Limits and Derivatives Test

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Limits and Derivatives Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \lim_{x\rightarrow \infty}\dfrac{x^3+1}{x^2+1}-(ax+b)=2$$, then

A
$$a=2$$ and $$b=-1$$

B
$$a = 1$$ and $$b = 1$$

C
$$a = 1$$ and $$b = -1$$

D
$$a = 1$$ and $$b = -2$$

Solution

Given $$\displaystyle \lim _{ x\rightarrow \infty }{ \frac { { x }^{ 3 }+1 }{ { x }^{ 2 }+1 } -(ax+b)=2 } $$

$$\Rightarrow \displaystyle \lim _{ x\rightarrow \infty }{ (\frac { { (1-a)x }^{ 3 }-b{ x }^{ 2 }-ax-b+1 }{ { x }^{ 2 }+1 } )=2 } $$

For the limit to exist, the coefficient of $$x^{3}$$ must be zero because if it is not zero then the limit is infinite
$$\Rightarrow a=1$$

So the given one will reduce to $$\displaystyle \lim _{ x\rightarrow \infty }{ (\frac { -b{ x }^{ 2 }-ax-b+1 }{ { x }^{ 2 }+1 } )=2 } $$

If we apply the limit , we get $$-b=2$$
$$\Rightarrow b=-2$$
Therefore the correct option is $$D$$
Question 2
1 / -0

The value of the constant $$\alpha$$ and $$\beta$$ such that $$\displaystyle \lim_{x\rightarrow \infty}\left(\displaystyle\frac{x^2+1}{x+1}-\alpha x-\beta\right)=0$$ are respectively.

A
$$(1, 1)$$

B
$$(-1, 1)$$

C
$$(1,-1)$$

D
$$(0, 1)$$

Solution

Simplifying the above expression gives us

$$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}+1-\alpha x(x+1)-\beta(x+1)}{x+1})$$

$$=\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}(1-\alpha)-x(\alpha +\beta)+1-\beta)}{x+1})$$
$$=0$$ implies that $$Numerator<Denominator$$.
Hence coefficient of $$x^{2}$$ will be 0. Therefore $$\alpha=1$$.
Hence
$$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-x(1 +\beta)+1-\beta)}{x+1})=0$$

Applying L'Hopital's Rule we get
$$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-(1 +\beta)}{1})=0$$
Or
$$\beta+1=0$$
Or
$$\beta=-1$$.
Hence
$$(\alpha,\beta)=1,-1$$
Question 3
1 / -0

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

A
$$2$$

B
$$3$$

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

D
$$\dfrac {1}{3}$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\dfrac {1 - \cos x}{x^{2}}=\lim_{x\to 0}\dfrac{2\sin^2\frac{x}{2}}{x^2}$$

$$=\displaystyle \lim_{x\rightarrow 0}\dfrac {2\sin^2\frac{x}{2}}{4\left(\frac{x}{2}\right)^{2}}=\dfrac{1}{2}\lim_{x\to 0}\left( \dfrac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\dfrac{1}{2}\cdot 1=\dfrac{1}{2}$$ (using basic limit formula )
Question 4
1 / -0

The limit of $$\left[\frac{1}{x^2}+\frac{(2013)^x}{e^x-1}-\frac{1}{e^x-1}\right]$$ as $$x\rightarrow 0$$

A
Approaches $$+\, \infty$$

B
Approaches $$- \,\infty$$

C
Is equal to $$log_e(2013)$$

D
Does not exist

Solution

$$\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { 1 }{ { x }^{ 2 } } +\cfrac { { \left( 2013 \right) }^{ x }-1 }{ { e }^{ x }-1 } \right] } $$
$$=\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { 1 }{ { x }^{ 2 } } +\cfrac { { \left( 2013 \right) }^{ x }-1 }{ x } \times \cfrac { x }{ { e }^{ x }-1 } \right] } $$

$$=\left( \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1 }{ { x }^{ 2 } } } \right) +\left( \log { 2013 } \right) \times 1\quad \quad \left[ \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { a }^{ x }-1 }{ x } } =\log _{ e }{ a } ;\quad \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { e }^{ x }-1 }{ x } } =1 \right] $$

$$=\infty +\log { 2013 } =\infty $$

as $$x\rightarrow 0\Rightarrow $$Limit $$\rightarrow \infty $$
Question 5
1 / -0

If the function $$f(x)$$ satisfies $$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$$, then $$\displaystyle \lim_{x\rightarrow 1}f(x)=$$

A
$$2$$

B
$$3$$

C
$$1$$

D
$$0$$

Solution

It is given that, $$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$$ exists and equal to $$1$$.

Clearly the denominator is zero at $$x=1$$, so for above limit to exist
numerator should also be zero, as x approaches $$1$$.
Therefore, $$\displaystyle \lim_{x\to 1}f(x)=2$$

Note: If $$\displaystyle \lim_{x\to 1}f(x) \neq 1$$, given limit will not exist
Question 6
1 / -0

$$f(x) = \log \left (e^{x} \left (\dfrac {x - 2}{x + 2}\right )^{\dfrac {3}{4}} \right ) \Rightarrow f'(0) =$$

A
$$\dfrac {1}{4}$$

B
$$4$$

C
$$\dfrac {-3}{4}$$

D
$$1$$

Solution

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

$$f(x) = x + \dfrac {3}{4} [\log (x - 2) - \log (x + 2)]$$

$$f'(x) = 1 + \dfrac {3}{4} \left [\dfrac {1}{x - 2} - \dfrac {1}{x + 2} \right ]$$

$$f'(x) = \left [1 + \dfrac {3}{x^{2} - 4}\right ]$$

$$f'(0) = 1 -\dfrac {3}{4} = \dfrac {1}{4}$$
Question 7
1 / -0

If $$f(x) = \sec (3x)$$, then $$f'\left (\dfrac {3\pi}{4}\right ) =$$

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

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

C
$$\dfrac {3}{2}$$

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

E
$$3\sqrt {2}$$

Solution

Given, $$f(x) = \sec(3x)$$
We get $$f'\left( x \right) =3\sec(3x)\tan(3x)$$
Therefore, we get $$f'\left( \dfrac { 3\pi }{ 4 } \right) =3\sec\left (\dfrac { 9\pi }{ 4 }\right )\tan\left (\dfrac { 9\pi }{ 4 } \right)$$
$$=3\sec\left (\dfrac { \pi }{ 4 } \right)\tan\left (\dfrac { \pi }{ 4 }\right )$$
$$=3\times \sqrt { 2 } \times 1$$
$$=3\sqrt 2$$
Question 8
1 / -0

If $$y = f(x^{2} + 2)$$ and $$f'(3) = 5$$, then $$\dfrac {dy}{dx}$$ at $$x = 1$$ is _____

A
$$5$$

B
$$25$$

C
$$15$$

D
$$10$$

Solution

$$y=f(x^2+2)$$
Using chain rule of differentiation
$$\dfrac{dy}{dx}=\dfrac{d}{d(x^2+2)}[f(x^2+2)]\cdot \dfrac{d}{dx}(x^2+2)$$
$$\quad =f'(x^2+2)\cdot 2x=2xf'(x^2+2)$$

Hence $$\dfrac{dy}{dx}\bigg|_{x=1}=2\cdot 1\cdot f'(1^2+2)=2\cdot f'(3)=2\cdot 5=10$$
Question 9
1 / -0

If f(x) is a function such that $$f^{\prime \prime}(x)+f(x)=0$$ and $$g(x)=[f(x)]^2+[f'(x)]^2$$ and g(3)=8, then $$g(8)= $$_____

A
$$0$$

B
$$3$$

C
$$5$$

D
$$8$$

Solution

We have, $$g(x)=[f(x)]^2+[f'(x)]^2$$
Differentiate the function g(x)
$$\Rightarrow g'(x)=2f(x)f'(x)+2f'(x)f''(x)$$, use chain rule
$$=2f'(x)[f(x)+f''(x)]=2f'(x)(0)=0$$, use the given condition
Hence $$g(x)$$ is a constant function
$$\Rightarrow g(x)=c$$, constant
But $$g(3)=8$$, so $$g(x)=8$$, for all real x
Hence $$g(8)=8$$
Question 10
1 / -0

The limit of $$\displaystyle \sum_{n=1}^{1000}(-1)^nx^n$$ as $$x\rightarrow \infty$$

A
does not exist

B
exists and equals to 0

C
exists and approaches $$+\infty$$

D
exists and approaches $$-\infty$$

Solution

$$\sum _{ n=1 }^{ 1000 }{ { \left( -1 \right) }^{ n }{ x }^{ n } } =-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }+....$$
It is in GP
$$S=\cfrac { -x\left( { (-x) }^{ 1000 }-1 \right) }{ (-x)-1 } $$

$$\displaystyle \lim _{ x\rightarrow \infty }{ \sum _{ n=1 }^{ 1000 }{ { \left( -1 \right) }^{ n }{ x }^{ n } } } =\displaystyle \lim _{ x\rightarrow \infty }{ S } =\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { x\left( { x }^{ 1000 }-1 \right) }{ x+1 } } $$

$$=\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \left( { x }^{ 1000 }-1 \right) }{ 1+\cfrac { 1 }{ x } } } =\cfrac { { \infty }^{ 1000 }-1 }{ 1+0 } =\infty $$

$$as \left( x\rightarrow \infty ;\cfrac { 1 }{ x } \rightarrow \infty \right) $$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 22

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Limits and Derivatives Test - 22

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SHARING IS CARING

Question 1

$$a=2$$ and $$b=-1$$

$$a = 1$$ and $$b = 1$$

$$a = 1$$ and $$b = -1$$

$$a = 1$$ and $$b = -2$$

Solution

Question 2

$$(1, 1)$$

$$(-1, 1)$$

$$(1,-1)$$

$$(0, 1)$$

Solution

Question 3

$$2$$

$$3$$

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

$$\dfrac {1}{3}$$

Solution

Question 4

Approaches $$+\, \infty$$

Approaches $$- \,\infty$$

Is equal to $$log_e(2013)$$

Does not exist

Solution

Question 5

$$2$$

$$3$$

$$1$$

$$0$$

Solution

Question 6

$$\dfrac {1}{4}$$

$$4$$

$$\dfrac {-3}{4}$$

$$1$$

Solution

Question 7

$$-3\sqrt {2}$$

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

$$\dfrac {3}{2}$$

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

$$3\sqrt {2}$$

Solution

Question 8

$$5$$

$$25$$

$$15$$

$$10$$

Solution

Question 9

$$0$$

$$3$$

$$5$$

$$8$$

Solution

Question 10

does not exist

exists and equals to 0

exists and approaches $$+\infty$$

exists and approaches $$-\infty$$

Solution

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