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CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 4 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 4
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
$$\displaystyle \dfrac {dy}{dx}=\dfrac {d}{dx}[\log x^3]+\dfrac {d}{dx}[3 \sin^{-1}x]+\dfrac {d}{dx}[kx^2]$$
$$=3\displaystyle \dfrac {d}{dx}[\log x]+3\dfrac {d}{dx}(\sin^{-1}x)+k\dfrac {d}{dx}(x^2)$$
$$=3\cdot \displaystyle \dfrac {1}{x}+3\cdot \dfrac {1}{\sqrt {1-x^2}}+k(2x)$$
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 $$y=e^{ax}\cdot \cos (bx+c)$$, then find $$\displaystyle \frac {dy}{dx}$$
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$$ae^{ax} \cos(bx+c)-be^{ax} \sin (bx+c)$$
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$$ae^{ax} \cos(bx+c)+be^{ax} \sin (bx+c)$$
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$$eae^{ax} \cos(bx+c)-be^{ax} \sin (bx+c)$$
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$$ae^{ax} \cos(bx+c)$$
Explanation
$$y=e^{ax}\cdot \cos (bx+c)$$
On differentiating, we get
$$\displaystyle \dfrac { dy }{ dx } ={ e }^{ ax }\dfrac { d }{ dx } \left( \cos { \left( bx+c \right) } \right) +\cos { \left( bx+c \right) } \dfrac { d }{ dx } { e }^{ ax }$$
$$=-{ e }^{ ax }\sin { \left( bx+c \right) \displaystyle \dfrac { d }{ dx } \left( bx+c \right) } +\cos { \left( bx+c \right) } { e }^{ ax }\displaystyle \dfrac { d }{ dx } (ax)$$
$$=-b{ e }^{ ax }\sin { \left( bx+c \right) } +a\cos { \left( bx+c \right) } { e }^{ ax }$$
$$=a{ e }^{ ax }\cos { \left( bx+c \right) -b{ e }^{ ax }\sin { \left( bx+c \right) } } \\ $$
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 $$\displaystyle y=e^{x \log a}+e^{a \log x}+e^{a \log a}$$, 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 $$\displaystyle y=\frac{1}{1+x^{\beta -\alpha}+x^{\gamma -\alpha}}+\frac{1}{1+x^{\alpha-\beta}+x^{\gamma -\beta }}+\frac{1}{1+x^{\alpha -\gamma }+x^{\beta-\gamma }}$$
then $$\displaystyle \frac{dy}{dx}$$ is equal to-
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$$0$$
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$$1$$
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$$\displaystyle
(a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$$
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None of these
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
$$f'(x)=\sin x+\sin 4x\cdot \cos x$$
$$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$$=$$\displaystyle \frac { d }{ dx } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) \left[ \sin { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) +\sin { 4 } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) .\cos { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) } } \right] $$
$$=4x\left[ \cos { 2{ x }^{ 2 }-\sin { 8 } { x }^{ 2 }.\sin { 2{ x }^{ 2 } } } \right] $$
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 $$e^{x \sin x}$$
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$$\displaystyle e^{x \sin x} (x \cos x-\sin x)$$
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$$\displaystyle e^{x \sin x} x \cos x$$
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$$\displaystyle e^{x \sin x} (-x \cos x+\sin x)$$
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$$\displaystyle e^{x \sin x} (x \cos x+\sin x)$$
Explanation
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
$$\displaystyle \dfrac { d }{ dx } \left( \log _{ 10 }{ x+\log _{ x }{ 10+\log _{ x }{ x+\log _{ 10 }{ 10 } } } } \right) $$
$$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } +\displaystyle \dfrac { \left( -\log _{ e }{ 10\left( \displaystyle \dfrac { 1 }{ x } \right) } \right) }{ { \left( \log_e { x } \right) }^{ 2 } } $$
$$=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } -\displaystyle \dfrac { \log _{ e }{ 10 } }{ x{ \left( \log_e { x } \right) }^{ 2 } } $$
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 $$x$$
$$\Rightarrow\displaystyle (ab+2ay)\frac{dy}{dx}=b\therefore \frac{dy}{dx}=\frac{b}{ab+2ay}$$
If
$$\displaystyle y=\frac { \sin { x } }{ 1+\displaystyle \frac { \cos { x } }{ 1+\displaystyle \frac { \sin { x } }{ 1+\displaystyle\frac { \cos { x } }{ 1+\displaystyle \frac { \sin { x } }{ 1+ .....\infty} } } } } $$,
then $$y'(0)$$ is
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equal to $$0$$
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equal to $$\frac{1}{2}$$
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equal to $$1$$
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non existent
Explanation
$$\displaystyle y=\frac { \sin { x } }{ 1+\frac { \cos { x } }{ 1+\frac { \sin { x } }{ 1+\frac { \cos { x } }{ 1+\frac { \sin { x } }{ 1+.....\infty } } } } } $$
here, $$y(0)=0$$
$$ \displaystyle \Rightarrow y=\frac { \sin { x } }{ 1+\frac { \cos { x } }{ 1+y } } $$
$$\displaystyle \Rightarrow y\left( 1+y+\cos { x } \right) =\left( 1+y \right) \sin { x } $$
differentiating wrt x
$$y'\left( 1+y+\cos { x } \right) +y\left( y'-\sin { x } \right) =\left( 1+y \right) \cos { x } +y'\sin { x } $$
$$ y'\left( 0 \right) \left( 1+y\left( 0 \right) +1 \right) +y\left( 0 \right) \left( y'\left( 0 \right) \right) =\left( 1+y\left( 0 \right) \right) $$
$$y'\left( 0 \right) =\dfrac { 1 }{ 2 } $$
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 $$0$$
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
$$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) +
\sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$$
$$y=\cos ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) +\sin ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) } } =\displaystyle \frac { \pi }{ 2 } $$ ............. $$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } } $$
$$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$$
Which of the following could be the sketch graph of $$y=\displaystyle \frac{d}{dx}(x\ln x)$$
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0%
0%
0%
Explanation
$$\displaystyle y=\frac { d }{ dx } \left( x\ln { x } \right) =\ln { x } +\frac { x }{ x } =\ln { x } +1$$
At $$x=0 ,y=1$$
At $$y=0, x={ e }^{ -1 }$$
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, $$x>0$$
The equation $$y^2e^{xy} =9e^{-3}.x^2$$ defines $$y$$ as a differentiable function of x. The value of $$\displaystyle \frac{dy}{dx}$$ for $$x=-1$$ and $$y= 3$$ is
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$$-\displaystyle \frac{15}{2}$$
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$$-\displaystyle \frac{9}{5}$$
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$$3$$
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$$15$$
Explanation
$$y^2e^{xy} =9e^{-3}.x^2$$
Differentiate both sides
$$\displaystyle \frac { d }{ dx } \left( { y }^{ 2 }{ e }^{ xy } \right) =\displaystyle \frac { d }{ dx } \left( 9{ e }^{ -3 }{ x }^{ 2 } \right) $$
$$2y\displaystyle \frac { dy }{ dx } { e }^{ xy }+{ y }^{ 2 }{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right) =9{ e }^{ -3 }\left( 2x \right) $$
On solving, we get
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { 18x{ e }^{ -3 }-{ y }^{ 3 }{ e }^{ xy } }{ 2y{ e }^{ xy }+{ y }^{ 2 }{ e }^{ xy }x } $$
putting $$x=-1,\quad y=3$$, we get
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { -18{ e }^{ -3 }-27{ e }^{ -3 } }{ 6{ e }^{ -3 }-9{ e }^{ -3 } } =\displaystyle \frac { -45{ e }^{ -3 } }{ -3{ e }^{ -3 } } =15$$
$$f:R\rightarrow R$$ 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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$$14$$
If $$y=x^{1/x}$$, the value of $$\displaystyle \frac{dy}{dx}$$ at $$x=e$$ is
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1
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0
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-1
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none of these
Explanation
Given, $$y=x^{1/x}$$
$$\displaystyle \log { y } =\frac { 1 }{ x } \log { x } $$
$$\Rightarrow \displaystyle x \log { y }= \log { x } $$
$$\displaystyle \frac { x }{ y } \frac { dy }{ dx } +\log { y } =\frac { 1 }{ x } $$
$$\Rightarrow \displaystyle \frac { dy }{ dx } =\frac { y(1-x\log { y } ) }{ { x }^{ 2 } } $$
$$\displaystyle \left( \frac { dy }{ dx } \right) _{ x=e }{ =\frac { e^{ 1/e }(1-e\log {e}^\frac {1}{e}) }{ { e }^{ 2 } } }=0$$
If for all $$x, y$$ the function $$f$$ is defined by $$f(x)+f(y)+f(x).f(y)=1$$ and $$f(x)>0$$ then
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$$f^{'}(x)$$ does not exist
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$$f^{'}(x)=0$$ for all $$x$$
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$$f^{'}(0)< f^{'}(1)$$
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none of these
Explanation
$$f(x)+f(y)+f(x).f(y)=1$$
put $$y = 0$$
$$\Rightarrow f(x)+f(0)+f(x).f(0)=1$$
$$\Rightarrow f(x) = \cfrac{1-f(0)}{1+f(0)} = C$$ (constant)
$$\therefore f'(x) = 0\forall x $$
If $${ S }_{ n }$$ denotes the sum of $$n$$ terms of a G.P. whose common ratio is $$r$$, then $$\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr } $$ is equal to
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$$\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }$$
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$$\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }$$
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$$\left( n-1 \right) { S }_{ n }$$
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None of these
Explanation
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
$$\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr } +{ S }_{ n }=na{ r }^{ n-1 }-0$$
$$\displaystyle \Rightarrow \left( r-1 \right) \frac { d{ S }_{ n } }{ dr } =na{ r }^{ n-1 }-{ S }_{ n }$$
$$=n[n$$ th term of $$g.p.]-{ S }_{ n }$$
$$=n\left( { S }_{ n }-{ S }_{ n-1 } \right) -{ S }_{ n }$$
$$=\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }.$$
If $$x^{y}=e^{x+y}$$ then $$\displaystyle \frac{dy}{dx}$$ at $$x=1$$ is equal to
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0%
$$0$$
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$$-2$$
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$$1$$
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none of these
Explanation
Given, $$x^y= e^{x+y}$$
Taking $$\log$$ on both sides, we get
$$y\log x=x+y$$
Taking derivative,
$$\displaystyle \frac{y}{x}+logx.y'=1+y'$$
At, $$x=1 => y=-1$$
So, $$y'=-2$$
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 $$\dfrac{1}{x}$$
$$\Rightarrow 5f\left ( \displaystyle \frac{1}{x} \right )+3f(x)=\dfrac{1}{x}+2\Rightarrow (2)$$
$$\Rightarrow 5\times(1)-3\times (2)\Rightarrow 16f(x) =5x-\dfrac{3}{x}+4=\dfrac{5x^2+4x-3}{x}$$
$$\therefore y = xf(x) =\dfrac{5x^2+4x-3}{16}$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{10x+4}{16}$$
$$\therefore \left(\dfrac{dy}{dx}\right)_{x=1}=\dfrac{10+4}{16}=\dfrac{7}{8}$$
$$y=\sqrt{\sin x+\sqrt{\sin x +\sqrt{\sin x+-\infty }}}$$ 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 $$y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}$$
$$2y\displaystyle \frac { dy }{ dx } =\cos { x } +\displaystyle \frac { dy }{ dx } $$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x } }{ 2y-1 } $$
If $$xe^{xy}-y=\sin ^{2}x$$ then $$\displaystyle \frac{dy}{dx}$$ at $$x=0$$ is
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$$0$$
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$$1$$
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$$-1$$
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None of these
Explanation
Given $$x{ e }^{ xy }=y+\sin ^{ 2 }{ x } $$ ...(1)
on putting $$x=0$$, we get $$0.{ e }^{ 0 }=y+0\Rightarrow y=0$$
on differentiating (1) both sides w.r.t $$x$$, we get
$$\displaystyle 1.{ e }^{ xy }+x.{ e }^{ xy }\left( x.\frac { dy }{ dx } +y \right) =\frac { dy }{ dx } +2\sin { x } \cos { x } $$
On putting $$x=0,y=0$$, we get
$$\displaystyle { e }^{ 0 }+0\left( 0+0 \right) =\left[ \frac { dy }{ dx } \right] +2\sin { 0 } $$
$$\Rightarrow \dfrac { dy }{ dx } =1$$
If $$x^{y}.y^{x}=16$$ then $$\frac{dy}{dx}$$ at (2, 2) is
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$$1$$
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$$-1$$
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$$0$$
0%
none of these
Explanation
$$x^{y}.y^{x}=16$$
$$\log { x^{ y } } +\log { y^{ x } } =\log { 16 } $$
$$y\log { x } +x\log { y } =\log { 16 } $$
Differentiating w.r.t x , we get
$$\displaystyle \frac { y }{ x } +\log { x } \frac { dy }{ dx } +\frac { x }{ y } \frac { dy }{ dx } +\log { y } =0$$
So, at x=2, y=2
$$\displaystyle 1+\log { 2 } \left( \frac { dy }{ dx } \right) _{ (2,2) }+1\left( \frac { dy }{ dx } \right) _{ (2,2) }+\log { 2 } =0$$
$$\Rightarrow \left( \frac { dy }{ dx } \right) _{ (2,2) }=-1$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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