Limits and Derivatives Test

Result

Limits and Derivatives Test - 26

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$ \lim _{ x\rightarrow 1 }{ \dfrac { \sqrt { 1-\cos { 2\left( x-1 \right) } } }{ x-1 } }$$

A
Exists and it equals $$\sqrt {2}$$

B
Exists and it equals $$-\sqrt {2}$$

C
Does not exist because $$x-1\rightarrow 0$$

D
Does not exist because left hand limit is not equal to right hand limit

Solution
Question 2
1 / -0

If $$f(x) = 2x^9 - 5x^8 + 7x^6 - 15x^4 + 5x + 7$$, then $$\underset{x \rightarrow 0}{\lim} \dfrac{f (1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$$ is

A
$$-\dfrac{35}{3}$$

B
$$\dfrac{35}{3}$$

C
$$\dfrac{37}{3}$$

D
$$-\dfrac{37}{3}$$
Question 3
1 / -0

If $$f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$$, then $$\underset{a \rightarrow 0}{\lim} \dfrac{f(1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$$ is

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

B
$$\dfrac{53}{3}$$

C
$$-\dfrac{55}{3}$$

D
$$\dfrac{55}{3}$$

Solution

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

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

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

$$=\dfrac{-53}{3}$$
Question 4
1 / -0

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

A
$$1$$

B
$$1/2$$

C
$$1/3$$

D
$$0$$

Solution

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.
Question 5
1 / -0

If $$y=|\cos x|+|\sin x|$$, then $$\dfrac{dy}{dx}$$ at $$x=\dfrac{2\pi}{3}$$ is

A
$$\dfrac{1-\sqrt{3}}{2}$$

B
$$0$$

C
$$\dfrac{\sqrt{3}-1}{2}$$

D
None of these

Solution

We have,

$$y=\left| \cos x \right|+\left| \sin x \right|$$

We know that the $$\cos x$$ is $$-ve$$ in $$\left( \dfrac{\pi }{2},\pi \right)$$ and $$\sin x$$ is $$+ve$$ in this domain.

Therefore,

$$y=-\cos x+\sin x$$

On differentiating both sides w.r.t $$x$$, we get

$$ \dfrac{dy}{dx}=-\left( -\sin x \right)+\cos x $$

$$ \dfrac{dy}{dx}=\sin x+\cos x $$

Since, $$x=\dfrac{2\pi }{3}$$

Therefore,

$$ {{\left. \dfrac{dy}{dx} \right|}_{x=\dfrac{2\pi }{3}}}=\sin \left( \dfrac{2\pi }{3} \right)+\cos \left( \dfrac{2\pi }{3} \right) $$

$$ {{\left. \dfrac{dy}{dx} \right|}_{x=\dfrac{2\pi }{3}}}=\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} $$

$$ {{\left. \dfrac{dy}{dx} \right|}_{x=\dfrac{2\pi }{3}}}=\dfrac{\sqrt{3}-1}{2} $$

Hence, this is the answer.
Question 6
1 / -0

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

A
1

B
-1

C
0

D
none of these

Solution

$$\lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - \cos{2x} \right)}}{x}}$$
$$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - \left( 1 - 2 \sin^{2}{x} \right) \right)}}{x}}$$
$$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - 1 + 2 \sin^{2}{x} \right)}}{x}}$$
$$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 2 \sin^{2}{x} \right)}}{x}}$$
$$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\sin^{2}{x}}}{x}}$$
$$= \lim_{x \rightarrow 0}{\cfrac{\sin{x}}{x}}$$
$$= 1 \; \left( \because \lim_{x \rightarrow 0}{\cfrac{\sin{x}}{x}} = 1 \right)$$
Question 7
1 / -0

If $$y=|\cos x|+|\sin x|$$, then $$\dfrac {dy}{dx}$$ at $$x=\dfrac {2\pi}{3}$$ is

A
$$\dfrac {1-\sqrt {3}}{2}$$

B
$$0$$

C
$$\dfrac {\sqrt {3}-1}{2}$$

D
$$\dfrac {\sqrt {3}+1}{2}$$

Solution

$$ y = |cos\,x|+|sin\,x| $$ $$ , x = \dfrac{2n}{3} $$
$$ y = -cos\,x+sin\,x $$ $$ \left \{ x\sum (\dfrac{\pi }{2},\pi ) \right \} $$
$$ y^{1} = sinx+cosx $$
$$ y^{1} = sin(\dfrac{2n}{3})+cos(\dfrac{2n}{3}) $$
$$ = \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} = \dfrac{\sqrt{3}-1}{2} $$ Option C is correct
Question 8
1 / -0

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

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
non existent

Solution

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,

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

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

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

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

$$=0$$
Question 9
1 / -0

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

A
$$1$$

B
$$0$$

C
$$2$$

D
$$\dfrac{1}{2}$$

Solution

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.
Question 10
1 / -0

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

A
$$1$$

B
$$0$$

C
$$2$$

D
$$1/2$$

Solution

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

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

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 26

Result

Limits and Derivatives Test - 26

Score

-

Rank

-

SHARING IS CARING

Question 1

Exists and it equals $$\sqrt {2}$$

Exists and it equals $$-\sqrt {2}$$

Does not exist because $$x-1\rightarrow 0$$

Does not exist because left hand limit is not equal to right hand limit

Solution

Question 2

$$-\dfrac{35}{3}$$

$$\dfrac{35}{3}$$

$$\dfrac{37}{3}$$

$$-\dfrac{37}{3}$$

Question 3

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

$$\dfrac{53}{3}$$

$$-\dfrac{55}{3}$$

$$\dfrac{55}{3}$$

Solution

Question 4

$$1$$

$$1/2$$

$$1/3$$

$$0$$

Solution

Question 5

$$\dfrac{1-\sqrt{3}}{2}$$

$$0$$

$$\dfrac{\sqrt{3}-1}{2}$$

None of these

Solution

Question 6

1

-1

0

none of these

Solution

Question 7

$$\dfrac {1-\sqrt {3}}{2}$$

$$0$$

$$\dfrac {\sqrt {3}-1}{2}$$

$$\dfrac {\sqrt {3}+1}{2}$$

Solution

Question 8

$$0$$

$$1$$

$$\dfrac{1}{2}$$

non existent

Solution

Question 9

$$1$$

$$0$$

$$2$$

$$\dfrac{1}{2}$$

Solution

Question 10

$$1$$

$$0$$

$$2$$

$$1/2$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.