+\text{h})^\text{n}}{1-(1+\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}+\frac{\text{n}(\text{n}-1)}{2!}\text{h}+\dots$
$=\text{n}$
$\text{f}(1)=\text{n}-1$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}( 1)$
Hence, funtion is discontinuous at x = 1
This is removable discotinuity.
View full question & answer→Question 525 Marks
Find the value of k if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ where
$\text{f}\text{(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, &\text{ x}\neq\frac{\pi}{2}\\3, &\text{ x}=\frac{\pi}{2}\end{cases}$
AnswerSince f(x) is continuous at $\text{x}=\frac{\pi}{2},$ L.H.Limit = R.H.Limit.
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi^+}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}=3$
$\Rightarrow\text{k}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}=3$
$\Rightarrow\text{k}=6$
View full question & answer→Question 535 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}\text{at x} = \pi$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}$
We have,
$(\text{LHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{k}(\pi-\text{h})+1=\text{k}\pi+1$
$(\text{RHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\cos(\pi+\text{h})=\cos\pi=-1$
If f(x) is continuous at $\text{x}=\pi,$ then
$\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}$
$\Rightarrow\text{k}\pi+1=-1$
$\Rightarrow\text{k}=\frac{-2}{\pi}$
View full question & answer→Question 545 Marks
Discuss the continuity of the function $\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
AnswerWhen x < 2, we have
f(x) = 2x - 1
We know that a polynomial function is everywhere continuous.
So, f(x) is continuous for each x < 2
When x > 2, we have
$\text{f(x)}=\frac{3\text{x}}{2}$
The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3\text{x}}{2}$ is continuous at each x > 2
Now,
Let us consider the point x = 2
$\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
We have
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2(2-\text{h})-1)=4-1=3$
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\frac{3(\text{h}+2)}{2}=3$
Also,
$\text{f}(2)=\frac{3(2)}{2}=3$
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\text{f}(2)$
Thus, f(x) is continuous at x = 2
Hence, f(x) is everywhere continuous.
View full question & answer→Question 555 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\sin\text{x}-\cos\text{x},&\text{if }\text{ x}\neq0\\-1,&\text{if }\text{ x}=0\end{cases}$
AnswerThe given function f is $\text{f(x)}=\begin{cases}\sin\text{x}-\cos\text{x},&\text{if }\text{ x}\neq0\\-1,&\text{if }\text{ x}=0\end{cases}$ It is evident that f is defind at all points of the real line. Let c be a real number.Case I:
If $\text{c}\neq0,$ then $\text{f(c)}=\sin\text{c}-\cos\text{c}$ $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(\sin\text{x}-\cos\text{x})=\sin\text{c}-\cos\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that $\text{x}\neq0$Case II:
If c = 0, then f(0) = -1 $\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}(\sin\text{x}-\cos\text{x})\\=\sin0-\cos0=0-1=-1$ $\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}(\sin\text{x}-\cos\text{x})\\=\sin0-\cos0=0-1=-1$ Therefore, f is continuousb at x = 0 From tha above observations, it can be continuous that f is continuous at every point of the real line. Thus, f is a continuous function.
View full question & answer→Question 565 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{if }\text{ x}<0\\2\text{x}+3,&\text{ x}\geq0\end{cases}$
AnswerWhen x < 0, we have $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ and the identity function continuous for x < 0, so the quotient function
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ is continuous for x < 0
When x > 0 f(x) = 2x + 3, which is a polynomial of degree 1. So, f(x) 2x + 3 is continuous for x > 0
Now, consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(-\text{h})}{-\text{h}}=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin\text{h}}{-\text{h}}=1$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
$\text{f}(0)=2\times0+3=3$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence, f(x) is discontinuous at x = 0
View full question & answer→Question 575 Marks
If $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}}$ for $\text{x}\neq\frac{\pi}{4},$ find the value which can be assigned to f(x) at $\text{x}=\frac{\pi}{4}$ so that the function f(x) becomes continuous every where in $\Big[0,\frac{\pi}{2}\Big]$
AnswerWhere $\text{x}\neq\frac{\pi}{4},\ \tan\big(\frac{\pi}{4}-\text{x}\big)$ and $\cot2\text{x}$ are continuous in $\Big[0,\frac{\pi}{2}\Big]$
Thus, the quotient function $\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}}$ is continuous in $\Big[0,\frac{\pi}{2}\Big]$ for each $\text{x}\neq\frac{\pi}{4}$
So, if f(x) is continuous at $\text{x}=\frac{\pi}{4},$ then it will be everywhere continuous in $\Big[0,\frac{\pi}{2}\Big]$
Now,
Let us consider the point $\text{x}=\frac{\pi}{4}$
Given, $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}},\text{x}\neq\frac{\pi}{4}$
We have
$\Big(\text{LHL at x}=\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\tan\big(\frac{\pi}{4}-\frac{\pi}{4}+\text{h}\big)}{\cot\big(\frac{\pi}{2}-2\text{h}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(\text{h})}{\tan(2\text{h})}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\frac{\tan(\text{h})}{\text{h}}}{\frac{2\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}\Bigg(\frac{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(\text{h})}{\text{h}}}{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}$
$\Big(\text{RHL at x}=\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{4}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\tan\big(\frac{\pi}{4}-\frac{\pi}{4}-\text{h}\big)}{\cot\big(\frac{\pi}{2}+2\text{h}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(-\text{h})}{-\tan(2\text{h})}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(\text{h})}{\tan(2\text{h})}\Big)=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\frac{\tan(\text{h})}{\text{h}}}{\frac{2\tan(2\text{h})}{2\text{h}}}\Bigg)\\=\frac{1}{2}\Bigg(\frac{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(\text{h})}{\text{h}}}{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}$
If f(x) is continuous at $\text{x}=\frac{\pi}{4},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\text{f}\Big(\frac{\pi}{4}\Big)$
$\therefore\ \text{f}\Big(\frac{\pi}{4}\Big)=\frac{1}{2}$
Hence, for $\text{f}\Big(\frac{\pi}{4}\Big)=\frac{1}{2},$ the function f(x) will be everywhere continuous in $\Big[0,\frac{\pi}{2}\Big]$
View full question & answer→Question 585 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{2^\text{z+2}-16}{4^\text{x}-16}, &\text{if x} \neq 2\\\text{k}, & \text{x} = 2\end{cases}$
is continuous at x = 2, Find k.
AnswerWe are given that the function is continuous at x = 2
$\therefore$ LHL = RHL = f(2)...(i)
Now,
f(2) = k...(A)
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{(2-\text{h)+2}}-16}{4^{(2-\text{h)}}-16}=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{2-\text{h}}-16}{4^{2-\text{h}}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^4-2^\text{-h}-16}{4^2.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16.2^\text{-h}-16}{16.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16\Big(2^\text{-h}-1\Big)}{16\Big(4^\text{-h}-1\Big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}\Big)-1^2}$ $\Big[\because2^{-2\text{h}}=\Big(2^{-\text{h}}\Big)^2=4^{-\text{h}}\Big]$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}-1\Big)\Big(2^\text{-h}+1\Big)}=\frac{1}{2}\dots(\text{B})$
$\therefore$ Using (i) from (A) & (B)
$\text{k}=\frac{1}{2}$
View full question & answer→Question 595 Marks
Discuss the continuity of the f(x) at the indicated points f(x) = |x| + |x - 1| at x = 0, 1.
AnswerGiven,
$\text{f(x)}=|\text{x}|+|\text{x}-1|$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|0-\text{h}|+|0-\text{h}-1|\big]=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|0+\text{h}|+|0+\text{h}-1|\big]=1$
Also, $\text{f}(1)=|1|+|1-1|=1+0=1$
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$ and $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\text{f}=1$
Hence, f(x) is continuous at x = 0, 1
View full question & answer→Question 605 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}2\text{x},&\text{ if}\text{ x}<0\\0,&\text{if }0\leq\text{x}\leq1\\4\text{x},&\text{if }\text{ x}>1\end{cases}$
AnswerThe given function is $\text{f(x)}=\begin{cases}2\text{x},&\text{ if}\text{ x}<0\\0,&\text{if }0\leq\text{x}\leq1\\4\text{x},&\text{if }\text{ x}>1\end{cases}$ The given function is defined at all points of the real line. Let c be a point on the real line.Case I:
If c < 0 then f(c) = 2c $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(2\text{x})=2\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x < 0Case II:
If c = 0, then f(c) = f(0) = 0 The left hand limit of f at x = 0 is, $\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^-}(2\text{x})=2\times0=0$ The right hand limit of at x = 0 is, $\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}(0)=0$ $\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\text{f}(0)$ Therefore, f is continuous at x = 0Case III:
If 0 < c < 1, then f(x) = 0 and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(0)=0$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f}(\text{c})$ Therefore, if is continuous at all points of the interval (0, 1)Case IV:
If c = 1, then f(c) = f(1) = 0 The left hand limit of at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(0)=0$ The right hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(4\text{x})=4\times1=4$ It is observes that the left and right hand limits of f at x = 1 do not coincide. Therefore, f is not continuous at x = 1Case V:
If c < 1, then f(c) = 4c and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}\text{(4x)}=4\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuouse at all points x such that x > 1 Hence, f is not continuous only at x = 1
View full question & answer→Question 615 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}, & \text{x} \neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$ is continuous at x = 0. find k.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}, & \text{x} \neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)\dots(\text{i})$
Consider:
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\Bigg(\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}\Bigg)=\lim\limits_{\text{x} \rightarrow 0}\Bigg(\frac{2\sin^2\frac{\text{kx}}{2}}{\text{x}\sin\text{x}}\Bigg)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{2\sin^2\frac{\text{kx}}{2}}{\text{x}^2\Big(\frac{\sin\text{x}}{\text{x}}\Big)}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\frac{2\text{k}^2}{4}\Big(\sin\frac{\text{kx}}{\text{x}}\Big)^2}{\Big(\frac{\text{kx}}{2}\Big)^2\Big(\frac{\sin\text{x}}{\text{x}}\Big)}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2\text{k}^2}{4}\begin{pmatrix}\frac{\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\sin\frac{\text{kx}}{2}\Big)^2}{\Big(\frac{\text{kx}}{2}\Big)^2}}{\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2\text{k}^2}{4}\times1=\frac{\text{k}^2}{2}$
From equation (i), we have
$\frac{\text{k}^2}{2}=\text{f}(0)$
$\Rightarrow\frac{\text{k}^2}{2}=\frac{1}{2}$
$\Rightarrow\text{k}=\pm1$
View full question & answer→Question 625 Marks
If $\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$ find whethe f is continuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$
We observe
$(\text{LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{e}^\frac{-1}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\bigg(\frac{1}{\text{e}^{\frac{1}{\text{h}}}}\bigg)=\frac{1}{\lim\limits_{\text{h} \rightarrow 0}\text{e}^{\frac{1}{\text{h}}}}=0$
$(\text{RHL at x}=0)=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}(\text{x})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0^+}\text{e}^\frac{1}{\text{h}}=\infty$
Given,
$\text{f}(0)=1$
It is known for a function f(x) to be continuous at x = a,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\text{f}\text{(a)}$
But here,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow \text{0}^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.
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