MCQ 11 Mark
Let $\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$ Then, $f(x)$ is continus at $x = 4$ when:
- A
$a = 0, b = 0$
- B
$a = 1, b = 1$
- C
$a = -1, b = 1$
- ✓
$a = 1, b = -1$
AnswerCorrect option: D. $a = 1, b = -1$
Given,
$\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$
We have
$(\text{LHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(4-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4- \text{h}-4}{|4-\text{h}-4|}+\text{a}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{-\text{h}}{|-\text{h}|}+\text{a}\Big)=\text{a}-1$
$(\text{RHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f(4+h)}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4+\text{h}-4}{|4+\text{h}-4|}+\text{b}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}}{|\text{h}|}+\text{b}\Big)=\text{b}+1$
Also,
$\text{f}(4)=\text{a+b}$
if $(x)$ is continuous at $x = 4,$ then
$\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\text{f}(4)$
$\Rightarrow\text{a}-1=\text{b}+1=\text{a + b}$
$\Rightarrow\text{a}-\text{1}=\text{a + b}$ and $\text{b}+1=\text{a + b} $
$\Rightarrow\text{b}=-1$ and $\text{a}=1$
View full question & answer→MCQ 21 Mark
If $\text{f(x)}=\frac{1-\sin\text{x}}{(\pi-2\text{x})^2},$ when $\text{x}\neq\frac{\pi}{2}=\lambda$ then $f(x)$ will be continuous function at $\text{x}=\frac{\pi}{2},$ where $\lambda=$
- ✓
$\frac{1}{8}$
- B
$\frac{1}{4}$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: A. $\frac{1}{8}$
If $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\text{x}}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\lim\frac{1-\sin\text{x}}{\text{x}\rightarrow\frac{\pi}{2}(\pi-2\text{x})^2}=\text{f}\Big(\frac{\pi}{2}\Big) \ ...(\text{i})$
suppose $\Big(\frac{\pi}{2}-\text{x}\Big)=\text{t},$ then
$\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\sin\Big(\frac{\pi}{2}-\text{t}\Big)}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big) [$From eq.$(i)]$
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\cos\text{t}}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\frac{1}{4}\lim\limits_{t\rightarrow0}\begin{bmatrix}\frac{2\sin^2\Big(\frac{\text{t}}{2}\Big)}{\text{t}^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow \frac{1}{4}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$
$\Rightarrow \frac{1}{8}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$
$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{\sin\Big(\frac{\text{t}}{2}\Big)}{\frac{\text{t}}{2}} \end{bmatrix}^2=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=\lambda=\frac{1}{8}$
View full question & answer→MCQ 31 Mark
If $\text{f(x)}=\begin{cases}\frac{1-\sin\text{x}}{(\pi-2\text{x}^2)}\times\frac{\log\sin\text{x}}{\log(1+\pi^2-4\pi\text{x}+4\text{x}^2)},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then $k =$
- A
$-\frac{1}{16}$
- B
$-\frac{1}{32}$
- ✓
$-\frac{1}{64}$
- D
$-\frac{1}{28}$
AnswerCorrect option: C. $-\frac{1}{64}$
if $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\text{f(x)}=\text{f}(\frac{\pi}{2})$
$\text{f }\frac{\pi}{2}-\text{x = t},$ then
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\text{f}\big(\frac{\pi}{2}-\text{t}\big)=\text{f}(\frac{\pi}{2})$
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{1-\sin(\frac\pi{2}-\text{t}{})}{4\text{t}^2}\times\frac{\log\sin(\frac{\pi}{2})}{\log\big(1+\pi^2-4\pi\big(\frac{\pi}{2}-\text{t}\big)+4\big(\frac{\pi}{2}-\text{t}\big)^2\big)}\Bigg)=\text{k}$
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log\cos\text{t}}{\log\big(1+\pi^2-2\pi^2+4\pi\text{t}+4\big(\frac{\pi^2}{4}+\text{t}^2-\pi\text{t}\big)}\Bigg)=\text{k}$
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log\cos\text{t}}{\log(1-\pi^2+4\pi\text{t})+(\pi^2+4\text{t}^2-4\pi\text{t}}\bigg)=\text{t}$
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log \cos\text{t}}{\log(1+4\text{t}^2)}\bigg)=\text{k}$
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{2\sin^2\frac{\text{t}}{2}}{16\times\frac{\text{t}^2}{4}}\times\frac{\log\cos\text{t}}{\log(1+4\text{t}^2)}\bigg)=\text{k}$
$\Rightarrow\frac{2}{16}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\bigg(\frac{\text{t}^2}{4}\bigg)}\times\frac{\log\cos\text{t}}{\bigg(\frac{4\text{t}^2\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$
$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{(\frac{\text{t}}{2})^2}\times\frac{\frac{\log\cos\text{t}}{4\text{t}^2}}{\bigg({\frac{\log(1+4\text{t}^2)}{4\text{t}}\bigg)}} \end{pmatrix}=\text{k}$
$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{(\frac{\text{t}}{2})}\times\frac{\frac{\log\sqrt{1-\sin^2\text{t}}}{4\text{t}^2}}{\bigg({\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)}} \end{pmatrix}=\text{k}$
$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\big(\frac{\text{t}}{2}\big)^2}\times\frac{\bigg(\frac{\log(1-\sin^2\text{t})}{(8\text{t}^2)}\bigg)}{\bigg({\frac{\log(1+4^2\text{t})}{4\text{t}^2}}\bigg)} \end{pmatrix}=\text{k}$
$\Rightarrow\frac{1}{64}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\big(\frac{\text{t}}{2}\big)}\times\frac{\bigg(\frac{\log(1-\sin^2\text{t})}{\text{t}^2}\bigg)}{\bigg(\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$
$\Rightarrow\frac{1}{64}\begin{pmatrix}\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{\sin\frac{\text{t}}{2}}{\big({\frac{\text{t}}{2}\big)}}\Bigg)^2\times\frac{\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\log(1-\sin^2\text{t})}{\text{t}^2}\bigg)}{\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$
$\Rightarrow\frac{1}{64}\bigg(1\times\lim\limits_{\text{t}\rightarrow0}\frac{(-\sin^2\text{t})\log(1-\sin^2\text{t})}{\text{t}^2(-\sin^2\text{t})}\bigg)=\text{k}$
$\Rightarrow\frac{-1}{64}\bigg(\lim\limits_{\text{t}\rightarrow0}\frac{(\sin^2\text{t})\log(1-\sin^2\text{t})}{\text{t}^2(-\sin^2\text{t})}\bigg)=\text{k}$
$\Rightarrow\frac{-1}{64}\Bigg(\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\sin\text{t}}{\text{t}}\bigg)^2\lim\limits_{\text{t}\rightarrow0}\frac{\log(1-\sin^2\text{t})}{(-\sin^2\text{t})}\Bigg)=\text{k}$
$\Rightarrow\frac{-1}{64}\Bigg(\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\sin\text{t}}{\text{t}}\bigg)^2\lim\limits_{\text{t}\rightarrow0}\frac{\log(1-\sin^2\text{t})}{(-\sin^2\text{t})}\Bigg)=\text{k}$
$\Rightarrow\text{k}=\frac{-1}{64}$ $\bigg[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1-\text{x})}{\text{x}}=1\bigg]$
View full question & answer→MCQ 41 Mark
If the function $\text{f(x)}=\frac{2\text{x}-\sin^{-1}\text{x}}{2\text{x}+\tan^{-1}\text{x}}$ is continuous at each point of its domain, then the value of $f(0)$ is:
- A
$2$
- ✓
$\frac{1}{3}$
- C
$-\frac{1}{3}$
- D
$\frac{2}{3}$
AnswerCorrect option: B. $\frac{1}{3}$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{2\times-\sin^{-1}\text{x}}{2\times+\tan^{-1}\text{x}}$
$\text{f}\text{(0)}=\lim\limits_{\text{x}\rightarrow0}\frac{2\times-\frac{\sin^{-1}\text{x}}{\text{x}}}{2+\frac{\tan^{-1}\text{x}}{\text{x}}}$
$\text{f}(0)=\frac{2-1}{2+1}=\frac{1}{3}$
View full question & answer→MCQ 51 Mark
If $\text{f(x)}=\begin{cases}\text{x}\sin\frac{\pi}{2}(\text{x}+1),&\text{x}\leq0\\\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3},&\text{x}>0\end{cases}$ is continuous at $x = 0,$ then a equals:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- D
$\frac{1}{6}$
AnswerCorrect option: A. $\frac{1}{2}$
Given, $\text{f(x)}=\begin{cases}\text{x}\sin\frac{\pi}{2}(\text{x}+1),&\text{x}\leq0\\\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3},&\text{x}>0\end{cases}$
We have,
$(\text{LHL at x}=0)=\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{a}\sin\Big(\frac{\pi}{2}(-\text{h+1})\Big)$
$=\text{a}\sin\Big(\frac{\pi}{2}\Big)=\text{a}$
$(\text{RHL at x}=0)=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}$
$=\lim\limits\frac{\tan\text{h}-\sin\text{h}}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\sin\text{h}}{\cos\text{h} }-\sin\text{h}}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\sin\text{h}}{\cos\text{h}}(1-\cos\text{h})}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos\text{h})\tan\text{h}}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\sin^2\frac{\text{h}}{2}\tan\text{h}}{4\times\frac{\text{h}^2}{4}\times\text{h}}$
$=\frac{2}{4}\lim\limits_{\text{h}\rightarrow0}\frac{\sin^2\frac{\text{h}}{2}\tan\text{h}}{\frac{\text{h}^2}{4}\times\text{h}}$
$=\frac{1}{2}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2\times\lim\limits_{\text{h}\rightarrow0}\frac{\tan\text{h}}{\text{h}}$
$=\frac{1}{2}\times1\times1$
$=\frac{1}{2}$
If $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\text{a}=\frac{1}{2}$
View full question & answer→MCQ 61 Mark
The value of $f(0)$, so that the function $\text{f(x)}=\frac{\sqrt{\text{a}^2+\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}{}}{\sqrt{\text{a+x}}-\sqrt{\text{a-x}}}$ becomes continuous for all $x$, given by:
AnswerCorrect option: C. $-\text{a}^{\frac{1}{2}}$
Given, $\text{f(x)}=\frac{\sqrt{\text{a}^2+\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}{}}{\sqrt{\text{a+x}}-\sqrt{\text{a-x}}}$
$\Rightarrow\text{f(x)}=\frac{\big(\sqrt{\text{a}^2-\text{ax+x}^2}-\sqrt{\text{a}^2+\text{ax+x}^2}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$
$\Rightarrow\text{f(x)}=\frac{\big(\text{a}^2-\text{ax+x}^2\big)-\big(\text{a}^2+\text{ax+x}^2\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$
$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a+x}}-\sqrt{\text{a-x}}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}$
$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\text{a+x-a+x}\big)\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2}+\text{ax+x}^2\big)}$
$\Rightarrow\text{f(x)}=\frac{(-2\text{ax})\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{(2\text{x})\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$
$\Rightarrow\text{f(x)}=\frac{-\text{a}\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a}^2-\text{ax+x}^2}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)}$
So, if $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f(0)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Bigg[\frac{-\text{a}\big(\sqrt{\text{a+x}}+\sqrt{\text{a-x}}\big)}{\big(\sqrt{\text{a}^2-\text{ax+x}}+\sqrt{\text{a}^2+\text{ax+x}^2}\big)} \Bigg]$
$\Rightarrow\bigg[\frac{-2\text{a}(\sqrt{\text{a}})}{(\sqrt{a}^2+\sqrt{\text{a}^2})}\bigg]=\text{f(0)}$
$\Rightarrow\begin{bmatrix}\frac{-2\text{a}(\sqrt{a})}{(\text{a+a})} \end{bmatrix}=\text{f(0)}$
$\Rightarrow\text{f}(0)$
$=-\sqrt{a}$
View full question & answer→MCQ 71 Mark
The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&0\leq\text{x}<1\\\text{a},&1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous for $0\leq\text{x}<\infty,$ then the most suitable values of $a$ and $b$ are:
- A
$\text{a}=1,\text{ b}=-1$
- B
$\text{a}=-1,\text{ b}=1+\sqrt{2}$
- ✓
$\text{a}=-1,\text{ b}=1$
- D
$\text{None of these}.$
AnswerCorrect option: C. $\text{a}=-1,\text{ b}=1$
Given, $f(x)$ is continuous for $0\leq\text{x}<\infty.$
This means that $f(x)$ is continuous for $\text{x}=1,\sqrt{2.}$
Now,
If $(x)$ is continuons at $x = 1,$ then
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})=\text{a}$
$\Rightarrow\frac{(1-\text{h})^2}{\text{a}}=\text{a}$
$\Rightarrow\frac{1}{\text{a}}=\text{a}$
$\Rightarrow\text{a}^2=1$
$\Rightarrow\text{a}=\pm1$
If $f(x)$ is continuous at $\text{x}={\sqrt{2}},$ then
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\text{f}(\sqrt{2})$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}-\text{h})=\frac{2\text{b}^2-4\text{b}}{2}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{a=b}^2-2\text{b}$
$\Rightarrow\text{a = b}^2-2\text{b}$
$\Rightarrow\text{b}^2-2\text{b - a}=0$
$\therefore$ For $a = -1,$ We have
$\text{b}^2-2\text{b}+1=0$
$\Rightarrow(\text{b}-1)^2=0$
$\Rightarrow\text{b}=1$
Thus, $a = -1$ and $b=1$
View full question & answer→MCQ 81 Mark
Let $\text{f(x)}=\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}},\text{ x}\neq\frac{\pi}{4}$ The value which should be assigned to $f(x)$ at $\text{x}=\frac{\pi}{4},$ so that it is continuous everywhere is:
AnswerCorrect option: B. $\frac{1}{2}$
$\text{f}\big(\frac{\pi}{2}\big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}}$
$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\frac{1-\tan\text{x}}{1+\tan\text{x}}}{\cot2\text{x}}$
$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\tan2\text{x}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$
$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{1-\tan^2\text{x}}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$
$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{(1-\tan\text{x})(1+\tan\text{x})}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$ $\begin{pmatrix}\because\tan\frac{\pi}{4}\rightarrow1\\1-\tan\frac{\pi}{4}\neq0 \end{pmatrix}$
$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{(1+\tan\text{x})^2}=\frac{2}{4}=\frac{1}{2}$
View full question & answer→MCQ 91 Mark
If $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{n}>\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then:
- A
$\text{m}=1,\text{ n}=0$
- B
$\text{m}=\frac{\text{n}\pi}{2}+1$
- ✓
$\text{n}=\frac{\text{m}\pi}{2}$
- D
$\text{m}=\text{n}=\frac{\pi}{2}$
AnswerCorrect option: C. $\text{n}=\frac{\text{m}\pi}{2}$
Here,
$\text{f}\Big(\frac{\pi}{2}\Big)=\frac{\text{m}\pi}{2}+1$
$\Big(\text{LHL}\text{ x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\text{m}\Big(\frac{\pi}{2}-\text{h}\Big)+1$
$=\frac{\text{m}\pi}{2}+1$
$\Big(\text{RHL}\text{ x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}$
$= \lim\limits_{\text{h}\rightarrow0}\text{f}(\frac{\pi}{2}+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\Big[\sin\Big(\frac{\pi}{2}+\text{h}\Big)+\text{n}\Big]$
$=\text{n}+1$
Thus,
if $f(x)$ is continuons at $\text{x}= \frac{\pi}{2},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}$
$\Rightarrow\frac{\text{m}\pi}{2}+1=\text{n}+1$
$\Rightarrow\frac{\text{m}\pi}{2}=\text{n}$
View full question & answer→MCQ 101 Mark
The value of $f(0),$ so that the function $\text{f(x)}=\frac{2-(256-7\text{x})^{\frac{1}{8}}}{(5\text{x}+32)^\frac{1}{5}-2},\text{ x}\neq0$ is continuous everywhere, is given by:
Answer$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{2-(256-7\text{x})^{\frac{1}{8}}}{(5\text{x}+32)^\frac{1}{5}-2}$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{-\bigg[(256-7\text{x})^\frac{1}{8}-256^\frac{1}{8}\bigg]}{(5\text{x+32})^\frac{1}{5}-32\frac{1}{5}}$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{-\bigg[(256-7\text{x})^{\frac{1}{8}}-256\bigg]}{-7\text{x}}}{\frac{(5\text{x}+32)^{\frac{1}{5}}-32^\frac{1}{5}}{5\text{x}}}$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{-\bigg[(256-7\text{x})^\frac{1}{8}-256^\frac{1}{8}\bigg]}{-7\text{x}}\text{x}-7}{\frac{(5\text{x}+32)^\frac{1}{5}-32^\frac{1}{5}}{5\text{x}}\times5}$
$\text{f}(0)=\frac{7}{5}\times\frac{\frac{1}{8}\times256^\frac{-7}{8}}{\frac{1}{5}\times32^\frac{-4}{5}}$
$\text{f}(0)=\frac{7}{5}\times\frac{5\times2^4}{8\times2^7}=\frac{7}{8}\times\frac{1}{8}=\frac{7}{64}$
View full question & answer→MCQ 111 Mark
The value of a for which the function $\text{f(x)}=\begin{cases}5\text{x}-4,&\text{if }0<\text{x}\leq1\\4\text{x}^2+3\text{ax},&\text{if }<\text{x}<2\end{cases}$ is continuous at every point of its domain, is:
- A
$\frac{13}{3}$
- B
$1$
- C
$0$
- ✓
$-1$
Answer$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}$
$\lim\limits_{\text{x}\rightarrow1}5\text{x}-4=\lim\limits_{\text{x}\rightarrow1}4\text{x}^2+3\text{ax}$
$1=4+3\text{a}$
$\text{a}=-1$
View full question & answer→MCQ 121 Mark
If $f(x)$ defind by $\text{f(x)}=\begin{cases}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}},&\text{x}\neq0,1\\1,&\text{x}=0\\-1,&\text{x}=1\end{cases}$ then $f(x)$ is continuse for all:
AnswerCorrect option: D. $x$ except at $x = 0$ and $x = 1$
Given function $\text{f(x)}=\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}$
Consider,
$\text{f}(0^+)=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x(x}-1)|}{\text{x(x}-1)}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x(x}-1)}{\text{x(x}-1)}=1$
$\text{f}(0^-)=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}(\text{x}-1)|}{\text{x}(\text{x}-1)}=\lim\limits_{\text{x}\rightarrow0}\frac{-\text{x}(\text{x}-1)}{\text{x}(\text{x}-1)}=-1$
Also, for $f(1^+)$ and $f(1^-)$ you can check.
View full question & answer→MCQ 131 Mark
If $\text{f(x)}=(\text{x+1})^{\cot\text{x}}$ be continuous at $x = 0,$ then $f(0)$ is equal to:
- A
$0$
- B
$\frac{1}{\text{e}}$
- ✓
$e$
- D
AnswerGiven, $\text{f(x)}=(\text{x+1})^{\cot\text{x}}$
$\log\text{f(x)}=(\cot\text{x})(\log(\text{x+1})) [$Taking log on both sides$]$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}(\cot\text{x})(\log(\text{x+1}))$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(\text{x+1})}{\tan\text{x}}\Big)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{\log(\text{x+1})}{\text{x}}\Big)}{\Big(\frac{\tan\text{x}}{\text{x}}\Big)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\log\text{f(x)}=\frac{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(\text{x+1})}{\text{x}}\Big)}{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\tan\text{x}}{\text{x}}\Big)}$
$\Rightarrow\log\Big(\lim\limits_{\text{x}\rightarrow0}\text{f(x)}\Big)=\frac{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(\text{x+1})}{\text{x}}\Big)}{\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\tan\text{x}}{\text{x}}\Big)}$ $[\because f(x)$ is continuous at $x =0]$
$\Rightarrow\log\Big(\lim\limits_{\text{x}\rightarrow0}\text{f(x)}\Big)=1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{e}$
$\Rightarrow\text{f(0)}=\text{e} $
$[\because f(x)$ is continuous at $x = 0]$
View full question & answer→MCQ 141 Mark
The function $\text{f(x)}=\tan\text{x}$ is discontinuous on the set:
- A
$\{\text{n}\pi:\text{n}\in\text{z}\}$
- B
$\{2\text{n}\pi:\text{n}\in\text{z}\}$
- ✓
$\{(2\text{n}+1)\frac{\pi}{2}:\text{n}\in\text{z}\}$
- D
$\Big\{\frac{\text{n}\pi}{2}:\text{n}\in\text{z}\Big\}$
AnswerCorrect option: C. $\{(2\text{n}+1)\frac{\pi}{2}:\text{n}\in\text{z}\}$
When $\tan(2\text{n}+1)\frac{\pi}{2}=\tan\Big(\text{n}\pi+\frac{\pi}{2}\Big)=-\cot\text{n}\pi,$ it is not defined at the integral points.
$[\text{n}\in\text{z}]$
Hence, $f(x)$ is discontinuous on the set $\{(2\text{n}+1)\frac{\pi}{2}:\text{n}\in\text{z}\}$
View full question & answer→MCQ 151 Mark
If $\text{f(x)}=\begin{cases}\text{ax}^2+\text{b},&0\leq\text{x}<1\\4,&\text{x}=1\\\text{x}+3,&1<\text{x}\leq2\end{cases}$ then the value of $(a, b)$ for which $f(x)$ cannot be continuous at $x = 1,$ is:
- A
$(2, 2)$
- B
$(3, 1)$
- C
$(4, 0)$
- ✓
$(5, 2)$
AnswerCorrect option: D. $(5, 2)$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\text{f}(1)$
$\lim\limits_{\text{x}\rightarrow1}\text{ax}^2+\text{b}=4$
$a + b = 4$
We have possible values as $(2, 2), (3, 1), (4, 0)$
But can not be $(5, 2).$
Function is can not be continuous at $(5, 2).$
View full question & answer→MCQ 161 Mark
The value of $k$ which makes $\text{f(x)}=\begin{cases}\sin\frac{1}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ continuous at $x = 0$, is:
AnswerIf $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\sin\frac{1}{\text{x}}\Big)=\text{k}$
But $\lim\limits_{\text{x}\rightarrow0}\Big(\sin\frac{1}{\text{x}}\Big)$ does not exist.
Thus, there dose not exist any $k$ thet makes $f(x)$ a continuous function.
View full question & answer→MCQ 171 Mark
Let $\text{f(x)=}\begin{cases}\frac{\text{x}^4-5\text{x}^2+4}{|(\text{x}-1)(\text{x-2})|},\text{x}\neq1,2\\6, \text{x}=1\\12,\text{x}=2\end{cases}$ Then $f(x)$ is continuous on the set:
- A
$R$
- B
$R - \{1\}$
- C
$R - \{2\}$
- ✓
$R - \{1, 2\}$
AnswerCorrect option: D. $R - \{1, 2\}$
Given:
$\text{f(x)=}\begin{cases}\frac{\text{x}^4-5\text{x}^2+4}{|(\text{x}-1)(\text{x-2})|},\text{x}\neq1,2\\6, \text{x}=1\\12,\text{x}=2\end{cases}$
Now,
$\Rightarrow\text{x}^4-5\text{x}^2+4=\text{x}^4-\text{x}^2-4\text{x}^2+4\\=\text{x}^2(\text{x}^2-1)-4(\text{x}^2-1)$
$=(\text{x}^2-1)(\text{x}^2-4)=(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)}{|(\text{x}-2)(\text{x}-1)|},&\text{x}\neq1,2\\6,& \text{x}=1\\12,&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}(\text{x}+1)(\text{x}+2),&\text{x}<1\\-(\text{x+1})(\text{x}+2),&1<\text{x}<2\$\text{x+1})(\text{x}+2),&\text{x}>2\\6,&\text{x=1}\\12,&\text{x}=2\end{cases}$
So,
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(1-\text{h}+1)(1-\text{h}+2)\\=2\times3=6$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(1+\text{h}+1)(1+\text{h}+2)\\=-2\times3=-6$
Also,
$\Rightarrow\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(2-\text{h}+1)(2-\text{h}+2)=-12$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2+\text{h}+1)(2+\text{h}+2)=12$
Thus,
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}$
$\therefore$ The only point of discontinuities of the function $f(x)$ are $x = 1$ and $x = 2.$
Hence, the given function is continuous on the set $R - \{1, 2\}.$
View full question & answer→MCQ 181 Mark
The points of discontinuity of the function $\text{f(x)}=\begin{cases}\frac{1}{5}(2\text{x}^2+3),&\text{x}\leq1\\6-5\text{x},&1<\text{x}<3\\\text{x}-3,&\text{x}\geq3\end{cases}$ is $($are$):$
- A
$x = 1$
- ✓
$x = 3$
- C
$x = 1, 3$
- D
AnswerCorrect option: B. $x = 3$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}\frac{1}{5}(2\text{x}^2+3)=1$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}6-5\text{x}=1$
Function is continuou at $x = 1$
$\lim\limits_{\text{x}\rightarrow3^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}6-5\text{x}=-9$
$\lim\limits_{\text{x}\rightarrow3^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}\text{x}-3=0$
Function is discontinuous at $x = 3$
View full question & answer→MCQ 191 Mark
The function $\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}$
- A
Discontinuous at only one point.
- B
Discontinuous exactly at two points.
- C
Discontinuous exactly at three points.
- ✓
AnswerGiven,
$\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}$
$\Rightarrow\text{f(x)}=\frac{4-\text{x}^2}{\text{x}(4-\text{x}^2)}$
$\Rightarrow\text{f(x)}=\frac{1}{\text{x}},\text{x}\neq0 $ and $4-\text{x}^2\neq0 $ or $ \text{x}\neq0,\pm2$
Clearly, $f(x)$ is defined and continuous for all real numbers except $\left\{0,\pm2\right\}$
Therefore, $f(x)$ is discontinuous exactly at three points.
View full question & answer→MCQ 201 Mark
If the fucnction $\text{f(x)}=\begin{cases}(\cos\text{x})^{\frac{1}{\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuouse at $x = 0,$ then the value of $k$ is:
Answer$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\text{x}}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(1+\cos \text{x}-1)^{\frac{1}{\text{x}}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\Big(1-2\sin^2\frac{\text{x}}{2}\Big)^\frac{1}{\text{x}}$
$\text{f(0)}=\lim\limits\Big(1-2\sin^2\frac{\text{x}}{1}\Big)^{\frac{1}{-2\sin^2\frac{\text{x}}{2}}\times\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$
$\text{f(0)}= \lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$
$\text{f(0)}=\lim \limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text{x}}{2}}{\text{x}}\times\sin\frac{\text{x}}{2}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text {x}}{2}}{\frac{\text{x}}{2}}\times\frac{1}{2}\sin\frac{\text{x}}{2}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{1\times\sin\frac{1}{2}}=\text{e}^0=1$
$\Rightarrow\text{k}=1$
View full question & answer→MCQ 211 Mark
The value of $b$ for which the function $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$ is continuous at every point of its domain, is:
- ✓
$-1$
- B
$0$
- C
$\frac{13}{3}$
- D
$1$
AnswerGiven, $f(x)$ is continuous at every point of its domain.
So, it is continuous at $x = 1.$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f}\text{(x)}=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Big(4(1+\text{h})^2+3\text{b}(1+\text{h})\Big)=5(1)-4$
$\Rightarrow4+3\text{b}=1$
$\Rightarrow-3=3\text{b}$
$\Rightarrow \text{b} = -1$
View full question & answer→MCQ 221 Mark
If $\text{f(x)}=\begin{cases}\frac{36^\text{x}-9^\text{x}-4\text{x}+1}{\sqrt{2}-\sqrt{1+\cos\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at $x = 0$, these k equals.
AnswerCorrect option: C. $16\sqrt{2}\text{ in }6\text{ in }3$
$\text{k}=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{ \text{x}}+1}{\sqrt2-\sqrt{1+\cos\text{x}}}$
consider,
$=\lim\limits_{\text{x}\rightarrow0}\frac{36^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{\sqrt{2}-\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(4\times9)^{\text{x}}-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(4^{\text{x}}\times9^{\text{x}})-9^{\text{x}}-4^{\text{x}}+1}{2-(1+\cos\text{x})}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\cos\text{x})}{1-\cos^2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(9^{\text{x}}-1)(4^{\text{x}}-1)(\sqrt{2}+\sqrt{1+\cos\text{x}})(1+\cos\text{x})}{\sin^2\text{x}}$
dividing by $x^2$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{9^{\text{x}}-1}{\text{x}}{\times\frac{4^{\text{x}}-1}{\text{x}}\times\big(\sqrt{2}+\sqrt{1+\cos\text{x}}\big)(1+\cos\text{x})}}{\frac{\sin^2\text{x}}{\text{x}^2}}$
$=(\log9)(\log4)\big(\sqrt{2}+\sqrt{1+1}\big)(1+1)$
$=4\sqrt{2}(\log9)(\log4)$
$=4\sqrt{2}(2\log3)(2\log2)$
$=16\sqrt{2}(\log3)(\log2)$
View full question & answer→MCQ 231 Mark
If $\text{f(x)}=\begin{cases}\frac{1-\cos10\text{x}}{\text{x}^2},&\text{x}<0\\\text{a},&\text{x}=0\\\frac{\sqrt{\text{x}}}{\sqrt{625+\sqrt{\text{x}}}-25},&\text{x}>0\end{cases}$ then the value of so that $f(x)$ may be continuous at $x = 0$ is:
AnswerIf $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(-10\text{h}))}{(-\text{h})^2}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(10\text{h}))}{\text{h}^2}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(2\sin^2(5\text{h}))}{\text{h}^2}=\text{a}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{2\times25(\sin^2(5\text{h}))}{25\text{h}^2}=\text{a}$
$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\frac{(\sin^2(5\text{h}))}{(5\text{h})^2}=\text{a}$
$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(5\text{h})}{5\text{h}}\Big)^2=\text{a}$
$\Rightarrow\text{a}=50$
View full question & answer→MCQ 241 Mark
If the function $f(x)$ defined by $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at $x = 0,$ then $k =$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If $f(x)$ is continuous at $x = 0,$ then $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f(0)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}}\Big)=\text{k}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3\log(1+3\text{x})}{3\text{x}}-\frac{2\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)-2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)+2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{-2\text{x}}\Big)=\text{k}$
$\Rightarrow3\times1+2\times1=\text{k}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$
$\Rightarrow\text{k}=3+2$
$\Rightarrow\text{k}=5$
View full question & answer→MCQ 251 Mark
If $\text{f(x)}=|\log_{10}\text{x}|\text{fx}=\log_{10}\text{x},$ then at $x = 1:$
- ✓
$f(x)$ is continuous and $\text{f}'(1^+)=\log_{10}\text{e}$
- B
$f(x)$ is continuous and $\text{f}'(1^+)=\log_{10}\text{e}$
- C
$f(x)$ is continuous and $\text{f}'(1^-)=-\log_{10}\text{e}$
- D
AnswerCorrect option: A. $f(x)$ is continuous and $\text{f}'(1^+)=\log_{10}\text{e}$
Given,
$\text{f(x)}=|\log_{10}\text{x}|=\bigg|\frac{\log_{\text{e}}\text{x}}{\log_{\text{e}}10}\bigg|$
$=|(\log_{\text{e}}\text{x})\times(\log_{10}\text{e})|$
$=(\log_{10}\text{e})|\log_{10}\text{x}|$
$\text{f}'(1^+)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{{10}}\text{e})|\log_{\text{e}}(1+\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1|}{\text{h}}$
$=(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$
$=(\log_{10}\text{e})\times1$
$=(\log_{10}\text{e})$
Also,
$\text{f'(1}^-)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{10}\text{e})|\log_{\text{e}}(1-\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1}{\text{h}}$
$=-(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$
$=(\log_{10}\text{e})\times1=-(\log_{10}\text{e})$
View full question & answer→MCQ 261 Mark
If $\text{f(x)}=\text{x}\sin\frac{1}{\text{x}},\text{ x}\neq0,$ then the value of the function at $x = 0,$ so that the function is continuous at $x = 0,$ is:
Answer$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\times\sin\frac{1}{\text{x}}$
$-1\leq\sin\frac{1}{\text{x}}\leq1$
$\text{x}\times(-1)\leq\text{x}\sin\frac{1}{\text{x}}\leq\text{x}$
$\lim\limits_{\text{x}\rightarrow0}-\text{x}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\frac{1}{\text{x}}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}$
$\text{f}(0)=0$
View full question & answer→MCQ 271 Mark
The values of the constants $a, b$ and for which the function $\text{f(x)}=\begin{cases}(1+\text{ax})^{\frac{1}{\text{x}}},&\text{x}>0\\\text{b},&\text{x}=0\\\frac{(\text{x}+\text{c})^{\frac{1}{2}}-1}{(\text{x}+1)^{\frac{1}{2}}-1},&\text{x}>0\end{cases}$ may be continuous at $x = 0,$ are:
- A
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=-\frac{2}{3},\text{ c}=1$
- B
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\frac{2}{3},\text{ c}=-1$
- ✓
$\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$
- D
AnswerCorrect option: C. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}(1+\text{ax})^{\frac{1}{\text{x}}}$
$\text{b}=\lim\limits_{\text{x}\rightarrow{\text{a}}}(1+\text{ax})^{\frac{1}{\text{ax}}\times\text{a}}$
$\text{b}=\text{e}^{\text{a}}$
$\text{a}=\log_{\text{e}}\text{b}$
$\text{f}(0)=\lim\limits_{\text{x}\rightarrow\text{a}^+}\frac{(\text{x}+\text{c})^{\frac{1}{3}}-1}{(\text{x}+1)^{\frac{1}{2}}-1}$
Here, $\text{c}=1$
$\text{x}+1=\text{y}$
$\text{x}\rightarrow0\Rightarrow\text{y}\rightarrow1$
$\text{f}(0)=\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}^{\frac{1}{2}}-1}$
$\text{b}=\lim\limits_{\text{y}\rightarrow1}\frac{\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}-1}}{\frac{\text{y}^{\frac{1}{2}}-1}{\text{y}-1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$
$\text{a}=\log\text{b}=\log\frac{2}{3}$
View full question & answer→MCQ 281 Mark
The function $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
AnswerCorrect option: B. is not continuous at $x = 0$
Given, $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},\text{x}\neq0\\0,\text{x}=0\end{cases}$
We have
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1}\Bigg)$
if $\text{e}^\frac{1}{\text{x}}=\text{t},$ then
$\text{x}\rightarrow0, \text{t}\rightarrow\infty$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{t}\rightarrow\infty}\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$
$=\lim\limits_{\text{t}\rightarrow\infty}\Bigg(\frac{1-\frac{1}{\text{t}}}{1+\frac{1}{\text{t}}}\Bigg)=\frac{1-0}{1+0}=1$
Also, $f(0) = 0$
$\because\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}\neq\text{f}(0)$
Hence, $f(x)$ is discontinuons at $x = 0.$
View full question & answer→MCQ 291 Mark
The function $\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$
AnswerCorrect option: C. Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and $x = 0$
Given function is
$\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$
Consider,
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^-}\text{f(x)}=\frac{1}{\text{n}^2}$
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}}1=1$
Hence, fnuction is discontinuous at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and $x = 0$
View full question & answer→MCQ 301 Mark
Let $f(x) = |x| + |x - 1|,$ then:
- ✓
$f(x)$ is continuous at $x = 0,$ as well as at $x = 1$
- B
$f(x)$ is continuous at $x = 0,$ but not at $x = 1$
- C
$f(x)$ is continuous at $x = 0,$ but not at $x = 0$
- D
AnswerCorrect option: A. $f(x)$ is continuous at $x = 0,$ as well as at $x = 1$
Since modulus function is everywhere continuous $|x|$ and $|x - 1|$ are also everywhere continuous.
Also,
It is known that if $f$ and $g$ are continuous functions,
then $f + g$ will also be continuous.
Thus, $|x| + |x - 1|$ is everywhere continuous.
Hence, $f(x)$ is continuous at $x = 0$ and $x = 1, x = 0$ and $x = 1.$
View full question & answer→MCQ 311 Mark
The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for $x = 2$. in order to make $f(x)$ continuous at $x = 2$, here $f(2)$ should be defined as:
AnswerHere,
$x^3+ x^2- 16x + 20$
$= x^3- 2x^2+ 3x^2- 6x - 10x + 20$
$= x^2(x - 2) + 3x(x - 2) - 10(x - 2)$
$= (x - 2)(x^2+ 3x - 10)$
$= (x - 2)(x - 2) (x - 5)$
$= (x - 2)^2(x + 5)$
So, the given function can be rewritten as
$\text{f(x)}=\frac{(\text{x}-2)^2(\text{x}+5)}{\text{x}-2}$
$\Rightarrow\text{f(x)}=(\text{x}-2)(\text{x}+5)$
If f(x) is continuous at $x = 2$, then
$\lim\limits_{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}(\text{x}-2)(\text{x}+5)=\text{f}(2)$
$\Rightarrow\text{f}(2)=0$
Hence, in order to make $f(x)$ continuous at $x = 2, f(2)$ should be defined as $0$.
View full question & answer→MCQ 321 Mark
If $\text{f(x)}=\begin{cases}\frac{\sin(\text{a}+1)}{\text{x}},&\text{x}<0\\\text{c},&\text{x}=0\\\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}},&\text{x}>0&\end{cases}$ is continuouse at $x = 0,$ then:
- A
$\text{a}=-\frac{3}{2},\text{b}=0,\text{c}=\frac{1}{2}$
- B
$\text{a}=-\frac{3}{2},\text{b}=1,\text{c}=-\frac{1}{2}$
- ✓
$\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
- D
$\text{None of these}.$
AnswerCorrect option: C. $\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}\times\frac{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c }$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x+bx}^2-\text{x}}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{bx}^2}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\sqrt{\text{x}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}\Big(1+\text{bx}^\frac{3}{2}\Big)}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}{\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\Bigg[\sqrt{\Big(1+\text{b}^\frac{3}{2}}\Big)+1\Bigg]}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{1}{\Bigg[\sqrt{\Big(1+\text{bx}^{\frac{3}{2}}\Big)}+1\Bigg]}=\text{c}$
$\text{c}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a+1})\text{x}+\sin\text{x}}{\text{x}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\bigg[\frac{(\text{a+1)x+x}}{2}\bigg]\cos\bigg(\frac{\text{ax}}{2}\bigg)}{\text{x}}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\Big[\frac{(\text{a+2)x}}{2}\Big]}{\frac{(\text{a+2)x}}{2}}\times\frac{(\text{a}+2)}{2}\cos\Big(\frac{\text{ax}}{2}\Big)=\frac{1}{4}$
$1\times\frac{(\text{a+2)}}{2}\times1=\frac{1}{4}$
$\text{a}+2=\frac{1}{2}$
$\text{a}=\frac{-3}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}$ exist if $\text{b}\neq0$
$\text{b }\in\text{ R}-(0)$
View full question & answer→MCQ 331 Mark
The function $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$ is continuous at $x = 0,$ then $k =$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$
If $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{\text{x}}=\text{f}(0)$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}=\frac{\text{k}}{2}$
$\Rightarrow3\times1=\frac{\text{k}}{2}$
$\Rightarrow\frac{\text{k}}{2}=3$
$\Rightarrow\text{k}=6$
View full question & answer→MCQ 341 Mark
If $\text{f(x)}=\begin{cases}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ and $f(x)$ is continous at $x = 0,$ then the value of $k$ is:
AnswerCorrect option: B. $a + b$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}}=\text{k}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{x}}-\frac{\log(1-\text{bx})}{\text{x}}=\text{k}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{ax}}\times\text{a} -\frac{\log(1-\text{bx})}{\text{-bx}}\times(-\text{b})=\text{k}$
$\text{a}+\text{b}=\text{k}$
View full question & answer→MCQ 351 Mark
The value of a for which the function $\text{f(x)}=\begin{cases}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}},&\text{x}\neq0\\12(\log4)^3,&\text{x}=0\end{cases}$ may be continuous at $x = 0$ is:
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^2}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^3}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{(4^\text{x}-1}{\text{x}^3}\Big)}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}}\text{a}\text{x}\ {\times}\frac{\frac{1}{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}}{\frac{\text{x}^3}{3}}\text{x}^3=12(\log4)^3$
$3(\log4)^3=12(\log4)^3$
$3\text{a}=12$
$\text{a}=12$
Note: The question is incorrect, so it has been modified.
View full question & answer→MCQ 361 Mark
If $\text{f(x)}=\frac{1}{1-\text{x}},$ then the set of points discontinuity of the function $f(f(f(x)))$ is:
- A
$\{1\}$
- ✓
$\{0,1\}$
- C
$\{-1, 1\}$
- D
AnswerCorrect option: B. $\{0,1\}$
Given, $\text{f}\text{(x)}=\frac{1}{1-\text{x}}$
Clearly, $\text{f}:\text{R}-\big\{1\big\}\rightarrow\text{R}$
Now,
$\text{f}\big(\text{f}\text{(x)}\big)=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\Bigg(\frac{1}{1-\big(\frac{1}{1-\text{x}}\big)}\Bigg)$
$=\Big(\frac{1-\text{x}}{-\text{x}}\Big)=\Big(\frac{\text{x}-1}{\text{x}}\Big)$
$\therefore\text{f}\text{o}\text{f}:\text{R}-\big\{0,1\big\}\rightarrow\text{R}$
Now,
$\text{f}\big(\text{f}\text{(x)}\big)=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\Bigg(\frac{1}{1-\big(\frac{1}{1-\text{x}}\big)}\Bigg)=\text{x}$
$\therefore\ \text{f}\text{o}\text{f}:\text{R}-\big\{0,1\big\}\rightarrow\text{R}$
Thus, $f(f(f(x)))$ is not defind at $x = 0, 1$
Hence, $f(f(f(x)))$ is discontinuous at $\{0, 1\}$
View full question & answer→MCQ 371 Mark
If $\text{f(x)}=\begin{cases}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then $k$ is equal to:
- ✓
$0$
- B
$\frac{1}{2}$
- C
$1$
- D
$-1$
Answer$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2}$
$\text{x}\rightarrow\frac{\pi}{2}+\text{h}$
$\text{x}\rightarrow\frac{\pi}{2}\Rightarrow\text{h}\rightarrow0$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\cos\big(\frac{\pi}{2}+\text{h}\big)\Big)-\cos\big(\frac{\pi}{2}+\text{h}\big)}{\Big(\pi-2\big(\frac{\pi}{2}+\text{h}\big)\Big)^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(-\sin\text{h})+\sin\text{h}}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{\frac{\text{h}-\sin\text{h}}{2}}\times\frac{\text{h}-\sin\text{h}}{2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\text{h}-\sin\text{h}}{2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times(\text{h}-\sin\text{h})$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{1}{4}\frac{\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{\text{h}^2}\times(\text{h}-\sin\text{h})=0$
$\text{k}=0$
View full question & answer→MCQ 381 Mark
If $\text{f(x)}=|\text{x}-\text{a}|\ \phi\ (\text{x}),$ where $\phi(\text{x})$ is continuous function, then:
- A
$\text{f}'(\text{a}^+)=\phi(\text{a})$
- ✓
$\text{f}'(\text{a}^-)=-\phi(\text{a})$
- C
$\text{f}'(\text{a}^+)=\text{f}'(\text{a}^-)$
- D
AnswerCorrect option: B. $\text{f}'(\text{a}^-)=-\phi(\text{a})$
Given that $\text{f(x)}=|\text{x}-\text{a}|\ \phi\ (\text{x}),$ where $\phi(\text{x})$ continuous function.
$|\text{x}-\text{a}|\Rightarrow\text{x}-\text{a}$ if $\text{x}-\text{a}>0$
$|\text{x}-\text{a}|\Rightarrow-(\text{x}-\text{a})$ if $\text{x}-\text{a}<0$
By definition of continuity,
$\text{f}'(\text{a})= \lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f(a)}}{\text{h}}$
Hence, $\text{f}(\text{a}^+)=\phi(\text{x})$ and $\text{f}'(\text{a}^-)=-\phi(\text{x})$
View full question & answer→MCQ 391 Mark
The value of $f(0)$, so that the function $\text{f(x)}=\frac{(27-2\text{x})^\frac{1}{3}-3}{9-3(243+5\text{x})^\frac{1}{5}}$ is continuous, is given by:
- A
$\frac{2}{3}$
- B
$6$
- ✓
$2$
- D
$4$
AnswerFor $f(x)$ to be continuous at $x = 0,$ we must have
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\frac{(27-2\text{x})^\frac{1}{3}-3}{9-3(243+5\text{x})^\frac{1}{5}}$
$\Rightarrow\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{3\Big(243^{\frac{1}{5}}-(243+5\text{x})^\frac{1}{5}\Big)}$
$=\frac{1}{3}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{\text{x}}}{\frac{\bigg(243^\frac{1}{5}-(243+5\text{x})^\frac{1}{5}\bigg)}{\text{x}}}$
$=\frac{-1}{3}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{\text{x}}}{\frac{\bigg((243+5\text{x})^{\frac{1}{5}}-243\frac{1}{5}\bigg)}{\text{x}}}$
$=\frac{2}{15}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{-2\text{x}}}{\frac{\bigg((243+5\text{x})^\frac{1}{5}-243^\frac{1}{5}\bigg)}{5\text{x}}}$
$=\frac{2}{15}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{27-2\text{x}-27}}{\frac{\bigg((243+5\text{x})^\frac{1}{5}{-243^{\frac{1}{5}}\bigg)}}{243+5\text{x}-243}}$
$=\frac{2}{15}\times\frac{\frac{1}{3}\times27^{\frac{-2}{3}}}{\frac{1}{5}\times243^\frac{-4}{5}}$
$=\frac{2}{15}\times\frac{\frac{1}{3}\times\frac{1}{27^{\frac{-2}{3}}}}{\frac{1}{5}\times\frac{1}{243^\frac{-4}{5}}}$
$=2$
View full question & answer→MCQ 401 Mark
If $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$ Then $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ if:
- A
$\text{a}=\frac{1}{3},\text{ b}=2$
- ✓
$\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
- C
$\text{a}=\frac{2}{3},\text{ b}=\frac{8}{3}$
- D
AnswerCorrect option: B. $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
Given, $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$
We have
$\Big(\text{LHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{1-\sin^2\big(\frac{\pi}{2}-\text{h}\big)}{3\cos^2\big(\frac{\pi}{2}-\text{h}\big)}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{1-\cos^2\text{h}}{3\sin^2\text{h}}\Big)$
$=\frac{1}{3}\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin^2\text{h}}{\sin^2\text{h}}\Big)$
$=\frac{1}{3}$
$\Big(\text{RHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\text{b}\big[1-\sin\big(\frac{\pi}{2}+\text{h})\big]}{\big[\text{x}-2\big(\frac{\pi}{2}+\text{h}\big)\big]^2}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{b}(1-\cos\text{h})}{[-2\text{h}]^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^{2}\frac{\text{h}}{2}}{4\text{h}^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^2\frac{\text{h}}{2}}{16\frac{\text{h}^2}{4}}\bigg)$
$=\frac{\text{b}}{8}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2$
$=\frac{\text{b}}{8}\times1$
$=\frac{\text{b}}{8}$
Also, $\text{f}\big(\frac{\pi}{2}\big)=\text{a}$
If $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\text{f}\big(\frac{\pi}{2}\big)$
$\Rightarrow\frac{1}{3}=\frac{\text{b}}{8}=\text{a}$
$\Rightarrow\text{a}=\frac{1}{3}$ and $\text{b}=\frac{8}{3}$
View full question & answer→MCQ 411 Mark
The points of discontinuity of the function $\text{f(x)}=\begin{cases}2\sqrt{\text{x}},&0\leq\text{x}\leq1\\4-2\text{x},&1<\text{x}<\frac{5}{2}\\2\text{x}-7,&\frac{5}{2}\leq\text{x}\leq4\end{cases}$ is $($are$)$:
- A
$\text{x}=1,\text{x}=\frac{5}{2}$
- ✓
$\text{x}=\frac{5}{2}$
- C
$\text{x}=1,\frac{5}{2},4$
- D
$\text{x}=0,4$
AnswerCorrect option: B. $\text{x}=\frac{5}{2}$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}2\sqrt{\text{x}}=2$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}4-2\text{x}=2$
Function is continuous at $x = 1$
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}4-2\text{x}=-1$
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}2\text{x}-7=-2$
Function is discontinuous at $\text{x}=\frac{5}{2}$
View full question & answer→MCQ 421 Mark
$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$ is continuous in the interval $[-1, 1],$ then $p$ is equal to:
- A
$-1$
- ✓
$-\frac{1}{2}$
- C
$\frac{1}{2}$
- D
$1$
AnswerCorrect option: B. $-\frac{1}{2}$
Given, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$
If $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f(0)}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}}{-\text{h}}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(1-\text{ph}-1-\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(-2\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(2\text{p})}{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\Big(\frac{(2\text{p})}{(2)}\Big)=\Big(\frac{1}{-2}\Big)$
$\Rightarrow\text{p}=\frac{-1}{2}$
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