Differentiation Test 16

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Differentiation Test 16

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Question 1
1 / -0

In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential.

A
True

B
False

C
Either

D
Neither

Solution

Combination of few things .
The order of selected things is not specified .
Combination is basically selection hence ordering in immaterial .
Permtation of few things the order of selected things is specified .
Permutation basically means selecting and arranging .
Hence ordering is essential .
Question 2
1 / -0

If $$y={(\tan{x})}^{\displaystyle{(\tan{x})}^{\displaystyle\tan{x}}}$$, then find $$\displaystyle\frac{dy}{dx}$$ at $$\displaystyle x=\frac{\pi}{4}$$.

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

Taking $$\log$$ on both sides, we get
$$\log{y}={(\tan{x})}^{\displaystyle\tan{x}}\log{\tan{x}}$$
Again taking $$\log$$ on both the sides
$$\log{\log{y}}=[\tan{x}\log{\tan{x}}]+\log{\log{\tan{x}}}$$
Differentiating w.r.t. $$x$$, we get
$$\displaystyle\frac{1}{y\log{y}}\frac{dy}{dx}=\log { \tan { x } \frac { d }{ dx } \left( \tan { x } \right) +\tan { x } \frac { d }{ dx } \left( \log { \tan { x } } \right) +\frac { d }{ dx } \left( \log { \log { \tan { x } } } \right) } $$

$$\displaystyle \frac{dy}{dx}=\log { \tan { x } .\sec ^{ 2 }{ x } +\tan { x } .\frac { \sec ^{ 2 }{ x } }{ \tan { x } } +\frac { \sec ^{ 2 }{ x } }{ \tan { x } .\log { \tan { x } } } } $$
$$\displaystyle \frac { dy }{ dx } =y\log { y.\sec ^{ 2 }{ x } \left[ \log { \tan { x } +1+\frac { 1 }{ \tan { x } .\log { \tan { x } } } } \right] } $$
At $$\displaystyle x=\frac{\pi}{4}$$, $$y=1$$ and $$\log{y}=0$$
So, putting this values in the equation of $$\displaystyle \frac { dy }{ dx } $$, we get
$$\displaystyle\therefore{\left(\frac{dy}{dx}\right)}_{\displaystyle x=\frac{\pi}{4}}=0$$
Question 3
1 / -0

There are $$6$$ boxes numbered $$1, 2 ....... 6$$. Each box is to be filled up either with a red or a green ball in such a way that at least $$1$$ box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

A
$$5$$

B
$$21$$

C
$$33$$

D
$$60$$

Solution

(B) The number of ways in which 1 green ball can be put $$=6$$ . The number of ways in which two green balls can be put such that the boxes are consecutive
$$=5$$ $$(i.e., (1, 2),(2, 3),(3, 4),(4, 5),(5, 6))$$

Similarly, the number of ways in which three green balls can be put
$$=4( i.e. (1, 2, 3),(2, 3, 4),(3, 4, 5),(4, 5, 6))$$
$$\cdots \cdots \cdots \cdots \cdots $$ and so on.
$$\therefore $$ Total number of ways of doing this
$$=6+5+4+3+2+1=21$$
Question 4
1 / -0

If $$\displaystyle y=5^{3-x^2}+(3-x^2)^5$$, then $$\displaystyle \frac{dy}{dx}=$$

A
$$-2x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$$

B
$$-x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$$

C
$$-2x\left \{5^{3-x^2}\cdot \log_e5+(3-x^2)^4\right \}$$

D
$$-2x\left \{5^{3-x^2}+5(3-x^2)^4\right \}$$

Solution

$$\displaystyle \frac { d }{ dx } \left( 5^{ 3-x^{ 2 } }+(3-x^{ 2 })^{ 5 } \right) $$
$$={ 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5\displaystyle \frac { d }{ dx } \left( 3-{ x }^{ 2 } \right) +5{ \left( 3-{ x }^{ 2 } \right) }^{ 4 }\displaystyle \frac { d }{ dx } { \left( 3-{ x }^{ 2 } \right) } } $$
$$={ 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5\left( -2x \right) } +5{ \left( 3-{ x }^{ 2 } \right) }^{ 4 }\left( -2x \right) $$
$$=-2x\left( { 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5+5\left( 3-{ x }^{ 2 } \right) } \right) $$
Question 5
1 / -0

If $$y=\log_{3}x+3 \log_{e} x+2 \tan x$$, then $$\displaystyle \frac{dy}{dx}=$$

A
$$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$$

B
$$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+ \sec^2 x$$

C
$$\displaystyle \frac {1}{\log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$$

D
$$\displaystyle \frac {1}{x \log_e 3}-\displaystyle \frac {3}{x}+2 \sec^2 x$$

Solution

$$\displaystyle \frac { d }{ dx } \left( \log _{ 3 }{ x } +3\log _{ e }{ x } +2\tan { x } \right) $$

$$=\displaystyle \frac { d }{ dx } \left( \displaystyle \frac { \log_e { x } }{ \log { 3 } } +3\log_e { x } +2\tan { x } \right) $$

$$=\displaystyle \frac { 1 }{ x\log _{ e }{ 3 } } +\displaystyle \frac { 3 }{ x } +2\sec ^{ 2 }{ x } $$
Question 6
1 / -0

If $$\displaystyle y=e^{x \log a}+e^{a \log x}+e^{a \log a}$$, then $$\displaystyle \frac{dy}{dx}=$$

A
$$a^x \log a+x^{a-1}$$

B
$$a^x \log a+ax$$

C
$$a^x \log a+ax^{a-1}$$

D
$$a^x \log a+ax^{a}$$

Solution

Let $$y=\displaystyle e^{x log a}+e^{a log x}+e^{a log a}$$
$$={ e }^{ \log _{ e }{ { a }^{ x } } }+{ e }^{ \log _{ e }{ { x }^{ a } } }+{ e }^{ \log _{ e }{ { a }^{ a } } }$$
$$={ a }^{ x }+{ x }^{ a }+{ a }^{ a }$$ ......... Since $${ e }^{ \log _{ e }{ x } }=x$$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left( { a }^{ x }+{ x }^{ a }+{ a }^{ a } \right) $$
$$={ a }^{ x }\log { a } +a.{ x }^{ a-1 }$$
Question 7
1 / -0

If $$f'(x)=\sin x+\sin 4x\cdot \cos x$$, then $$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$$ is

A
$$4x\left \{\cos(2x^2)-sin 8x^2\cdot \sin 2x^2\right \}$$

B
$$4x\left \{\cos(2x^2)+\sin 8x^2\cdot \sin 2x^2\right \}$$

C
$$\left \{\cos (2x^2)-\sin 8x\cdot \sin 2x^2\right \}$$

D
none of the above

Solution

$$f'(x)=\sin x+\sin 4x\cdot \cos x$$
$$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$$=$$\displaystyle \frac { d }{ dx } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) \left[ \sin { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) +\sin { 4 } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) .\cos { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) } } \right] $$
$$=4x\left[ \cos { 2{ x }^{ 2 }-\sin { 8 } { x }^{ 2 }.\sin { 2{ x }^{ 2 } } } \right] $$
Question 8
1 / -0

If $$y=|\cos x|+|\sin x|$$, then $$\displaystyle \dfrac {dy}{dx}$$ at $$x=\dfrac {2\pi}{3}$$ is

A
$$\displaystyle \dfrac {1}{2}(\sqrt 3+1)$$

B
$$2(\sqrt 3-1)$$

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

D
none of these

Solution

$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left( |cosx|+|sinx| \right) $$
$$=-\sin { x } +\cos { x } $$ at $$x=\displaystyle \frac { 2\pi }{ 3 }$$
$$=\left| -\sin { \displaystyle \frac { 2\pi }{ 3 } } \right| +\left| \cos { \displaystyle \frac { 2\pi }{ 3 } } \right| $$
$$=\displaystyle \frac { \sqrt { 3 } }{ 2 } -\displaystyle \frac { 1 }{ 2 } $$
$$=\displaystyle \frac { 1 }{ 2 } \left( \sqrt { 3 } -1 \right) $$
Question 9
1 / -0

Find the derivative of $$\sec^{-1}\left (\displaystyle \frac {x+1}{x-1}\right )+\sin^{-1}\left (\displaystyle \frac {x-1}{x+1}\right )$$

A
$$0$$

B
$$1$$

C
$$-1$$

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

Solution

Let $$y=\sec^{-1}\left (\displaystyle \frac {x+1}{x-1}\right )+\sin^{-1}\left (\displaystyle \frac {x-1}{x+1}\right )$$
$$y=\cos ^{ -1 }{ \left( \displaystyle \frac { x-1 }{ x+1 } \right) +\sin ^{ -1 }{ \left( \displaystyle \frac { x-1 }{ x+1 } \right) } } =\displaystyle \frac { \pi }{ 2 } $$
$$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$$
$$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } } $$
Question 10
1 / -0

If $$y=\log_{10}x+\log_x 10+\log_xx+\log_{10} 10$$, then $$\displaystyle \frac{dy}{dx}=$$

A
$$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$$

B
$$\displaystyle \frac {1}{\log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$$

C
$$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x^2(\log_ex)^2}$$

D
None of these

Solution

$$\displaystyle \dfrac { d }{ dx } \left( \log _{ 10 }{ x+\log _{ x }{ 10+\log _{ x }{ x+\log _{ 10 }{ 10 } } } } \right) $$
$$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } +\displaystyle \dfrac { \left( -\log _{ e }{ 10\left( \displaystyle \dfrac { 1 }{ x } \right) } \right) }{ { \left( \log_e { x } \right) }^{ 2 } } $$
$$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } -\displaystyle \dfrac { \log _{ e }{ 10 } }{ x{ \left( \log_e { x } \right) }^{ 2 } } $$

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Re-Attempt Weekly Quiz Competition

Differentiation Test 16

Result

Differentiation Test 16

Score

-

Rank

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SHARING IS CARING

Question 1

True

False

Either

Neither

Solution

Question 2

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 3

$$5$$

$$21$$

$$33$$

$$60$$

Solution

Question 4

$$-2x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$$

$$-x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$$

$$-2x\left \{5^{3-x^2}\cdot \log_e5+(3-x^2)^4\right \}$$

$$-2x\left \{5^{3-x^2}+5(3-x^2)^4\right \}$$

Solution

Question 5

$$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$$

$$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+ \sec^2 x$$

$$\displaystyle \frac {1}{\log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$$

$$\displaystyle \frac {1}{x \log_e 3}-\displaystyle \frac {3}{x}+2 \sec^2 x$$

Solution

Question 6

$$a^x \log a+x^{a-1}$$

$$a^x \log a+ax$$

$$a^x \log a+ax^{a-1}$$

$$a^x \log a+ax^{a}$$

Solution

Question 7

$$4x\left \{\cos(2x^2)-sin 8x^2\cdot \sin 2x^2\right \}$$

$$4x\left \{\cos(2x^2)+\sin 8x^2\cdot \sin 2x^2\right \}$$

$$\left \{\cos (2x^2)-\sin 8x\cdot \sin 2x^2\right \}$$

none of the above

Solution

Question 8

$$\displaystyle \dfrac {1}{2}(\sqrt 3+1)$$

$$2(\sqrt 3-1)$$

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

none of these

Solution

Question 9

$$0$$

$$1$$

$$-1$$

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

Solution

Question 10

$$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$$

$$\displaystyle \frac {1}{\log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$$

$$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x^2(\log_ex)^2}$$

None of these

Solution

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