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CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 3
lf $$\mathrm{y}=\log(\log(x+\sqrt{1+x^{2}}))$$ then $$\displaystyle \frac{dy}{dx}=$$
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$$\displaystyle \frac{1}{x+\sqrt{1+x^{2}}}$$
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$$\displaystyle \frac{x}{\log(x+\sqrt{1+x^{2}})}$$
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$$\displaystyle \frac{-1}{\log(x+\sqrt{1+x^{2}})}$$
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$$\displaystyle \frac{1}{\sqrt{1+x^{2}}\log(x+\sqrt{1+x^{2}})}$$
Explanation
We have, $$y =\log(\log(x+\sqrt{1+x^2}))$$
Using chain rule of differentiation,
$$\displaystyle \frac{dy}{dx}=\frac{1}{\log(x+\sqrt{1+x^2})}\times \frac{1}{(x+\sqrt{1+x^2})}\times \left( 1+\frac{2x}{2\sqrt{1+x^2}}\right)$$
$$\displaystyle =\frac{1}{\log(x+\sqrt{1+x^2})}\times \frac{1}{(x+\sqrt{1+x^2})}\times \left(\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^2}}\right)$$
$$=\dfrac{1}{\sqrt{1+x^2}\log(x+\sqrt{1+x^2})}$$
Hence,
$$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1+x^2}\log(x+\sqrt{1+x^2})}$$
lf $$\mathrm{a}^{\mathrm{y}}=\log_{\mathrm{a}}(\mathrm{x}^{2}+\mathrm{x}+1)$$, then $$\displaystyle \frac{dy}{dx}=$$
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$$\displaystyle \frac{\log_{a}e.(2x+1)}{(x^{2}+x+1)\log(x^{2}+x+1)}$$
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$$\displaystyle \frac{(2x+1)}{(x^{2}+x+1)\log(x^{2}+x+1)}$$
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$$\displaystyle \frac{1}{(x^{2}+x+1)\log(x^{2}+x+1)}$$
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$$\displaystyle \frac{-1}{(x^{2}+x+1)\log(x^{2}+x+1)}$$
Explanation
We have, $$a^y =\log_a(x^2+x+1) \Rightarrow (1)$$
Now differentiating both sides, w.r.t $$x$$
$$\Rightarrow a^y(\log a)\dfrac{dy}{dx} =\log_ae \cdot \dfrac{2x+1}{x^2+x+1}$$
$$\Rightarrow \dfrac{dy}{dx} =\dfrac{\log_ae\cdot (2x+1)}{(x^2+x+1)\cdot \log(x^2+x+1)}$$, using (1)
If $$\sqrt{x+y}+\sqrt{y-x}=c$$, then $$\dfrac{d^2y}{dx^2}$$ is
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$$\displaystyle \frac{2}{c}$$
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$$\displaystyle -\frac{2}{c^2}$$
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$$\displaystyle \frac{2}{c^2}$$
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$$\displaystyle -\frac{2}{c}$$
Explanation
$$\sqrt{x+y}=c-\sqrt{y-x}$$
On squaring we get,
$$2x-c^2=-2c\sqrt{y-x}$$
Again squaring we get,
$$y = \dfrac{(2x-c^2)^2}{4c^2}+x$$
$$y'=\dfrac{4(2x-c^2)}{4c^2}+1$$
$$y''=\dfrac{8}{4c^2}=\dfrac{2}{c^2}$$
Option C
If $$f(x)=(ax+b)\cos x + (cx+d)\sin x$$ and $$f^{'}(x)=x \cos x$$, for all values of $$x\in R$$, then $$a,b,c,d$$ are given by
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$$a = b = c = d$$
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$$0, 1, -1, 0$$
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$$1, 0, -1, 0$$
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$$0, 1, 1, 0$$
Explanation
Given, $$f(x) = (ax+b)\cos x + (cx+d)\sin x$$
Thus $$f'(x) = (ax+b)(-\sin x) + a\cos x + c\sin x + (cx+d)\cos x$$
$$= (-ax-b+c)\sin x + (a+cx+d)\cos x$$
Also given, $$f'(x) = x\cos x$$
Comparing the coefficients, we get
$$a =0, b=1, c =1, d=0$$
The set onto which the derivative of the function $$f(x)=x(\log x -1)$$ maps the ray $$[1, \infty)$$ is ?
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$$[1, \infty)$$
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$$(10, \infty)$$
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$$[0,\infty)$$
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$$(0,0)$$
Explanation
Given function $$f\left( x \right)=x\left( \log { x } -1 \right) $$
differentiate w.r.t $$x$$
we get, $$f^{ \prime }\left( x \right) =\cfrac { x }{ x } +\log { x } -1\\ f^{ \prime }\left( x \right) =\log { x } $$
Clearly, $$\log x$$ is defined in the given region.
as $$x\rightarrow 1,\log { x } =0\\ x\rightarrow \infty ,\log { x } =\infty $$
The set onto which the derivative of the function maps is [$$0,\infty $$).
lf $$y=\displaystyle \frac{\log(x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}$$
then $$(1+x^{2})y_{1}+xy=$$
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$$y$$
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$$-y$$
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$$0$$
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$$1$$
Explanation
$$y\sqrt{1+x^{2}}=\log(x+\sqrt{1+x^{2}})$$
Differentiate both sides:
$$y_1\sqrt{1+x^{2}}+\dfrac{yx}{\sqrt{1+x^{2}}}=\dfrac{1+\dfrac{x}{\sqrt{1+x^{2}}}}{x+\sqrt{1+x^{2}}}$$
$$\Rightarrow y_1\sqrt{1+x^{2}}+\dfrac{yx}{\sqrt{1+x^{2}}}=\dfrac{1}{\sqrt{1+x^{2}}}$$
$$\Rightarrow y_1(1+x^{2})+yx=1$$
lf $$f(x)=\displaystyle \frac{x^{2}}{x+a}$$ then $$f^{'}(a)=$$
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4
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$$\displaystyle \frac{3}{8}$$
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$$\displaystyle \frac{3}{4}$$
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8
Explanation
Given, $$f(x)=\displaystyle \frac{x^{2}}{x+a}$$
using division rule,
$$f'(x)=\displaystyle \frac{(x+a)(2x)-x^2(1)}{(x+a)^2}$$
$$=\dfrac{x(2x+2a-x)}{(x+a)^2}$$
$$=\dfrac{x(x+2a)}{(x+a)^2}$$
$$\therefore f'(a)=\displaystyle \frac{a(a+2a)}{(a+a)^2}=\frac{3}{4}$$
If $${ e }^{ y }+xy=e,$$ then $$\displaystyle{ \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } }_{ x=0 }^{ }$$ is
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$$\displaystyle \frac { 1 }{ { e }^{ 2 } } $$
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$${e}^{-1}$$
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$$e$$
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None of these
Explanation
We have $${ e }^{ y }+xy=e$$
Differentiating w.r.t $$x$$, we get
$$\displaystyle{ e }^{ y }\frac { dy }{ dx } +y+x\frac { dy }{ dx } =0$$ -----------(1)
Differentiating again w.r.t $$x$$, we get
$$\displaystyle { e }^{ y }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +e^y{ \left( \frac { dy }{ dx } \right) }^{ 2 }+2\frac { dy }{ dx } +x\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0$$ -------(2)
Put $$x=0$$ in $${ e }^{ y }+xy=e$$, we get $$y=1$$
Putting $$x=0,y=1$$ in (1) we get $$\displaystyle e\frac { dy }{ dx } +1=0\Rightarrow \frac { dy }{ dx } =-\frac { 1 }{ e } $$
Putting $$x=0,y=1,\displaystyle \frac { dy }{ dx } =-\frac { 1 }{ e } $$ in (2) we get $$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { 1 }{ { e }^{ 2 } } $$
If $$x^{4}+y^{4}-a^{2}xy=0$$ defines $${y}$$ implicitly as function of $$x$$ , then $$ \displaystyle \frac{dy}{dx}=$$
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$$\displaystyle \frac{4x^{3}-a^{2}y}{4y^{3}-a^{2}x}$$
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$$-\left(\displaystyle \frac{4x^{3}-a^{2}y}{4y^{3}-a^{2}x}\right)$$
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$$\displaystyle \frac{4x^{3}}{4y^{3}-a^{2}x}$$
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$$\displaystyle \frac{-4x^{3}}{4y^{3}-a^{2}x}$$
Explanation
Given, $$x^{4}+y^{4}-a^{2}xy=0$$
$$x^4+y^4=a^2xy$$
Differentiate both sides w.r.t $$x$$ we get,
$$4x^3+4y^3\dfrac{dy}{dx}$$
$$=a^2x\dfrac{dy}{dx}+a^2y$$
$$\therefore \dfrac{dy}{dx}=\dfrac{a^2y-4x^3}{4y^3-a^2x}$$
$$=-\left(\dfrac{4x^3-a^2y}{4y^3-a^2x} \right)$$
lf $$\mathrm{y}=(\mathrm{x}^{2}+1)^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}$$, then $$\mathrm{y}^{'}(0)$$ is equal to
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$$\dfrac{1}{2}$$
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$$e^{2}$$
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$$0$$
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$$\dfrac{3}{2}$$
Explanation
$$\mathrm {y'(x)}=\left(\mathrm{x}^{2}+1)^{\sin \mathrm{x}}(\cos \mathrm{x } log(1+\mathrm {x^{2}})+\sin \mathrm { x}\left ( \dfrac{2{\mathrm{x}}}{1+\mathrm{x^{2}}} \right ) \right)$$
$$\mathrm{y'(0)}=1(0+0)$$
$$=0$$
If $$f(x)=e$$ $$^{x}$$ $$g(x),$$
$$g(0)=1,g'(0)=3$$, then$$ f' (0)$$ is
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$$0$$
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$$4$$
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$$3$$
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$$7$$
Explanation
$${f}'(x)=e^{^{x}}g(x)+e^{^{x}}{g}'\left ( x \right )$$
For x=0:
$$f'(0)= g\left ( 0 \right )+{g}'\left ( 0\right )=4$$
If $$ax^{2}+2hxy+by^{2}+2gx+2fy +c=0$$ then $$\displaystyle \frac{dy}{dx}=$$
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$$-(\displaystyle \frac{ax+hy+g}{hx+by+f})$$
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$$-(\displaystyle \frac{ax+hy+g}{bx+hy+f})$$
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$$-(\displaystyle \frac{hx+by+f}{ax+hy+g})$$
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$$-(\displaystyle \frac{hx+by+f}{hx+ay+g})$$
Explanation
Given: $$ax^{2}+2hxy+by^{2}+2gx+2fy +c=0$$
$$\displaystyle2ax+2hy+2hx\frac{dy}{dx}+2by\frac{dy}{dx}+2g+2f\frac{dy}{dx}=0$$
$$\dfrac{-(2ax+2hy+2g)}{2hx+2by+2f}=\dfrac{dy}{dx}$$
If $$u=\displaystyle \tan^{-1}\left (\frac{x^{2}+y^{2}}{x+y}\right)$$,
then $$x\displaystyle \frac{d u}{d x}+y\frac{d u}{d y}=$$
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$$\sin 2u$$
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$$\dfrac {1}{2}\sin 2u$$
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$$\dfrac {1}{3}\sin 2u$$
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$$2\sin 2u$$
Explanation
Given, $$u=\tan^{-1}\left(\dfrac {x^2+y^2}{x+y}\right)$$
$$\Rightarrow \tan u=\dfrac{x^{2}+y^{2}}{(x+y)}$$
On differentiating both the sides w.r.t. $$x$$ and $$y$$ respectively, we get
$$\sec^{2}u\dfrac{du}{dx}=\dfrac{2x(x+y)-(x^{2}+y^{2})}{(x+y)^{2}}$$ ....(1)
and $$\sec^{2}u\dfrac{du}{dy}=\dfrac{2y(x+y)-(x^{2}+y^{2})}{(x+y)^{2}}$$ ....(2)
Multiplying $$x$$ with equation (1) and $$y$$ to equation (2).
Therefore, $$x\sec^2u\dfrac {du}{dx}=\dfrac {x^3+2x^2y-xy^2}{(x+y)^2}$$ ....(3)
and $$y\sec^2u\dfrac {du}{dy}=\dfrac {y^3+2xy^2-x^2y}{(x+y)^2}$$ ....(4)
Adding equations (3) and (4), we get
$$\sec^{2}u\left ( x\dfrac{du}{dx}+y\dfrac{du}{dy} \right )=\dfrac{x^{3}+2x^{2}y-xy^{2}+y^{3}+2xy^{2}-x^{2}y}{(x+y)^{2}}$$
$$=\dfrac {x^3+x^2y+y^3+xy^2}{(x+y)^2}$$
$$=\dfrac {(x+y)(x^2+y^2)}{(x+y)^2}$$
$$=\dfrac {x^2+y^2}{x+y}$$
$$=\tan u$$
Thus $$x\dfrac{du}{dx}+y\dfrac{du}{dy}=\sin u \cos u$$
$$ = \dfrac{1}{2} \sin 2u$$
If $$y=\sqrt{\sin{x}+y}$$, then $$\displaystyle\frac{dy}{dx}$$ is
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$$\displaystyle\frac{\cos{x}}{2y-1}$$
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$$\displaystyle\frac{\cos{x}}{1-2y}$$
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$$\displaystyle\frac{\cos{x}}{2y+1}$$
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$$\displaystyle\frac{\sin{x}}{2y-1}$$
Explanation
We have,
$$y=\sqrt{\sin{x}+y}$$
On squaring both the sides, we get
$$y^2=\sin(x)+y$$
Differentiating w.r.t. $$x$$,
$$\displaystyle 2y\frac{dy}{dx}=\cos{x}+\frac{dy}{dx}$$
$$2y\displaystyle \frac { dy }{ dx } -\frac { dy }{ dx } =\cos { x } $$
$$\displaystyle \frac { dy }{ dx } \left( 2y-1 \right) =\cos { x } $$
$$\displaystyle\frac{dy}{dx}=\frac{\cos{x}}{2y-1}$$
If $$\sqrt{x}+\sqrt{y}=4$$, then find $$\displaystyle\frac{dx}{dy}$$ at $$y=1$$.
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$$3$$
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$$4$$
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$$-3$$
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$$-4$$
Explanation
$$\sqrt{x}+\sqrt{y}=4$$
Differentiating both sides of the given equation w.r.t. y, we get
$$\displaystyle\frac{1}{2\sqrt{x}}\frac{dx}{dy}+\frac{1}{2\sqrt{y}}=0$$
or $$\displaystyle\frac{dx}{dy}=-\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{y}-4}{\sqrt{y}}$$ ........................................Since $$\sqrt{x}+\sqrt{y}=4$$
or $$\displaystyle{\left[\frac{dx}{dy}\right]}_{\displaystyle y=1}=\frac{1-4}{1}=-3$$
If $$x^3+y^3=3axy$$, then $$\displaystyle\frac{dy}{dx}$$ is
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$$\displaystyle\frac{ay-y^2}{x^2-ax}$$
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$$\displaystyle\frac{ay-x^2}{y^2-ax}$$
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$$\displaystyle\frac{by-x^2}{y^2-bx}$$
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None of these
Explanation
Given that
$$x^3+y^3=3axy$$
We have to find $$\displaystyle \frac { dy }{ dx } $$
Differentiate w r to $$x$$
$$\displaystyle \frac { d }{ dx } \left( { x }^{ 3 }+{ y }^{ 3 }-3axy \right) =0$$
$$3{ x }^{ 2 }+3{ y }^{ 2 }\displaystyle \frac { dy }{ dx } -3a\left( x\displaystyle \frac { dy }{ dx } +y \right) =0$$
$$3{ x }^{ 2 }+3{ y }^{ 2 }\displaystyle \frac { dy }{ dx } -3ax\displaystyle \frac { dy }{ dx } -3ay=0$$
Dividing throughout by $$3$$
$$ \displaystyle \frac { dy }{ dx } \left( { y }^{ 2 }-ax \right) =ay-{ x }^{ 2 }$$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { ay-{ x }^{ 2 } }{ { y }^{ 2 }-ax } $$
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
$$Given:\cfrac { d }{ dx } (x^{ n })=nx^{ n-1 }\\ \therefore for\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...............+\cfrac { x^{ 2 } }{ 2 } +1\\ f'(x)=\cfrac { 100x^{ 99 } }{ 100 } +99\cfrac { x^{ 98 } }{ 99 } +.............+\cfrac { 2x }{ 2 } +1\\ Here\; f'(1)=1+1+.........to\; 100term=\; 100\\ Hence\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...........+\cfrac { x^{ 2 } }{ 2 } +x+1=100$$
If $$ y=x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\cdots}}}$$ then $$\displaystyle\frac{dy}{dx}=$$
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$$\displaystyle\frac{dy}{dx}=\frac{x}{2x-y}$$
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$$\displaystyle\frac{dy}{dx}=\frac{y}{2y-x}$$
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$$\displaystyle\frac{dy}{dx}=\frac{2y}{2x-y}$$
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None of these
Explanation
We have
$$ y=x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\cdots}}}$$
$$=x+\displaystyle\frac{1}{y}$$
or $$y^2=xy+1$$
or $$2y\displaystyle\frac{dy}{dx}=y+x\frac{dy}{dx}+0$$ $$[$$Differentiating both sides w.r.t. $$x]$$
or $$\displaystyle\frac{dy}{dx}(2y-x)=y$$
or $$\displaystyle\frac{dy}{dx}=\frac{y}{2y-x}$$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
Explanation
Let $$f(x) = (px+q) \displaystyle \left ( \cfrac{r}{x}+s \right )$$
We have, $$\cfrac{d(uv)}{dx} = u'v + uv'$$
$$\therefore \displaystyle f'(x) = \left [ \cfrac{d}{dx}(px+q) \right ] \left ( \cfrac{r}{x} +s \right ) +(px+q) \cfrac{d}{dx} \left ( \cfrac{r}{x} +s \right)$$
$$=\displaystyle p\left ( \cfrac{r}{x} + s \right ) +(px+q)\left ( \cfrac{-r}{x^2} \right )$$
$$= \displaystyle \cfrac{pr}{x}+ps -\cfrac{pr}{x}-\cfrac{qr}{x^2}=ps - \cfrac{qr}{x^2}$$
Hence, assertion is incorrect but reason is correct.
If $$\displaystyle x^2+y^2=t-\frac{1}{t}$$ and $$\displaystyle x^4+y^4=t^2+\frac{1}{t^2}$$, then $$\displaystyle x^3y\frac{dy}{dx}=$$
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$$0$$
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$$1$$
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$$-1$$
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none of these
Explanation
We have $$\displaystyle x^2+y^2=t-\frac{1}{t}$$ ..........(1)
$$\displaystyle x^4+y^4=t^2+\frac{1}{t^2}$$ ................(2)
On squaring equation (1), we get
$$\displaystyle {(x^2+y^2)}^2=t^2+\frac{1}{t^2}-2$$
$${ x }^{ 4 }+2{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 4 }=x^4+y^4-2$$ ..............from (2)
$$\therefore 2x^2y^2=-2$$
$$x^2y^2=-1$$
$$\displaystyle y^2=-\frac{1}{x^2}$$
Differentiating both the sides,
$$\displaystyle 2y\frac{dy}{dx}=\frac{2}{x^3}$$
$$\displaystyle x^3y\frac{dy}{dx}=1$$
If $$xy+y^2=\tan x + y$$, then $$\displaystyle\frac{dy}{dx}$$ is equal to
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$$\displaystyle\frac{\sec^{2}{x}-y}{(x+2y-1)}$$
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$$\displaystyle\frac{\cos^{2}{x}+y}{(x+2y-1)}$$
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$$\displaystyle\frac{\sec^{2}{x}-y}{(2x+y-1)}$$
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$$\displaystyle\frac{\cos^{2}{x}+y}{(2x+2y-1)}$$
Explanation
The given relation is $$xy+y^2=\tan x + y$$.
Differentiating both sides with respect to $$x$$, we get
$$\displaystyle\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(\tan x)+\frac{dy}{dx}$$
or $$\displaystyle\left[y.1+x.\frac{dy}{dx}\right]+2y\frac{dy}{dx}=\sec^{2}{x}+\frac{dy}{dx}$$
or $$\displaystyle(x+2y-1)\frac{dy}{dx}=\sec^{2}{x}-y$$
$$\displaystyle\therefore\frac{dy}{dx}=\frac{\sec^{2}{x}-y}{(x+2y-1)}$$
If $$y^x=x^y$$, then find $$\displaystyle\frac{dy}{dx}$$.
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$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$
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$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$
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$${y(x\log{y}-y)}$$
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$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$
Explanation
$$y^x=x^y$$
Taking log on both sides
$$\Rightarrow$$ $$\log{y^x}=\log{x^y}$$
$$\Rightarrow$$ $$x\log{y}=y\log{x}$$
$$\Rightarrow$$ $$\displaystyle x\frac{1}{y}\frac{dy}{dx}+\log{y}\times 1=\frac{y}{x}+\log{x}\frac{dy}{dx}$$
$$\Rightarrow$$ $$\displaystyle\left(\frac{x}{y}-\log{x}\right)\frac{dy}{dx}=\frac{y}{x}-\log{y}$$
$$\Rightarrow$$ $$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{y}{x}-\log{y}}{\displaystyle\frac{x}{y}-\log{x}}=\frac{y(y-x\log{y})}{x(x-y\log{x})}=\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$
Differentiate $$\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$ with respect to $$x$$.
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$$\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}-\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$$
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$$\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-2}+\frac{1}{x-3}+\frac{1}{x-4}+\frac{1}{x-5}\right]$$
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$$\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$$
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None of these
Explanation
Let $$y=\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$
Taking logarithm on both sides, we get
$$\log{y}=log{\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}}$$
$$\displaystyle =\frac{1}{2}[\log{(x-1)}+\log{(x-2)}-\log{(x-3)}-\log{(x-4)}-\log{(x-5)}]$$ ....................(Since $$\log { \displaystyle \frac { a }{ b } =\log { a } } -\log { b }$$ and $$\log { ab=\log { a } +\log { b } } $$)
Differentiating both sides w.r.t. $$x$$, we get
$$\displaystyle\frac{1}{y}\frac{dy}{dx}=\frac { 1 }{ 2 } \left( \frac { 1 }{ x-1 } \frac { d }{ dx } \left( x-1 \right) +\frac { 1 }{ x-2 } \frac { d }{ dx } \left( x-2 \right) -\frac { 1 }{ x-3 } \frac { d }{ dx } \left( x-3 \right) -\frac { 1 }{ x-4 } \frac { d }{ dx } \left( x-4 \right) -\frac { 1 }{ x-5 } \frac { d }{ dx } \left( x-5 \right) \right) $$
$$\displaystyle\therefore\frac{dy}{dx}=\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$$
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 $$y=e^x \sin x$$, then find $$\displaystyle \frac {dy}{dx}$$
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$$e^x(\sin x+\cos x)$$
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$$e^x(\sin x-\cos x)$$
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$$e^x \sin x$$
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None of these
Explanation
Given that
$$y={ e }^{ x }.\sin { x } $$
On differentiating, we get
$$\displaystyle \frac { dy }{ dx } =\sin { x } \frac { d }{ dx } { e }^{ x }+{ e }^{ x }\frac { d }{ dx } \sin { x } $$
$$\displaystyle \frac { dy }{ dx } =\sin { x } { e }^{ x }+{ e }^{ x }\cos { x } $$
$$\therefore \displaystyle \frac { dy }{ dx } ={ e }^{ x }\left( \sin { x } +\cos { x } \right) $$
$$\displaystyle\frac{dy}{dx}$$ for $$y=x^x$$ is
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$$x^x(1-\log{x})$$
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$$x^x(1-\log{y})$$
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$$x^x(1+\log{y})$$
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$$x^x(1+\log{x})$$
Explanation
$$y=x^x$$
Taking $$\log $$ on both the sides
$$\log { y } =\log { \left( { x }^{ x } \right) } $$
$$\log { y } =x\log { x } $$
Differentiate w.r to $$x$$
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =x\frac { d }{ dx } \left( \log { x } \right) +\left( \log { x } \right) \frac { d }{ dx } \left( x \right) $$
$$\displaystyle \frac { dy }{ dx } =y\left[ \frac { x }{ x } +\log { x } \right] $$
$$\displaystyle \frac { dy }{ dx } ={ x }^{ x }\left[ 1+\log { x } \right] $$
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 $$y=\sqrt{\sin{x}+\sqrt{\sin{x}+\sqrt{\sin{x}+\cdots\:to\:\infty}}}$$, then $$\displaystyle\frac{dy}{dx}$$ is
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$$\displaystyle\frac{\cos{x}}{1+2y}$$
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$$\displaystyle -\frac{\sin{x}}{1-2y}$$
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$$\displaystyle\frac{\cos{x}}{1-2y}$$
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$$\displaystyle\frac{\cos{x}}{2y-1}$$
Explanation
We have
$$y=\sqrt{\sin{x}+\sqrt{\sin{x}+\sqrt{\sin{x}+\cdots\:to\:\infty}}}$$
The given series may be written as
$$y=\sqrt{\sin{x}+y}$$
Squaring both the sides, we get
$$y^2=\sin{x}+y$$
Differentiating both sides w.r.t. $$x$$
$$\displaystyle 2y\frac{dy}{dx}=\cos{x}+\frac{dy}{dx}$$
$$2y\displaystyle \frac { dy }{ dx } -\frac { dy }{ dx } =\cos { x } $$
$$\displaystyle\frac{dy}{dx}(2y-1)=\cos{x}$$
$$\displaystyle\frac{dy}{dx}=\frac{\cos{x}}{2y-1}$$
If $$y=e^x \tan x + x\cdot \log_ex$$, then find $$\displaystyle \frac {dy}{dx}$$
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$$\displaystyle \frac {dy}{dx}=e^x \tan x+(\log x+1)$$
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$$\displaystyle \frac {dy}{dx}=e^x(\tan x+\sec^2x)+(\log x+x)$$
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$$\displaystyle \frac {dy}{dx}=e^x \tan x+1$$
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$$\displaystyle \frac {dy}{dx}=e^x(\tan x+\sec^2x)+(\log x+1)$$
Explanation
$$y=e^x \tan x+x\cdot \log_ex$$
On differentiating, we get
$$\displaystyle \frac {dy}{dx}=\frac {d}{dx}(e^x \tan x)+\frac {d}{dx}(x \log x)$$
$$=\tan { x } \displaystyle \frac { d }{ dx } { e }^{ x }+{ e }^{ x }\frac { d }{ dx } \left( \tan { x } \right) +x\frac { d }{ dx } \left( \log { x } \right) +\log { x } \frac { d }{ dx } x$$
$$=e^x\cdot \tan x+e^x\cdot \sec^2x+1\cdot \log x+x\cdot \displaystyle \frac {1}{x}$$
Hence, $$\displaystyle \frac {dy}{dx}=e^x(\tan x+\sec^2x)+(\log x+1)$$
If $$y={(\tan{x})}^{\displaystyle{(\tan{x})}^{\displaystyle\tan{x}}}$$, then find $$\displaystyle\frac{dy}{dx}$$ at $$\displaystyle x=\frac{\pi}{4}$$.
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$$0$$
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$$1$$
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$$-1$$
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$$2$$
Explanation
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
$$\displaystyle\therefore{\left(\frac{dy}{dx}\right)}_{\displaystyle x=\frac{\pi}{4}}=0$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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