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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 4 - MCQExams.com
CBSE
Class 11 Engineering Maths
Limits And Derivatives
Quiz 4
lf $$[\mathrm{x}]$$ denotes the greatest integer contained in $$\mathrm{x},$$ then for 4 $$<\mathrm{x}<5,\ \displaystyle \frac{d}{dx}\{[x]\}=$$
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$$[x - 4, 5]$$
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$$[x]$$
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$$0$$
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$$1$$
Explanation
$$[\ ]$$ denote greatest integer function ,i.e $$[n+f]=n$$,
where $$n=integer\ $$and $$f= fraction$$,
If $$4<x<5$$ , then $$[x] =4 =$$ constant
Hence $$\dfrac{d}{dx}\{[x]\} =\dfrac{d}{dx}(4)=0$$
For the function $$f(x) = \displaystyle \frac{x^{100}}{100} + \frac{x^{99}}{99} + ........... + \frac{x^2}{2} + x+1$$, $$f'(1) =$$
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$$x^{100}$$
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$$100$$
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$$101$$
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None of these
Explanation
$$Given:\cfrac { d }{ dx } (x^{ n })=nx^{ n-1 }\\ \therefore for\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...............+\cfrac { x^{ 2 } }{ 2 } +1\\ f'(x)=\cfrac { 100x^{ 99 } }{ 100 } +99\cfrac { x^{ 98 } }{ 99 } +.............+\cfrac { 2x }{ 2 } +1\\ Here\; f'(1)=1+1+.........to\; 100term=\; 100\\ Hence\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...........+\cfrac { x^{ 2 } }{ 2 } +x+1=100$$
lf $$f(x)=\left\{\begin{matrix}\displaystyle \frac{1-\cos x}{x} &x\neq 0 \\ 0 & x=0\end{matrix}\right.$$,
then $$f^{'}(0)=$$
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$$\dfrac{1}{2}$$
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$$\dfrac {1}{4}$$
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$$\dfrac{3}{4}$$
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Does not exist
Explanation
Using fundamental theorem,
$$\displaystyle f'(0) =\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$
$$=\displaystyle \lim_{h\to 0}\frac{\dfrac{1-\cos h}{h}-0}{h} $$
$$=\displaystyle \lim_{h\to 0}\frac{1-\cos h}{h^2}$$
$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{1}{2}.\frac{\sin^2 \frac{h}{2}}{\frac{h^2}{4}}= \dfrac{1}{2}$$
Derivative of which function is $$f'(x) = x \sin x$$?
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$$x \sin x + \cos x$$
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$$x \cos x+ \sin x$$
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$$x\sin \left ( \dfrac{\pi}{2}-x \right ) + \cos \left ( \dfrac{\pi}{2}-x \right )$$
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$$x \cos \left ( \dfrac{\pi}{2}-x \right ) + \sin \left ( \dfrac{\pi}{2}-x \right )$$
Explanation
We know $$\frac { d }{ dx } \left( f.g \right) =\frac { df }{ dx } .g+\frac { dg }{ dx } .f\\ \Rightarrow \frac { d }{ dx } \left( x.sin(x) \right) =\frac { d(x) }{ dx } .sin(x)+\frac { d(sin(x)) }{ dx } .x\\ \Rightarrow 1\times sin(x)+cos(x).x=x.cos(x)+sin(x)\\$$
We know that $$ cos(x)=sin\left( \frac { \pi }{ 2 } -x \right) sin(x)=cos\left( \frac { \pi }{ 2 } -x \right) \\\Rightarrow x.sin(\left( \frac { \pi }{ 2 } -x \right) +cos\left( \frac { \pi }{ 2 } -x \right) $$
Both B and C are correct
If $$y=x^{-\tfrac12}+\log_5x+\displaystyle \frac {\sin x}{\cos x}+2^x$$, then find $$\dfrac {dy}{dx}$$
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$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$
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$$\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$
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$$-\displaystyle \frac {3}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$
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$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\cos^2x+2^x\log 2$$
Explanation
Here, we have function $$y=x^{-1/2}+\log _5x+\tan x+2^x$$
On differentiating w.r.t x, we get
$$\dfrac {dy}{dx}=\dfrac {d}{dx}(x)^{-1/2}+\dfrac {d}{dx}(\log _5x)+\dfrac {d}{dx} \tan x+\dfrac {d}{dx}(2^x)$$
$$=-\dfrac {1}{2}(x)^{-1/2-1}+\dfrac {1}{x \log _e5}+\sec ^2x+2^x \log 2$$
$$=-\dfrac {1}{2}x^{-3/2}+\dfrac {1}{x\log _e5}+\sec ^2x+2^x\log 2$$
If $$\displaystyle y=5^{3-x^2}+(3-x^2)^5$$, then $$\displaystyle \frac{dy}{dx}=$$
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$$-2x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$$
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$$-x\left \{5^{3-x^2}\cdot \log_e5+5(3-x^2)^4\right \}$$
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$$-2x\left \{5^{3-x^2}\cdot \log_e5+(3-x^2)^4\right \}$$
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$$-2x\left \{5^{3-x^2}+5(3-x^2)^4\right \}$$
Explanation
$$\displaystyle \frac { d }{ dx } \left( 5^{ 3-x^{ 2 } }+(3-x^{ 2 })^{ 5 } \right) $$
$$={ 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5\displaystyle \frac { d }{ dx } \left( 3-{ x }^{ 2 } \right) +5{ \left( 3-{ x }^{ 2 } \right) }^{ 4 }\displaystyle \frac { d }{ dx } { \left( 3-{ x }^{ 2 } \right) } } $$
$$={ 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5\left( -2x \right) } +5{ \left( 3-{ x }^{ 2 } \right) }^{ 4 }\left( -2x \right) $$
$$=-2x\left( { 5 }^{ 3-{ x }^{ 2 } }\log _{ e }{ 5+5\left( 3-{ x }^{ 2 } \right) } \right) $$
If $$y=\log_{3}x+3 \log_{e} x+2 \tan x$$, then $$\displaystyle \frac{dy}{dx}=$$
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$$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$$
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$$\displaystyle \frac {1}{x \log_e 3}+\displaystyle \frac {3}{x}+ \sec^2 x$$
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$$\displaystyle \frac {1}{\log_e 3}+\displaystyle \frac {3}{x}+2 \sec^2 x$$
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$$\displaystyle \frac {1}{x \log_e 3}-\displaystyle \frac {3}{x}+2 \sec^2 x$$
Explanation
$$\displaystyle \frac { d }{ dx } \left( \log _{ 3 }{ x } +3\log _{ e }{ x } +2\tan { x } \right) $$
$$=\displaystyle \frac { d }{ dx } \left( \displaystyle \frac { \log_e { x } }{ \log { 3 } } +3\log_e { x } +2\tan { x } \right) $$
$$=\displaystyle \frac { 1 }{ x\log _{ e }{ 3 } } +\displaystyle \frac { 3 }{ x } +2\sec ^{ 2 }{ x } $$
If $$y=x^2+sin^{-1}x+log_ex$$, find $$\dfrac {dy}{dx}$$
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$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$
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$$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$
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$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$$
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$$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$
Explanation
$$y=x^2+sin^{-1}+log_ex$$
On differentiating, we get
$$\dfrac {dy}{dx}=\dfrac {d}{dx}(x^2)+\dfrac {d}{dx}sin^{-1}x)+\dfrac {d}{dx}(log_ex)$$
or $$\dfrac {dy}{dx}=2(x)^{2-1}+\dfrac {1}{\sqrt {1-x^2}}+\dfrac {d}{dx}(log_ex)$$
$$\dfrac {dy}{dx}=2x+\dfrac {1}{\sqrt {1-x^2}}+\dfrac {1}{x}$$
If $$\displaystyle y=e^{x \log a}+e^{a \log x}+e^{a \log a}$$, then $$\displaystyle \frac{dy}{dx}=$$
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$$a^x \log a+x^{a-1}$$
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$$a^x \log a+ax$$
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$$a^x \log a+ax^{a-1}$$
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$$a^x \log a+ax^{a}$$
Explanation
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 $${ e }^{ \log _{ e }{ x } }=x$$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left( { a }^{ x }+{ x }^{ a }+{ a }^{ a } \right) $$
$$={ a }^{ x }\log { a } +a.{ x }^{ a-1 }$$
If $$y=|\cos x|+|\sin x|$$, then $$\displaystyle \dfrac {dy}{dx}$$ at $$x=\dfrac {2\pi}{3}$$ is
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$$\displaystyle \dfrac {1}{2}(\sqrt 3+1)$$
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$$2(\sqrt 3-1)$$
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$$\displaystyle \dfrac {1}{2}(\sqrt 3-1)$$
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none of these
Explanation
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left( |cosx|+|sinx| \right) $$
$$=-\sin { x } +\cos { x } $$ at $$x=\displaystyle \frac { 2\pi }{ 3 }$$
$$=\left| -\sin { \displaystyle \frac { 2\pi }{ 3 } } \right| +\left| \cos { \displaystyle \frac { 2\pi }{ 3 } } \right| $$
$$=\displaystyle \frac { \sqrt { 3 } }{ 2 } -\displaystyle \frac { 1 }{ 2 } $$
$$=\displaystyle \frac { 1 }{ 2 } \left( \sqrt { 3 } -1 \right) $$
Find the derivative of $$\sec^{-1}\left (\displaystyle \frac {x+1}{x-1}\right )+\sin^{-1}\left (\displaystyle \frac {x-1}{x+1}\right )$$
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$$0$$
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$$1$$
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$$-1$$
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$$\displaystyle \frac{x+1}{x-1}$$
Explanation
Let $$y=\sec^{-1}\left (\displaystyle \frac {x+1}{x-1}\right )+\sin^{-1}\left (\displaystyle \frac {x-1}{x+1}\right )$$
$$y=\cos ^{ -1 }{ \left( \displaystyle \frac { x-1 }{ x+1 } \right) +\sin ^{ -1 }{ \left( \displaystyle \frac { x-1 }{ x+1 } \right) } } =\displaystyle \frac { \pi }{ 2 } $$
$$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$$
$$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } } $$
If $$y=\log_{10}x+\log_x 10+\log_xx+\log_{10} 10$$, then $$\displaystyle \frac{dy}{dx}=$$
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$$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$$
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$$\displaystyle \frac {1}{\log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}$$
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$$\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x^2(\log_ex)^2}$$
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None of these
Explanation
$$\displaystyle \dfrac { d }{ dx } \left( \log _{ 10 }{ x+\log _{ x }{ 10+\log _{ x }{ x+\log _{ 10 }{ 10 } } } } \right) $$
$$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } +\displaystyle \dfrac { \left( -\log _{ e }{ 10\left( \displaystyle \dfrac { 1 }{ x } \right) } \right) }{ { \left( \log_e { x } \right) }^{ 2 } } $$
$$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } -\displaystyle \dfrac { \log _{ e }{ 10 } }{ x{ \left( \log_e { x } \right) }^{ 2 } } $$
$$\displaystyle \lim_{x\to0}\left( x^{-3}\sin{3x} + ax^{-2} + b \right)$$ exists and is equal to 0, then
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$$a = -3$$ and $$b = \dfrac{9}{2}$$
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$$a = 3$$ and $$b = \dfrac{9}{2}$$
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$$a = -3$$ and $$b = -\dfrac{9}{2}$$
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$$a = 3$$ and $$b = -\dfrac{9}{2}$$
Explanation
$$\displaystyle\lim_{x\to\infty}\displaystyle\frac{\sin3x}{x^3}+\displaystyle\frac{a}{x^2}+b
= \lim_{x\to0}\displaystyle\frac{\sin3x+ax+bx^3}{x^3}$$
$$=\displaystyle\lim_{x\to0}\displaystyle\frac{3\displaystyle\frac{\sin3x}{3x}+a+bx^2}{x^2}$$
$$\mbox{For existence, }$$
$$(3+a)=0 \mbox{ or } a=-3$$
$$\therefore L=\displaystyle\lim_{x\to0}\displaystyle\frac{\sin3x-3x+bx^3}{x^3}$$
$$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\sin t - t}{t^3}+b = 0 ........ (3x=t ) $$
The left hand side reduces to $$\dfrac{0}{0}$$ form by substituting the limit
Then, using L'Hospital's rule
$$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\cos t - 1}{3t^2}+b = 0$$
Again, applying
L'Hospital's rule
$$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{-\sin t}{6t}+b = 0$$
$$\Rightarrow -\displaystyle\frac{27}{6}+b=0 \mbox{ or } b=\frac{9}{2}$$
If $$y=logx^3+3 sin^{-1}x+kx^2$$, then find $$\displaystyle \frac {dy}{dx}$$
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$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$
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$$3\cdot \displaystyle \frac {1}{x^3}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$
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$$3\cdot \displaystyle \frac {1}{x}-3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$
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$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+2x$$
Explanation
Here, $$y=\log x^3+3 \sin^{-1} x+kx^2$$
On differentiating we get
$$\displaystyle \dfrac {dy}{dx}=\dfrac {d}{dx}[\log x^3]+\dfrac {d}{dx}[3 \sin^{-1}x]+\dfrac {d}{dx}[kx^2]$$
$$=3\displaystyle \dfrac {d}{dx}[\log x]+3\dfrac {d}{dx}(\sin^{-1}x)+k\dfrac {d}{dx}(x^2)$$
$$=3\cdot \displaystyle \dfrac {1}{x}+3\cdot \dfrac {1}{\sqrt {1-x^2}}+k(2x)$$
The value of $$\displaystyle\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$$ for $$(a>1)$$ is equal to?
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$$1$$
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$$0$$
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$$\displaystyle\frac{\pi}{2}$$
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Does not exist
Explanation
$$\mathrm{L} \displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$$
$$\displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{\log{x}}{\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log a}{\log x}}}}=\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{1}{a\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log^2 a. x}{1}}}}=\frac{\cot^{-1}0}{\sec^{-1} \infty}=\frac{\pi/2}{\pi/2}=1$$
Note: L-Hospital's rule has been used in step two in both numerator and denominator alone.
The value of
$$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 \sin^2 x + \sin x-1}{2 \sin^2 x - 3 \sin x + 1} $$
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$$3$$
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$$-3$$
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$$6$$
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$$0$$
Explanation
We have
$$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 sin^2 x + sin x -1}{2 sin^2 x-3 sin x + 1}$$
$$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{(2 sin x - 1)(sin x + 1)}{(2 sin x - 1)(sin x - 1)}$$
$$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{sin x + 1}{sin x - 1} = - 3$$
If $$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) + \sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$$, then $$\displaystyle\frac{dy}{dx}$$ equals
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$$1$$
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$$0$$
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$$\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}$$
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$$\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}$$
Explanation
$$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) +
\sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$$
$$y=\cos ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) +\sin ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) } } =\displaystyle \frac { \pi }{ 2 } $$ ............. $$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } } $$
$$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$$
If $$2^x+2^y=2^{x+y}$$, then $$\displaystyle \frac {dy}{dx}$$ has the value equal to
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$$\displaystyle -\frac {2^y}{2^x}$$
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$$\displaystyle \frac {1}{1-2^x}$$
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$$\displaystyle 1-2^y$$
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$$\displaystyle \frac {2^x(1-2^y)}{2^y(2^x-1)}$$
Explanation
$$2^x+2^y=2^{x+y}$$
Differentiating both the sides
$$\Rightarrow \displaystyle \dfrac { d }{ dx } \left( 2^{ x }+2^{ y } \right) =\displaystyle \dfrac { d }{ dx } \left(2^{ x+y }\right)$$
$$\Rightarrow { 2 }^{ x }\log { 2+{ 2 }^{ y }\log { 2\displaystyle \dfrac { dy }{ dx } ={ 2 }^{ x+y }\log { 2 } \left( 1+\displaystyle \dfrac { dy }{ dx } \right) } } $$
$$\Rightarrow \log { 2\left( { 2 }^{ x }+{ 2 }^{ y }\displaystyle \dfrac { dy }{ dx } \right) =\log { 2 } \left( { 2 }^{ x+y }+{ 2 }^{ x+y }\displaystyle \dfrac { dy }{ dx } \right) } $$
$$\Rightarrow \left( { 2 }^{ y }-{ 2 }^{ x+y } \right)\displaystyle \dfrac { dy }{ dx } ={ 2 }^{ x+y }-{ 2 }^{ x }$$
$$\Rightarrow \displaystyle \dfrac { dy }{ dx } =\displaystyle \dfrac { { 2 }^{ x }{ 2 }^{ y }-{ 2 }^{ x } }{ { 2 }^{ y }-{ 2 }^{ x }{ 2 }^{ y } } =\displaystyle \dfrac { { 2 }^{ x }\left( { 2 }^{ y }-1 \right) }{ { 2 }^{ y }\left( 1-{ 2 }^{ x } \right) } $$
$$=\displaystyle \dfrac { { 2 }^{ x }\left( 1-{ 2 }^{ y } \right) }{ { 2 }^{ y }\left( { 2 }^{ x }-1 \right) } $$
If $$f'(x)=\sin x+\sin 4x\cdot \cos x$$, then $$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$$ is
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$$4x\left \{\cos(2x^2)-sin 8x^2\cdot \sin 2x^2\right \}$$
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$$4x\left \{\cos(2x^2)+\sin 8x^2\cdot \sin 2x^2\right \}$$
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$$\left \{\cos (2x^2)-\sin 8x\cdot \sin 2x^2\right \}$$
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none of the above
Explanation
$$f'(x)=\sin x+\sin 4x\cdot \cos x$$
$$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$$=$$\displaystyle \frac { d }{ dx } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) \left[ \sin { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) +\sin { 4 } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) .\cos { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) } } \right] $$
$$=4x\left[ \cos { 2{ x }^{ 2 }-\sin { 8 } { x }^{ 2 }.\sin { 2{ x }^{ 2 } } } \right] $$
The solution set of $${f}'(x)>{g}'(x)$$ where $$f(x)=\displaystyle \frac{1}{2}(5^{2x+1})$$ & $$g(x)= 5^x+4x(\ln 5)$$ is
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$$x>1$$
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$$0< x< 1$$
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$$x \leq 0$$
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$$x>0$$
Explanation
Given, $$f(x)= \dfrac {1}{2}(5^{2x+1})$$
$$f'(x)= 5^{2x+1}\times \log 5$$
$$g'(x)=5^{x}l\log 5 +4\ln 5=\ln 5(5^{x}+4)$$
$$f'(x)>g'(x)$$
$$=>5\times 5^{2x}>(5^{x}+4)$$
Let $$t=5^{x}$$
$$=> 5t^{2}-t-4>0$$
$$=>(t-1)(t+\frac{4}{5})>0$$$$=>t>1$$ or $$ t>-\frac{4}{5}$$
$$=>5^{x}>5^{0} =>x>0$$
so, $$x>0$$
$$f:R\rightarrow R$$ and $$\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}$$, then $$f(x)$$ is
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one-one ito
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many-one onto
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one-one onto
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many-one into
Explanation
$$\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}$$
$$\displaystyle \quad \quad =\frac {x(x^4+1)(x+1)+(x^4+1)+1}{x^2+x+1}$$
$$\displaystyle \quad \quad =\frac {(x^4+1)(x^2+x+1)+1}{x^2+x+1}$$
$$\displaystyle \quad \quad =(x^4+1)+\frac {1}{x^2+x+1}$$
$$\displaystyle f'(x) =4x^3-\frac {2x+1}{(x^2+x+1)^2}=$$ not always positive or negative
Thus, $$f$$ is many one.
Also range and co-domain of $$f$$ are not same,
Hence is many-one into function
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
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$$19$$
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$$9$$
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$$17$$
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$$14$$
Which one of the following statements is true?
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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.
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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.
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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.
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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.
Explanation
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.
If $$\displaystyle y=\frac { x }{ a+\displaystyle\frac { x }{ b+\displaystyle\frac { x }{ a+\displaystyle\frac { x }{ b+.....\infty } } } } $$, then $$\cfrac{dy}{dx} =$$
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$$\displaystyle\frac{a}{ab+2ay}$$
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$$\displaystyle\frac{b}{ab+2by}$$
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$$\displaystyle\frac{a}{ab+2by}$$
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$$\displaystyle\frac{b}{ab+2ay}$$
Explanation
$$\displaystyle y = \frac{x}{a+\frac{x}{b+y}}$$
$$\Rightarrow \displaystyle y =\frac{x(b+y)}{ab+ay+x}$$
$$\Rightarrow aby+ay^2+xy=bx+xy\Rightarrow aby+ay^2=bx$$
Differentiating both sides w.r.t $$x$$
$$\Rightarrow\displaystyle (ab+2ay)\frac{dy}{dx}=b\therefore \frac{dy}{dx}=\frac{b}{ab+2ay}$$
Given : $$f(x)=4x^3-6x^2\cos2a+3x \sin 2a.\sin 6a+\sqrt{\ln (2a-a^2)}$$ then
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$$f(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$
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$${f}'(\displaystyle \frac{1}{2})<0$$
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$$f'(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$
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$${f}'(\displaystyle \frac{1}{2})>0$$
Explanation
$$f'(x)=12x^{2}-12xcos2a+3sin2a.sin6a+0$$
$$f'(1/2)=\frac{12}{4}-\frac{12}{2}cos2a+3sin2a.sin6a$$
$$=3-6cos2a+1.5(cos4a-cos8a)$$
So, it will alwyas be greater than $$0$$
Let $$\displaystyle f\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\frac { f\left( { x }_{ 1 } \right) +f\left( { x }_{ 2 } \right) +...+f\left( { x }_{ n } \right) }{ n } $$ where all $${ x }_{ i }\in R$$ are independent to each other and $$n\in N$$. if $$f(x)$$ is differentiable and $$f'\left( 0 \right) =a,f\left( 0 \right) =b$$ and $$f'\left( x \right) $$ is equal to
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$$a$$
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$$0$$
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$$b$$
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None of these
Explanation
Differentiating the given equation w.r.t. $${ x }_{ 1 },$$ we get
$$\displaystyle \dfrac { 1 }{ n } f'\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\dfrac { f'\left( { x }_{ 1 } \right) }{ n } $$
$$[$$Since all $${ x }_{ i }'s$$ are independent to each other, $$\displaystyle \therefore \dfrac { d{ x }_{ i } }{ { dx }_{ j } } =0$$ if $$i\neq j$$ and $$\displaystyle \dfrac { { dx }_{ i } }{ { dx }_{ j } } =1$$ if $$(i=j)]$$
On putting $${ x }_{ 1 }={ x }_{ 2 }=...={ x }_{ n-1 }=0$$ and $${ x }_{ n }=x,$$ we get $$\displaystyle f'\left( \dfrac { x }{ n } \right) =f'\left( 0 \right) =a.$$
On integrating, we get $$\displaystyle nf'\left( \dfrac { x }{ n } \right) =ax+c$$
Since $$f(0)=b,$$ we have $$c=nb$$
$$\displaystyle \therefore nf'\left( \dfrac { x }{ n } \right) =ax+nb\Rightarrow nf'\left( x \right) =nax+nb\Rightarrow f\left( x \right) =ax+b.$$
$$\displaystyle \therefore f'\left( x \right) =a,\forall x\in R$$
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 ?
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$$14$$
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$$\displaystyle \frac{7}{8}$$
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$$1$$
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none of these
Explanation
$$5f(x)+3f\left ( \displaystyle \frac{1}{x} \right )=x+2\Rightarrow (1)$$
Replace $$x$$ by $$\dfrac{1}{x}$$
$$\Rightarrow 5f\left ( \displaystyle \frac{1}{x} \right )+3f(x)=\dfrac{1}{x}+2\Rightarrow (2)$$
$$\Rightarrow 5\times(1)-3\times (2)\Rightarrow 16f(x) =5x-\dfrac{3}{x}+4=\dfrac{5x^2+4x-3}{x}$$
$$\therefore y = xf(x) =\dfrac{5x^2+4x-3}{16}$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{10x+4}{16}$$
$$\therefore \left(\dfrac{dy}{dx}\right)_{x=1}=\dfrac{10+4}{16}=\dfrac{7}{8}$$
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
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$$\displaystyle \frac{1}{2\sqrt{x}}$$
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{4\sqrt{x}}$$
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$$\displaystyle \frac{2\sqrt{x}+1}{4\sqrt{x}}$$
Explanation
Let $$y=f\left( x \right) =\sqrt { 1+\sqrt { x } } $$
Then, $${ y }^{ 2 }=1+\sqrt { x } $$
Differentiating w.r.t. x, we get,
$$\displaystyle 2y\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$$
$$\Rightarrow f(x)f'(x)=\displaystyle \frac{1}{4\sqrt{x}}$$
$$y=\sqrt{\sin x+\sqrt{\sin x +\sqrt{\sin x+-\infty }}}$$ then $$\displaystyle \frac{dy}{dx}$$ equals:$$(\sin x> 0)$$
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$$\displaystyle \frac{\cos x}{2y-1}$$
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$$\displaystyle \frac{y}{2\tan x+y\sec x}$$
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$$\displaystyle \frac{1}{\sqrt{1+4\sin x}}$$
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$$\displaystyle \frac{2\cos x}{\sin x+2y}$$
Explanation
$$y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}$$
Taking square on both the sides, we get
$${ y }^{ 2 }=\sin { x } +\sqrt { \sin x\sqrt { \sin x\sqrt { \sin x+-\infty } } } $$
$${ y }^{ 2 }=\sin { x } +y$$ ............ Since $$y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}$$
$$2y\displaystyle \frac { dy }{ dx } =\cos { x } +\displaystyle \frac { dy }{ dx } $$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x } }{ 2y-1 } $$
$$f\left( x \right)=\begin{cases} \sin { x } \qquad ;\qquad x\neq n\pi ,n=0,\pm 1,\pm 2,\pm 3..... \\ 2\qquad \qquad ;\qquad otherwise \end{cases}$$ 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
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$$1$$
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$$4$$
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$$5$$
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non-existent
Explanation
$$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right) \right) } =g\left( f\left( 0 \right) \right) $$
$$=g(2)={ 2 }^{ 2 }+1=5$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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