Limits and Continuity Test 31

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Limits and Continuity Test 31

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Weekly Quiz Competition

Question 1
1 / -0

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

A
0

B
1

C
19

D
20

Solution

$\because \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n} x=\begin{cases}1, x=r \pi, r \in I \\ 0, x \neq r \pi, r \in I\end{cases}$
Here, for $x=10, \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n}(x-10)=1$
and in all other cases it is zero.
$\therefore \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{\infty} \cos ^{2 n}(x-10)=1$
Question 2
1 / -0

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

A
2

B
-2

C
1

D
-1

Solution

$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{In(cos(2x^2-x))}$
$$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{log \left(1-2 sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}$$
$=\displaystyle \lim_{x \to 0} \dfrac{sin(x^2)x^2}{\dfrac{x^2log \left(1-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}{-2sin^2 \left(\dfrac{2x^2-x}{2}\right)}\left[-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right]}$
$=\displaystyle \lim_{x \to 0} \dfrac{x^2}{\dfrac{2sin^2 \left(\dfrac{2x^2-x}{2}\right)}{\left(\dfrac{2x^2-x}{2}\right)^2}\left(\dfrac{2x^2-x}{2}\right)^2}$
$\displaystyle \lim_{x \to 0}-\dfrac{2x^2}{(2x^2-x)^2}=\displaystyle \lim_{x \to 0}-\dfrac{2}{(2x-1)^2}=-2$
Question 3
1 / -0

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

A
2

B
-2

C
1/2

D
-1/2

Solution

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

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

A
$e^{\sin ^{2} y}$

B
$\sin 2 y e^{\sin ^{2} y}$

C
0

D
None of these

Solution

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

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

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

$=e^{\sin ^{2} y}\left[1+\dfrac{d y}{d x}-\dfrac{d y}{d x}\right]=e^{\sin ^{2} y}$
Question 5
1 / -0

If $y^{r}=\dfrac{n !^{n+r-1} C_{r-1}}{r^{n}},$ where $n=k r(k \text { is constant }),$ then $\operatorname{lim}_{r\rightarrow\infty} y$ is equal to

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

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

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

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

Solution

$y^{r}=\left(1+\dfrac{1}{r}\right)\left(1+\dfrac{2}{r}\right)\left(1+\dfrac{3}{r}\right) \cdots\left(1+\dfrac{n-1}{r}\right)\\$
$\Rightarrow \log y=\dfrac{1}{r} \sum_{p=1}^{n-1} \log \left(1+\dfrac{p}{r}\right)\\$
$\Rightarrow \lim _{n \rightarrow \infty} y=\lim _{r \rightarrow \infty} y=\int_{-\infty}^{k} \log (1+x) d x=(k-1) \log _{e}(1+k)-k$
Question 6
1 / -0

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

A
1/6

B
-1/3

C
1/2

D
1

Solution

$\cos (\tan x)-\cos x=2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)$
$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{\cos (\tan x)-\cos x}{x^{4}}=\lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}}$
$=\displaystyle \lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}\left(\dfrac{x+\tan x}{2}\right)\left(\dfrac{x-\tan x}{2}\right)}\left(\dfrac{x^{2}-\tan ^{2} x}{4}\right)$
$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\tan ^{2} x}{x^{4}}$
$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\left(x+\dfrac{x^{3}}{3}+\dfrac{2}{15} x^{5}+\cdots\right)^{2}}{x^{4}}$
$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x^{2}}\left(1-\left(1+\dfrac{x^{2}}{3}+\dfrac{2}{15} x^{4}+\dots\right)^{2}\right)=-\dfrac{1}{3}$
Question 7
1 / -0

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

A
$\dfrac{2 a}{\pi}$

B
$-\dfrac{2 a}{\pi}$

C
$\dfrac{4 a}{\pi}$

D
$-\dfrac{4 a}{\pi}$

Solution

$\displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cdot \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$
$=\displaystyle \lim _{x \rightarrow a} \dfrac{\sqrt{a^{2}-x^{2}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}$
$=\dfrac{2}{\pi} \displaystyle \lim _{x \rightarrow a} \dfrac{\dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}(a+x)=\dfrac{4 a}{\pi}$
Question 8
1 / -0

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

A
1

B
0

C
e-1

D
e+1

Solution

$\quad \displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1}{n}+\dfrac{e^{1 / n}}{n}+\dfrac{e^{2 / n}}{n}+\dots+\dfrac{e^{(n-1) / n}}{n}\right]$
$=\displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1+e^{1 / n}+\left(e^{1 / n}\right)^{2}+\cdots+\left(e^{1 / n}\right)^{n-1}}{n}\right]$
$=\displaystyle \lim _{n \rightarrow \infty} \dfrac{1 \cdot\left[\left(e^{1 / n}\right)^{n}-1\right]}{n\left(e^{1 / n}-1\right)}=(e-1) \lim _{n \rightarrow \infty} \dfrac{1}{\left(\dfrac{e^{1 / n}-1}{1 / n}\right)}$
$=(e-1) \times 1=(e-1)$
Question 9
1 / -0

If function $f(x)=\dfrac{x^2-9}{x-3}$ is continuous at $x=3$ , then value of $(3)$ will be:

A
$6$

B
$3$

C
$1$

D
$0$

Solution

$f(x)=\dfrac{x^2-9}{x-3}$
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.
Question 10
1 / -0

If $f(x)=\begin{cases} \begin{matrix} \dfrac{\log (1+mx)- \log (1-nx)}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} k; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$
is continuous at $x=0$ then the value of $k$ will be:

A
$0$

B
$m+n$

C
$m-n$

D
$m.n$

Solution

$\therefore$ 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.

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 31

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Limits and Continuity Test 31

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SHARING IS CARING

Question 1

0

1

19

20

Solution

Question 2

2

-2

1

-1

Solution

Question 3

2

-2

1/2

-1/2

Solution

Question 4

esin⁡2ye^{\sin ^{2} y}esin2y

sin⁡2yesin⁡2y\sin 2 y e^{\sin ^{2} y}sin2yesin2y

0

None of these

Solution

Question 5

(k−1)log⁡e(1+k)−k(k-1) \log _{e}(1+k)-k(k−1)loge​(1+k)−k

(k+1)log⁡e(k−1)+k(k+1) \log _{e}(k-1)+k(k+1)loge​(k−1)+k

(k+1)log⁡e(k−1)−k(k+1) \log _{e}(k-1)-k(k+1)loge​(k−1)−k

(k−1)log⁡e(k−1)+k(k-1) \log _{e}(k-1)+k(k−1)loge​(k−1)+k

Solution

Question 6

1/6

-1/3

1/2

1

Solution

Question 7

2aπ\dfrac{2 a}{\pi} π2a​

−2aπ-\dfrac{2 a}{\pi} −π2a​

4aπ \dfrac{4 a}{\pi} π4a​

−4aπ-\dfrac{4 a}{\pi} −π4a​

Solution

Question 8

1

0

e-1

e+1

Solution

Question 9

666

333

111

000

Solution

Question 10

000

m+nm+nm+n

m−nm-nm−n

m.nm.nm.n

Solution

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$e^{\sin ^{2} y}$

$\sin 2 y e^{\sin ^{2} y}$

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

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

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

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

$\dfrac{2 a}{\pi}$

$-\dfrac{2 a}{\pi}$

$\dfrac{4 a}{\pi}$

$-\dfrac{4 a}{\pi}$

$6$

$3$

$1$

$0$

$0$

$m+n$

$m-n$

$m.n$