MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 6

${ x }_{ n }={ \left( 1-\cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( 1-\cfrac { 1 }{ 6 } \right) }^{ 2 }{ \left( 1-\cfrac { 1 }{ 10 } \right) }^{ 2 }...{ \left( 1-\cfrac { 1 }{ \cfrac { n(n+1) }{ 2 } } \right) }^{ 2 },n\ge 2$ . Then the value of

$\displaystyle\lim _{ n\rightarrow \infty }{ { x }_{ n } }$

0%
$1/3$
0%
$1/9$
0%
$1/81$
0%
$0$ (zero)

$\displaystyle \lim_{n \rightarrow \infty}\dfrac {1^{2}+2^{2}+3^{2}+....+n^{2}}{n^{3}}$ is equal to

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

Explanation

Solution

$\displaystyle \lim_{x\rightarrow \infty } \dfrac{1^2 + 2^2 +--- n^2}{n^3}$

we know that

$1^2 + 2^2 + --- n^2 = \dfrac{n (n+1)(2n+1)}{6}$

$\displaystyle =\lim_{x\rightarrow \infty }\frac{n(n+1)(2n+1)}{6n^3}$

$\displaystyle= \lim_{x\rightarrow \infty }\frac{(n+1)(2n+1)}{6n^2} = \lim_{x\rightarrow \infty } \frac{2n^2 +3n +1}{6n^2}$

Apply L-Hospital rule twice.

$\displaystyle = \lim_{x\rightarrow \infty } \dfrac{4n+3}{12n}$

$= \dfrac{4}{12} = \dfrac{1}{3}$

C is correct.

1063331_1116651_ans_f9e176fd269b4ce5ac81647f9adf3277.jpg

The value of k which makes

$f(x)=\left\{\begin{matrix} \sin\dfrac{1}{x}, x\neq 0\\ k, x=0\end{matrix}\right.$ continuous at

$x=0$ is?

0%
$8$
0%
$1$
0%
$-1$
0%
None

Explanation

Then k=?

$\begin{matrix} \Rightarrow f\left( 0 \right) =k \\ \Rightarrow { f\left( { { 0^{ + } } } \right) }_{ \lim \, \, x\to { 0^{ + } } }={ f\left( { { 0^{ +h } } } \right) }_{ \lim \, \, h\to 0 }={ \sin }_{ \lim \, \, h\to 0 }\dfrac { 1 }{ { (o+h) } } =\sin 0 \\ \Rightarrow { f\left( { { 0^{ - } } } \right) }_{ \lim \, \, x\to { 0^{ - } } }={ f\left( { { 0^{ -h } } } \right) }_{ \lim \, \, h\to 0 }={ \sin }_{ \lim \, \, h\to 0 }\dfrac { 1 }{ { (o-h) } } =\sin 0 \\ \Rightarrow f\left( 0 \right) =f\left( { { 0^{ + } } } \right) =f\left( { { 0^{ - } } } \right) =f\left( 0 \right) =k=0 \\ \end{matrix}$

the value of

$k$ is zero

$f(x) = 2x^9 - 5x^8 + 7x^6 - 15x^4 + 5x + 7$ , then

$\underset{x \rightarrow 0}{\lim} \dfrac{f (1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$ is

0%
$-\dfrac{35}{3}$
0%
$\dfrac{35}{3}$
0%
$\dfrac{37}{3}$
0%
$-\dfrac{37}{3}$

The value of

$\displaystyle \lim _{ x\rightarrow 0 } \dfrac{1+\sin{x}-\cos{x}+\log{(1-x)}}{x^{3}}$ , is

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

$f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$ , then

$\underset{a \rightarrow 0}{\lim} \dfrac{f(1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$ is

0%
$-\dfrac{53}{3}$
0%
$\dfrac{53}{3}$
0%
$-\dfrac{55}{3}$
0%
$\dfrac{55}{3}$

Explanation

$\lim_{\alpha \rightarrow 0}\dfrac{f(1-\alpha )-f(1)}{\alpha ^3+3\alpha }$

Apply L-Hospital's rule

$=\lim_{\alpha \rightarrow 0}\dfrac{f'(1-\alpha )-0}{3\alpha ^2+3 }$

$=\dfrac{f'(1 )}{3 }$

Considering, f(x)

$f(x )=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$

$f'(x )=30x^{9}-56x^7+30x^5-63x^2+6x$

Calculating required limit value:

$=\dfrac{f'(1 )}{3 }$

$=\dfrac{30(1)^{9}-56(1)^7+30(1)^5-63(1)^2+6(1)}{3 }$

$=\dfrac{-53}{3}$

Let

$f(x) = \underset{0}{\overset{x}{\int}} |2t - 3| dt$ , then

$f$ is

0%
continuous at $x = \dfrac{3}{2}$
0%
continuous at $x = 3$
0%
differentiable at $x = \dfrac{3}{2}$
0%
differentiable at $x = 0$

The value of

$\lim_{x\rightarrow o}\dfrac{\sqrt{\dfrac{1}{2}(1-cos 2 x)}}{x}$

0%
1
0%
-1
0%
0
0%
none of these

Explanation

$\lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - \cos{2x} \right)}}{x}}$

$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - \left( 1 - 2 \sin^{2}{x} \right) \right)}}{x}}$

$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - 1 + 2 \sin^{2}{x} \right)}}{x}}$

$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 2 \sin^{2}{x} \right)}}{x}}$

$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\sin^{2}{x}}}{x}}$

$= \lim_{x \rightarrow 0}{\cfrac{\sin{x}}{x}}$

$= 1 \; \left( \because \lim_{x \rightarrow 0}{\cfrac{\sin{x}}{x}} = 1 \right)$

solve the limit

$\mathop {\lim }\limits_{x \to 3} \dfrac{2}{{x - 3}}$

0%
$2$
0%
$3$
0%
$4$
0%
Does not exist

Explanation

$\displaystyle \lim_{x\rightarrow 3} \dfrac{2}{x-3}$

put

$h = x - 3$

$\displaystyle \lim_{h\rightarrow 0} \dfrac{2}{h}$

this is

$\dfrac{2}{0}$ that is not defined.

so limit does not exist.

The value of

$\underset { x\longrightarrow \infty }{ Lim } \dfrac{d}{dx}\overset { \sqrt { 3 } }{ \underset { -\sqrt { 3 } }{ \int } } \dfrac{r^3}{(r+1)(r-1)}$ dr,is

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

Explanation

Given,

$\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{r^3}{(r+1)(r-1)}dr$

using long division process, we get,

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{r^3}{r^2-1}dr$

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}r+\dfrac{r}{r^2-1}dr$

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\left [ \dfrac{r^2}{2}+\dfrac{1}{2}\ln \left|r^2-1\right| \right ]_{-\sqrt{3}}^{\sqrt{3}}$

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx} (0)$

$=0$

$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ is equal to

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

Explanation

$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$

can be written as

$\mathop {\lim }\limits_{x \to 0} \dfrac{\sin^{-1}(\dfrac{2x}{1+x^2})}{x}$

it is of the form

$\dfrac00$

So, we will apply L Hospital rule until we will get determinate form

Differentiating numerator and denominator we will get,

$=\lim_{x\to 0}\dfrac{\dfrac{1}{\sqrt{1-(\dfrac{2x}{1+x^2})^2}}.\dfrac{2(1+x^2)-4x^2}{(1+x^2)^2}}{1}=\lim_{x\to0}\dfrac{1}{\sqrt{1-(\dfrac{2x}{1+x^2})^2}}.\dfrac{2(1+x^2)-4x^2}{(1+x^2)^2}=2$

$\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{\cot x - \cos x}{\left(\dfrac{\pi}{2} -x \right)^3} =$

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

$\mathop {\lim }\limits_{x \to 0} \,\dfrac{1}{x}\,{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ is equal to

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

Explanation

Solution -

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

we know that

$\displaystyle sin^{-1}x+sin^{-1}y = sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})$

$\displaystyle sin^{-1}(\frac{2x}{1+x^{2}})= sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})+sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})$

$\displaystyle = 2sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})-\pi$

$\displaystyle \lim_{x\rightarrow 0} \frac{2sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})-\pi }{x}$

from triangle.

$\displaystyle \lim_{x\rightarrow 0} \frac{2tan^{-1}(\frac{1}{x})-\pi }{x}$

Apply L Hospital rule-

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

$\displaystyle \lim_{x\rightarrow 0} \frac{2x^{2}}{1+x^{2}}$

$= 0.$

B is correct.

1092192_1174617_ans_0efb6bf82d704d8fbc4c1a54fc5d5961.png

Solve

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x}}{{\tan 3x}}$

0%
$-\dfrac{5}{3}$
0%
$\dfrac{5}{3}$
0%
$\dfrac{7}{3}$
0%
None of these

Explanation

Let

$L=\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}$

$= \displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}\times \dfrac {3x}{\tan x}\times \dfrac {5}{3}$

we know that

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}=1$ and

$\displaystyle \lim_{x\rightarrow 0}\dfrac {3x}{\tan 3x}=1$

$L=1\times 1\times \dfrac {5}{3}$

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}= \dfrac {5}{3}$

$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{2{x^2} + x - 3}} =$

0%
$\dfrac{1}{{10}}$
0%
$\dfrac{{ - 1}}{{10}}$
0%
$\dfrac{2}{5}$
0%
$- \dfrac{2}{5}$

Explanation

Now,

$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{2{x^2} + x - 3}}$

$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(2{x^2} +3 x-2x - 3)}}$

$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(x-1)(2x + 3)}}$

$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(\sqrt x-\sqrt 1)(\sqrt x+\sqrt 1)(2x + 3)}}$

$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)}}{{(\sqrt x+\sqrt 1)(2x + 3)}}$

$=\dfrac{-1}{2\times5}$

$=-\dfrac{1}{10}$ .

$\mathrm { L } = \lim _ { \mathrm { x } ^ { 2 } \rightarrow \mathrm { a } } \frac { \mathrm { b } - \cos \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) } { \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) \sin \left( \mathrm { cx } ^ { 2 } - \mathrm { a } \right) }$ is non-
zero finite

$( \mathrm { a } > 0 ) ,$ then-

0%
L = 2 , b = 1 , c = 1
0%
$L = \frac { 1 } { 2 } , b = 1 , c = 1$
0%
L = 4 , b = - 1 , c = - 1
0%
$L = \frac { 1 } { 4 } , b = - 1 , c = - 1$

${z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}}$ , where

$r= 1, 2, 3, ....n$ , then

$\mathop {\lim }\limits_{n \to \infty } \left( {{z_1}.{z_2}.....{z_n}} \right)$ is equal to

0%
$\cos \alpha + i\sin \alpha$
0%
$\sin \frac{\alpha }{2} + i\cos \frac{\alpha }{2}$
0%
${e^{i\alpha /2}}$
0%
$\sqrt {{e^{i\alpha }}}$

Explanation

Solution -

$z_{r} = \dfrac{cosr\alpha }{n^{2}}+i\dfrac{sinr\alpha }{n^{2}}$

Comparing with

$cos \theta +isin \theta$

$\theta =\dfrac {r\alpha}{n^2}$

$\therefore z_r= e^{\dfrac{ir\alpha }{n^{2}}}$

$\lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = e^{\dfrac{i\alpha }{n^{2}}}e^{\dfrac{2i\alpha }{n^{2}}} e^{\dfrac{3i\alpha }{n^{2}}}...e^{\dfrac{ni\alpha }{n^{2}}}$

$= \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{1}{n^{2}}+\dfrac{2}{n^{2}}+...)}$

$= \lim_{n\rightarrow \infty } e^{i\alpha \dfrac{1}{n^2}(1+2+3)}$

$= \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{n^{2}+n}{2n^{^{2}}})} \quad \dots S_{n}=\dfrac{n}{2}\left [ 2a+(n-1)d \right ]$

$n\rightarrow \infty$

$= e^{i\alpha }\times \dfrac{1}{2}$

$\lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = \sqrt{e^{i\alpha }}$

Solve:

$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}dx=$

0%
$\dfrac{4}{3}(\sqrt{2}+1)$
0%
$\dfrac{4}{3}(\sqrt{2}-1)$
0%
$\dfrac{3}{4}(\sqrt{2}-1)$
0%
$\dfrac{3}{4}(\sqrt{2}-2)$

Explanation

$\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}}$

$=\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}\times\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}}$

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

$=\displaystyle\int_{0}^{1}{\left(\sqrt{x+1}+\sqrt{x}\right)dx}$

$=\left[\dfrac{{\left(x+1\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+\dfrac{{\left(x\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{1}$

$=\left[2\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+2\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$

$=2\left[\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$

$=\dfrac{2}{3}\left[\left({2}^{\frac{3}{2}}-1\right)-\left(1-0\right)\right]$

$=\dfrac{2}{3}\left[2\sqrt{2}-2\right]$

$=\dfrac{4}{3}\left(\sqrt{2}-1\right)$

$f(x)$ is the integral of

$\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0$ . Find

$\lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right) }$ , where

$f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}$

0%
$\dfrac{1}{2}$
0%
$1$
0%
$\dfrac{1}{3}$
0%
$2$

Explanation

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

$x\neq 0$

$f'(x)=\dfrac{df(x)}{dx}=\dfrac{2\sin x-\sin 2x}{x^3}$

$=\dfrac{2\sin x}{x}\left(\dfrac{1-\cos x}{x^2}\right)$

$\Rightarrow \displaystyle \lim_{x\rightarrow 0}{f'(x)}=\displaystyle \lim_{x\rightarrow 0}{2\left(\dfrac{\sin x}{x}\right)}\left( \dfrac{2\sin^2(x/2)}{x^2}\right)$

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

$=1$

$\lim_{x\to \infty} \dfrac{\sqrt{x^2 + sin^2x}}{x+cosx}$ equals

0%
$1$
0%
$0$
0%
$\infty$
0%
does not exist

Explanation

Given,

$\lim _{x\to \infty \:}\left(\dfrac{\sqrt{x^2+\sin ^2\left(x\right)}}{x+\cos \left(x\right)}\right)$

dividing by highest denominator power

$=\lim _{x\to \infty \:}\left(\dfrac{\sqrt{1+\frac{\sin ^2\left(x\right)}{x^2}}}{1+\frac{\cos \left(x\right)}{x}}\right)$

$=\dfrac{\lim _{x\to \infty \:}\left(\sqrt{1+\frac{\sin ^2\left(x\right)}{x^2}}\right)}{\lim _{x\to \infty \:}\left(1+\frac{\cos \left(x\right)}{x}\right)}$

$=\dfrac{\sqrt{1+0}}{1+0}$

$=1$

Find the value of limit

$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x-1 } }{ 2\sin ^{ 2 }{ x } -3\sin { x+1 } } = }$ .

0%
$0$
0%
$3$
0%
$-3$
0%
$1$

Explanation

$\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } +1 } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin ^{ 2 }{ x } +2\sin { x } -\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -2\sin { x } -\sin { x } +1 } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { 2\sin { x } \left( \sin { x } +1 \right) -1\left( \sin { x } +1 \right) }{ 2\sin { x } \left( \sin { x } -1 \right) -1\left( \sin { x } -1 \right) } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { \left( 2\sin { x } -1 \right) \left( \sin { x } +1 \right) }{ \left( 2\sin { x } -1 \right) \left( \sin { x } -1 \right) } } \\ =\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \frac { \sin { x } +1 }{ \sin { x } -1 } } \\ =\frac { \frac { 1 }{ 2 } +1 }{ \frac { 1 }{ 2 } -1 } \\ =-3$

$\lim\limits_{x\rightarrow0}\dfrac{x\tan 2x-2x\tan x}{\left(1-\cos 2x\right)^{2}}=$

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

Explanation

$\underset{x\rightarrow 0}{\lim}\dfrac{x\tan2x-2x\tan x}{(1-\cos 2x)^2}$

$\underset{x\rightarrow 0}{\lim}\dfrac{x[\tan2x-2\tan x]}{(1-\cos 2x)^2}$

$\underset{x\rightarrow 0}{\lim}x\dfrac { \left[ \frac { 2\tan { x } }{ 1-{ \tan { x } }^{ 2 } } -2\tan { x } \right] }{ { \left( 2{ \sin { x } }^{ 2 } \right) }^{ 2 } }$

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

$\underset{x\rightarrow 0}{\lim}\dfrac { \dfrac { x\tan { x } }{ \left( 1-{ \tan { x } }^{ 2 } \right) } \left[ 1-1+{ \tan { x } }^{ 2 } \right] }{ 2{ \sin { x } }^{ 4 } }$

$\underset{x\rightarrow 0}{\lim} {\dfrac { x{ \tan^3 { x } }^{ 3 } }{ ( 1-{ \tan^2 { x } )}.2{ \sin^4 { x } } } }$

Divide num and den by

$x^4$

$\underset{x\rightarrow 0}{\lim}\dfrac{\dfrac{\sin^3 x}{\cos^3 x}.\dfrac{x}{x^4}}{2\dfrac{\sin^4x}{x^4}(1-\tan^2 x)}$

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

Apply limit

${ \dfrac { { \left( \dfrac { \sin { 1 } }{ 1 } \right) }^{ 3 }.\dfrac { 1 }{ { \cos ^{ 3 }{1 } } } }{ 2{ \left( \dfrac { \sin { 1 } }{ 1 } \right) }^{ 4 }\left( 1-{ \tan^2 { 1 } }\right) } }=\dfrac{1.1}{2.1(1-0)}$

$=\dfrac{1.1}{2.1(1-0)}$

$=\dfrac{1}{2}$

$\therefore \underset{x\rightarrow 0}{\lim}\dfrac{x \tan 2x- 2x\tan x}{(1-\cos 2x)^2}=\dfrac{1}{2}$

$\displaystyle\lim_{h\rightarrow 0}\dfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}=$ __________.

0%
$\cos\sqrt{x}$
0%
$\dfrac{1}{2\sin \sqrt{x}}$
0%
$\dfrac{\cos \sqrt{x}}{2\sqrt{x}}$
0%
$\sin\sqrt{x}$

Explanation

$Lt_{x\rightarrow 0}\cfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{(x+h)-x}$

Let

$f(x)=\sin\sqrt{x}$

$Lt_{h\rightarrow 0}\cfrac{\sin{x+h}-\sin\sqrt{x}}{(x+h)-x}=Lt_{x\rightarrow 0}f\cfrac{(x+h)-f(x)}{h}=f^1(x)\\ \Rightarrow \cos\sqrt x\times\cfrac{1}{2\sqrt x}\\ \Rightarrow \cfrac{\cos\sqrt x}{2\sqrt x}$

The value of

$\displaystyle lim_{x\to 0} \dfrac{cos (sin x) - cos x}{x^4}$ is equal to

0%
$1/5$
0%
$1/6$
0%
$1/4$
0%
$1/2$

Explanation

Given,

$\lim _{x\to \:0}\left(\dfrac{\cos \left(\sin \left(x\right)\right)-\cos \left(x\right)}{x^4}\right)$

apply L-Hospital's rule

$=\lim _{x\to \:0}\left(\dfrac{-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)+\sin \left(x\right)}{4x^3}\right)$

again apply L-Hospital's rule

$=\lim _{x\to \:0}\left(\dfrac{-\cos ^2\left(x\right)\cos \left(\sin \left(x\right)\right)+\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos \left(x\right)}{12x^2}\right)$

once again apply L-Hospital's rule

$=\lim _{x\to \:0}\left(\dfrac{\frac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2\cos ^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{2}}{24x}\right)$

upon simplification, we get,

$=\lim _{x\to \:0}\left(\dfrac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2\cos ^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{48x}\right)$

again applying L-Hospital's rule,

$=\lim _{x\to \:0}\left(\dfrac{3\left(2\cos \left(2x\right)\cos \left(\sin \left(x\right)\right)-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)\sin \left(2x\right)\right)+2\left(-3\cos ^2\left(x\right)\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos ^4\left(x\right)\cos \left(\sin \left(x\right)\right)\right)+2\left(-\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos ^2\left(x\right)\cos \left(\sin \left(x\right)\right)\right)-2\cos \left(x\right)}{48}\right)$

substituting

$x=0$

$\dfrac{3\left(2\cos \left(2\cdot \:0\right)\cos \left(\sin \left(0\right)\right)-\sin \left(\sin \left(0\right)\right)\cos \left(0\right)\sin \left(2\cdot \:0\right)\right)+2\left(-3\cos ^2\left(0\right)\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+\cos ^4\left(0\right)\cos \left(\sin \left(0\right)\right)\right)+2\left(-\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+\cos ^2\left(0\right)\cos \left(\sin \left(0\right)\right)\right)-2\cos \left(0\right)}{48}$

we get,

$=\dfrac{1}{6}$

The value of

$\lim_{x \rightarrow 1} \sec \dfrac{\pi}{2x} \log x$ is-

0%
$\pi /2$
0%
$2 /\pi$
0%
$-\pi/2$
0%
$-2/ \pi$

The value of

$\displaystyle \lim_{\theta \rightarrow 0^{+}} \dfrac {\sin \sqrt {\theta}}{\sqrt {\sin \theta}}$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
$4$

For

$x>y$ ,

$\displaystyle\lim_{x\rightarrow 0}{\left[\left(\sin{x}\right)^{1/x}+\left(\cfrac{1}{x}\right)^{\sin{x}}\right]}$ is :

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

$\underset{x\rightarrow 0}{lim} \dfrac{\sqrt{a^2 -ax+x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} -\sqrt{a-x}}$ is equal to (a > 0)

0%
$\sqrt{a}$
0%
$-\sqrt{a}$
0%
$a \sqrt{a}$
0%
$-a\sqrt{a}$

Explanation

$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})(\sqrt{a+x}+\sqrt{a-x})}{(a+x)-(a-x)}$

$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\sqrt{a}}{2x}(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})$

$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a}}{x}\left(\dfrac{(a^2-ax+x^2)-(a^2+ax+x^2)}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\right)$

$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a}}{x}\times \dfrac{(-2ax)}{2\sqrt{a}}=-\sqrt{a}$

$\Rightarrow (B)$ .

1165369_1245288_ans_1b8e3f98ee4a423c9cabe059818fac18.jpg

The function

$f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$ is

0%
discontinuous at only one point
0%
discontinuous exactly at two points
0%
discontinuous exactly at three points
0%
None of these

Explanation

$f\left(x\right)= \dfrac{4-x^2}{4x-x^3}$

$=\dfrac{4-x^2}{x\left(4-x^2\right)}$

$for 4-x^2\neq 0$

$x\neq\pm 2$

$f\left(x\right)=\dfrac{1}{x}$

so this function will be discontinue at

$x=0$

$x=\pm2$

three Points

$\displaystyle\lim _{ x\rightarrow \dfrac { x }{ 2 } }{ \dfrac { \cot { x } -\cos { x } }{ \left( \pi -2x \right) ^{ 3 } } }$ is equal to

0%
$\dfrac{1}{24}$
0%
$\dfrac{1}{16}$
0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{4}$

$\lim _ { x \rightarrow 1 } \{ 1 - x + [ x + 1 ] + [ 1 - x ] \} , \text { where } [ x ]$ denotes greatest integer function is

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

$\underset { x\rightarrow 0 }{ lim } \cfrac { 1+cos\left( \pi x \right) }{ \left( 1-x \right)^ 2 }$ is equal to :

0%
$\cfrac { \pi} {2}$
0%
$- \cfrac { \pi ^2} {2}$
0%
$\cfrac { \pi ^2} {2}$
0%
$\cfrac { \pi} {3}$

Evaluate:

$\underset{x\rightarrow 0}{lim} \dfrac{e^{1/x} - 1}{e^{1/x }+ 1}$

0%
$0$
0%
$1$
0%
$-1$
0%
does not exist

Explanation

As it is

$\dfrac{\infty }{\infty }$ form, applying L'hospital

$L = \lim_{x\rightarrow 0}\dfrac{{e^{1/x}(-1/x^{2})}}{{e^{1/x}(-1/x^{2})}} = 1 \Rightarrow (B)$

1167307_1247005_ans_8b306a676d304837b4c557dcd861bdb8.jpg

$\underset { x\rightarrow { 0 } }{ lim } \dfrac { \left( ax+b \right) -\sqrt { 4+\sin x } }{ \tan\quad x } =\dfrac { 27 }{ 4 } ~where ~a,b\in R$ then the value of

0%
$a = 2~ and~ b = 7$
0%
$a = -2~and~ b = 7$
0%
$a = 7~ and~ b = 2$
0%
$a=7~ and ~b = -2$

Evaluate:

$\underset { x\rightarrow 0 }{ \lim } \dfrac { x\tan2x-2x\tan x }{ \left( 1-\cos2x \right) ^{ 2 } }$

0%
$\dfrac { 1 }{ 4 }$
0%
$1$
0%
$\dfrac { 1 }{ 2 }$
0%
$-\dfrac { 1 }{ 2 }$

Explanation

$\underset{x\rightarrow 0}{lt}\dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}$

$=\underset{x\rightarrow 0}{lt}\dfrac{x\dfrac{2\tan x}{1-\tan^2x}-2x\tan x}{(1-1+2\sin^2x)^2}$ (since

$\tan 2\theta =\dfrac{2\tan \theta}{1-\tan^2\theta}$ )

$=\underset{x\rightarrow 0}{lt}\dfrac{2x\tan x(1-1+\tan^2x)}{(1-\tan^2x)y\sin^4x}=\underset{x\rightarrow 0}{lt}\dfrac{x\tan^3x}{2\sin^4x}\underset{x\rightarrow 0}{lt}\dfrac{1}{1-\tan^2x}$

$=\underset{x\rightarrow 0}{lt}\dfrac{\sin^3x/\cos^3x}{2\sin^3x(\sin x/x)}\times 1=\underset{x\rightarrow 0}{lt}\dfrac{1}{2\cos^3x}\underset{x\rightarrow 0}{lt}\dfrac{1}{\sin x/x}=\dfrac{1}{2}\times 1$

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

$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } }$ equals:

0%
$\frac { 1 } { 24 }$
0%
$\frac { 1 } { 16 }$
0%
0
0%
$\frac { 1 } { 4 }$

Evaluate:

$\underset { x\rightarrow { 0 } }{ lim } \dfrac { x\tan 2x-2x \tan\ x\quad }{ (1-\cos2x)^{ 2 } }$

0%
$\dfrac { 1 }{ 4 }$
0%
$1$
0%
$\dfrac { 1 }{ 2 }$
0%
$-\dfrac { 1 }{ 2 }$

Explanation

$\rightarrow \underset{x \rightarrow 0}{lt} \dfrac{x (\tan 2x - 2 \tan x)}{(1 - \cos 2x)^2}$

$= \underset{x \rightarrow 0}{lt} \dfrac{x\left[\dfrac{2 \tan x}{1 - \tan^2 x} - 2 \tan x\right]}{(2 \sin^2 x)^2}$

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

$= \underset{x \rightarrow 0}{lt} \dfrac{\dfrac{x \tan x}{1 - \tan^2 x} [x - 1 + \tan^2 x]}{2 \sin^4x}$

$= \underset{x \rightarrow 0}{lt} \dfrac{x \tan^3 x \left(\dfrac{\sin^3 x}{\cos^3 x} \right) / x^4}{(1 - \tan^2 x). 2 \sin^4 x/ xy}$

$= \underset{x \rightarrow 0}{lt} \dfrac{\left(\dfrac{\sin x}{x} \right)^3 . \dfrac{1}{\cos^3x}}{2 \left(\dfrac{\sin x}{x} \right)^4 (1 - \tan^2 x)} = \dfrac{1.1}{2.1 (1 - 0)} = \dfrac{1}{2}$

Given

$\underset{x \rightarrow 0}{lt} \dfrac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} = \dfrac{1}{2}$

$\underset {x\rightarrow0}{ lim } \dfrac{x \tan 2 x - 2 x \tan x}{(1 - \cos 2 x)^2}$ equals

0%
$\dfrac{1}{4}$
0%
1
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$

Solve:

$\underset{x \rightarrow 2}{Lt} \dfrac{x^{\sqrt{2}} - 2^{\sqrt{2}}}{x - 2} =$

0%
$\sqrt{2}.2^{\sqrt{2}}$
0%
$2^{\sqrt{2}- 1}$
0%
$2^{\sqrt{2} - \dfrac{1}{2}}$
0%
$2^{\sqrt{2}}$

Explanation

$\underset{x \rightarrow 2}{Lt} \dfrac{x^{\sqrt{2}} - 2^{\sqrt{2}}}{x - 2} \Rightarrow$

L' Hospital -

$\underset{x \rightarrow 2}{Lt}\frac{\sqrt{2}.x^{\sqrt{2}-1}}{1}$

$= \sqrt{2}\times (2)^{\sqrt{2}-1}$

$= \frac{\sqrt{2}\times 2^{\sqrt{2}}}{2}$

$= (2)^{\sqrt{2}-\frac{1}{2}}$

1117297_1275460_ans_7e2d0b450cd244d7a22ce935f836b052.JPG

Let

$f(x)=\begin{cases} { x }^{ 2 }+k,\quad \quad when\quad x\ge 0 \\ -{ x }^{ 2 }-k,\quad \quad when \quad x<0 \end{cases}$ . If the function

$f(x)$ be continous at

$x=0$ , then

$k=$

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

Explanation

We have,

${{x}^{2}}+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,when,\,x\ge 0\,$

$-{{x}^{2}}-k\,\,\,\,\,\,\,\,\,\,\,\,\,when,x\le 0$

Given that,

$L.H.L=R.H.L=f\left( x \right)$

At point $x=0$

${{x}^{2}}+k=-{{x}^{2}}-k=0$

$\Rightarrow {{x}^{2}}+k=0$

$\Rightarrow k=-{{x}^{2}}$

$\Rightarrow k=0$

Hence, this is the answer.

$2f(\sin x)+\sqrt {2}f(\cos x)=\tan x,\ (x> 0)$ , then

$\displaystyle \lim _{ x\rightarrow 1 }{ \sqrt { 1-x } f\left( x \right) = }$

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

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow a}\dfrac{x-a}{\sqrt{x}-\sqrt{a}}$ .

0%
$2\sqrt{a}$
0%
$2{a}$
0%
$2{a^{\frac 13}}$
0%
None of these

Explanation

$\displaystyle \lim_{x\rightarrow a}{\dfrac{x-a}{\sqrt x-\sqrt a}}$

$x\rightarrow a$ , expression

$\rightarrow \dfrac{0}{0}$ , an indetermined form

using factorisation

$\displaystyle \lim_{x\rightarrow a}{\dfrac{(\sqrt{x}-\sqrt a)(\sqrt x+\sqrt a)}{(\sqrt x-\sqrt a)}}=2\sqrt a$

$Lt_{x\to 0} \dfrac{sin x - x+\dfrac{x^3}{6}}{x^5}=$ _________

0%
$\dfrac{1}{120}$
0%
$\dfrac{-1}{120}$
0%
$0$
0%
$\dfrac{1}{6}$

The value of

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

$f(x)=\dfrac {\cos (\sin x)-\cos x}{x^{4}}$ , is

0%
$2$
0%
$\dfrac {1}{6}$
0%
$\dfrac {2}{3}$
0%
$-\dfrac {1}{3}$

Explanation

$\mathop {f\left( x \right)}\limits_{\lim \,\,x \to 0} = \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}$

$= \frac{{ - \sin \left( {\sin x} \right)\cos x - \sin x}}{{4{x^3}}}$

$= \frac{{ - \left( {\sin x} \right){{\cos }^2}x + \sin \left( {\sin x} \right)\sin x - \cos x}}{{12{x^2}}}$

$= \frac{{\sin \left( {\sin x} \right){{\cos }^3}x + \cos \left( {\sin x} \right)\cos 2x + \cos \left( {\sin x} \right)\sin x + \sin \left( {\sin x} \right)\cos x + \sin x}}{{24x}}$

$= \frac{{1 + 1 + Q + 1 + 1}}{{24}} = \frac{1}{6}$

$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } \int _{ 0 }^{ x }{ \left( \sqrt { { t }^{ 2 }+5t } -t \right) dt } }$

0%
$0$
0%
$1$
0%
$\dfrac{5}{2}$
0%
$5$

The value of

$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } } \left( a>b \right)$ is

0%
$\dfrac {1}{4a}$
0%
$\dfrac {1}{a\sqrt {a-b}}$
0%
$\dfrac {1}{2a\sqrt {a-b}}$
0%
$\dfrac {1}{4a\sqrt {a-b}}$

Explanation

We have,

$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } }$

This is the

$\dfrac{0}{0}$ form.

Apply L-hospital rule,

$\lim_{x\to a}\dfrac{\dfrac{1}{2\sqrt {x-b}}-0}{2x-0}$

$\lim_{x\to a}\dfrac{1}{4x\sqrt {x-b}}$

$=\dfrac{1}{4a\sqrt {a-b}}$

Hence, this is the answer.

$\underset { x\rightarrow 1 }{ Lim } \left[ { \left[ \frac { 4 }{ { x }^{ 2 }-{ x }^{ -1 } } -\frac { { 1-3x+x }^{ 2 } }{ { 1-x }^{ 3 } } \right] }^{ -1 }+\frac { 3\left( { x }^{ 4 }-1 \right) }{ { x }^{ 3 }-{ x }^{ -1 } } \right] =$

0%
$\frac { 1 }{ 3 }$
0%
3
0%
$\frac { 1 }{ 2 }$
0%
$\frac { 3 }{ 2 }$

Explanation

$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4}{x^2-x^{-1}}-\dfrac{1-3x+x^2}{1-x^3}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$

$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4x}{x^3-1}+\dfrac{1-3x+x^2}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$

$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{x^2+x+1}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)x}{x^4-1}\right]$

$\displaystyle\lim_{x\rightarrow 0}\left[\dfrac{(x-1)(x^2+x+1)}{x^2+x+1}+3x\right]$ as

$(x^3-1)2(x-1)(x^2+x+1)$

$\displaystyle\lim_{x\rightarrow 1}[(x-1)+3x]=(1-1)+3(1)=3$

Answer

$=3$ .

1182108_1301374_ans_e0d4cf58fd6f414f8639427665c36517.jpg

The value of

$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x }{ 5 } } \left[ \frac { x }{ 2 } \right]$ (where

$[.]$ denotes the greatest integer function) is

0%
$\dfrac{2}{5}$
0%
$-\dfrac{2}{5}$
0%
$0$
0%
$\infty$

Explanation

Given,

$\lim _{x\to 0}\left(\dfrac{x}{5}\left[\dfrac{x}{2}\right]\right)$

$\lim _{x\to \:0}\left(\dfrac{x}{5}\cdot \dfrac{x}{2}\right)$

put the value

$x=0$

$=\dfrac{0}{5}\cdot \dfrac{0}{2}$

$=0$

The value of

$\displaystyle \lim_{x\rightarrow 0} \dfrac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}$ is

0%
$10/3$
0%
$3/10$
0%
$6/5$
0%
$5/6$

Explanation

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right) \cdot \sin 5x}}{{{x^2}\sin 3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{ 2{{\sin }^2}x \cdot \sin 5x}}{{{x^2}\sin 3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\left( {\dfrac{{\sin x}}{x}} \right)}^2} \cdot \left( {\dfrac{{\sin 5x}}{{5x}}} \right) \cdot 5x}}{{\left( {\dfrac{{\sin 3x}}{{3x}}} \right) \cdot 3x}}$

$= \dfrac{{10}}{3}$ .

Let

$U_{ { n } }=\dfrac { n! }{ (n+2)! }$ where

$n \in N .$ If

$S_{ { n } }=\sum _{ { n-1 } }^{ { n } } U_{ { n } }$ then

$\lim _ { n \rightarrow \infty } \mathrm { S } _ { n }$ equals :

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

Explanation

$\begin{array}{l} { U_{ n } }=\frac { { n! } }{ { \left( { n+2 } \right) ! } } =\frac { 1 }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } =\frac { { \left( { n+1 } \right) -\left( { n+1 } \right) } }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } \\ =\frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } \\ { S_{ n } }=\sum _{ n=1 }^{ n }{ { U_{ n } } } =\left[ { \left( { \frac { 1 }{ 2 } -\frac { 1 }{ 3 } } \right) +\left( { \frac { 1 }{ 3 } -\frac { 1 }{ 4 } } \right) +............+\left( { \frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } } \right) } \right] \\ { S_{ n } }=\frac { 1 }{ 2 } -\frac { 1 }{ { n+2 } } \\ { \lim }_{ n\to \infty }{ S_{ n } }=\frac { 1 }{ 2 } \\ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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