MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 8

$\displaystyle \lim _{x \rightarrow 0} \dfrac{\cos (\tan x)-\cos x}{x^{4}}$ is equal to

0%
1/6
0%
-1/3
0%
1/2
0%
1

Explanation

$\cos (\tan x)-\cos x=2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)$

$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{\cos (\tan x)-\cos x}{x^{4}}=\lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}}$

$=\displaystyle \lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}\left(\dfrac{x+\tan x}{2}\right)\left(\dfrac{x-\tan x}{2}\right)}\left(\dfrac{x^{2}-\tan ^{2} x}{4}\right)$

$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\tan ^{2} x}{x^{4}}$

$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\left(x+\dfrac{x^{3}}{3}+\dfrac{2}{15} x^{5}+\cdots\right)^{2}}{x^{4}}$

$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x^{2}}\left(1-\left(1+\dfrac{x^{2}}{3}+\dfrac{2}{15} x^{4}+\dots\right)^{2}\right)=-\dfrac{1}{3}$

The value of

$\displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1}{n}+\dfrac{e^{1 / n}}{n}+\dfrac{e^{2 / n}}{n}+\ldots .+\dfrac{e^{(n-1) / n}}{n}\right]$ is

0%
1
0%
0
0%
e-1
0%
e+1

Explanation

$\quad \displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1}{n}+\dfrac{e^{1 / n}}{n}+\dfrac{e^{2 / n}}{n}+\dots+\dfrac{e^{(n-1) / n}}{n}\right]$

$=\displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1+e^{1 / n}+\left(e^{1 / n}\right)^{2}+\cdots+\left(e^{1 / n}\right)^{n-1}}{n}\right]$

$=\displaystyle \lim _{n \rightarrow \infty} \dfrac{1 \cdot\left[\left(e^{1 / n}\right)^{n}-1\right]}{n\left(e^{1 / n}-1\right)}=(e-1) \lim _{n \rightarrow \infty} \dfrac{1}{\left(\dfrac{e^{1 / n}-1}{1 / n}\right)}$

$=(e-1) \times 1=(e-1)$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}$ is equal to

0%
1
0%
0
0%
2
0%
None of these

Explanation

$\dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}$

$=\dfrac{x^{4}\left(1+\tan ^{2} x+\tan ^{4} x\right)}{\tan ^{4} x\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\dfrac{x^{4}}{\tan ^{4} x}, x \neq 0$

$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}}{\tan ^{4} x}=1$

$\displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{20} \cos ^{2 n}(x-10)$ is equal to

0%
0
0%
1
0%
19
0%
20

Explanation

$\because \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n} x=\begin{cases}1, x=r \pi, r \in I \\ 0, x \neq r \pi, r \in I\end{cases}$
Here, for

$x=10, \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n}(x-10)=1$
and in all other cases it is zero.

$\therefore \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{\infty} \cos ^{2 n}(x-10)=1$

The value of

$\displaystyle \underset { n\rightarrow \infty }{ lim } \left( \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 6n } \right)$ is

0%
$\displaystyle \log { 2 }$
0%
$\displaystyle \log { 6 }$
0%
$\displaystyle 1$
0%
$\displaystyle \log { 3 }$

A point where function

$f(x)$ is not continuous where

$f(x)=\left[ \sin { \left[ x \right] } \right]$ in

$\left( 0,2\pi \right)$ ; is (

$\left[ \ast \right]$ denotes greatest integer

$\le x$ )

0%
$(3,0)$
0%
$(2,0)$
0%
$(1,0)$
0%
$(4,-1)$

$\underset { x\rightarrow \pi /4 }{ Lim } \dfrac { 2\sqrt { 2 } \left( cosx+sinx \right) ^{ 3 } }{ 1-sin2x } =2$ is equal to

0%
$\dfrac{3\sqrt{2}}{2}$
0%
$2\sqrt{2}$
0%
$\dfrac{4\sqrt{2}}{3}$
0%
$does \not \exist$ $

$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin \left(x^{2}\right)}{\ln \left(\cos \left(2 x^{2}-x\right)\right)}$ is equal to

0%
2
0%
-2
0%
1
0%
-1

Explanation

$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{In(cos(2x^2-x))}$
$$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{log \left(1-2 sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}$$

$=\displaystyle \lim_{x \to 0} \dfrac{sin(x^2)x^2}{\dfrac{x^2log \left(1-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}{-2sin^2 \left(\dfrac{2x^2-x}{2}\right)}\left[-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right]}$

$=\displaystyle \lim_{x \to 0} \dfrac{x^2}{\dfrac{2sin^2 \left(\dfrac{2x^2-x}{2}\right)}{\left(\dfrac{2x^2-x}{2}\right)^2}\left(\dfrac{2x^2-x}{2}\right)^2}$

$\displaystyle \lim_{x \to 0}-\dfrac{2x^2}{(2x^2-x)^2}=\displaystyle \lim_{x \to 0}-\dfrac{2}{(2x-1)^2}=-2$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}$ is equal to

0%
2
0%
-2
0%
1/2
0%
-1/2

Explanation

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{4 \sin ^{4} x}$

$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x}{4 \sin ^{4} x}\left[\dfrac{2 \tan x}{1-\tan ^{2} x}-2 \tan x\right]$

$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan ^{3} x}{2 \sin ^{4} x\left(1-\tan ^{2} x\right)}$

$=\dfrac{1}{2} \lim _{x \rightarrow 0} \dfrac{x}{\sin x} \dfrac{1}{\cos ^{3} x} \dfrac{1}{1-\tan ^{2} x}$

$=\dfrac{1}{2} \times 1 \times \dfrac{1}{1^{3}} \times \dfrac{1}{1-0}=\dfrac{1}{2}$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ is equal to

0%
$e^{\sin ^{2} y}$
0%
$\sin 2 y e^{\sin ^{2} y}$
0%
0
0%
None of these

Explanation

$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t+\int_{a}^{x+y} e^{\sin ^{2} t} d t\right]=\lim _{x \rightarrow 0} \dfrac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t \\$

$\therefore\left(\dfrac{0}{0} \text { form }\right)\\$

Apply L'Hopital Rule

$\displaystyle =\lim _{x \rightarrow 0} \dfrac{e^{\sin ^{2}(x+y)}\left(1+\dfrac{d y}{d x}\right)-e^{\sin ^{2} y} \dfrac{d y}{d x}}{1} \\$

$=e^{\sin ^{2} y}\left[1+\dfrac{d y}{d x}-\dfrac{d y}{d x}\right]=e^{\sin ^{2} y}$

$y^{r}=\dfrac{n !^{n+r-1} C_{r-1}}{r^{n}},$ where

$n=k r(k \text { is constant }),$ then

$\operatorname{lim}_{r\rightarrow\infty} y$ is equal to

0%
$(k-1) \log _{e}(1+k)-k$
0%
$(k+1) \log _{e}(k-1)+k$
0%
$(k+1) \log _{e}(k-1)-k$
0%
$(k-1) \log _{e}(k-1)+k$

Explanation

$y^{r}=\left(1+\dfrac{1}{r}\right)\left(1+\dfrac{2}{r}\right)\left(1+\dfrac{3}{r}\right) \cdots\left(1+\dfrac{n-1}{r}\right)\\$

$\Rightarrow \log y=\dfrac{1}{r} \sum_{p=1}^{n-1} \log \left(1+\dfrac{p}{r}\right)\\$

$\Rightarrow \lim _{n \rightarrow \infty} y=\lim _{r \rightarrow \infty} y=\int_{-\infty}^{k} \log (1+x) d x=(k-1) \log _{e}(1+k)-k$

The value of

$\displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$ is

0%
$\dfrac{2 a}{\pi}$
0%
$-\dfrac{2 a}{\pi}$
0%
$\dfrac{4 a}{\pi}$
0%
$-\dfrac{4 a}{\pi}$

Explanation

$\displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cdot \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$

$=\displaystyle \lim _{x \rightarrow a} \dfrac{\sqrt{a^{2}-x^{2}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}$

$=\dfrac{2}{\pi} \displaystyle \lim _{x \rightarrow a} \dfrac{\dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}(a+x)=\dfrac{4 a}{\pi}$

If function

$f(x)=\dfrac{x^2-9}{x-3}$ is continuous at

$x=3$ , then value of

$(3)$ will be:

0%
$6$
0%
$3$
0%
$1$
0%
$0$

Explanation

$f(x)=\dfrac{x^2-9}{x-3}$

Left hand limit

$f(3+0)=\displaystyle \lim_{h \rightarrow 0}f(3+h)$

$=\displaystyle \lim_{h \rightarrow 0} \dfrac{(3+h)^2-9}{3+h-3}$

$=\displaystyle \lim_{h \rightarrow 0} \dfrac{h(6+h)}{h}$

$=\displaystyle \lim_{h \rightarrow 0} (6+h)$

$=6$

Function is continous at

$x=3$ , so

$f(3)=f(3+0)$

$f(3)=6$

Hence, option

$(a)$ is correct.

$f(x)=\begin{cases} \begin{matrix} \dfrac{\log (1+mx)- \log (1-nx)}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} k; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$
is continuous at

$x=0$ then the value of

$k$ will be:

0%
$0$
0%
$m+n$
0%
$m-n$
0%
$m.n$

Explanation

$\therefore$ Function is continous at

$x=0$

$\therefore \displaystyle \lim_{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{\log (1+mx)- \log (1-nx)}{x}=k$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{ [mx-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3} - ...]- [nx-\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} - ...]}{x}=k$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{x \left[ m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... \right]}{x}=k$

$\Rightarrow \displaystyle \lim_{x \rightarrow 0} m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... =k$

$\Rightarrow m+n=k$

$\Rightarrow k=m+n$

Hence, option

$(b)$ is correct.

If function

$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$
is continuous at

$x=2$ then value of

$m$ will be:

0%
$3$
0%
$1/3$
0%
$1$
0%
$0$

Explanation

$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$

$f(0)=m$

$f(0+0)=\displaystyle \lim_{h \rightarrow 0}f(0+h)$

$\displaystyle \lim_{h \rightarrow 0} \dfrac{\sin 3 (0+h)}{0+h}$

$\displaystyle \lim_{h \rightarrow 0}3 \dfrac{\sin 3 h}{3h}$

$=3 \times 1$

$=3$

$x=a$ function is continuous.

So,

$f(0)=f(0+0)$

$m=3$

Hence, option

$(a)$ is correct.

The value of

$\displaystyle \lim_{x\rightarrow 0} \dfrac {xe^{x} - \log_{e} (1 + x)}{x^{2}}$ is

0%
$2/3$
0%
$1/3$
0%
$1/2$
0%
$3/2$

Explanation

$\displaystyle \lim_{x \rightarrow 0} \dfrac {xe^{x} - \log_{e} (1 + x)}{x^{2}}$

$x \left [1 + x + \dfrac {x^{2}}{2!} + \dfrac {x^{3}}{3!} + ....\right ]$

$= \displaystyle \lim_{x \rightarrow 0} \dfrac {-\left [x - \dfrac {x^{2}}{2} + \dfrac {x^{3}}{3} - \dfrac {x^{4}}{4} + ....\right ]}{x^{2}}$

$= \displaystyle \lim_{x \rightarrow 0} \dfrac {\left [x + x^{2} + \dfrac {x^{3}}{2!} + \dfrac {x^{4}}{3!} + .... - x + \dfrac {x^{2}}{2} - \dfrac {x^{3}}{3} + \dfrac {x^{4}}{4} - ....\right ]}{x^{2}}$

$= \displaystyle \lim_{x \rightarrow 0} \dfrac {x^{2} \left [1 + \dfrac {x}{2!} + \dfrac {x^{2}}{3!} + ... + \dfrac {1}{2} - \dfrac {x}{3} + \dfrac {x^{2}}{4} - .... \right ]}{x^{2}}$

$= \left [1 + 0 + 0 + ... + \dfrac {1}{2} - 0 + 0 ... \right ] = 1 + \dfrac {1}{2} = \dfrac {3}{2}$
Hence, option (D) is correct.

$f(x)=\begin{cases} \dfrac { 1-\sqrt { 2 } \sin { x } }{ \pi -4x } ,\quad \quad ifx\neq \dfrac { \pi }{ 4 } \\ a\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ,\quad \quad ifx=\dfrac { \pi }{ 4 } \end{cases}$ is continous at

$x=\dfrac {\pi}{4}$ then

$a=$

0%
$4$
0%
$2$
0%
$1$
0%
$1/4$

The value of

$\displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin x}{x}$ is

0%
$0$
0%
$\infty$
0%
$1$
0%
$-1$

Explanation

$\displaystyle \lim_{x \rightarrow \infty} \dfrac {\sin x}{x}$
Let

$x = 1/y$ or

$y = \dfrac {1}{x}$
Som

$x\rightarrow \infty \Rightarrow y \rightarrow 0$

$\therefore \displaystyle \lim_{x \rightarrow \infty} \left (\dfrac {\sin x}{x}\right ) = \displaystyle \lim_{y \rightarrow 0} y \sin \left (\dfrac {1}{y}\right )$

$= \displaystyle \lim_{y \rightarrow 0} y \cdot \displaystyle \lim_{y \rightarrow 0} \sin \dfrac {1}{y}$

$= 0\times \text {(any variable quantity)}$

$= 0$
Hence, option (A) is correct.

The value of

$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$ is

0%
$0$
0%
$1/2$
0%
$-1/2$
0%
$-1$

Explanation

$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$

$= \displaystyle \lim_{x\rightarrow 0} \dfrac {1 - 1 + 2\sin^{2} x/2}{x^{2}}$

$= \displaystyle \lim_{x\rightarrow 0} \dfrac {2\sin^{2} x/2}{x^{2}}$

$= \displaystyle \lim_{x\rightarrow 0} 2\left (\dfrac {\sin x/2}{x/2}\right )^{2} \times \dfrac {1}{4}$

$= \dfrac {1}{2}\times \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin x/2}{x/2}\right )^{2}$

$= \dfrac {1}{2} \times 1 = \dfrac {1}{2}$
Hence option (B) is correct.

The value of

$\displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{\tan x}\right )^{4}$ is

0%
$0$
0%
$81$
0%
$4$
0%
$1$

Explanation

$\displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{\tan x}\right )^{4}$

$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\times 3x\right )^{4}\times \dfrac {1}{\tan^{4} x}$

$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\right )^{4} \times \dfrac {3^{4}}{\dfrac {\tan^{4}x}{x^{4}}}$

$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\right )^{4} \times \dfrac {1}{\left (\dfrac {\tan x}{x}\right )^{4}} \times 81$

$= 1\times 1\times 81 = 81$
Hence, option (B) is correct.

The value of

$\displaystyle \lim_{x\rightarrow \infty} \sin \dfrac {\pi}{4x} \cos \dfrac {\pi}{4x}$ is

0%
$\pi/4$
0%
$\pi/2$
0%
$0$
0%
$\infty$

Explanation

$\displaystyle \lim_{\rightarrow \infty} x \sin \dfrac {\pi}{4x}\cos \dfrac {\pi}{4x}$

$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {2\sin \dfrac {\pi}{4x} \cos \dfrac {\pi}{4x}}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {1}{2} \times \dfrac {\pi}{2x}$

$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (2\times \dfrac {\pi}{4x}\right )}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {\pi}{4}$

$= \dfrac {\pi}{4}\times \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (\dfrac {\pi}{2x}\right )}{\left (\dfrac {\pi}{2x}\right )} = \dfrac {\pi}{4}\times 1 = \dfrac {\pi}{4}$
Hence, option (A) is correct.

Let

$f(x)=\displaystyle \frac{(256+ax)^{1/8}-2}{(32+bx)^{1/5}-2}$ . If

$f$ is continuous at

$x = 0$ , then the value of

$a / b$ is:

0%
$\displaystyle \frac{8}{5}f(0)$
0%
$\displaystyle \frac{32}{5}f(0)$
0%
$\displaystyle \frac{64}{5}f(0)$
0%
$\displaystyle \frac{16}{5}f(0)$

Explanation

$f(x)=\displaystyle \frac{(256+ax)^{1/8}-2}{(32+bx)^{1/5}-2}$
Given ,

$f$ is continuous at

$x=0$ .

$\displaystyle \lim _{ x\rightarrow 0 } f(x)=f(0)$
Now,

$\displaystyle \lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ \frac { (256+ax)^{ 1/8 }-2 }{ (32+bx)^{ 1/5 }-2 } }$
It is of the form

$\displaystyle \frac{0}{0}$ , so applying L-Hospital's rule

$\displaystyle =\lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ 8 } a(256+ax)^{ -7/8 } }{ \frac { 1 }{ 5 } b(32+bx)^{ -4/5 } } }$

$\displaystyle = \frac { 5a }{ 64b }$
Hence,

$\displaystyle \frac { 5a }{ 64b }=f(0)$

$\Rightarrow \displaystyle \frac { a }{ b }=\frac { 64 }{ 5 }f(0)$

Let

$\alpha(a)$ and

$\beta(a)$ be the roots of the equation

$(\sqrt[3]{1+a}-1)x^{2}+(\sqrt{1+a}-1){x}+(\sqrt[6]{1+a}-1)=0$ where

$a>-1$ . Then

$\underset{a\rightarrow 0^{+}}{\lim}\alpha(a)$ and

$\underset{a\rightarrow 0^{+}}{\lim}\beta(a)$ are

0%
$-\displaystyle \frac{5}{2}$ and 1
0%
$-\displaystyle \frac{1}{2}$ and $-1$
0%
$-\displaystyle \frac{7}{2}$ and 2
0%
$-\displaystyle \frac{9}{2}$ and 3

Explanation

Let

$1+a=y$ .

The equation becomes

$(y^{\frac {1}{3}}-1)x^2+(y^{\frac {1}{2}}-1)x+(y^{\frac {1}{6}}-1)=0$

Dividing by

$y-1$ , we get

$\displaystyle \left(\dfrac {y^{\frac {1}{3}}-1}{y-1}\right)x^2+\left(\dfrac {y^{\frac {1}{2}}-1}{y-1}\right)x+\left(\dfrac {y^{\frac {1}{6}}-1}{y-1}\right)=0$

When

$a\rightarrow 0, y\rightarrow 1$ .

Hence, taking

$\underset {y\rightarrow 1}{\lim}$ on both the sides,

$\Rightarrow \displaystyle \underset {y\rightarrow 1}{\lim} \left(\dfrac {y^{\frac {1}{3}}-1}{y-1}\right)x^2+\left(\dfrac {y^{\frac {1}{2}}-1}{y-1}\right)x+\left(\dfrac {y^{\frac {1}{6}}-1}{y-1}\right)=0$

$\Rightarrow \dfrac {1}{3}x^2+\cfrac {1}{2}x+\cfrac {1}{6}=0$ ...[Using

$\underset {x\rightarrow a}{\lim}\cfrac {x^n-a^n}{x-a}=na^{n-1}]$

$\Rightarrow 2x^2+3x+1=0$

Solving the quadratic equation, we get

$\Rightarrow x=-1$ or

$x=\dfrac {-1}{2}$ .

Since

$\alpha (a)$ and

$\beta (a)$ are roots of the given equation, we get

$\underset {a\rightarrow 0^+}{\lim}\alpha (a)=-1$ and

$\underset {a\rightarrow 0^+}{\lim}\beta (a)=\dfrac {-1}{2}$ .

$f(x)=\left\{\begin{matrix}(cos x)^{1/sinx} &for &x\neq 0 \\ k & for & x=0 \end{matrix}\right.$
Then the value of

$k$ , so that

$f$ is continuous at

$x=0$ is

0%
$0$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$2$

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =k$

$\displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cos { x } \right) }^{ \cfrac { 1 }{ \sin { k } } } } =k$

${ 1 }^{ \infty }$ form

${ e }^{\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { \cos { x } -1 }{ \sin { x } } } }={ e }^{\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { -\sin { x } }{ \cos { x } } } }\quad \left( \cfrac { 0 }{ 0 } form \right)$

$={ e }^{ -\cfrac { 0 }{ 1 } }=1=k$

The function

$f$ :

$R/\{0\}\rightarrow R$ given by

$f(x)=\displaystyle \frac{1}{x}-\frac{2}{e^{2x}-1}$ can be made continuous at

$x=0$ by defining

$f(0)$ as -

0%
$2$
0%
$-1$
0%
$0$
0%
$1$

Explanation

$f(x) = \dfrac { 1 }{ x } -\dfrac { 2 }{ { e }^{ 2x }-1 } = \dfrac { { e }^{ 2x }-1-2x }{ x{ (e }^{ 2x }-1) }$

Using expansion of

$e^x$

$f(x) = \dfrac { \left(1+2x+\dfrac { { (2x) }^{ 2 } }{ 2! } +\dfrac { { (2x) }^{ 3 } }{ 3! } +...\right)-1-2x }{ x{ (e }^{ 2x }-1) }$

$f(x)=\dfrac { \dfrac { { (2x) }^{ 2 } }{ 2! } +\dfrac { { (2x) }^{ 3 } }{ 3! } }{ x{ (e }^{ 2x }-1) } =\quad \dfrac { 2x\left(\dfrac { { (2x) } }{ 2! } +\dfrac { { (2x) }^{ 2 } }{ 3! }\right) }{ x\times{ (e }^{ 2x }-1) }$

$f(x)=\dfrac { 2x }{ { (e }^{ 2x }-1) }\times\dfrac {\left(\dfrac { { (2x) } }{ 2! } +\dfrac { { (2x) }^{ 2 } }{ 3! } \right) }{ x }$
Now,

$\displaystyle\lim_{ x\rightarrow 0}\dfrac { 2x }{ { (e }^{ 2x }-1) }\times\dfrac { \dfrac { { (2x) } }{ 2! } +\dfrac { { (2x) }^{ 2 } }{ 3! } +... }{ x } =1\times1=1$

Hence, option D.

$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} \frac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } -1 }{ \sqrt { { x }^{ 2 }+4 } -2} , & x\neq 0 \end{matrix} \\ \begin{matrix} a, & x=0 \end{matrix} \end{cases}$ then the value of

$a$ in order that

$f(x)$ may be continuous at

$x=0$ is

0%
$-8$
0%
$8$
0%
$-4$
0%
$4$

Explanation

Given:

$\displaystyle f\left( x \right) =\begin{cases} \begin{matrix} \dfrac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x-1 } }{ \sqrt { { x }^{ 2 }+4 } -2 } , & \left( x\neq 0 \right) \end{matrix} \\ \begin{matrix} a, & \left( x\neq 0 \right) \end{matrix} \end{cases}$ is continuous at

$x=0$ , then

$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =\lim _{ x\rightarrow 0 }{ \dfrac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x-1 } }{ \sqrt { { x }^{ 2 }+4 } -2 } }$

$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { \cos { 2x-1 } }{ \sqrt { { x }^{ 2 }+4 } -2 } }$

As th function is of the

$\dfrac{0}{0}$ form,applying L-Hospital's rule,

$\displaystyle =\lim _{ x\rightarrow 0 }{ \left[ \dfrac { \dfrac { -2\sin { 2x } }{ x } }{ \sqrt { { x }^{ 2 }+4 } } \right] }$

$\displaystyle =\lim _{ x\rightarrow 0 }{ 2\frac { -\sin { 2x } \sqrt { { x }^{ 2 }+4 } }{ x } }$

$\displaystyle =-\lim _{ x\rightarrow 0 }{ 2\left[ 2\cos { 2x } \sqrt { { x }^{ 2 }+4 } +\sin { 2x } \frac { x }{ \sqrt { { x }^{ 2 }+4 } } \right] }$

$\displaystyle =-\left[ 4\sqrt { 4 } \right] =-8$

Thus,

$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =-8$

For function to be continuous at

$x=0$ .

$\displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ f\left( x \right) }$

$\displaystyle \Rightarrow a=-8$

$\displaystyle f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^{5}};x\neq 0$ is continuous at

$x=0$ , then

0%
$\displaystyle {A}+{B}=2$
0%
$\displaystyle {A}+{B}=1$
0%
$\displaystyle {A}+{B}=0$
0%
$\displaystyle {A}-{B} =1$

Explanation

$f(x)=\displaystyle \frac { sin3x+Asin2x+Bsinx }{ x^{ 5 } } ;x\neq 0$

$\displaystyle \lim _{ x\rightarrow 0 }{ f(x)= } \lim _{ x\rightarrow 0 }{\displaystyle \frac { sin3x+Asin2x+Bsinx }{ x^{ 5 } } }$
It is of the form

$\displaystyle \frac{0}{0}$ , so applying L-Hospital's rule

$=\displaystyle \lim _{ x\rightarrow 0 }{ \displaystyle\frac { 3\cos { 3x } +2A\cos { 2x } +B\cos { x } }{ 5x^{ 4 } } }$
As

$x\to 0 , D^{r} \to 0 \Rightarrow N^{r} \to 0$

$\displaystyle \lim _{ x\rightarrow 0 }{ 3\cos { 3x } +2A\cos { 2x } +B\cos { x } =0 }$

$\Rightarrow 3+2A+B=0$ .....(i)
Again

$\displaystyle \lim _{ x\rightarrow 0 }{\displaystyle \frac { 3\cos { 3x } +2A\cos { 2x } +B\cos { x } }{ 5x^{ 4 } } }$ is of the form

$\displaystyle \frac {0}{0}$

$=\displaystyle \lim _{ x\rightarrow 0 }{ \displaystyle\frac { -9\sin { 3x } -4A\sin { 2x } -B\sin { x } }{ 20x^{ 3 } } }$
Again of the form

$\displaystyle \frac {0}{0}$

$=\displaystyle \lim _{ x\rightarrow 0 }{\displaystyle \frac { -27\cos { 3x } -8A\cos { 2x } -B\cos { x } }{ 60x^{ 2 } } }$
As

$x\to 0 , D^{r} \to 0 \Rightarrow N^{r} \to 0$

$\displaystyle \lim _{ x\rightarrow 0 }{ -27\cos { 3x } -8A\cos { 2x } -B\cos { x } } =0$

$\Rightarrow 27+8A+B=0$ .....(ii)
Solving (i) and (ii), we get

$A=-4, B=5$
Thus,

$A+B=1$

$f(x)=\left\{\begin{array}{ll}\dfrac{(x+bx^{2})^{1/_{2}}-x^{1/2}}{bx^{3/2}} & x>0\\c & x=0\\\dfrac{\sin(a+1)x+\sin x}{x} & x<0\end{array}\right.$ is continuous at

${x}=0$ , then

0%
$a=\displaystyle \frac{-3}{2},b=0,c=\frac{1}{2}$
0%
$a=\displaystyle \frac{-3}{2},b\neq 0,c=\frac{1}{2}$
0%
$a=\displaystyle \frac{3}{2},b\neq 0,c=\frac{1}{2}$
0%
$a=\displaystyle \frac{3}{2},b\neq 0,c=-\dfrac{1}{2}$

Explanation

Given:

$f(x)=\left\{\begin{array}{ll}\dfrac{(x+bx^{2})^{1/_{2}}-x^{1/2}}{bx^{3/2}} & x>0\\c & x=0\\\dfrac{\sin(a+1)x+\sin x}{x} & x<0\end{array}\right.$ is continuous at

${x}=0$

Multiply and divide by

$((x+bx^{2})^{1/_{2}}+x^{1/2})$

$\displaystyle\lim_{x\rightarrow 0^{+}} =\lim_{x\to0}\dfrac{(x+bx^2) + (x)}{bx^{\frac{3}{2}}((x+bx^2)^{\frac{1}{2}} + x^{\frac{1}{2}})}$

$\displaystyle = \lim_{x\to 0}\dfrac{x^{\frac{1}{2}}}{{(x+bx^{2})^{\frac{1}{2}}+x^{\frac{1}{2}}}}$

$\displaystyle = \lim_{x\to0}\dfrac{1}{(1+bx)\dfrac{1}{2}+1} = \dfrac12$

$\displaystyle\lim_{x\rightarrow 0} \dfrac{(a+1)sin(a+1)x}{(a+1)x}+\dfrac{sin x}{x}$

$(a+1)+1 =\dfrac{1}{2}$

$a=\dfrac{-3}{2}$

$c=\dfrac{1}{2}$ (By continuity)

The graph of the function

$y = f (x)$ has a unique tangent at the point

$(e^{a} ,0)$ through which the graph passes then

$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-sinf(x)}{3f(x)}$ is

0%
$1$
0%
$2$
0%
$0$
0%
$-1$

Explanation

Given

$y=f(x)$ has a unique tangent at the point

$(e^a,0)$ .
So, as

$x\rightarrow e^{a}$ ,

$f(x)\rightarrow 0$

Now,

$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-\sin f(x)}{3f(x)}$

$\displaystyle \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 3f(x) } -\frac { \sin f(x) }{ 3f(x) }$

$\displaystyle =\frac { 7 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 7f(x) } -\frac { 1 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { \sin f(x) }{ f(x) }$

$=\dfrac{7-1}{3}=2$

Assertion(A):

$f(x)=\left\{\begin{array}{ll}x^{2}\sin(\frac{1}{x}) , & x\neq 0\\0, & x=0\end{array}\right.$ is continuous at

${x}=0$
Reason(R): Both

$h(x)=x^{2},g(x)= \left\{\begin{array}{ll}\sin(\frac{1}{x}) , & x\neq 0\\0, & x=0\end{array}\right.$ are continuous at

$x = 0$

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true and R is not the correct explanation of A
0%
A is true but R is false
0%
R is true but A is false

Explanation

Assertion

$f(0)=0$

$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ 2 }\sin { \left( \cfrac { 1 }{ x } \right) } } ={ 0 }^{ 2 }\times$ (finite value)

$=0$

$\therefore$ It is continuous at

$x=0$

Reason:

$h\left( x \right) ={ x }^{ 2 }$ is continuous

but

$g\left( x \right)$ is not continuous

$\displaystyle \lim _{ x\rightarrow 0 }{ \sin { \left( \cfrac { 1 }{ x } \right) } } =$ not defined (value oscillates)

$\displaystyle \lim _{ x\rightarrow 0 }{ \sin { \left( \cfrac { 1 }{ x } \right) } } \neq 0$

$\therefore$ not continuous.

Let

$a, b, c \epsilon R^+$ and

$\displaystyle\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\displaystyle \frac{n}{(k+an)(k+bn)}=\displaystyle \frac{A}{a-b}ln\displaystyle \frac{a(b+B)}{b(a+C)}, a \neq b,$ then

$(A + B + C)$ is equal to

0%
2
0%
3
0%
4
0%
5

Explanation

Let

$a, b, c$

$\epsilon R^+$ and

$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\dfrac{n}{n^{2}\left ( a+\dfrac{k}{n} \right )\left ( b+\dfrac{k}{n} \right )}$

$=\lim_{n\rightarrow \infty }\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\left ( a+\dfrac{k}{n} \right )\left ( b+\dfrac{k}{n} \right )}$

$=\int_{0}^{1}\dfrac{1}{(a+x)(b+x)}dx=\dfrac{1}{a-b}\int_{0}^{1}\dfrac{(a+x)-(b+x)}{(a+x)(b+x)}dx$

$=\dfrac{1}{a-b}\int_{0}^{1}\left ( \dfrac{1}{b+x}-\dfrac{1}{a+x} \right )dx$

$=\dfrac{1}{a-b}\left ( ln(b+x)-ln(a+x) \right )_{0}^{1}$

$=\dfrac{1}{a-b}\left ( ln\dfrac{(b+x)}{(a+x)} \right )_{0}^{1}=\dfrac{1}{a-b}ln\dfrac{(b+1)a}{(a+1)b}$

$\therefore A + B + C = 3$
Hence, option 'B' is correct.

If f (x)

$=\left \{ \displaystyle \frac{|x+2|}{tan^{-1}(_{2}x+2)} \right.x\neq -2$

$x=-2,$ then f(x) is

0%
continuous at $x = - 2$
0%
not continuous at $x = - 2$
0%
differentiable at $x = - 2$
0%
continuous but not differentiable at $x = - 2$

Explanation

L.H.L

$=\lim_{x\rightarrow -2^{-}}f(x)=\lim_{h\rightarrow 0^{+}}f(-2-h)$

$=\lim_{h\rightarrow 0^{+}}\dfrac{|-2-h+2|}{tan^{-1}(-2-h+2)}$

$\lim_{h\rightarrow 0^{+}}\dfrac{h}{tan^{-1}(-h)}$

$\lim_{h\rightarrow 0^{+}}\dfrac{h}{-tan^{-1}h}$

& R.H.L.

$= \lim_{x\rightarrow -2^{+}}f(x)=\lim_{h\rightarrow 0^{+}}f_-2+h$

$=\lim_{h\rightarrow 0^{+}}\dfrac{|-2+h+2|}{tan^{-1}(-2+h+2)}$

$=\lim_{h\rightarrow 0^{+}}\dfrac{h}{tan^{-1}(h)}=1$

$\therefore \lim_{x\rightarrow -2^{-}}f(x)\neq \lim_{x\rightarrow -2^{+}}f(x)$

so, f(x) is not continuous & not differentiable at x

$=- 2$

The function

$f(x) = x - |x - x^2|, -1 \leq x \leq 1$ is continuous on the interval

0%
$[-1, 1]$
0%
$[-1, 2]$
0%
$[-1, 1]- \{0\}$
0%
$(-1, 1)- \{0\}$

Explanation

Given,

$f(x)=x-|x-x^2|=x-|x(1-x)|$

$\therefore$ continuity is to be checked at

$x=0$ and

$x=1$

At

$x=0$

$LHL =\displaystyle\lim_{h \rightarrow 0}f(0-h)=\displaystyle\lim_{h \rightarrow 0}[-h-|-h(1+h)|]=0$

$RHL =\displaystyle\lim_{h \rightarrow 0}f(0+h)=\displaystyle\lim_{h \rightarrow 0}[h-|h(1-h)|]=0$

Also,

$f(0)=0$

Since

$LHL = RHL=f(0)$

$\therefore f(x)$ is continuous at

$x=0$

At

$x=1$

$LHL = \displaystyle\lim_{h \rightarrow 0}f(1-h)=\displaystyle\lim_{h \rightarrow 0}[(1-h)-|(1-h)(1-1+h)|]=1$

$RHL =\displaystyle\lim_{h \rightarrow 0}f(1+h)=\displaystyle\lim_{h \rightarrow 0}[(1+h)-|(1+h)(1-1-h)|]=1$

Also,

$f(1)=1$

$\therefore f(x)$ is continuous at

$x=1$

Hence

$f(x)$ is continuous for all

$x \in [-1,1]$

$\displaystyle \lim_{x\to\infty}{\displaystyle \frac{2\sqrt{x}+3\sqrt [\Large 3]{x}+4\sqrt [\Large 4]{x}+...+n\sqrt [\Large n]{x}}{\sqrt{(2x-3)}+\sqrt[\Large 3]{(2x-3)}+\sqrt[\Large 4]{(2x-3)}+...+\sqrt[\Large n]{(2x-3)}}}$ is equal to

0%
$1$
0%
$\infty$
0%
$\sqrt{2}$
0%
None of these

Explanation

$\displaystyle \lim_{x\to\infty}{\displaystyle \dfrac{2\sqrt{x}+3\sqrt [\Large 3]{x}+4\sqrt [\Large 4]{x}+...+n\sqrt [\Large n]{x}}{\sqrt{(2x-3)}+\sqrt[\Large 3]{(2x-3)}+\sqrt[\Large 4]{(2x-3)}+...+\sqrt[\Large n]{(2x-3)}}}$

Dividing numerator and denominator by

$\sqrt{x}$

$\displaystyle=\lim _{ x\to \infty }{ \displaystyle\dfrac { 2+3{ x }^{ -\frac { 1 }{ 6 } }+4{ x }^{ -\frac { 1 }{ 4 } }+...+n{ x }^{ \frac { (2-n) }{ 2n } } }{ \sqrt { (2-\dfrac { 3 }{ x } ) } +\dfrac { \sqrt [ 3 ]{ (2x-3) } }{ \sqrt { x } } +...+\dfrac { \sqrt [ n ]{ (2x-3) } }{ \sqrt { x } } } }$

$\displaystyle =\lim _{ x\to \infty }{ \dfrac { 2+\dfrac { 3 }{ { x }^{ 1/6 } } +\dfrac { 4 }{ { x }^{ 1/4 } } +...+\dfrac { n }{ { x }^{ \frac { (n-2) }{ 2n } } } }{ \sqrt { (2-\dfrac { 3 }{ x } ) } +\sqrt [ 6 ]{ \dfrac { { (2x-3) }^{ 2 } }{ { x }^{ 3 } } } +...+\sqrt [ 2n ]{ \dfrac { { (2x-3) }^{ 2 } }{ { x }^{ n } } } } }$

$\displaystyle=\dfrac { 2 }{ \sqrt { 2 } } =\sqrt { 2 }$

$f(x) \displaystyle =\frac{3x^2 + ax + a + 1}{x^2 + x - 2}$ , then which of the following can be correct?

0%
$\displaystyle \lim_{x \rightarrow 1} f(x)$ exists $\Rightarrow a = - 2$
0%
$\displaystyle \lim_{x \rightarrow -2} f(x)$ exists $\Rightarrow a = 13$
0%
$\displaystyle \lim_{x \rightarrow 1} f(x) = \dfrac{4}{3}$
0%
$\displaystyle \lim_{x \rightarrow -2} f(x) = -\dfrac{1}{3}$

Explanation

$f(x) \displaystyle = \frac{3x^2 + ax + a + 1}{(x+2)(x-1)}$
As

$x \rightarrow 1,$ Denominator

$\rightarrow 0$ . Hence as

$x \rightarrow 1,$ Numerator

$\rightarrow 0$ .
Therefore,

$3 + 2a + 1 = 0$ or

$a=-2$

As

$x \rightarrow -2,$ Denominator

$\rightarrow 0$ .Hence as

$x \rightarrow - 2,$ Numerator

$\rightarrow 0$ .
Therefore,

$12 - 2a + a + 1 = 0$ or

$a = 13$

Now,

$\lim_{x \rightarrow 1} f(x) = \displaystyle \lim_{x \rightarrow1} \frac{3x^2 - 2x - 1}{(x + 2)(x-1)}$

$\displaystyle = \lim_{x \rightarrow 1} \frac{(3x + 1)(x-1)}{(x+2)(x-1)} = \frac{4}{3}$

Now,

$\displaystyle \lim_{x \rightarrow -2} \frac{3x^2 + 13x +14}{(x + 2)(x-1)}$

$\displaystyle = \lim_{x \rightarrow -2} \frac{(3x + 7)(x+2)}{(x + 2)(x-1)} = -\frac{1}{3}$

If the function

$f(x)= \left\{\begin{matrix} (1+\left | \tan x \right |)^{ \displaystyle \frac{p}{\left| \tan {x} \right|}} &, -\frac{\pi }{3}< x< 0 \\ \\ q& x=0\\ \\ e^{ \displaystyle \frac{\sin {3x}}{\sin {2x}}},& 0\: < \, x\, < \frac{\pi }{3} \end{matrix}\right.$
is continuous at

$x=0$ , then

0%
$\displaystyle \mathrm{p}=\frac{3}{2}$
0%
$\displaystyle \mathrm{p}=\frac{2}{3}$
0%
$\log_{e}\mathrm{q}=\mathrm{p}$
0%
$\mathrm{q}=2$

Explanation

Since f(x) is continuous at

$x=0$

$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=f(0)=\lim_{x\rightarrow 0^{+}}f(x)$ ...(1)

Now ,

$LHL= \displaystyle \lim _{ x\rightarrow 0^{ - } } f(x)=\lim _{ x\rightarrow 0 } (1-tan\, x)^{ -\frac { p }{ tan\, x } }$

Since

$\displaystyle \lim _{ x\rightarrow 0 } (1+x)^{ 1/x }=e$

$\Rightarrow LHL=e^{ p }$ ...(2)

Now,

$RHL \displaystyle =\lim_{x\rightarrow 0^{+}}f(x) =\lim_{x\rightarrow 0}e^{\displaystyle \frac{\sin {3x}}{\sin {2x}}}$

$\displaystyle =\lim _{ x\rightarrow 0 } e^{ \displaystyle \frac { 3 }{ 2 } (\displaystyle \frac { \sin { 3x } }{ 3x } )(\frac { 2x }{ \sin { 2x } } ) }$

$RHL=e^{3/2}$ ....(3)
From (1), (2) and (3)

$e^{ p }=e^{ 3/2 }=q$

$\Rightarrow p=\frac { 3 }{ 2 } =\log _{ e } q$

Let

$\tan \alpha .x+\sin \alpha .y=\alpha$ and

$\alpha \ \text{cosec} \alpha .x+\cos \alpha .y=1$ be two variable straight line,

$\alpha$ being the parameter. Let

$P$ be the point of intersection of the lines. In the limiting position when

$\alpha \rightarrow 0$ , the point

$P$ lies on the line

0%
$x=2$
0%
$x=-1$
0%
$y+1=0$
0%
$y=2$

Explanation

Solving

$\tan \alpha .x+\sin \alpha .y=\alpha$ and

$\alpha \ \text{cosec}\alpha .x+\cos \alpha .y=1$ , we get

$x=\cfrac { \alpha \cos { \alpha } -\sin { \alpha } }{ \sin { \alpha } -\alpha }$ and

$y=\cfrac { \alpha -x\tan { \alpha } }{ \sin { \alpha } }$

$\displaystyle \lim _{ \alpha \rightarrow 0 }{ x } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { \cos { \alpha } -\alpha \sin { \alpha } -\cos { \alpha } }{ \cos { \alpha } -1 } } \\ =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { \alpha \sin { \alpha } }{ 2\sin ^{ 2 }{ \cfrac { \alpha }{ 2 } } } } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { 4{ \left( \cfrac { \alpha }{ 2 } \right) }^{ 2 }\cfrac { \sin { \alpha } }{ \alpha } }{ { \left( \sin { \cfrac { \alpha }{ 2 } } \right) }^{ 2 }2 } } =2\\ \displaystyle \lim _{ \alpha \rightarrow 0 }{ y } =\displaystyle \lim _{ \alpha \rightarrow \alpha }{ \cfrac { \alpha -x\tan { \alpha } }{ \sin { \alpha } } } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \left( \cfrac { \alpha }{ \sin { \alpha } } -\cfrac { x }{ \cos { \alpha } } \right) } =1-2=-1$
Hence

$P=\left( 2,-1 \right)$

$f\left( x \right) = \left\{ {\matrix{ {\cos \left[ x \right],} & {x \ge 0} \cr {\left| x \right| + a,} & {x < 0} \cr } } \right\}$ Find
the value of a , given that

$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ exists,
where[.] denotes

0%
-1
0%
2
0%
1
0%
0

$f(x) = \left\{\begin{matrix}(3/x^{2})\sin 2x^{2} & if x M 0 \\\dfrac {x^{2} + 2x + c}{1 - 3x^{2}} & if\ x \geq 0, x \neq \dfrac {1}{\sqrt {3}}\\ 0 & x = 1/ \sqrt {3}\end{matrix}\right.$ then in order that

$f$ be continuous at

$x = 0$ , the value of

$c$ is

0%
$2$
0%
$4$
0%
$6$
0%
$8$

$\displaystyle\underset{x\rightarrow 0}{Lt}\left(cosec x-\dfrac{1}{x}\right)=?$

0%
$0$
0%
$1/2$
0%
$1$
0%
Does not exits

For each

$t\in R$ , let [t] be the greatest integer less than or equal to t. Then,

$\underset { { x\rightarrow 0 }^{ + } }{ lim } x([\frac { 1 }{ x } ]+[\frac { 2 }{ x } ]+...+[\frac { 15 }{ x } ])$

0%
is equal to 0
0%
is equal to 15
0%
is equal to 120
0%
does not exist (in R)

$\phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$ , then

0%
$\phi (x) = g(x)$ for all x $\in$ R
0%
$\phi (x) = f(x)$ for all x $\in$ R
0%
$\left\{\begin{matrix}g(x) & for -1 < x < 1\\ f(x) & for |x| \geq 1\end{matrix}\right.$
0%
$\left\{\begin{matrix}g(x) & for |x| < 1\\ f(x) & for |x| > 1 \\\displaystyle \frac{f(x) + g(x)}{2} & for |x| = 1\end{matrix}\right.$

Explanation

We have,

$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = \left\{\begin{matrix}0,& if |x| < 1\\ \infty, & if |x| > 1 \\ 1, & if |x| = 1\end{matrix}\right.$
Thus, we have the following cases:
CASE I When

$- 1 < x < 1$
In this case, we have

$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = 0$

$\therefore \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}} = g(x)$
CASE II When

$|x| > 1$
In this case, we have

$\displaystyle \lim_{n \rightarrow \infty} \displaystyle \frac{1}{x^{2n}} = 0$

$\therefore \phi (x) = \displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$

$\Rightarrow \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{f(x) + \displaystyle \frac{g(x)}{x^{2n}}}{1 + \displaystyle \frac{1}{x^{2n}}} = \displaystyle \frac{f(x) + 0}{1 + 0} = f(x)$
CASE III When

$|x| = 1$
In this case, we have

$x^{2n} = 1\Rightarrow \displaystyle \lim_{n \rightarrow \infty } x^{2n}=1$ .

$\therefore \phi (x) = \displaystyle \frac{f(x) + g(x)}{2}$

$\displaystyle f(x) = \frac{x - e^x + cos 2x}{x^2}, x \neq 0$ , is continuous at

$x = 0$ ,
where [x] and {x} denotes the greatest integer and fractional part functions, respectively.

Then which of the following is correct?

0%
$f(0) = 5/2$
0%
$[f(0)] = - 2$
0%
$\{ f(0) \} = - 0.5$
0%
$[f(0)]\{ f(0) \} = - 1.5$

Explanation

$\displaystyle \lim_{h \rightarrow 0} \frac{x- e^x + 1 - (1 - cos 2x)}{x^2}$

$\displaystyle = \lim_{h \rightarrow 0} \left [ \frac{x - e^x + 1}{x^2} - \frac{(1 - cos 2x)}{x^2} \right ]$

$= \displaystyle\lim_{h \rightarrow 0} \left [ \frac{x + 1 - \left (1 + x + \frac{x^2}{2} \right )}{x^2} - \frac{2sin^2 x}{x^2} \right ]$ (Using expansion of

$e^x$ )

$\displaystyle = - \frac{1}{2} - 2 = - \frac{5}{2}$
Hence, for continuity,

$f(0) = - \displaystyle \frac{5}{2}$

Now,

$[f(0)] = - 3; \{ f(0) \} = \displaystyle \left \{ - \frac{5}{2} \right \} = \frac{1}{2}$ .
Hence,

$[f(0)] \{ f(0) \} = \displaystyle - \frac{3}{2} = - 1.5$
Hence, option D is correct.

Let

$f(x) \displaystyle = \frac{x^2 - 9x + 20}{x -[x]}$ where [x] is the greatest integer not greater than

$x$ , then

0%
$\displaystyle \lim_{x \rightarrow 5^-} f(x) = 0$
0%
$\displaystyle \lim_{x \rightarrow 5^+} f(x) = 1$
0%
$\displaystyle \lim_{x \rightarrow 5} f(x)$ does not exists
0%
$none\ of\ these$

Explanation

$\displaystyle \lim_{x \rightarrow 5^-} f(x)$

$=\lim_{x \rightarrow 5^-} \displaystyle \dfrac{x^2 - 9x + 20}{x - [x]}$

$=\displaystyle \lim_{x \rightarrow 5^-} \dfrac{(x - 5)(x - 4)}{x - 4}$

$=\displaystyle \lim_{x \rightarrow 5^-} (x-5) = 0$

Hence, option A is correct.

Now,

$\displaystyle \lim_{x \rightarrow 5^-} f(x) = 0$

$\displaystyle \lim_{x \rightarrow 5^+} \dfrac{x^2 - 9x + 20}{x - [x]}$

$=\displaystyle \lim_{x \rightarrow 5^+} \dfrac{(x- 5)(x-4)}{x - 5}$

$=\displaystyle \lim_{x \rightarrow 5^+} (x - 4) = 1$

Hence, option B is correct.

Since,

$LHL \ne RHL$

Hence, limit does not exist.

${ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },..,{ x }_{ n }$ are the roots of the equation

$x^n+ax+b=0,$ then the value of

$\left( { x }_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) \left( { x }_{ 1 }-{ x }_{ 4 } \right) ...\left( { x }_{ 1 }-{ x }_{ n } \right)$ is equal to

0%
$n{ { x }_{ 1 } }^{ n-1 }+a$
0%
$n{ { x }_{ 1 } }^{ n-1 }$
0%
$nx-1+b$
0%
$n{ { x }_{ 1 } }^{ n-1 }+b$

Explanation

Given:

${ x }_{ 1, }{ x }_{ 2, }{ x }_{ 3 }.....{ x }_{ n }$ are the roots of

${ x }^{ n }+ax+b=0$

To find:

$\left( { x }_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) ......\left( { x }_{ 1 }-{ x }_{ n } \right) =?$

Sol:

$\left( { x }_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) ......\left( { x }_{ 1 }-{ x }_{ n } \right)$

$=\dfrac { \left( { x }_{ 1 }-{ x }_{ 1 } \right) \left( { x }_{ 1 }-{ x }_{ 2 } \right) }{ ({ x }_{ 1 }-{ x }_{ 1 } )} ........\left( { x }_{ 1 }-{ x }_{ 2 } \right)$

$=\lim _{ x\rightarrow { x }_{ 1 } }{ \dfrac { \left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right) .........\left( x-{ x }_{ n } \right) }{ \left( x-{ x }_{ 1 } \right) } }$

$=\lim _{ x\rightarrow { x }_{ 1 } }{ \dfrac { { x }^{ n }+ax+b }{ \left( x-{ x }_{ 1 } \right) } }$

Using l'Hospital rule, we get

$\lim _{ x\rightarrow { x }_{ 1 } }{ \dfrac { n.{ x }^{ n-1 }+a }{ 1 } } ={ nx }_{ 1 }^{ n-1 }+a$

Hence, correct answer is

${ nx }_{ 1 }^{ n-1 }+a$

STATEMENT-1 :

$\displaystyle \lim_{x \rightarrow 0} [x] \left \{ \frac{e^{1/x} - 1}{e^{1/x} + 1} \right \}$ (where [.] represents the greatest integer function) does not exist.
STATEMENT-2 :

$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$ does not exists.

0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is False
0%
STATEMENT-1 is False, STATEMENT-2 is True

Explanation

$\displaystyle \lim_{x \rightarrow 0} [x] \left \{ \frac{e^{1/x} - 1}{e^{1/x} + 1} \right \}$

$RHL=\displaystyle \lim_{x \rightarrow 0^+} [x] \left ( \frac{e^{1/x} - 1}{e^{1/x}+1} \right)$

$=\lim _{ h\rightarrow 0 } [h]\left(\displaystyle \frac { e^{ 1/h }-1 }{ e^{ 1/h }+1 } \right)$

$= \lim_{h \rightarrow 0} [h] \left ( \displaystyle\frac{1 - e^{-1/h}}{1 + e^{-1/h}} \right)$

$= 0 \times 1 = 0$

$LHL=\displaystyle \lim_{x \rightarrow 0^-} [x] \left ( \frac{e^{1/x}- 1}{e^{1/x} + 1} \right )$

$= \lim_{h \rightarrow 0} [-h] \left ( \displaystyle\frac{e^{-1/h} - 1}{e^{-1/h} + 1} \right)$

$= -1 \times (-1) = 1$
Here,

$LHL \ne RHL$
Thus, given limit does not exist.

II)

$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$

$RHL=\displaystyle \lim_{x \rightarrow 0^+} \left ( \frac{e^{1/x} - 1}{e^{1/x}+1} \right)$

$=\lim _{ h\rightarrow 0 }\left(\displaystyle \frac { e^{ 1/h }-1 }{ e^{ 1/h }+1 } \right)$

$= \lim_{h \rightarrow 0} \left ( \displaystyle\frac{1 - e^{-1/h}}{1 + e^{-1/h}} \right)$

$RHL= 1$

$LHL=\displaystyle \lim_{x \rightarrow 0^-} \left ( \frac{e^{1/x}- 1}{e^{1/x} + 1} \right )$

$= \lim_{h \rightarrow 0} \left ( \displaystyle\frac{e^{-1/h} - 1}{e^{-1/h} + 1} \right)$

$LHL= -1$

So,

$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right)$ does not exist, but this cannot be taken as only reason for non-existence of

$\displaystyle \lim_{x \rightarrow 0} [x] \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$ .

$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$ and

$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$ ,

then the value of

$\displaystyle\lim_{x\rightarrow a}{f(x)g(x)}$ is?

0%
Does not exist
0%
Exists and is $\displaystyle\frac{3}{4}$
0%
Exists and is $\displaystyle-\frac{3}{4}$
0%
Exists and is $\displaystyle\frac{4}{3}$

Explanation

Given,

$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$ and

$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$ ,

Lets, assume

$F(x) = (f(x) + g(x))$ and

$G(x) = (f(x) - g(x))$

$\because$ Limits of both

$F(x)$ and

$G(x)$ exists as

$x\rightarrow a$ , we can say that

$\displaystyle\lim_{x\rightarrow a}{(F(x)+G(x))}$ and

$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))}$ also exists.

$\displaystyle\therefore\lim_{x\rightarrow a}{(F(x)+G(x))} = 3$

$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{f(x)} = 3$

$\Rightarrow \lim_{x\rightarrow a}{f(x)} = \dfrac{3}{2}$
Similarly,

$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))} = 1$

$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{g(x)} = 1$

$\Rightarrow \lim_{x\rightarrow a}{g(x)} = \dfrac{1}{2}$
Hence,

$\displaystyle\lim_{x\rightarrow a}{(f(x)\cdot g(x))} = \lim_{x\rightarrow a}{f(x)}\cdot\lim_{x\rightarrow a}{g(x)} = \dfrac{3}{2}\times \dfrac{1}{2} = \dfrac{3}{4}$

Hence, option B.

$x$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$4$	$3$	$7$	$1$	$3$

The function f is continuous on the closed interval

$[1, 5]$ and values of the function are shown in the table above. If the values in the table are used to calculate a trapezoidal sum, the approximate value of

$\int_{1}^{5}f(x)dx$ is

0%
$14$
0%
$14.5$
0%
$15$
0%
$29$

$\quad \lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \sum _{ r=1 }^{ 2n }{ \cfrac { r }{ \sqrt { { n }^{ 2 }+{ r }^{ 2 } } } } }$ equal to:

0%
$1+\sqrt { 5 }$
0%
$-1+\sqrt { 5 }$
0%
$1+\sqrt { 2 }$
0%
$1+\sqrt { 2 }$

The value of

$\lim _{ x\rightarrow 0 }{ \left( { \left( \sin { x } \right) }^{ 1/x }+{ \left( 1+x \right) }^{ \sin { x } } \right) }$ whre

$x> 0$ is

0%
$0$
0%
$-1$
0%
$1$
0%
2

Explanation

Given

$\lim _{ x\rightarrow 0 }{ \left( { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+{ \left( 1+x \right) }^{ sinx } \right) }$

$=\lim _{ x\rightarrow 0 }{ { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } }$

$={ \left( sin0 \right) }^{ { \frac { 1 }{ 0 } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } }$

$={ \left( 0 \right) }^{ { \infty } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } }$

$=0+{ e }^{ \lim _{ x\rightarrow 0 }{ \left[ log{ \left( 1+x \right) }^{ sinx } \right] } }$

$\left( \because { e }^{ loga }=a \right)$

$={ e }^{ \lim _{ x\rightarrow 0 }{ \left[ sinx.log{ \left( 1+x \right) } \right] } }$

$={ e }^{ \lim _{ x\rightarrow 0 }{ sinx } \times \lim _{ x\rightarrow 0 }{ log\left( 1+x \right) } }$

$={ e }^{ sin\left( 0 \right) \times log\left( 1+0 \right) }$

$={ e }^{ 0\times log\left( 1 \right) }$

$={ e }^{ 0\times 0 }={ e }^{ 0 }$

$=1$

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.