Limits and Derivatives Test

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Limits and Derivatives Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Suppose the function $$f(x)-f(2x)$$ has the derivative $$5$$ at $$x=1$$ and derivative $$7$$ at $$x=2$$.The derivative of the function $$f(x)-f(4x)$$ at $$x=1$$, has the value equal to

A
$$19$$

B
$$9$$

C
$$17$$

D
$$14$$

Solution

The derivative will be,$$f'(x)-2f'(2x)$$
$$f'(x)-2f'(2x)|_{1}=5=f'(1)-2f'(2)...........(1)$$
$$f'(x)-2f'(2x)|_{2}=7=f'(2)-2f'(4)$$
Hence, $$ 14=2f'(2)-4f'(4)$$ ........................(2)
Adding (1) and (2),
$$19=2f'(2)-4f'(4)+f'(1)-2f'(2)$$
$$\Rightarrow f'(1)-4f'(4)=19$$
Question 2
1 / -0

Which one of the following statements is true?

A
If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exists.

B
If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exist.

C
If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exists.

D
If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exist.

Solution

For option A and B, take $$f(x)=x$$ and $$g(x)=\dfrac { 1 }{ x }$$.
Limit of $$g(x)$$ doesn't exist at $$x=0$$, but for $$\displaystyle \lim _{ x\to 0} f(x).g(x)$$ and $$f(x)=x$$ limit exists.

For option C,
Lets assume $$h(x) = f(x)+g(x)$$,
Take limit on both sides as $$x\to c$$
$$\lim_{x\to c}h(x) = \lim_{x\to c}(f(x)+g(x))$$
And it is given that $$\lim_{x\to c}f(x)$$ exists.
Using sum law of limits:
If $$\lim_{x\to c}F(x)$$ and $$\lim_{x\to c}G(x)$$ exists, then $$\lim_{x\to c}(F(x)\pm G(x))$$ also exists.

$$\therefore \lim_{x\to c}(h(x) - f(x)) = \lim_{x\to c}(f(x) + g(x) - f(x)) = \lim_{x\to c}g(x)$$ exists.

For option D, take $$f(x)=-\dfrac{cosx}{x^2}$$ and $$g(x)=\dfrac { 1 }{ x^2 }$$.
Limit of $$g(x)$$ and $$f(x)=-\dfrac{cosx}{x^2}$$ doesn't exist at $$x=0$$, but for $$\displaystyle \lim _{ x\to 0} f(x)+g(x)$$ limit exists.

Hence option C.
Question 3
1 / -0

If $$\displaystyle y=\frac { x }{ a+\displaystyle\frac { x }{ b+\displaystyle\frac { x }{ a+\displaystyle\frac { x }{ b+.....\infty } } } } $$, then $$\cfrac{dy}{dx} =$$

A
$$\displaystyle\frac{a}{ab+2ay}$$

B
$$\displaystyle\frac{b}{ab+2by}$$

C
$$\displaystyle\frac{a}{ab+2by}$$

D
$$\displaystyle\frac{b}{ab+2ay}$$

Solution

$$\displaystyle y = \frac{x}{a+\frac{x}{b+y}}$$
$$\Rightarrow \displaystyle y =\frac{x(b+y)}{ab+ay+x}$$
$$\Rightarrow aby+ay^2+xy=bx+xy\Rightarrow aby+ay^2=bx$$
Differentiating both sides w.r.t $$x$$
$$\Rightarrow\displaystyle (ab+2ay)\frac{dy}{dx}=b\therefore \frac{dy}{dx}=\frac{b}{ab+2ay}$$
Question 4
1 / -0

Let $$f(x)=\sin x$$, $$g(x)=\left [ x+1 \right ]$$ and $$g(f(x))=h(x)$$, where [.] is the greatest integer function. Then $$h^+\left ( \displaystyle \dfrac{\pi }{2} \right )$$ is

A
non existent

B
$$1$$

C
$$-1$$

D
none of these

Solution

$$f(x) =\sin(x)$$ and $$g(x) = [1+x]$$

$$\Rightarrow g(f(x)) = [1+\sin x]=h(x)$$

$$h^{+}\left( \cfrac{\pi}{2}\right) =[1+\sin(\pi/2)]= 1$$
Question 5
1 / -0

Find the solution of $$\displaystyle \frac{dy}{dx}= \frac{2x+2y-2}{3x+y-5}.$$

A
$$\displaystyle \left ( 2x+y-3 \right )= k\left ( x-y-3 \right )^{4}$$

B
$$\displaystyle \left ( 2x-y-3 \right )= k\left ( x-y-3 \right )^{4}$$

C
$$\displaystyle \left ( 2x+y+3 \right )= k\left ( x-y-3 \right )^{4}$$

D
$$\displaystyle \left ( 2x+y-3 \right )= k\left ( 2x-y-3 \right )^{4}$$
Question 6
1 / -0

Given : $$f(x)=4x^3-6x^2\cos2a+3x \sin 2a.\sin 6a+\sqrt{\ln (2a-a^2)}$$ then

A
$$f(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

B
$${f}'(\displaystyle \frac{1}{2})<0$$

C
$$f'(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

D
$${f}'(\displaystyle \frac{1}{2})>0$$

Solution

$$f'(x)=12x^{2}-12xcos2a+3sin2a.sin6a+0$$
$$f'(1/2)=\frac{12}{4}-\frac{12}{2}cos2a+3sin2a.sin6a$$
$$=3-6cos2a+1.5(cos4a-cos8a)$$
So, it will alwyas be greater than $$0$$
Question 7
1 / -0

$$\displaystyle \lim_{n\to\infty }\frac{n^{p}\sin ^{2}\left ( n! \right )}{n+1}$$, $$0<p<1$$, is equal to

A
$$0$$

B
$$\infty $$

C
$$1$$

D
$$none\ of\ these$$

Solution

We have,
$$\displaystyle \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n + 1} = \lim_{n \rightarrow \infty} \dfrac{sin^2 (n!)}{1 + \dfrac{1}{n}} \times n^{p - 1}$$
$$\Rightarrow \displaystyle \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n + 1} = \lim_{n \rightarrow \infty} \dfrac{sin^2 (n!)}{n^{1 - p}} \times \dfrac{1}{1 + \dfrac{1}{n}}$$
$$\displaystyle \Rightarrow \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n+1} = \dfrac{\text{An oscillating number}}{\infty} \times 1 = 0 \times 1 = 0$$.
Hence, option 'A' is correct.
Question 8
1 / -0

$$f\left( x \right)=\begin{cases} \sin { x } \qquad ;\qquad x\neq n\pi ,n=0,\pm 1,\pm 2,\pm 3..... \\ 2\qquad \qquad ;\qquad otherwise \end{cases}$$ and $$g\left( x \right) =\begin{cases} { x }^{ 2 }+1\qquad ;\qquad x\neq 0 \\ 4\qquad \qquad ;\qquad x=0 \end{cases}.$$
Then $$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right)\right)} $$ is

A
$$1$$

B
$$4$$

C
$$5$$

D
non-existent

Solution

$$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right) \right) } =g\left( f\left( 0 \right) \right) $$
$$=g(2)={ 2 }^{ 2 }+1=5$$
Question 9
1 / -0

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\left ( 1-\tan \dfrac{x}{2} \right )\left ( 1-\sin x \right )}{\left ( 1+\tan \dfrac{x}{2} \right )\left ( \pi -2x \right )^{3}}$$ is

A
$$0$$

B
$$\displaystyle \frac{1}{32}$$

C
$$\infty $$

D
$$\displaystyle \frac{1}{8}$$

Solution

$$\because \tan\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right) = \dfrac{1 - \tan\dfrac{x}{2}}{1+\tan\dfrac{x}{2}}$$
$$\therefore \displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )\left ( 1-\sin x \right )}{4\left ( \dfrac{\pi -2x}{4} \right )\left ( \pi -2x \right )^{2}}$$

$$\therefore \displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{2}-\dfrac{x}{2} \right )\left(1-\cos \left ( \dfrac{\pi }{2}-x \right )\right)}{4.\left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )\left ( \pi -2x \right )^{2}}$$

$$\therefore \displaystyle\lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{2}-\dfrac{x}{2}\right )\cdot2\cdot\sin ^{2}\left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )}{4\cdot\left (\dfrac{\pi }{4}-\dfrac{x}{2} \right )\cdot4^{2}\cdot\left ( \dfrac{\pi -2x}{4} \right )^{2}}$$

$$\displaystyle =\frac{1}{4}\times \frac{2}{16}=\frac{1}{32}$$
Question 10
1 / -0

Let $$f\left ( x \right )=\begin{cases}\sin x, x\neq n\pi
\\ 2, x=n\pi \end{cases}$$, where $$n\epsilon \mathbb{Z}$$ and
$$g\left ( x \right )=\begin{cases}x^{2}+1, x\neq 2 \\
3, x=2 \end{cases}$$.
Then $$\displaystyle \lim_{x\to 0}g\left ( f\left ( x \right ) \right )$$ is

A
$$0$$

B
$$1$$

C
$$3$$

D
none of these

Solution

$$\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ g\left( f\left( x \right) \right) } =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ g\left( \sin { x } \right) } \\ =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \left( \sin ^{ 2 }{ x } +1 \right) } =0+1=1\\ \displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ g\left( f\left( x \right) \right) } =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ g\left( \sin { x } \right) } \\ =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \left( \sin ^{ 2 }{ x } +1 \right) }=1$$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 19

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Limits and Derivatives Test - 19

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SHARING IS CARING

Question 1

$$19$$

$$9$$

$$17$$

$$14$$

Solution

Question 2

If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exists.

If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exist.

If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exists.

If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exist.

Solution

Question 3

$$\displaystyle\frac{a}{ab+2ay}$$

$$\displaystyle\frac{b}{ab+2by}$$

$$\displaystyle\frac{a}{ab+2by}$$

$$\displaystyle\frac{b}{ab+2ay}$$

Solution

Question 4

non existent

$$1$$

$$-1$$

none of these

Solution

Question 5

$$\displaystyle \left ( 2x+y-3 \right )= k\left ( x-y-3 \right )^{4}$$

$$\displaystyle \left ( 2x-y-3 \right )= k\left ( x-y-3 \right )^{4}$$

$$\displaystyle \left ( 2x+y+3 \right )= k\left ( x-y-3 \right )^{4}$$

$$\displaystyle \left ( 2x+y-3 \right )= k\left ( 2x-y-3 \right )^{4}$$

Question 6

$$f(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

$${f}'(\displaystyle \frac{1}{2})<0$$

$$f'(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

$${f}'(\displaystyle \frac{1}{2})>0$$

Solution

Question 7

$$0$$

$$\infty $$

$$1$$

$$none\ of\ these$$

Solution

Question 8

$$1$$

$$4$$

$$5$$

non-existent

Solution

Question 9

$$0$$

$$\displaystyle \frac{1}{32}$$

$$\infty $$

$$\displaystyle \frac{1}{8}$$

Solution

Question 10

$$0$$

$$1$$

$$3$$

none of these

Solution

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