Limits and Derivatives Test

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Limits and Derivatives Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The limit of $\left\{\frac{1}{x}\sqrt{1-x}-\sqrt{1+\frac{1}{x^2}}\right\}$ as $x\rightarrow 0$

A
Does not exist

B
Is equal to $\frac{1}{2}$

C
Is equal to 0

D
Is equal to 1

Solution

Let $p(x)=\cfrac { 1 }{ x } \sqrt { 1-x } ,q(x)=\sqrt { 1+\cfrac { 1 }{ { x }^{ 2 } } }$
$\lim _{ x\rightarrow { 0 }^{ + } }{ p\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { 1 }{ x } \sqrt { 1-x } } =\infty \times \sqrt { 1-10 }$
$=\infty$
$\lim _{ x\rightarrow { 0 }^{ - } }{ p\left( x \right) } =-\infty \times \sqrt { 1-0 } =-\infty$
as $x\rightarrow { 0 }^{ + }\quad \cfrac { 1 }{ x } \rightarrow +\infty$
$x\rightarrow { 0 }^{ - }\quad \cfrac { 1 }{ x } \rightarrow -\infty$
$\Rightarrow p(x)$ is discontinuous as $x\rightarrow 0$
Let $f\left( x \right) =p\left( x \right) -q\left( x \right)$
Since $p\left( x \right)$ is discontinuous $f\left( x \right)$ is also discontinuous by Algebra of continuous functions.
Question 2
1 / -0

If $y = \tan^{-1} \left (\dfrac {1}{1 + x + x^{2}}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 3x + 2}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 5x + 6}\right ) + .... +$ upto $n$ terms then $\dfrac {dy}{dx}$ at $x = 0$ and $n = 1$ is equal to

A
$\dfrac {1}{2}d$

B
$-\dfrac {1}{2}$

C
$0$

D
$\dfrac {1}{3}$

Solution

$y= \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) + \tan^{-1}\left (\dfrac {1}{1 + (x + 1)(x + 2)}\right ) +\\ \quad\tan^{-1} \left (\dfrac {1}{1 + (x + 2)(x + 3)}\right ) + ..... + \tan^{-1} \left (\dfrac {1}{1 + (x + n - 1) (x + n)}\right )$

when $n = 1$
$y = \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) = \tan^{-1} \left (\dfrac {(x + 1) - x}{1 + x(x + 1)}\right ) = \tan^{-1} (x + 1) - \tan^{-1} x$

$\Rightarrow \dfrac {dy}{dx} = \dfrac {1}{1 + (x + 1)^{2}} - \dfrac {1}{1 + x^{2}} \\ \therefore \dfrac {dy}{dx}]_{x = 0} = \dfrac {1}{1 + 1} - \dfrac {1}{1} = \dfrac {1}{2} - 1 = -\dfrac {1}{2}$
Question 3
1 / -0

What is $\displaystyle \lim_{x \rightarrow 0 } x^2 \sin \left(\frac{1}{x}\right)$ equal to ?

A
0

B
1

C
1/2

D
Limit does not exist.

Solution

$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ 2 }\sin { \left( \cfrac { 1 }{ x } \right) } }$
$={ 0 }^{ 2 }\times \sin { \left( \cfrac { 1 }{ 0 } \right) } \quad \left[ -1\le \sin { \left( \cfrac { 1 }{ x } \right) } \le 1 \right]$
$=0$
Question 4
1 / -0

If $f\left( x \right) =\begin{vmatrix} \sin { x } & \cos { x } & \tan { x } \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}$ , then $\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) }{ { x }^{ 2 } } }$ is

A
$-1$

B
$3$

C
$1$

D
Zero

Solution

$f(x)=\begin{vmatrix} sinx & cosx & tanx \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}\\ \Longrightarrow f(x)=sinx({ x }^{ 2 }-x)-cosx({ x }^{ 3 }-2{ x }^{ 2 })+tanx({ x }^{ 3 }-2{ x }^{ 3 })\\ \Longrightarrow f(x)={ x }^{ 2 }sinx-xsinx-{ x }^{ 3 }cosx+2{ x }^{ 2 }cosx-{ x }^{ 3 }tanx\\ \Longrightarrow \dfrac { f(x) }{ { x }^{ 2 } } =sinx-xcosx+2cosx-xtanx-\dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =\underset { x\longrightarrow 0 }{ lim } sinx-\underset { x\longrightarrow 0 }{ lim } xcosx+\underset { x\longrightarrow 0 }{ lim } 2cosx-\underset { x\longrightarrow 0 }{ lim } xtanx-\underset { x\longrightarrow 0 }{ lim } \dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =0-0+2-0-1\\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =1\\ \\$
Question 5
1 / -0

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$ is equal to :

A
$\pi+2$

B
$\pi-2$

C
e

D
$e^{-1}$

Solution

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$
$=\pi - 2\left(\dfrac{\pi}{2}\right)^{\cos \frac{\pi}{2}}$
$=\pi-2\left(\dfrac{\pi}{2}\right)^{0}$
$=\pi-2$
Question 6
1 / -0

The derivative of $ln(x+\sin x)$ with respect to $(x+\cos x)$ is

A
$\dfrac{1+\cos x}{(x+\sin x)(1-\sin x)}$

B
$\dfrac{1-\cos x}{(x+\sin x)(1+\sin x)}$

C
$\dfrac{1-\cos x}{(x-\sin x)(1+\cos x)}$

D
$\dfrac{1+\cos x}{(x-\sin x)(1-\cos x)}$

Solution

Let $u=ln(x+\sin x)$ and $v=x+\cos x$
Now,
$\cfrac{du}{dx}=\cfrac1{x+\sin x}(1+\cos x)$ .... $(i)$
and $\cfrac{dv}{dx}=1-\sin x$ ..... $(ii)$
Now, we can find derivative of $u$ with respect to $v$ by dividing $(i)$ by $(ii)$ , we get
$\cfrac{du/dx}{dv/dx}=\cfrac{(1+\cos x)}{(x+\sin x)(1-\sin x)}$
or, $\cfrac{du}{dv}=\cfrac{1+\cos x}{(x+\sin x)(1-\sin x)}$
Hence, A is the correct option.
Question 7
1 / -0

Let $C(\theta)=\displaystyle\sum _{n=0}^{\infty}\dfrac{\cos(n\theta)}{n!}$
Which of the following statements is FALSE?

A
$C(0).C(\pi)=1$

B
$C(0).C(\pi) > 2$

C
$C(\theta) > 0$ for all $\theta \in R$

D
$C(\theta) \neq 0$ for all $\theta \in R$

Solution

$C(0)=e$
$C(\pi)=\dfrac{1}{e}$
Hence, option A and B are correct

$C'(\theta)= -\displaystyle \sum_{n=0}^{\infty }\dfrac{\sin\theta }{(n-1)!}$
$C'(\theta)=0$
Therefore, option D is false because $C'(\theta) =0$ for $n=0$
Question 8
1 / -0

$\displaystyle\lim_{x\rightarrow\frac{\pi}{6}}\frac{\sin\left(x-\displaystyle\frac{\pi}{6}\right)}{\sqrt{3-2cos x}}$ is equal to :

A
$0$

B
$\displaystyle\frac{1}{(\sqrt{3}-2)}$

C
$1$

D
$\infty$

Solution

$\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \dfrac { \sin { (x-\cfrac { \pi }{ 6 } } ) }{ \sqrt { 3-2\cos { x } } } }$
at $\cfrac { \pi }{ 6 }$ we get the numerator as $\sin { (x-\cfrac { \pi }{ 6 } ) } = 0$
and the denominator as $\sqrt { 3-2\cos { x } } =\sqrt { 3-\sqrt { 3 } }$
So,it is not an indeterminant form as we have $\dfrac{0}{\sqrt{3 - \sqrt{3}} }$
Hence the answer will be zero.
Question 9
1 / -0

$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x}$ is equal to

A
$1$

B
$0$

C
$-2$

D
$-1$

Solution

$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x} = \displaystyle \lim_{h\rightarrow 0}\dfrac {\tan \left (\dfrac {\pi}{4} + h\right ) - 1}{\cos 2\left (\dfrac {\pi}{4} + h\right )}$
$\left [\because x = \dfrac {\pi}{4} + h\right ]$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {\left (\dfrac {1 + \tan h}{1 - \tan h}\right ) - 1}{\cos \left (\dfrac {\pi}{2} + 2h\right )}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {1 + \tan h - 1 + \tan h}{-\sin 2h (1 - \tan h)}$
$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-2\tan h}{2\sin h \cos h (1 - \tan h)}$
$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-1}{\cos^{2} h (1 - \tan h)} = -1$
Hence, option D is correct.
Question 10
1 / -0

The value of $\displaystyle \lim _{ x\rightarrow \pi /6 }{ \cfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } }$ is

A
$3$

B
$-3$

C
$6$

D
$0$

Solution

$L=\lim _{ x\rightarrow { \pi }/{ 6 } }{ \dfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } } \\ L=\dfrac { 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } +\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 }{ 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } -3\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 } \\ L=\dfrac { 2.\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 2 } -1 }{ 2.\dfrac { 1 }{ 4 } -3.\dfrac { 1 }{ 2 } -1 } =\dfrac { 0 }{ -2 }= 0$

So option $D$ is correct

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 23

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Limits and Derivatives Test - 23

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Question 1

Does not exist

Is equal to 12\frac{1}{2}21​

Is equal to 0

Is equal to 1

Solution

Question 2

12d\dfrac {1}{2}d21​d

−12-\dfrac {1}{2}−21​

000

13\dfrac {1}{3}31​

Solution

Question 3

0

1

1/2

Limit does not exist.

Solution

Question 4

−1-1−1

333

111

Zero

Solution

Question 5

π+2\pi+2π+2

π−2\pi-2π−2

e

e−1e^{-1}e−1

Solution

Question 6

1+cos⁡x(x+sin⁡x)(1−sin⁡x)\dfrac{1+\cos x}{(x+\sin x)(1-\sin x)}(x+sinx)(1−sinx)1+cosx​

1−cos⁡x(x+sin⁡x)(1+sin⁡x)\dfrac{1-\cos x}{(x+\sin x)(1+\sin x)}(x+sinx)(1+sinx)1−cosx​

1−cos⁡x(x−sin⁡x)(1+cos⁡x)\dfrac{1-\cos x}{(x-\sin x)(1+\cos x)}(x−sinx)(1+cosx)1−cosx​

1+cos⁡x(x−sin⁡x)(1−cos⁡x)\dfrac{1+\cos x}{(x-\sin x)(1-\cos x)}(x−sinx)(1−cosx)1+cosx​

Solution

Question 7

C(0).C(π)=1C(0).C(\pi)=1C(0).C(π)=1

C(0).C(π)>2C(0).C(\pi) > 2C(0).C(π)>2

C(θ)>0C(\theta) > 0C(θ)>0 for all θ∈R\theta \in Rθ∈R

C(θ)≠0C(\theta) \neq 0C(θ)=0 for all θ∈R\theta \in Rθ∈R

Solution

Question 8

000

1(3−2)\displaystyle\frac{1}{(\sqrt{3}-2)}(3​−2)1​

111

∞\infty∞

Solution

Question 9

111

000

−2-2−2

−1-1−1

Solution

Question 10

333

−3-3−3

666

000

Solution

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Is equal to $\frac{1}{2}$

$\dfrac {1}{2}d$

$-\dfrac {1}{2}$

$0$

$\dfrac {1}{3}$

$-1$

$3$

$1$

$\pi+2$

$\pi-2$

$e^{-1}$

$\dfrac{1+\cos x}{(x+\sin x)(1-\sin x)}$

$\dfrac{1-\cos x}{(x+\sin x)(1+\sin x)}$

$\dfrac{1-\cos x}{(x-\sin x)(1+\cos x)}$

$\dfrac{1+\cos x}{(x-\sin x)(1-\cos x)}$

$C(0).C(\pi)=1$

$C(0).C(\pi) > 2$

$C(\theta) > 0$ for all $\theta \in R$

$C(\theta) \neq 0$ for all $\theta \in R$

$0$

$\displaystyle\frac{1}{(\sqrt{3}-2)}$

$1$

$\infty$

$1$

$0$

$-2$

$-1$

$3$

$-3$

$6$

$0$