Question 13 Marks
Determine the value of the constant k so that the function $\text{f(x)}=\begin{cases}\text{kx}^2,&\text{if }\text{ x}\leq2\\3,&\text{if }\text{ x}>2\end{cases}$ is continuous at x = 2.
AnswerGiven, $\text{f(x)}=\begin{cases}\text{kx}^2,&\text{if }\text{ x}\leq2\\3,&\text{if }\text{ x}>2\end{cases}$ If f(x) is continuous at x = 2, then $\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}=\text{f}(2)\ ...(\text{i})$ Now, $\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\text{k}(2-\text{h})^2=4\text{k}$ And, f(2) = 3 From (i) we have, $4\text{k}=3$$\Rightarrow\text{k}=\frac{3}{4}$
View full question & answer→Question 23 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\text{kx}+5,&\text{if }\text{ x}\leq2\\\text{x}-1,&\text{if }\text{ x}>2\end{cases}$
Answer$\text{f(x)}=\begin{cases}\text{kx}+5,&\text{if }\text{ x}\leq2\\\text{x}-1,&\text{if }\text{ x}>2\end{cases}$
It is given that the function is continuous
$\therefore\ \text{LHL}=\text{RHL}=\text{f}(2)\ ...(\text{i})$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{k}(2-\text{h})+5=2\text{k}+5$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})=\lim_\limits{\text{h}\rightarrow0}(2+\text{h})-1=1$
Thus, using (i) we get
$2\text{k}+5=1$
$\text{k}=-2$
View full question & answer→Question 33 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}\text{at x}=0$
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\cos2\text{kx}}{\text{x}^2}=8$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{2\text{k}^2\sin^2\text{kx}}{\text{k}^2\text{x}^2}=8$
$\Rightarrow2\text{k}^2\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin\text{kx}}{\text{kx}}\Big)^2=8$
$\Rightarrow2\text{k}^2\times1=8$
$\Rightarrow\text{k}^2=4$
$\Rightarrow\text{k}=\pm2$
View full question & answer→Question 43 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}\text{x}^2,&\text{x}\geq1\\4,&\text{x}<1\end{cases}\text{at x} =1$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}\text{x}^2,&\text{x}\geq1\\4,&\text{x}<1\end{cases}$
We have,
$(\text{LHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$\lim_\limits{\text{h}\rightarrow0}4=4$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$\lim_\limits{\text{h}\rightarrow0}\text{k}(1+\text{h})^2=\text{k}$
If f(x) is continuous at x = 1, then
$\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}$
$\Rightarrow\text{k}=4$
View full question & answer→Question 53 Marks
Find the value of k for which the function $\text{f(x)}=\begin{cases}\frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} - 2},&\text{x}\neq2\\\text{k},&\text{x} = {2}\end{cases}$ is continues at x = 2.
AnswerGiven
$\text{f(x)} = \frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} - 2}$
Continuity
$\text{x} = 2$
$\lim\limits_{\text{x} \rightarrow 2} \frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} -2} = \text{k}$
$\lim\limits_{\text{x}\rightarrow 2} \frac{\text{x}^{2} + 5\text{x} - 2\text{x} - 10}{\text{x} - 2 } = \text{k}$
$\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x} (\text{x} + 5) - 2 (\text{x} + 5)}{\text{x} - 2} = \text{k}$
$\lim\limits_{\text{x} \rightarrow 2} \frac{(\text{x} - 2) (\text{x} + 5)}{\text{(x} - 2)} = \text{k}$
When x = 2
x + 5 = k
k = 5 + 2 = 7
k = 7
View full question & answer→Question 63 Marks
Test the continuity of the function on f(x) at the origin:
$\text{f}\ (\text{x})=\begin{cases}\frac{\text{x}}{\text{|x|}},& \text{x}\neq0\\1, & \text{x} = 0\end{cases}$
AnswerGiven,
$\text{f}\ (\text{x})=\text{x},\text{ x}\neq0$
$\text{f}\ (\text{x})=1,\text{ x}=0$
We observe
$\text{(LHL at x}= 0)$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f} \ (0-\text{h})$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (-h)}=\lim\limits_{\text{h} \rightarrow 0} \frac{\text{-h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}-1=-1$
$\text{(RHL at x}=0)$
$\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\ \text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\ \text{(h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}1=1$
Hence, f(x) is discontinuous at the origin.
View full question & answer→Question 73 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+3\text{x}),&\text{if }\text{ x}<0\\\cos2\text{x},&\text{if }\text{ x}\geq0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+3\text{x}),&\text{if }\text{ x}<0\\\cos2\text{x},&\text{if }\text{ x}\geq0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}{(-\text{h)}}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Big(\text{k}\big((-\text{h})^2-3\text{h}\big)\Big)=\lim_\limits{\text{h}\rightarrow0}(\cos2\text{h})$
$\Rightarrow0=1$ [It is not possible]
Hence, there does not exist any value of k, which can make the given function continuous.
View full question & answer→Question 83 Marks
For what value of k is the following function continuous at x = 2?
$\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}2\text{x}+1,&\text{if }\text{ x}<2\\\text{k},&\text{x}=2\\3\text{x}-1,&\text{x}>2\end{cases}$
We have,
$(\text{LHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2(2-\text{h})+1)=5$
$(\text{RHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})=\lim_\limits{\text{h}\rightarrow0}3(2+\text{h})-1=5$
Also, $\text{f}(2)=\text{k}$
If f(x) is continuous at x = 2, then
$=\lim_\limits{\text{x}\rightarrow2^-}\text{f}(\text{x})=\lim_\limits{\text{x}\rightarrow2^+}\text{f}(\text{x})=\text{f}(2)$
$\Rightarrow5=5=\text{k}$
Hence, for k = 5, (fx) is continuous at x = 2
View full question & answer→Question 93 Marks
Find f(x) is continuse at x = 0, then $\text{f(x)}=\frac{\text{x}}{1-\sqrt{1-\text{x}}}$ becomes continuous at x = 0.
AnswerIf f(x) is continuous at x = 0, then $\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\text{f}(0)\ ....(\text{i})$
Given, $\text{f(x)}=\frac{\text{x}}{1-\sqrt{1-\text{x}}}$
$\Rightarrow\text{f(x)}=\frac{\text{x}\big(1+\sqrt{1-\text{x}}\big)}{\big(1-\sqrt{1-\text{x}}\big)\big(1+\sqrt{1-\text{x}}\big)}$
$\Rightarrow\text{f(x)}=\frac{\text{x}\big(1+\sqrt{1-\text{x}}\big)}{1-(1-\text{x})}$
$\Rightarrow\text{f(x)}=\big(1+\sqrt{1-\text{x}}\big)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}=\big(1+\sqrt{1-\text{x}}\big)=\text{f}(0)$ [From eq. (i)]
$\Rightarrow\text{f}(0)=2$
So, for f(0) = 2, the function f(x) becomes continuous x = 0
View full question & answer→Question 103 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq5\\3\text{x}-5,&\text{if}\text{ x}>5\end{cases}\text{at x} =5$
Answer$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq5\\3\text{x}-5,&\text{if}\text{ x}>5\end{cases}\text{at x} =5$We have given that function is continuous at x = 5
$\therefore\ \text{LHL}=\text{RHL}=\text{f}(5)\ ....(\text{i})$
$\text{f}(5)=5\text{k}+1$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(5+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}3(5+\text{h})-5=10$
Thus, using (i) we get,
$5\text{k}+1=10$
$5\text{k}=9$
$\text{k}=\frac{9}{5}$
View full question & answer→Question 113 Marks
Write the value of b which $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$ is continuous at x = 1.
AnswerGiven, $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\text{f}(1)\ ...(\text{i})$
Now,
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}5(1-\text{h})-4=5-4=1$
$\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1+\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}4(1+\text{h})^2+3\text{b}(1+\text{h})=4+3\text{b}$
Also,
$\text{f}(1)=5(1)-4=1$
$=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x)}=\text{f}(1)$ [From eq. (i)]
$\Rightarrow1=4+3\text{b}=1$
$\Rightarrow1=4+3\text{b}$
$\Rightarrow-3=3\text{b}$
$\Rightarrow\text{b}=-1$
Thus, for b = -1, the function f(x) is continuous at x = 1.
View full question & answer→Question 123 Marks
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$ continuous at x = 0.
AnswerGiven, $\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\times1=\text{k}$
$\Rightarrow\text{k} =2$
View full question & answer→Question 133 Marks
Given the funcation $\text{f(x)}=\frac{1}{\text{x}+2}.$ Find the points of discontinuity of the function f(f(x)).
Answer$\text{f}\big[\text{f(x)}\big]=\frac{1}{\frac{1}{\text{x}+2}+2}=\frac{\text{x}+2}{2\text{x}+5}$
So, f[f(x)] is not defind at x + 2 = 0 and 2x + 5 = 0
If x + 2, then x = - 2
If 2x + 5 = 0, then $\text{x}=-\frac{5}{2}$
Hence, the function is dicontinuous at $\text{x}=-\frac{5}{2}$ and -2
View full question & answer→Question 143 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}},&\text{if }\text{ x}\neq0\\3\text{k},&\text{if }\text{ x}=0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}},&\text{if }\text{ x}\neq0\\3\text{k},&\text{if }\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\Rightarrow\lim_\limits{\text{x}\rightarrow0}\frac{\sin2\text{x}}{5\text{x}}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow0}\frac{2\sin2\text{x}}{2\times5\text{x}}=\text{f}(0)$
$\Rightarrow\frac{2}{5}\lim_\limits{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}=\text{f}(0)$
$\Rightarrow\frac{2}{5}=3\text{k}$
$\Rightarrow\text{k}=\frac{2}{15}$
View full question & answer→Question 153 Marks
For what value of k is the following function continuous at x = 1
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{\text{x} \rightarrow 1}\text{f}\text{(x)}=\text{f}(1)$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^2-1}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{(x}-1)(\text{x}+1)}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}(\text{x}+1)=\text{k}$
$\text{k}=2$
View full question & answer→Question 163 Marks
Determine the value of the constant k so that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-3\text{x}+2}{\text{x}-1}, &\text{if}\text{ x}\neq1\\\text{k}, &\text{if}\text{ x}=1\end{cases}$ is continuous at x = 1
AnswerWe have that the function is continuous at x = 1
$\therefore$ LHL = RHL = f(1) ....(1)
Now,
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1-\text{h})^2-3(1-\text{h})+2}{(1-\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}-\text{h}-1=-1$
$\text{f}(1)=\text{k}$
from (1), We get,
k = -1
View full question & answer→Question 173 Marks
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0?
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{3\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{5\sin5\text{x}}{3\times5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\times1=\text{k}$
$\Rightarrow\text{k}=\frac{5}{3}$
View full question & answer→Question 183 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}\text{at x} = 0$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{k}(\text{h}^2+2\text{h})=0$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\cos\text{h}=1$
$\therefore\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\neq\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
Thus, no value of k exists for which f(x) is continuous at x = 0.
View full question & answer→Question 193 Marks
If $\text{f(x)}=\frac{2\text{x}+3\sin\text{x}}{3\text{x}+2\sin\text{x}},\text{ x}\neq0$ is continuous at x = 0, then find f(0).
AnswerIt is given that the function is continuous at x = 0
$\therefore\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow 0}=\lim_\limits{\text{h}\rightarrow 0}(0-\text{h})=\lim_\limits{\text{h}\rightarrow 0}\frac{2(-\text{h})+3\sin(-\text{h})}{3(-\text{h})+2\sin(-\text{h})}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{-2\text{h}-3\sin\text{h}}{-3\text{h}-2\sin\text{h}}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{\frac{2\text{h}+3\sin\text{h}}{\text{h}}}{\frac{3\text{h}+2\sin\text{h}}{\text{h}}}$
$=\lim_\limits{\text{x}\rightarrow 0}\frac{2+3\frac{\sin\text{h}}{\text{h}}}{3+2\frac{\sin\text{h}}{\text{h}}}$
$=\frac{2+3}{3+2}=1$ $\Big[\because=\lim_\limits{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1\Big]$
Using (i) we get
$=\text{f}(0)=1$
View full question & answer→Question 203 Marks
Discuss the continuity of the function f(x) at the point $\text{x}=\frac{1}{2}$ where
$\text{f}\text{(x)}=\begin{cases}\text{x}, & 0\leq\text{x} < \frac{1}{2}\\\frac{1}{2},&\text{x}=\frac{1}{2}\\1-\text{x}, &\frac{1}{2}< \text{x}\leq 1\end{cases}$
AnswerWe want to discuss the continuity of the function at $\text{x}=\frac{1}{2}.$
$\text{LHL}=\lim\limits_{\text{x} \rightarrow \frac{1}{2}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}-\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}\frac{1}{2}-\text{h}=\frac{1}{2}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow \Big(\frac{1}{2}^-\Big)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}+\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}1-\Big(\frac{1}{2}+\text{h}\Big)=\frac{1}{2}$
$\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Thus, $\text{LHL}=\text{RHL}=\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Hence, the function is continuous at $\text{x}=\frac{1}{2}.$
View full question & answer→Question 213 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\text{|x|}\cos\Big(\frac{1}{\text{x}}\Big), & \text{ x}\neq 0\\0 &\text{ x} = 0\end{cases}\text{at x}=0$
AnswerGiven,
$\text{f}\text{(x)}=\text{x}\cos\Big(\frac{1}{\text{x}}\Big),\text{x}\neq0$
$\text{f}\text{(x)}=0,\ \text{x}=0$
We observe
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{x}\cos\Big(\frac{1}{\text{x}}\Big)$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{x}\lim\limits_{\text{x} \rightarrow 0}\cos\Big(\frac{1}{\text{x}}\Big)$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=0\times\lim\limits_{\text{x} \rightarrow 0}\cos\Big(\frac{1}{\text{x}}\Big)$
$=0$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
Hence, f(x) is continuous at x = 0.
View full question & answer→Question 223 Marks
Determine that value of the constant 'k' so that function $\text{f(x)}=\begin{cases}\frac{\text{kx}}{|\text{x}|},&\text{if }\text{ x}<0\\3,&\text{if }\text{ x}\geq0\end{cases}$ is continuous at x = 0.
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{kx}}{|\text{x}|},&\text{if }\text{ x}<0\\3,&\text{if }\text{ x}\geq0\end{cases}$
Since, the function is continuous at x = 0, therefore,
$\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow0^+}\text{f(x})=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\frac{-\text{kx}}{\text{x}}=\lim\limits_{{\text{x}}\rightarrow0}3=3$
$\Rightarrow-\text{k}=3$
$\Rightarrow\text{k}=-3$
View full question & answer→Question 233 Marks
A function f(x) is defined as, $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-9}{\text{x}-3},&\text{if }\text{x}\neq3\\6,&\text{if }\text{ x}=3\end{cases}$ show that f(x) is continuous that x = 3
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-9}{\text{x}-3},&\text{if } \text{ x}\neq3\\6 ,&\text{if }\text{ x}=3\end{cases}$
We observe
$(\text{LHL at x }= 3)=\lim\limits_{\text{x} \rightarrow 3^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3-\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{(3-\text{h})^2-9}{(3-\text{h})-3}=\lim\limits_{\text{h} \rightarrow 0}\frac{3^2-\text{h}^2-6\text{h}-9}{3-\text{h}-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+6\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}(\text{h}-6)}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}(6-\text{h})=6$
$\text{(RHL at x = 3)}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}(3-\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{(3+\text{h})^2-9}{3+\text{h}-3}=\lim\limits_{\text{h} \rightarrow 0}\frac{3^2+\text{h}^2+6\text{h}-9}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+6\text{h}}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}(6+\text{h})}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}(6+\text{h})=6$
Given:
$\text{f}(3)=6$
$\therefore$ $\lim\limits_{\text{x} \rightarrow 3^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}(\text{x})=\text{f}(3)$
Hence, f(x) is continuous at x = 3
View full question & answer→Question 243 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{\text{|x}^2-1|}{\text{x}-1},\text{for} & \text{x} \neq1\\2, &\text{for} \text{ x} = 1\end{cases} \text{at x}=1$
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{|x}^2-1|}{\text{x}-1}, & \text{x} \neq1\\2, &\text{ x} = 1\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\text{x}+1, & \text{x}< -1\\-\text{x}-1, & -1\leq \text{x} <0\\\text{x}+1,& \text{x}>1\\2,&\text{x}=1\end{cases}$
We obseve
$(\text{LHL at x}=1)=\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(1-h)}$
$=\lim\limits_{\text{h} \rightarrow 0}-\text{(1-h)}-1=\lim\limits_{\text{h} \rightarrow 0}-2+\text{h}=-2$
And, $\text{f}(1)=2$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}\neq\text{f}(1)$
Hence, f(x) is discontinuous at x = 1
View full question & answer→Question 253 Marks
Discuss the continuity of the function $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$
$\text{f(x)}=\frac{\text{x}}{|\text{x}|}=\begin{cases}\frac{-\text{x}}{\text{x}}=-1;&\text{x}<0\\\frac{\text{x}}{|\text{x}|}=1;&\text{x}>0\end{cases}$
So, f(x) is a constant function when $\text{x}\neq0,$
Hence, is continuous for all x < 0 and x > 0
Now, Consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{|-\text{h}|}=-1$
$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{|\text{h}|}=1$
So, $\text{LHL}\neq\text{RHL}$
Hence, function is discontinuous at x = 0
View full question & answer→Question 263 Marks
Discuss the continuity of $\text{f}\text{(x)}=\begin{cases}2\text{x}-1, & \text{x} < 0\\2\text{x}+1, & \text{x} \geq 0\end{cases}\text{at}\text{ x}=0$
Answer$\text{f}\text{(x)}=\begin{cases}2\text{x}-1, & \text{x} < 0\\2\text{x}+1, & \text{x} \geq 0\end{cases}\text{at}\text{ x}=0$
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=2(0)-1=-1$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=2(0)+1=1$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 273 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\text{e}^{\text{x}}}{\log_\text{e}(1+2\text{x})},&\text{if }\text{ x}\neq0\\7,&\text{if }\text{ x}=0\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{e}^{\text{x}}}{\log_\text{e}(1+2\text{x})},&\text{if }\text{ x}\neq0\\7,&\text{if }\text{ x}=0\end{cases}$
We have,
$\lim_\limits{\text{x}\rightarrow0}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{e}^{\text{x}}}{\log_\text{e}(1+2\text{x})}=\lim_\limits{\text{x}\rightarrow0}\frac{\big(\frac{\text{e}^{\text{x}}-1}{\text{x}}\Big)}{\Big(\frac{2\log_\text{e}(1+2\text{x})}{2\text{x}}\Big)}$
$=\frac{1}{2}\times\frac{\big(\frac{\text{e}^{\text{x}}-1}{\text{x}}\Big)}{\Big(\frac{2\log_\text{e}(1+2\text{x})}{2\text{x}}\Big)}=\frac{1}{2}$
It is given that f(0) = 7
$\lim_\limits{\text{x}\rightarrow0}\text{f(x)}\neq\text{f}(0)$
Hence, the given function is discontinuous at x = 0 and continuous elsewhere.
View full question & answer→Question 283 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},& \text{when}\text{ x}\neq0 \\1,&\text{when} \text{ x}=0\end{cases}$ Find whether f(x) is continuous at x = 0.
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 293 Marks
Determine the value of the constant k so that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0.
AnswerWe have given that the funtion is continuous at x = 0
So, LHL = RHL = f(0) ....(i)
Now,
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin2(-\text{h})}{5(-\text{h})}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin2\text{h}}{-5\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin2\text{h}}{2\text{h}}\times\frac{2\text{h}}{5\text{h}}=\frac{2}{5}$
$\text{f}(0)=\text{k}$
Using(i), $\text{k}=\frac{2}{5}$
View full question & answer→Question 303 Marks
If $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{2},&\text{if }0\leq\text{ x}\leq1\\2\text{x}^2-3\text{x}+\frac{3}{2},&\text{if }1<\text{x}\leq2\end{cases}$ Show that f is continuous at x = 1.
AnswerWe want to discuss the continuity of the function at x = 1
We need to prove that
$\text{LHL}=\text{RHL}=\text{f}(1)$
$\text{f}(1)=\frac{1^2}{2}=\frac{1}{2}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{(1-\text{h})^2}{2}=\frac{1}{2}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(1+\text{h})=\lim_\limits{\text{h}\rightarrow0}2(1+\text{h}^2)-3(1+\text{h})+\frac{3}{2}$
$=2-3+\frac{3}{2}=\frac{1}{2}$
Thus, $\text{LHL}=\text{RHL}=\text{f}(1)=\frac{1}{2}$
Hence, function is continuous at x = 1
View full question & answer→Question 313 Marks
Discuss the continuity of the function f(x) at the point x = 0, where$\text{f}\text{(x)}=\begin{cases}\text{x}, & \text{x} > 0\\1,&\text{x}=0\\\text{-x}, & \text{x} > 0\end{cases}$
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\text{x}, & \text{x} > 0\\1,&\text{x}=0\\\text{-x}, & \text{x} > 0\end{cases}$
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}=\lim\limits_{\text{h} \rightarrow 0}-(-\text{h)}=0$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(h)}=0$
And, $\text{f}(0)=1$
$\therefore\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}\neq\text{f}(0).$
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 323 Marks
Show that $\text{f}\text{(x)}=\begin{cases}1+\text{x}^2,&\text{if } 0\leq\text{x}\leq 1\\2-\text{x},&\text{if }\text{x} > 1\end{cases}$ is discontinuous at x = 1.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}1+\text{x}^2,&\text{if } 0\leq\text{x}\leq 1\\2-\text{x},&\text{if }\text{x} > 1\end{cases}$
We observe
$\text{(LHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(1+1-\text{h)}^2=\lim\limits_{\text{h} \rightarrow 0}(2+\text{h}^2-\text{2h)}=2$
$\text{(RHL at x}=1)\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(2-(1+\text{h))}=\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=1$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}$
Thus, f(x) is discontinuous at x = 1.
View full question & answer→Question 333 Marks
Find the value of k for which $\text{f(x)}=\begin{cases}\frac{1-\cos4\text{x}}{8\text{x}^2},&\text{when x}\neq0\\\text{k},&\text{when x}=0\end{cases}$ is continous at x = 0.
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{1-\cos4\text{x}}{8\text{x}^2},&\text{when x}\neq0\\\text{k},&\text{when x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\cos4\text{x}}{8\text{x}^2}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{2\sin^22\text{x}}{8\text{x}^2}=\text{f}(0)$
$\Rightarrow\frac{2}{2}\lim_\limits{\text{x}\rightarrow 0}\frac{\sin^22\text{x}}{4\text{x}^2}=\text{f}(0)$
$\Rightarrow\frac{2}{2}\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin2\text{x}}{2\text{x}^2}\Big)^2=\text{f}(0)$
$\Rightarrow1\times1=\text{f}(0)$
$\Rightarrow\text{k}=1$ $(\because\text{f}(0)=\text{k})$
View full question & answer→Question 343 Marks
A function f(x) is defined as,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-\text{x}-6}{\text{x}-3}&; &\text{if} \text{x}\neq3\\5 &;&\text{if}\text{ x}=3\end{cases}$
show that f(x) is continuous that x = 3.
AnswerWe have, to check the continuity at x = 3.
$\text{L.H.L}=\lim\limits_{\text{x} \rightarrow 3^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\frac{(3-\text{h})^2-(3-\text{h})-6}{(3-\text{h})-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2-5\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}-\text{h}+5=5$
$\text{R.H.L}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3\text{+h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(3+\text{h})^2-(3+\text{h})-6}{(3+\text{h})-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+5\text{h}}{\text{h}}=\lim\limits_{\text{h}\rightarrow 0}\text{h}+5=5$
$\text{f}(3)=5$
Thus, We have, LHL = RHL = f(3) = 5
So,The function is continus at x = 3
View full question & answer→Question 353 Marks
If $\text{f(x)}=\begin{cases}2\text{x}^2+\text{k},&\text{if }\text{ x}\geq0\\-2\text{x}^2+\text{k},&\text{if }\text{ x}<0\end{cases},$ then what should be the value of k so that f(x) is continuous at x = 0.
AnswerIt is given that function is continous at x = 0 then,
$\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
Now, $\text{f}(0)=2\times0+\text{k}=\text{k}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\lim_\limits{\text{h}\rightarrow0}=\lim_\limits{\text{h}\rightarrow0}-2(-\text{h})^2+\text{k}=\text{k}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}2(\text{h}^2)+\text{k}=\text{k}$
Thus, the function will be continuous for any $\text{k}\in\text{R}$
View full question & answer→Question 363 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2-25}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}\text{at x} =5$
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{\text{x}^2-25}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{(\text{x}-5)(\text{x}+5)}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
If f(x) is continuous at x = 5, then,
$\lim_\limits{\text{x}\rightarrow5}\text{f(x)}=\text{f}(5)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow5}\text{(x}+5)=\text{k}$
$\Rightarrow\text{k}=5+5=10$
View full question & answer→Question 373 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}(\text{x}-\text{a}){\sin}\Big(\frac{1}{\text{x}-\text{a}}\Big) & \text{x} \neq \text{a}\\\ 0, & \text{ x} = \text{a}\end{cases}\text{at x}=\text{a}$
AnswerWe want, to check the continuity at x = 0.
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{ h)}=\lim\limits_{\text{x} \rightarrow 0}(-\text{h)}^2$
$\sin\Big(\frac{1}{\text{-h}}\Big)=0$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{x} \rightarrow 0}\text{h}^2$
$\sin\Big(\frac{1}{\text{h}}\Big)=0$
$\text{f}(0)=0$
Thus, LHL = RHL = f(0) = 0
Hence, the function is continuous at x = 0.
View full question & answer→Question 383 Marks
show that $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-|\text{x}|}{2}, & \text{when} \text{ x}\neq 0\\2, & \text{when}\text{ x} = 0\end{cases}$ is discontinuous at x = 0.
AnswerWe want, to check the continuty of the function at x = 0.
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-|-\text{h|}}{2}=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-\text{h}}{2}=0$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}-\text{(|h|)}}{2}=0$
$\text{f}(0)=2$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence,The function is discotinuous at x = 0
This is rem ovable discontinuty.
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