CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 15 - MCQExams.com

$$f(x)=(kx+e^{x})h(x)$$

Given:

Let $$f\left( x \right) ={ \left( x-n \right)  }^{ m }+{ \left( x-{ n }^{ 2 } \right)  }^{ m }+{ \left( x-{ n }^{ 3 } \right)  }^{ m }+...+{ \left( x-{ n }^{ m } \right)  }^{ m }$$

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ at $$(\sqrt 2, \sqrt 2))$$
$$2x+2y\displaystyle \frac { dy }{ dx } =0$$
$$\displaystyle \frac { dy }{ dx } =-\displaystyle \frac { x }{ y } =-\displaystyle \frac { \sqrt { 2 }  }{ \sqrt { 2 }  } =-1$$
$$ B: \displaystyle \frac {d}{dx}(\sin y+\sin x=\sin x\cdot \sin y)$$ at $$(\pi, \pi)$$
$$\cos

{ y\displaystyle \frac { dy }{ dx }  } +\cos { x } =\sin { x } \cos {

y\displaystyle \frac { dy }{ dx } +\sin { y } \cos { x }  } $$
$$\displaystyle \frac { dy }{ dx } \left( \cos { y } -\sin { x } \cos { y }  \right) =\sin { y\cos { x-\cos { x }  }  } $$
$$\displaystyle

\frac { dy }{ dx } =\displaystyle \frac { \cos { x\left( \sin { y-1 } 

\right)  }  }{ \cos { y\left( 1-\sin { x }  \right)  }  } $$
$$B=\displaystyle \frac { -1\left( 0-1 \right)  }{ -1\left( 1-0 \right)  } =-1$$
$$C: \displaystyle \frac {d}{dx}(2e^{xy}+e^xe^y-e^x-{ e }^{ y }=e^{xy+1})$$ at $$(1, 1)$$
$$\left[

2{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right) 

\right] +{ e }^{ x }{ e }^{ y }\displaystyle \frac { dy }{ dx } +{ e }^{

y }{ e }^{ x }-{ e }^{ x } -{ e }^{ y }\frac { dy }{ dx } ={ e }^{ xy+1 }\left( x\displaystyle \frac {

dy }{ dx } +y \right) $$
$$\displaystyle \frac { dy }{ dx }

=\displaystyle \frac { y{ e }^{ xy+1 }-2y{ e }^{ xy }-{ e }^{ x+y }+{ e

}^{ x } }{ 2x{ e }^{ xy }+{ e }^{ x+y }-x{ e }^{ xy+1 }-{ e }^{ y } } $$..........

since $${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$$
$$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 }-{ e }^{ 1 } }=-{1} $$
$$A+B+C=-1-1-1=-3$$

Taking log on both sides,

Given $$g\left( x \right)=f\left( { e }^{ x } \right) { e }^{ f\left( x \right) }$$

Given, $$g(x)=f(x)+f^{'}(x)+f^{"}(x)$$     ....(1)
Let $$f(x)=ax^{2}+bx+c  a\ne 0$$
$$f'(x)=2ax+b$$
$$\Rightarrow f''(x)=2a$$

Given, $$f(x-y), f(x),f(y)$$ and $$f(x+y)$$ are in AP
$$\Rightarrow f(x-y)+f(x+y)=2f(x)f(y)$$           ....(1)
Substitute $$x=0, y=0$$ in (1)
$$\Rightarrow 2f(0)=2f(0)f(0)$$
$$\Rightarrow f(0)(f(0)-1)=0$$
$$\Rightarrow f(0)=1          (\because f(0)\ne 0$$ given) 

Substitute $$x=0, y=x$$ in (1), we get
$$f(-x)+f(x)=2f(0)f(x)$$
$$\Rightarrow f(-x)=f(x)$$        ....(2)
$$\Rightarrow f(-2)=f(2), f(-3)=f(3)$$
Differentiating (2) w.r.t x
$$f'(-x)(-1)=f'(x)$$
$$f'(-x)+f'(x)=0$$
$$f'(-2)+f'(2)=0$$ and $$f'(-3)+f'(3)=0$$

$$y^{3}-3y+x=0$$
Or 
$$3y^{2}.y'-3y'+1=0$$
Or 
$$3y'(y^{2}-1)=-1$$
Or 
$$y'=\dfrac{1}{3(1-y^{2})}$$
Now 
$$y"=\dfrac{1}{3(1-y^{2})^{2}}[-2y'y]$$
$$=3(y')^{2}(-2y'y)$$
$$=-6yy'^{3}$$
Now 
$$y'=\dfrac{1}{3(1-y^{2})}$$ at $$y=2\sqrt{2}$$
$$=\dfrac{1}{3(-7)}$$
Hence
$$y"=-12\sqrt{2}\dfrac{1}{3^{3}.7^{3}}$$

$$=\dfrac{-4\sqrt{2}}{3^{2}.7^{3}}$$

$$x|x|=x.\sqrt{x^{2}}$$
Its derivative will be,
$$\displaystyle =\frac{x^{2}}{\sqrt{x^{2}}}+\sqrt{x^{2}}$$
$$\displaystyle =\frac{2x^{2}}{\sqrt{x^{2}}}$$
$$=2|x|$$

$$\displaystyle \sqrt{y+x}+\sqrt{y-x}=c$$
Differentiating,
$$\displaystyle \frac{y'+1}{2\sqrt{y+x}}+\frac{1-y'}{2\sqrt{y-x}} =0 $$
$$\Rightarrow y'\sqrt{y-x}+\sqrt{y-x}=\sqrt{y+x}-y'\sqrt{y+x}$$
$$\displaystyle y'=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y-x}+\sqrt{y+x}}$$multiplying and dividing by $$\sqrt{y-x}-\sqrt{y+x},$$
$$\displaystyle =\frac{y+x+y-x-2\sqrt{y+x}\sqrt{y-x}}{y-x-y-x}$$$$=\frac{y-\sqrt{y+x}\sqrt{y-x}}{-x}$$
$$\displaystyle =\frac{y-\sqrt{y^{2}-x^{2}}}{x}$$
multiplying and dividing by $$\sqrt{y-x}+\sqrt{y+x},$$
$$\displaystyle =\frac{y+x-y+x}{y-x+y+x+2\sqrt{y^{2}-x^{2}}}$$
$$\displaystyle =\frac{x}{y+\sqrt{y^{2}-x^{2}}}$$
We know,
 $$\displaystyle y+x+y-x+2\sqrt{y^{2}-x^{2}}=c^{2}$$
$$\displaystyle y+\sqrt{y^{2}-x^{2}}=\frac{c^{2}}{2}$$
so, the differentiation would be, $$\displaystyle \frac{2x}{c^{2}}$$

$$\displaystyle \dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{-cosxsinx}{|cosx|}+\dfrac{cosxsinx}{|sinx|}$$

$$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { f\left( x \right)  } -a }{ \sqrt { \phi \left( x \right)  } -b }  } =\lim _{ x\rightarrow a }{ \frac { f\left( x \right) -{ a }^{ 2 } }{ \phi \left( x \right) -{ b }^{ 2 } } \times \frac { \sqrt { \phi \left( x \right)  } +b }{ \sqrt { f\left( x \right)  } +a }  } $$

$$\displaystyle =\lim _{ x\rightarrow a }{ \frac { f\left( x \right) -f\left( a \right)  }{ \phi \left( x \right) -\phi \left( a \right)  } \times \frac { \sqrt { \phi \left( x \right)  } +b }{ \sqrt { f\left( x \right)  } +a }  } $$

$$\displaystyle =\frac { f'\left( a \right)  }{ \phi' \left( a \right)  } \times \frac { \sqrt { \phi \left( a \right)  } +b }{ \sqrt { f\left( a \right)  } +a } =\frac { 3b }{ a } $$

To find the approximation of the value of $$\sqrt{104}$$
The nearest number to $$104$$ whose square root can be taken is $$100$$
so let us consider that $$x=100$$ and $$\delta x= dx= 4$$
Now consider,
$$y=\sqrt{x}$$
Differentiating Equation(i) with respect to x we have
$$\cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( \sqrt { x }  \right) $$
$$\cfrac { dy }{ dx } =\cfrac { 1 }{ 2\sqrt { x }  } \rightarrow \left( i \right) $$
Taking the differential of eq(i) we get,
$$ dy=\cfrac { 1 }{ 2\sqrt { x }  } dx$$
using values $$x=100$$ and $$dx=4$$ we have
$$dy=\cfrac { 1 }{ 2\sqrt { 100 }  } \left( 4 \right) =\cfrac { 4 }{ 20 } $$
$$dy=0.2$$
$$\therefore \sqrt { 104 } =y+dy=\sqrt { x } +dy$$
$$=\sqrt { 100 } +0.2=10+0.2$$
$$\sqrt { 102 } =10.2$$

We have, 

$$\displaystyle { 40 }^{ o }=\frac { \pi  }{ 4 } -\frac { \pi  }{ 36 } $$ radians 

Given, $$y={ x }^{ 2 }\cos x\\ u=\cos x\quad and\quad v={ x }^{ 2 }\\ { y }_{ 8 }={ u }_{ 8 }v+{ 8C }_{ 1 }{ u }_{ 7 }{ v }_{ 1 }+{ 8C }_{ 2 }{ u }_{ 6 }{ v }_{ 2 }+........+{ 8C }_{ 7 }{ u }{ v }_{ 7 }+{ 8C }_{ 8 }{ u }{ v }_{ 8 }$$
except  $${ v }_{ 2 }=2$$ all other derivatives of v are zero at x = 0 i.e
$${ v }(0)={ v }_{ 1 }(0)=.......{ v }_{ n }(0)=0\quad \Rightarrow { y }_{ 8 }={ 8C }_{ 2 }{ u }_{ 6 }{ v }_{ 2 }\\ { u }_{ n }=\cos (x+\frac { n\pi  }{ 2 } )\quad \quad \quad \quad { u }_{ 6 }=\cos(x+3\pi )=-\cos x\\ { u }_{ 6 }(0)=-1\\ { y }_{ 8 }(0)=28(2)(-1)=-56$$

Let $$\displaystyle f(x)=\begin{Bmatrix} \int _{ 0 }^{ x }{ \{ 1+|1-t| \} dt }

 &  x>2 \\ 5x-7 &  x\le 2 \end{Bmatrix}$$

Let $$\displaystyle y=\int _{ 0 }^{ x }{ \{ 1+|1-t| \} dt } $$ as $$x>2$$

$$\displaystyle y=\int _{ 0 }^{ 1 }{ (1+1-t)dt } +\int _{ 1 }^{ x }{ tdt } $$

$$\displaystyle y=2-\frac { 1 }{ 2 } +\frac { { (x }^{ 2 }-1) }{ 2 } $$

$$\displaystyle y=1+\frac{ { x }^{ 2 } }{ 2 } $$

$$\displaystyle f(x)=\begin{Bmatrix} 1+\frac { { x }^{ 2 } }{ 2 } ,\quad \quad  & x>2 \\ 5x-7,\quad \quad  & x\le 2 \end{Bmatrix}$$

$$\displaystyle f({ 2 }^{ + })=1+\frac { 4 }{ 2 } =3$$

$$\displaystyle f({ 2 }^{ - })=10-7=3$$

Differentiating with respect to $$x$$ we get

$$\displaystyle { f }^{ 1 }(x)=\begin{Bmatrix} x; & x>2 \\ 5; & x\le 2 \end{Bmatrix}$$

We can see that it is not differentiable at $$x=2$$ but continuous  at $$x=2$$

$$x^{2x}-2x^{x}\cot y-1=0$$ ........ (i)

$$y={ x }^{ 4 }{e}^{2x}\\ u={e}^{2x}\quad and\quad v={ x }^{ 4 }\\ { y }_{ 10 }={ u }_{ 10 }v+{ 10C }_{ 1 }{ u }_{ 9 }{ v }_{ 1 }+{ 10C }_{ 2 }{ u }_{ 8 }{ v }_{ 2 }+........+{ 10C }_{ 9 }{ u }{ v }_{ 9 }+{ 10C }_{ 10 }{ u }{ v }_{ 10 }$$
except  $${ v }_{ 4 } = 4! = 24$$ all other derivatives of v are zero at x = 0 i.e
$${ v }(0)={ v }_{ 1 }(0)=.......{ v }_{ n }(0)=0\quad \Rightarrow { y }_{ 10 }={ 10C }_{ 4 }{ u }_{ 6 }{ v }_{ 4 }\\ { u }_{ n }={2}^{n}{e}^{2x}\quad \quad \quad \quad { u }_{ 6 }={2}^{6}{e}^{2x}\\ { u }_{ 6 }(0)={2}^{6}\\ { y }_{ 10 }(0)={ 10C }_{ 4 } \times 24 \times {2}^{6}$$
$${ y }_{ 10 }(0)=\frac{10 \times 9 \times 8 \times 7}{2 \times 3 \times 4} \times 24 \times {2}^{6}=315 \times {2}^{10}$$

$$\displaystyle V=\frac { 4 }{ 3 } \pi { r }^{ 3 }\Rightarrow dV=4\pi { r }^{ 2 }h$$
when $$r=10$$ and $$h=0.02$$ 
We have $$dV=4\pi { \left( 10 \right)  }^{ 2 }\left( 0.02 \right) =8\pi $$

The given equation is:

$$\displaystyle y\left ( n \right )=e^{x+x^{2}+...+x^{n}}=e^{\dfrac{x\left ( 1-x^{n} \right )}{1-x}}$$
So $$\displaystyle \dfrac{dy\left ( n \right )}{dx}=e^{\dfrac{x\left ( 1-x^{n} \right )}{1-x}}\times \dfrac{d}{dx}\left ( \dfrac{x\left ( 1-x^{n} \right )}{1-x} \right )$$
$$\displaystyle \lim_{n\rightarrow \infty }\dfrac{dy\left ( n \right )}{dx}=\lim_{n}e^{\dfrac{x}{1-x}-\dfrac{x^{n+1}}{1-x}}\dfrac{d}{dx}\left [ \lim_{n\rightarrow \infty }\dfrac{x}{1-x}-\dfrac{x^{n+1}}{1-x} \right ]$$
   $$\displaystyle =e^{\dfrac{x}{1-x}}\dfrac{d}{dx}\left ( \dfrac{x}{1-x} \right )$$
   $$\displaystyle =e^{\dfrac{x}{1-x}}\dfrac{1}{\left ( 1-x \right )^{2}}$$
$$\displaystyle \lim_{n\rightarrow \infty }\dfrac{dy\left ( n \right )}{dx}|_{x=\frac{1}{2}}=4e$$

$$\displaystyle \int { f\left( x \right) dx } =\dfrac { 3 }{ 55 } \sqrt [ 3 ]{ { \left( \tan { x }  \right)  }^{ 5 } } \left( 5\tan ^{ 2 }{ x+11 }  \right) $$

$$\displaystyle f\left( x \right) =\dfrac { d }{ dx } \left( \dfrac { 3 }{ 55 } \sqrt [ 3 ]{ { \left( \tan { x }  \right)  }^{ 5 } } \left( 5\tan ^{ 2 }{ x+11 }  \right)  \right) $$

$$\displaystyle =\dfrac { 3 }{ 55 } \dfrac { d }{ dx } \left( 5\tan ^{ \tfrac { 11 }{ 3 }  }{ x } +11\tan ^{ \tfrac { 5 }{ 3 }  }{ x }  \right) \sec ^{ 2 }{ x } $$

$$\displaystyle =\dfrac { 3 }{ 55 } \left( \dfrac { 55 }{ 3 } \tan ^{ \tfrac { 8 }{ 3 }  }{ x } +\dfrac { 55 }{ 3 } \tan ^{ \tfrac { 2 }{ 3 }  }{ x }  \right) \sec ^{ 2 }{ x } $$

$$\displaystyle =\left( \tan ^{ \tfrac { 8 }{ 3 }  }{ x } +\tan ^{ \tfrac { 2 }{ 3 }  }{ x }  \right) \sec ^{ 2 }{ x } =\tan ^{ \tfrac { 2 }{ 3 }  }{ x } \left( { \left( 1+\tan ^{ 2 }{ x }  \right)  }^{ 2 } \right) \\ =\sqrt [ 3 ]{ \sin ^{ 2 }{ x } \cos ^{ -14 }{ x }  } $$

$$ (x^{2}+ax+1)f(x)=x^{2}-ax+1$$
Differentiating we obtain $$\displaystyle (2x+a)f(x)+(x^{2}+ax+1) f'(x)=2x-a$$   ...(i)
Putting $$x=1$$ and noting that $$ \displaystyle f(1)=\frac{2-a}{2+a},$$ we obtain
$$ (2+a)f(1)+(2+a)f'(1)=2-a\Rightarrow f'(1)=0 $$
Differentiating (i) again
$$ 2f(x)+(2x+a)f'(x)+(2x+a)f'(x) +(x^{2}+ax+1)f''(x)=2$$   ...(ii)
Putting $$x=1$$, we have $$ \displaystyle 2\cdot \frac{2-a}{2+a}+(2+a) f''(1)=2$$ $$\displaystyle \Rightarrow f''(1)=\frac{4a}{(2+a)^{2}}$$
Putting $$x=-1$$ in (i) and noting that $$\displaystyle f(-1)=\frac{2+a}{2-a} \Rightarrow (-2+a)f(-1)+(2-a) f'(-1)=-(2+a)\Rightarrow f'(-1)=0$$
Putting $$x=-1 $$ in (ii), we obtain $$ \displaystyle f''(-1)=\frac{-4a}{(2-a)^{2}}$$