Limits and Derivatives Test

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Limits and Derivatives Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \lim_{x\rightarrow 0}(f(x)\:g(x))$$ exists for any functions $$f$$ and $$g$$ then

A
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

B
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

C
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ may not exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

D
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

Solution

Take $$\displaystyle f(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$$ and $$\displaystyle g(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$$. Then $$f(x)\:g(x)=1$$ for $$x\neq 0$$. Hence $$\displaystyle \displaystyle \lim_{x\rightarrow 0}f(x)$$ and $$\displaystyle \lim_{x\rightarrow 0}g(x)$$ does not exist but $$\displaystyle\lim_{x\rightarrow 0}f(x)\:g(x)=1$$
Hence, option 'D' is correct.
Question 2
1 / -0

$$\displaystyle \lim_{x\rightarrow\infty}\left(\frac{\sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+ \sqrt(1 - \cos x)+...\infty) - 1}}}{x^2}\right)$$ equals to

A
0

B
$$\displaystyle\frac{1}{2}$$

C
1

D
2

Solution

Let $$\sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+...\infty)}}}=y$$

$$\Rightarrow y = \sqrt{(1 - \cos x) + y}$$

$$\Rightarrow y^2 - y + (\cos x - 1) = 0$$

$$\Rightarrow y = \displaystyle\frac{1 + \sqrt{5 - 4 \cos x}}{2}$$

Now required limit is $$\lim_{x\rightarrow0}$$ $$\displaystyle\frac{1 + \sqrt{5 - 4 cosx - 1}}{2x^2}$$ $$=\displaystyle\frac{1}{2}$$
Question 3
1 / -0

$$\underset{x\rightarrow0}{lim}\displaystyle\frac{1-cos^{3}x+sin^{3}x+\ell n(1+x^{3})+\ell n(1+cos\,\,x)}{x^{2}-1+2\,cos^{2}x+tan^{4}x+sin^{3}x}$$ is equal to -

A
$$\,\,\displaystyle\frac{3}{4}$$

B
$$\,\,ln2$$

C
$$\,\,\displaystyle\frac{ln2}{4}$$

D
$$\,\,3/2$$

Solution

$$\displaystyle \underset { x\rightarrow 0 }{ lim } \frac { 1-cos^{ 3 }x+sin^{ 3 }x+\ell n(1+x^{ 3 })+\ell n(1+cos\, \, x) }{ x^{ 2 }-1+2\, cos^{ 2 }x+tan^{ 4 }x+sin^{ 3 }x } $$

$$\displaystyle =\frac { 1-1+0+\ln { \left( 1+0 \right) } +\ln { \left( 1+1 \right) } }{ 0-1+2+0+0 } =\ln { 2 } $$
Question 4
1 / -0

let a, b, c are non zero constant number then $$\lim_{r\rightarrow\infty}\displaystyle\frac{cos\displaystyle\frac{a}{r}-cos\displaystyle\frac{b}{r}cos\displaystyle\frac{c}{r}}{sin\displaystyle\frac{b}{r}sin\displaystyle\frac{c}{r}}$$ equals to

A
$$\displaystyle\frac{a^2 + b^2 - c^2}{2bc}$$

B
$$\displaystyle\frac{c^2 + a^2 - b^2}{2bc}$$

C
$$\displaystyle\frac{b^2 + c^2 - a^2}{2bc}$$

D
independent of a, b and c

Solution

$$\displaystyle \lim _{ r\rightarrow \infty }{ \dfrac { \cos { \dfrac { a }{ r } -\cos { \dfrac { b }{ r } } \cos { \dfrac { c }{ r } } } }{ \sin { \dfrac { b }{ r } \sin { \dfrac { c }{ r } } } } } $$

$$\displaystyle =\lim _{ r\rightarrow \infty }{ \dfrac { \cos { \dfrac { a }{ r } -\dfrac { 1 }{ 2 } \left( \cos { \left( \dfrac { b+c }{ r } \right) } +\cos { \left( \dfrac { b-c }{ r } \right) } \right) } }{ \dfrac { 1 }{ 2 } \left( \cos { \left( \dfrac { b-c }{ r } \right) -\cos { \left( \dfrac { b+c }{ r } \right) } } \right) } } $$

By Applying L Hospital,s Rule

$$\displaystyle =\lim _{ r\rightarrow \infty }{ \dfrac { -a\sin { \dfrac { a }{ r } -\dfrac { 1 }{ 2 } \left( -\left( b+c \right) \sin { \left( \dfrac { b+c }{ r } \right) } -\left( b-c \right) \sin { \left( \dfrac { b-c }{ r } \right) } \right) } }{ \dfrac { -1 }{ 2 } { \left( b-c \right) }\sin { \left( \dfrac { b-c }{ r } \right) } +\dfrac { 1 }{ 2 } { \left( b+c \right) }\sin { \left( \dfrac { b+c }{ r } \right) } } } $$ $$(here\,\,\,\dfrac{-1}{r^2}$$

$$would\,\, cancel \,\,out\,\, from\,\, numerator \,\,and\,\, denominator)$$

$$\displaystyle =\lim _{ r\rightarrow \infty }{ \dfrac { { -a }^{ 2 }\cos { \dfrac { a }{ r } +\dfrac { 1 }{ 2 } \left( { \left( b+c \right) }^{ 2 }\cos { \left( \dfrac { b+c }{ r } \right) +{ \left( b-c \right) }^{ 2 }\cos { \left( \dfrac { b-c }{ r } \right) } } \right) } }{ \dfrac { -1 }{ 2 } { \left( b-c \right) }^{ 2 }\cos { \left( \dfrac { b-c }{ r } \right) } +\dfrac { 1 }{ 2 } { \left( b+c \right) }^{ 2 }\cos { \left( \dfrac { b+c }{ r } \right) } } } $$

$$\displaystyle =\dfrac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 2bc } $$
Question 5
1 / -0

$$ \displaystyle f^{ ' }\left( x \right) =g\left( x \right) $$ and $$ \displaystyle g^{ ' }\left( x \right) =-f\left( x \right)$$ for all real x and $$ \displaystyle f\left( 5 \right) =2=f^{ ' }\left( 5 \right) $$ then $$ \displaystyle f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right) $$ is -

A
$$2$$

B
$$4$$

C
$$8$$

D
None of these

Solution

Given, $$ \displaystyle f^{ ' }\left( x \right) =g\left( x \right)$$ and $$g^{ ' }\left( x \right) =-f\left( x \right)$$
Now $$ \displaystyle \frac { d }{ dx } \left[ f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right) \right] =2f\left( x \right) f^{ ' }\left( x \right) +2g\left( x \right) g^{ ' }\left( x \right) $$
$$ \displaystyle =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0$$
$$ \displaystyle \therefore \quad f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right)$$ =constant
$$ \displaystyle f^{ 2 }\left( 5 \right) +g^{ 2 }\left( 5 \right) = 4 + 4 = 8$$
$$ \displaystyle \therefore \quad f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right) $$ = 8
Question 6
1 / -0

Evaluate $$\displaystyle \lim_{n\rightarrow \infty }\left [ \frac{n!}{n^{n}} \right ]^{1/n}$$.

A
$$\displaystyle\frac{1}{e}$$

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

C
$$\displaystyle\frac{1}{n}$$

D
$$none\ of\ above$$

Solution

Let $$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \frac{n!}{n^{n}} \right )^{1/n}$$

$$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \frac{1.2.3.4...n}{n.nnn....n} \right )^{1/n}$$

$$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \left ( \frac{1}{n} \right )\left ( \frac{2}{n} \right )\left ( \frac{3}{n} \right )...\left ( \frac{n}{n} \right ) \right )^{1/n}$$

$$\displaystyle\Rightarrow ln P=\lim_{n\rightarrow \infty }\frac{1}{n}\left ( \log \left ( \frac{1}{n} \right )+\log \left ( \frac{2}{n} \right )+...\log \left ( \frac{n}{n} \right ) \right )$$

$$\displaystyle P=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{n}\log \frac{r}{n}$$

$$=\int_{0}^{1}lnxdx=\left [ xlnx-x \right ]^{1}_{0}$$

$$=\displaystyle \left ( 0-1 \right )-\lim_{x\rightarrow 0 }\left ( xlnx \right )+0$$

$$=\displaystyle -1-\lim_{x\rightarrow 0 }\frac{lnx}{1/x}=-1-\lim_{x\rightarrow 0 }\frac{1/x}{\left ( -1/x^{2} \right )}$$

$$=\displaystyle -1-\lim_{x\rightarrow 0 }x=-1+0=-1\Rightarrow lnP=-1$$

$$\displaystyle P=e^{-1}=1/e$$
Question 7
1 / -0

$$\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{a-x}{1+ax} \right ) \right )$$ equals if ax > -1

A
$$\displaystyle \frac{a}{1+x^{2}}$$

B
$$\displaystyle \frac{1}{1+x^{2}}$$

C
$$\displaystyle- \frac{a}{1+x^{2}}$$

D
$$\displaystyle -\frac{1}{1+x^{2}}$$

Solution

Let, $$\displaystyle y=\tan ^{-1}\left ( \frac{a-x}{1+ax} \right )$$
$$\displaystyle y=\tan ^{-1}(a)-\tan ^{-1}(x)$$
$$\therefore \displaystyle \frac{dy}{dx}=-\frac{1}{1+x^2}$$
Question 8
1 / -0

If $$f(x) = \displaystyle \left | \cos x-\sin x \right |$$ then $$\displaystyle f'\left ( \dfrac{\pi}4 \right )$$ is equal to-

A
$$\displaystyle \sqrt{2}$$

B
$$\displaystyle -\sqrt{2}$$

C
$$0$$

D
$$Does\ not\ exist$$

Solution

$$f\left( x \right) =\left| \cos { x } -\sin { x } \right| $$
$$\displaystyle =\begin{cases} \cos { x-\sin { x } \quad \quad x<\dfrac { \pi }{ 4 } } \\ -\cos { x+\sin { x } } \quad x>\dfrac { \pi }{ 4 } \end{cases}$$
$$\displaystyle f'\left( x \right) =\begin{cases} -\sin { x } -\cos { x } \quad x<\dfrac { \pi }{ 4 } \\ +\sin { x+\cos { x } \quad x>\dfrac { \pi }{ 4 } } \end{cases}$$
Hence $$\displaystyle f'\left( \dfrac { \pi }{ 4 } \right) $$ does not exists.
Question 9
1 / -0

If $$\displaystyle y=\frac{1}{1+x^{\beta -\alpha}+x^{\gamma -\alpha}}+\frac{1}{1+x^{\alpha-\beta}+x^{\gamma -\beta }}+\frac{1}{1+x^{\alpha -\gamma }+x^{\beta-\gamma }}$$
then $$\displaystyle \frac{dy}{dx}$$ is equal to-

A
$$0$$

B
$$1$$

C
$$\displaystyle

(a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$$

D
None of these

Solution

$$\displaystyle y=\frac{1}{1+x^{\beta -\alpha}+x^{\gamma -\alpha}}+\frac{1}{1+x^{\alpha-\beta}+x^{\gamma -\beta }}+\frac{1}{1+x^{\alpha -\gamma }+x^{\beta-\gamma }}$$
$$\displaystyle y=\frac{1}{\displaystyle 1+\frac{x^{\beta }}{x^{\alpha}}+\frac{x^{\gamma }}{x^{\alpha} }}+\frac{1}{{\displaystyle 1+\frac{x^{\alpha }}{x^{\beta}}+\frac{x^{\gamma }}{x^{\beta} }}}+\frac{1}{\displaystyle 1+\frac{x^{\alpha}}{x^{\gamma} }+\frac{x^{\beta} }{x^{\gamma }}}$$
$$\displaystyle y=\frac{x^{\alpha }+x^{b}+x^{\gamma }}{x^{\alpha }+x^{\beta }+x^{\gamma }}=1$$
$$\therefore \cfrac{dy}{dx}=0$$
Question 10
1 / -0

If $${ S }_{ n }$$ denotes the sum of $$n$$ terms of $$g.p$$. whose common ratio is $$r$$, then $$\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr } $$ is equal to

A
$$\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }$$

B
$$\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }$$

C
$$\left( n-1 \right) { S }_{ n }$$

D
None of these

Solution

We have, $$\displaystyle {S}_{n}=\frac { a\left( { r }^{ n }+1 \right) }{ r-1 } $$
$$\Rightarrow \left( r-1 \right) { S }_{ n }={ ar }^{ n }-a$$
Differentiating both sides with respect to $$r,$$ we get
$$\displaystyle \left( r-1 \right) \frac { { dS }_{ n } }{ dr } +{ S }_{ n }=na{ r }^{ n-1 }-0$$
$$\displaystyle \Rightarrow \left( r-1 \right) \frac { { dS }_{ n } }{ dr } =na{ r }^{ n-1 }-{ S }_{ n }$$
$$=n[n$$th term from of $$G.P.]$$ $$-{S}_{n}$$
$$=n\left( { S }_{ n }-{ S }_{ n-1 } \right) -{ S }_{ n }$$
$$=\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }.$$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 20

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Limits and Derivatives Test - 20

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SHARING IS CARING

Question 1

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ may not exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

Solution

Question 2

0

$$\displaystyle\frac{1}{2}$$

1

2

Solution

Question 3

$$\,\,\displaystyle\frac{3}{4}$$

$$\,\,ln2$$

$$\,\,\displaystyle\frac{ln2}{4}$$

$$\,\,3/2$$

Solution

Question 4

$$\displaystyle\frac{a^2 + b^2 - c^2}{2bc}$$

$$\displaystyle\frac{c^2 + a^2 - b^2}{2bc}$$

$$\displaystyle\frac{b^2 + c^2 - a^2}{2bc}$$

independent of a, b and c

Solution

Question 5

$$2$$

$$4$$

$$8$$

None of these

Solution

Question 6

$$\displaystyle\frac{1}{e}$$

$$\displaystyle\frac{1}{t}$$

$$\displaystyle\frac{1}{n}$$

$$none\ of\ above$$

Solution

Question 7

$$\displaystyle \frac{a}{1+x^{2}}$$

$$\displaystyle \frac{1}{1+x^{2}}$$

$$\displaystyle- \frac{a}{1+x^{2}}$$

$$\displaystyle -\frac{1}{1+x^{2}}$$

Solution

Question 8

$$\displaystyle \sqrt{2}$$

$$\displaystyle -\sqrt{2}$$

$$0$$

$$Does\ not\ exist$$

Solution

Question 9

$$0$$

$$1$$

$$\displaystyle (a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$$

None of these

Solution

Question 10

$$\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }$$

$$\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }$$

$$\left( n-1 \right) { S }_{ n }$$

None of these

Solution

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$$\displaystyle

(a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$$