Differentiation Test 18

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Differentiation Test 18

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle y^{x}=x^{\sin y} $$, find $$\cfrac{dy}{dx}$$.

A
$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

B
$$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$$

C
$$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

D
$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$$

Solution

Given $$\displaystyle y^{x}\:=x^{\sin \:y}$$.

Take log both sides.

$$\displaystyle x \log y = \sin y \: \log \:x.$$

Differentiate w.r.t. x

$$\displaystyle \log y+x.\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \sin y+ (\log x)\cos y .\frac{dy}{dx}$$

$$\displaystyle \left ( \log y - \frac{\sin y}{x} \right )=\frac{dy}{dx}\left [ \cos y \log x-\frac{x}{y} \right ]$$

$$\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}\left [ \frac{x \log y-\sin y}{y \log x \cos y-x}\right ]$$
Question 2
1 / -0

The numbers 1, 2, ..., 100 are arranged in the squares of an table in the following way: the numbers 1, ... , 10 are in the bottom row in increasing order, numbers 11, ... ,20 are in the next row in increasing order, and so on. One can choose any number and two of its neighbors in two opposite directions (horizontal, vertical, or diagonal). Then either the number is increased by 2 and its neighbors are decreased by 1, or the number is decreased by 2 and its neighbors are increased by 1. After several such operations the table again contains all the numbers 1, 2, ... , 100. Prove that they are in the original order.

A
$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 appear in their original order.

B
$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 appear in their original order.

C
$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

D
$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

Solution

Label the table entry in the i th row and j th column by $$a_{ ij }$$
where the bottom-left corner is in the first row and first column.
Let $$\displaystyle b_{ij}= 10(i-1)+j$$ be the number originally in the i th row and j th column. Observe that $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
is invariant. Indeed, every time entries
$$\displaystyle a_{mn}, a_{pq}, a_{rs}$$ are changed (with m+r= 2p and n+s= 2q), P increases or decreases by $$\displaystyle b_{mn}-2b_{pq}+b_{rs},$$
But this equals $$\displaystyle 10\left ( \left ( m-1 \right )+ \left ( r-1 \right )-2\left ( p-1 \right )+\left ( n+s-2q \right )\right )= 0.$$
In the beginning $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$ at the end, the entires $$a_{ij}$$ equal the $$b_{ij}$$
In some order,we now have $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
By the rearrangement inequality, this is at least $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
with equality only when each $$a_{ij}=b_{ij}$$
The equality does occur since P is invariant. Therefore the $$a_{ij}$$ do indeed equal the $$b_{ij}$$
in the same order, and thus the entries $$1, 2, ... , 100$$ appear in their original order.
Question 3
1 / -0

$$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$$.Find dy/dx

A
$$\sin x(\cot x)^{ \sin x-1 }(-cosec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(-sinx) } $$

B
$$\sin x(\cot x)^{ \sin x-1 }(-co\sec ^{ 2 } x)+\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } $$

C
$$\sin x(\cot x)^{ \sin x-1 }(-sec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x+1 } }co\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(sinx) } $$

D
None of these

Solution

Given $$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$$

Let $$u=(\cot x)^{\sin x}$$

Taking log on both sides

$$\log u=\sin x \log(\cot x)$$

Differentiating w.r.t. x, we get,

$$\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{\sin x}{\cot x}(-cosec^2 x)+\cos x \log \cot x$$

$$\Rightarrow \dfrac{du}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x$$

Now, $$v=(\tan x)^{\cos x}$$

Taking log on both sides

$$\log v=\cos x \log(\tan x)$$

Differentiating w.r.t. x, we get,

$$\dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{\cos x}{\tan x}(\sec^2 x)-\sin x \log \tan x$$

$$\Rightarrow \dfrac{dv}{dx}=\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$$

Since, $$y=u+v$$

$$\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$$

$$\Rightarrow \dfrac{dy}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x+$$
$$ \cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$$
Question 4
1 / -0

$$\displaystyle \frac{d}{dx}(\log_{e}\left ( \frac{1+x}{1-x} \right )^{1/4}-\frac{1}{2}\tan^{-1}x.)$$

A
$$\displaystyle \frac{x^{2}}{1-x^{4}}.$$

B
$$\displaystyle \frac{x^{3}}{1-x^{4}}.$$

C
$$\displaystyle \frac{x^{4}}{1-x^{4}}.$$

D
$$-\displaystyle \frac{x^{2}}{1-x^{4}}.$$

Solution

Let $$\displaystyle y=\frac{1}{4}\left [ \log\left ( 1+x \right )-\log\left ( 1-x \right ) \right ]-\frac{1}{2}\tan^{-1}x.$$
$$\Rightarrow \displaystyle

\frac{dy}{dx}=\frac{1}{4}\left [ \frac{1}{1+x}-\frac{1}{1-x}\left ( -1

\right ) \right ]-\frac{1}{2}.\frac{1}{1+x^{2}}$$
$$\displaystyle =\frac{1}{2\left ( 1-x^{2} \right )}-\frac{1}{2\left ( 1+x^{2} \right )}=\frac{x^{2}}{1-x^{4}}.$$
Question 5
1 / -0

$$\displaystyle \frac{d}{dx}\tan ^{-1}\left(\frac{a \cos x-b\sin x}{b\cos x+a\sin x}\right)$$

A
$$-1$$

B
$$-2$$

C
$$1$$

D
$$2$$

Solution

Put $$\displaystyle a= r\cos \alpha , b= r\sin \alpha $$

$$\displaystyle \therefore r= \sqrt{a^{2}+b^{2}},\tan \alpha = \dfrac{b}{a}$$

$$\displaystyle \therefore y= \tan ^{-1}\frac{r\cos\left ( x+\alpha \right )}{r\sin\left ( x+\alpha \right )}=\tan^{-1}\cot \left ( x+\alpha \right )$$

$$\displaystyle =\tan^{-1}\tan\left [ \dfrac{\pi}{2} -\left ( x+\alpha \right )\right ]$$

$$\displaystyle \therefore y=\dfrac{\pi} {2}-x-\alpha $$

$$\therefore \dfrac{dy}{dx}=-1$$
Question 6
1 / -0

$$\displaystyle \frac{d}{dx}(\tan ^{-1}\frac{\sin x+\cos x}{\cos x-\sin x})$$

A
$$-1$$

B
$$-2$$

C
$$1$$

D
$$2$$

Solution

Let $$\displaystyle y = \tan ^{-1}\frac{\cos x+\sin x}{\cos x-\sin x}=\tan ^{-1}\frac{1+\tan x}{1-\tan x}$$

$$\displaystyle \Rightarrow y =\tan ^{-1}\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}=\tan^{-1}\tan(\frac{\pi}{4}+x)=\frac{\pi}{4}+x$$

$$\therefore \cfrac{dy}{dx}=1$$
Question 7
1 / -0

If $$5f(x)+3f\left ( \displaystyle \frac{1}{x} \right )=x+2$$ and $$y=xf(x)$$ then $$\left (\displaystyle \frac{dy}{dx} \right )_{x=1}$$ is equal to ?

A
$$14$$

B
$$\displaystyle \frac{7}{8}$$

C
$$1$$

D
none of these

Solution

$$5f(x)+3f\left ( \displaystyle \frac{1}{x} \right )=x+2\Rightarrow (1)$$
Replace $$x$$ by $$\dfrac{1}{x}$$
$$\Rightarrow 5f\left ( \displaystyle \frac{1}{x} \right )+3f(x)=\dfrac{1}{x}+2\Rightarrow (2)$$
$$\Rightarrow 5\times(1)-3\times (2)\Rightarrow 16f(x) =5x-\dfrac{3}{x}+4=\dfrac{5x^2+4x-3}{x}$$
$$\therefore y = xf(x) =\dfrac{5x^2+4x-3}{16}$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{10x+4}{16}$$
$$\therefore \left(\dfrac{dy}{dx}\right)_{x=1}=\dfrac{10+4}{16}=\dfrac{7}{8}$$
Question 8
1 / -0

If $$\displaystyle f\left( x \right) =\sqrt { 1+\sqrt { x } } , x > 0,$$ then $$\displaystyle f\left ( x \right )\cdot f'\left ( x \right )$$ is equal to

A
$$\displaystyle \frac{1}{2\sqrt{x}}$$

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

C
$$\displaystyle \frac{1}{4\sqrt{x}}$$

D
$$\displaystyle \frac{2\sqrt{x}+1}{4\sqrt{x}}$$

Solution

Let $$y=f\left( x \right) =\sqrt { 1+\sqrt { x } } $$

Then, $${ y }^{ 2 }=1+\sqrt { x } $$

Differentiating w.r.t. x, we get,

$$\displaystyle 2y\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$$

$$\Rightarrow f(x)f'(x)=\displaystyle \frac{1}{4\sqrt{x}}$$
Question 9
1 / -0

Rajdhani Express going from Bombay to Delhi stops at five inter-mediate stations, $$10$$ passengers enter the train during the journey with $$10$$ different ticket of two classes. The number of different sets of tickets they may have is

A
$$^{15}C_{10}$$

B
$$^{20}C_{10}$$

C
$$^{30}C_{10}$$

D
None of these

Solution

For a particular class, the total number of different tickets from first intermediate station is $$5.$$
Similarly, number of different tickets from second intermediate station is $$4.$$
So the total number of different tickets is $$5+4+3+2+1=15$$.
And same number of tickets for another class is equal to total number of different tickets,
which is equal to $$30$$ and number of selection is $$^{30}C_{10}$$.
Question 10
1 / -0

$$\displaystyle \dfrac{d}{dx}\tan ^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)$$

A
$$-\displaystyle \dfrac{1}{2}$$

B
$$-\displaystyle \dfrac{1}{4}$$

C
$$-\displaystyle \dfrac{1}{8}$$

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

Solution

Let $$\displaystyle y = \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$$

$$\Rightarrow \displaystyle y = \tan ^{-1}\left(\dfrac{(\cos^2 \dfrac{x}{2}-\sin^2\dfrac{x}{2})}{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin \dfrac{x}{2}.\cos\dfrac{x}{2}}\right)$$

$$\displaystyle y=\tan ^{-1}\left(\dfrac{\cos\left ( x/2 \right )-\sin \left ( x/2 \right )}{\cos \left ( x/2 \right )+\sin \left ( x/2 \right )}\right)$$

$$\displaystyle =\tan^{-1}\left(\frac{1-\tan\left ( x/2 \right )}{1+\tan \left ( x/2 \right )}\right)$$

$$\displaystyle =\tan^{-1}\left(\tan \left ( \pi /4-x/2 \right )\right)=\pi/4-x/2$$

$$\displaystyle \therefore \dfrac{dy}{dx}=-\dfrac{1}{2}$$

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Differentiation Test 18

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Differentiation Test 18

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SHARING IS CARING

Question 1

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$$

$$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$$

Solution

Question 2

$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 appear in their original order.

$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 appear in their original order.

$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

Solution

Question 3

$$\sin x(\cot x)^{ \sin x-1 }(-cosec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(-sinx) } $$

$$\sin x(\cot x)^{ \sin x-1 }(-co\sec ^{ 2 } x)+\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } $$

$$\sin x(\cot x)^{ \sin x-1 }(-sec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x+1 } }co\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(sinx) } $$

None of these

Solution

Question 4

$$\displaystyle \frac{x^{2}}{1-x^{4}}.$$

$$\displaystyle \frac{x^{3}}{1-x^{4}}.$$

$$\displaystyle \frac{x^{4}}{1-x^{4}}.$$

$$-\displaystyle \frac{x^{2}}{1-x^{4}}.$$

Solution

Question 5

$$-1$$

$$-2$$

$$1$$

$$2$$

Solution

Question 6

$$-1$$

$$-2$$

$$1$$

$$2$$

Solution

Question 7

$$14$$

$$\displaystyle \frac{7}{8}$$

$$1$$

none of these

Solution

Question 8

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

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

$$\displaystyle \frac{1}{4\sqrt{x}}$$

$$\displaystyle \frac{2\sqrt{x}+1}{4\sqrt{x}}$$

Solution

Question 9

$$^{15}C_{10}$$

$$^{20}C_{10}$$

$$^{30}C_{10}$$

None of these

Solution

Question 10

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

$$-\displaystyle \dfrac{1}{4}$$

$$-\displaystyle \dfrac{1}{8}$$

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

Solution

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