Limits and Derivatives Test

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Limits and Derivatives Test - 17

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Question 1
1 / -0

lf $$[\mathrm{x}]$$ denotes the greatest integer contained in $$\mathrm{x},$$ then for 4 $$<\mathrm{x}<5,\ \displaystyle \frac{d}{dx}\{[x]\}=$$

A
$$[x - 4, 5]$$

B
$$[x]$$

C
$$0$$

D
$$1$$

Solution

$$[\ ]$$ denote greatest integer function ,i.e $$[n+f]=n$$,
where $$n=integer\ $$and $$f= fraction$$,
If $$4<x<5$$ , then $$[x] =4 =$$ constant
Hence $$\dfrac{d}{dx}\{[x]\} =\dfrac{d}{dx}(4)=0$$
Question 2
1 / -0

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

A
$$-3$$

B
$$3$$

C
$$0$$

D
$$-1$$

Solution

Given that $$y=3\cos x$$
$$ \Rightarrow \dfrac { dy }{ dx } =-3\sin x$$ at $$ x=\dfrac { \pi }{ 2 } $$
$$\Rightarrow \dfrac { dy }{ dx } =-\sin\dfrac { \pi }{ 2 } $$
$$ =-3$$
Question 3
1 / -0

lf $$f(x)=\left\{\begin{matrix}\displaystyle \frac{1-\cos x}{x} &x\neq 0 \\ 0 & x=0\end{matrix}\right.$$, then $$f^{'}(0)=$$

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

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

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

D
Does not exist

Solution

Using fundamental theorem,
$$\displaystyle f'(0) =\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$

$$=\displaystyle \lim_{h\to 0}\frac{\dfrac{1-\cos h}{h}-0}{h} $$

$$=\displaystyle \lim_{h\to 0}\frac{1-\cos h}{h^2}$$

$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{1}{2}.\frac{\sin^2 \frac{h}{2}}{\frac{h^2}{4}}= \dfrac{1}{2}$$
Question 4
1 / -0

If $$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$$ is non-zero finite, then $$n$$ must be equal

A
4

B
1

C
2

D
3

Solution

$$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$$

$$\mbox{For n = 0,we have }\displaystyle\lim_{x\to0}\displaystyle\frac{1-1}{x-1}=0$$

$$\mbox{For n = 1,}\displaystyle\lim_{x\to0}\displaystyle\frac{x-\sin x}{x- \sin x}=1\\$$

$$\mbox{For n = 2,}\displaystyle\lim_{x\to0}\displaystyle\frac{x^2-\sin^2x}{x-\sin^2x}=\displaystyle\lim_{x\to0}\displaystyle\frac{1-\displaystyle\frac{\sin^2x}{x^2}}{\displaystyle\frac{1}{x}-\displaystyle\frac{\sin^2x}{x^2}}.\\$$
$$\mbox{This does not exist.}\\ $$

$$\displaystyle { For\ \ n=3, }\lim _{ x\to 0 } \dfrac { x^{ 3 }-\sin ^{ 3 } x }{ x-\sin ^{ 3 } x } =\lim _{ x\to 0 } \dfrac { 1-\dfrac { \sin ^{ 3 } x }{ x^{ 3 } } }{ \dfrac { 1 }{ x^{ 2 } } -\dfrac { \sin ^{ 3 } x }{ x^{ 3 } } } $$
For $$n = 3$$, also given limit does not exist.
Hence, $$n = 1$$
Question 5
1 / -0

If $$y=x^{-\tfrac12}+\log_5x+\displaystyle \frac {\sin x}{\cos x}+2^x$$, then find $$\dfrac {dy}{dx}$$

A
$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

B
$$\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

C
$$-\displaystyle \frac {3}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

D
$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\cos^2x+2^x\log 2$$

Solution

Here, we have function $$y=x^{-1/2}+\log _5x+\tan x+2^x$$
On differentiating w.r.t x, we get
$$\dfrac {dy}{dx}=\dfrac {d}{dx}(x)^{-1/2}+\dfrac {d}{dx}(\log _5x)+\dfrac {d}{dx} \tan x+\dfrac {d}{dx}(2^x)$$

$$=-\dfrac {1}{2}(x)^{-1/2-1}+\dfrac {1}{x \log _e5}+\sec ^2x+2^x \log 2$$

$$=-\dfrac {1}{2}x^{-3/2}+\dfrac {1}{x\log _e5}+\sec ^2x+2^x\log 2$$
Question 6
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 7
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 8
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 9
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

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

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 17

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Limits and Derivatives Test - 17

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SHARING IS CARING

Question 1

$$[x - 4, 5]$$

$$[x]$$

$$0$$

$$1$$

Solution

Question 2

$$-3$$

$$3$$

$$0$$

$$-1$$

Solution

Question 3

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

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

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

Does not exist

Solution

Question 4

4

1

2

3

Solution

Question 5

$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

$$\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

$$-\displaystyle \frac {3}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\cos^2x+2^x\log 2$$

Solution

Question 6

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

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

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

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

Solution

Question 7

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

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

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

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

Solution

Question 8

$$0$$

$$1$$

$$-1$$

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

Solution

Question 9

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

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

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

None of these

Solution

Question 10

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

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

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

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

Solution

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