CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 14 - MCQExams.com

$$y=x^{{\displaystyle(\log{x})}^{\displaystyle\log{(\log{x})}}}$$

Let $$y=\displaystyle\frac{\sqrt{x}{(x+4)}^{\tfrac{3}{2}}}{{(4x-3)}^{\tfrac{4}{3}}}$$

$$\displaystyle\frac{d}{dx}\left\{f_n(x)\right\}=\frac{d}{dx}\left\{e^{\displaystyle f_{n-1}(x)}\right\}$$

$$\Rightarrow e^{\displaystyle f_{n-1}(x)}\frac{d}{dx}\left\{f_{n-1}(x)\right\}=f_n(x)\frac{d}{dx}\left\{f_{n-1}(x)\right\}$$

$$\Rightarrow f_n(x).\frac{d}{dx}\left\{e^{\displaystyle f_{n-2}(x)}\right\}=f_n(x).e^{\displaystyle f_{n-2}(x)}\frac{d}{dx}\left\{f_{n-2}(x)\right\}$$

$$\Rightarrow f_n(x)f_{n-1}(x)\frac{d}{dx}\left\{f_{n-2}(x)\right\}$$
$$\cdots$$
$$\Rightarrow f_n(x)f_{n-1}(x)\cdots f_2(x)\frac{d}{dx}\left\{f_1(x)\right\}$$

$$\Rightarrow f_n(x)f_{n-1}(x)\cdots f_2(x)\frac{d}{dx}\left\{e^{\displaystyle f_0(x)}\right\}$$

$$\Rightarrow f_n(x)f_{n-1}(x)\cdots f_2(x)e^{\displaystyle f_0(x)}\frac{d}{dx}\left\{f_0(x)\right\}$$

Use $$e^{\displaystyle f_0(x)}=f_1(x)$$ and $$f_0(x)=x$$
Hence, 
$$\displaystyle\frac{d}{dx}\left\{f_n(x)\right\} = f_n(x)f_{n-1}(x) ... f_2(x)f_1(x)$$
Hence, option C is correct.

Given, $$\displaystyle f(x)=\Pi_{n=1}^{100}(x-n)^{n(101-n)}$$
$$\Rightarrow log f(x)= \left \{n(101-n)\left \{\Pi_{n=1}^{100}log(x-n)\right \}\right \}$$
$$\left \{Here, \Pi \text {changes to} \sum \text {when taken log}\right \}$$
$$\displaystyle \Rightarrow log f(x)=\sum_{n=1}^{100} n(101-n) log (x-n)$$
Differentiating both the sides, we get
$$\displaystyle \frac {f'(x)}{f(x)}=\sum_{n=1}^{100}n(101-n)\cdot \frac {1}{x-n}$$
$$\displaystyle \therefore \frac {f'(101)}{f(101)}= \sum_{n=1}^{100}\frac {n(101-n)}{(101-n)}=\sum_{n=1}^{100}n=5050$$

$$y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\dots\infty}}}}$$
$$=\sqrt{x+\sqrt{y+y}}$$
or $$y^2=x+\sqrt{2y}$$
Differentiating w.r.t. $$x$$ , we get
or $$\displaystyle 2y\frac{dy}{dx}=1+\frac{1}{\sqrt{2y}}\times\frac{dy}{dx}$$
or $$\displaystyle\frac{dy}{dx}\left[2y-\frac{1}{\sqrt{2y}}\right]=1$$
or $$\displaystyle\frac{dy}{dx}=\frac{\sqrt{2y}}{2y\sqrt{2y}-1}$$
$$\displaystyle =\frac{y^2-x}{2y^3-2xy-1}$$

$$\lim_{x\rightarrow 0}\frac{f(x)}{x}=\frac{f(x+y)-f(y)-x^{2}y-xy^{2}}{x}$$
Applying L'Hospital's Rule,
$$=\lim_{x\rightarrow 0}\frac{f(x)}{x}=\frac{f'(x+y)-2xy-y^{2}}{1}=1(given)$$
$$=\frac{f'(y)-y^{2}}{1}=1(given)$$
Let $$y=3,$$
$$f'(y)-y^{2}=1$$
$$f'(3)-3^{2}=1$$
$$f'(3)=10$$

Here, $$f(x)+f(y)+f(x)\cdot f(y)=1$$ .....(i)
Substitute $$x=y=0$$ , we get

$$2f(0)+\left \{f(0)\right \}^2=1\Rightarrow \left \{f(0)\right \}^2+2f(0)-1=0$$

$$f(0)=\frac {-2\pm \sqrt {4+4}}{2}=-1-\sqrt 2$$ and $$-1+\sqrt 2$$

As $$f(0) > 0\Rightarrow f(0)=\sqrt 2-1$$ [neglecting $$f(0)=-1-\sqrt 2$$ as f(0) is positive]

Again, putting $$y=x$$ is Eq, (i), $$2f(x)+\left \{f(x)\right \}^2=1$$

On differentiating w.r.t. x, $$2f'(x)+2f(x)f'(x)=0$$
$$2f'(x)\left \{1+f(x)\right \}=0\Rightarrow f'(x)=0$$  because $$f(x) > 0$$
Thus, $$f'(x)=0$$ when $$f(x) > 0$$

Let $$y=e^x+x$$
on differentiating w.r.t y,
$$1=(e^x+1)\dfrac {dx}{dy}\Rightarrow \dfrac {dx}{dy}=\dfrac {1}{1+e^x}$$
$$\therefore \left (\dfrac {dx}{dy}\right )_{x=log_e2}=\dfrac {1}{1+e^{log 2}}=\dfrac {1}{3}$$
or $$\left [\dfrac {d}{dx}(f^{-1}(x))\right ]_{x=log 2}=\dfrac {1}{3}$$

$$f(0)=0$$
So, $$\displaystyle\lim _{x\rightarrow 0}{\frac{f(x)}{x}}=f^\prime(0)=1$$ (Using L'Hospital's rule)
$$f(x+y)=f(x)+f(y)+x^2y+xy^2$$
Differentiating w.r.t. $$x$$ , keeping $$y$$ constant, we get
$$f^\prime(x+y)=f^\prime(x)+2xy+x^2$$
Substitute $$x=0$$ . Then
$$f^\prime(y)=y^2+1$$ or $$f^\prime(x)=x^2+1$$
$$\displaystyle\therefore f(x)=\frac{x^3}{3}+x+c$$
$$f(0)=0=c$$
$$\displaystyle\therefore f(x)=\frac{x^3}{3}+x$$
$$f^\prime(3)=10$$
and $$f(9)=243+9=252$$

$$f(x)=-\displaystyle \frac {x^3}{3}+x^2 \sin 1.5 a-x \sin a\cdot \sin 2a-5 arc \sin (a^2-8a+17)$$
$$f(x)=\displaystyle \frac { -{ x }^{ 3 } }{ 3 } +{ x }^{ 2 }\sin { 6 } -x\sin { 4\sin { 8 } -5\sin ^{ -1 }{ \left( { \left( a-4 \right)  }^{ 2 }+1 \right)  }  }$$
$$f'(x)=-{ x }^{ 2 }+2x\sin { 6-\sin { 4.\sin { 8 }  }  }$$ .......( $$a=4$$ considering the domain of inverse sine function) 
$$f'\left( \sin { \quad 8 }  \right) =-\sin ^{ 2 }{ 8 } +2\sin { 6 } \sin { 8 } -\sin { 4 } \sin { 8 }$$
$$=\sin { 8\left( -\sin { 8+2\sin { 6-\sin { 4 }  }  }  \right)  }$$
$$=-\sin { 8 } \left( \sin { 8 } +\sin { 4-2\sin { 6 }  }  \right)$$
$$=-\sin { 8\left( 2\sin { 6\cos { 2-2\sin { 6 }  }  }  \right)  }$$
$$=2\sin { 8 } \sin { 6\left( 1-\cos { 2 }  \right)  }$$
Now, 
$$sin 8 > 0$$           (Since, $$2\pi  <  8 < 3\pi$$ )
$$sin 6 < 0$$          (Since, $$\pi < 6 < 2\pi$$ )
$$(1-\cos 2) > 0$$      
Hence, $$f'(\sin { 8 } )<0$$

$$h'(x) = 2f(x) \times f'(x) + 2 g(x) g'(x)$$

Let $$y={(x\cos{x})}^x$$ .
Taking logarithm on both sides, we get
$$\log{y}=\log{{(x\cos{x})}^x}$$
$$=x\log{(x\cos{x})}$$
$$=x\log{x}+x\log{\cos{x}}$$
Differentiating both sides with respect to $$x$$ , we get

$$\displaystyle\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x\log{x})+\frac{d}{dx}(x\log{\cos{x}})$$
$$\displaystyle \frac { dy }{ dx } =y\left[ \left( 1+\log { x }  \right) +\left( \log { \cos { x\quad +\displaystyle \frac { x }{ \cos { x }  } \displaystyle \frac { d }{ dx } \left( \cos { x }  \right)  }  }  \right)  \right]$$
$$\therefore \displaystyle\frac{dy}{dx}={(x\cos{x})}^x\left[(\log{x}+1)+\left\{\log{\cos{x}}+\frac{x}{\cos{x}}.(-\sin{x})\right\}\right]$$

We can write $$|x|$$ as $$\sqrt{x^{2}}$$
so, $$\frac{\mathrm{d} (\sqrt{x^{2}})}{\mathrm{d} x}=\frac{2x}{2\sqrt{x^{2}}}=\frac{x}{|x|}$$
and, $$\frac{\mathrm{d} (a_{0}x^{n})}{\mathrm{d} x}=na_{0}x^{n-1}$$
Similarly, $$\frac{\mathrm{d} (a_{n-1}x)}{\mathrm{d} x}=a_{n-1}$$ and $$\frac{\mathrm{d} (a_{n})}{\mathrm{d} x}=0$$
Hence, option A is the correct option. 

Equation $$x^n-1=0, n > 1, n\epsilon N$$ , has roots $$1, a_2, \dots a_n$$ .
We have to find the value of $$\displaystyle \sum_{r=2}^{n}\frac {1}{2-a_r}$$
$${ x }^{ n }-1=\left( x-1 \right) \left( x-a_{ 2 } \right)\dots\left( x-{ a }_{ n } \right)\rightarrow$$ {i}
Taking log on both the sides
$$\log { \left( { x }^{ n }-1 \right) =\log { \left( x-1 \right)} +\log { \left( x-{ a }_{ 2 } \right) +\dots+ } \log { \left( x-{ a }_{ n } \right)  }  }$$
Differentiating both the sides, we get
$$\displaystyle \frac { n{ x }^{ n-1 } }{ { x }^{ n }-1 } =\displaystyle \frac { 1 }{ x-1 } +\displaystyle \frac { 1 }{ x-{ a }_{ 2 } } +\dots+\displaystyle \frac { 1 }{ x-{ a }_{ n } } \rightarrow$$ {ii}
Putting $$x=2$$ in eqn {ii}
$$\displaystyle \frac { n{ 2 }^{ n-1 } }{ { 2 }^{ n }-1 } =1+\displaystyle \frac { 1 }{ 2-{ a }_{ 2 } } +\dots+\displaystyle \frac { 1 }{ 2-{ a }_{ n } }$$
$$\displaystyle \frac { 1 }{ 2-{ a }_{ 2 } } +\dots+\displaystyle \frac { 1 }{ 2-{ a }_{ n } } =\displaystyle \frac { n{ 2 }^{ n-1 } }{ { 2 }^{ n }-1 } -1=\displaystyle \frac { n{ 2 }^{ n-1 }-{ 2 }^{ n }+1 }{ { 2 }^{ n }-1 }$$
$$\displaystyle \frac { 1 }{ 2-{ a }_{ 2 } } +\dots+\displaystyle \frac { 1 }{ 2-{ a }_{ n } } =\displaystyle \frac { { 2 }^{ n-1 }\left( n-2 \right) +1 }{ { 2 }^{ n }-1 }$$

We have,
$$f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}, 1 < x < 26$$

We have, $$\lambda =\sqrt { { y }^{ 2 }+{ z }^{ 2 } }$$

$$f(x)=\left( kx+{ e }^{ x } \right) .h(x)$$
$$h(0)=5,h'(0)=-2,f'(0)=18$$
$$f'(x)=\displaystyle \dfrac { d }{ dx } \left[ \left( kx+{ e }^{ x } \right) h(x) \right]$$
$$\Rightarrow f'(x)=\left( kx+{ e }^{ x } \right) \displaystyle \frac { d }{ dx } h(x)+h(x)\frac { d }{ dx } \left( kx+{ e }^{ x } \right)$$
$$\Rightarrow f'(x)=\left( kx+{ e }^{ x } \right) h'(x)+h(x)\left( k+{ e }^{ x } \right)$$
put $$x=0$$ , we get
$$f'(0)=\left( 0+1 \right) h'(0)+h(0)\left( k+{ e }^{ 0 } \right)$$
$$f'(0)=h'(0)+h(0)\left( k+1 \right)$$
$$18=-2+5\left( k+1 \right)$$
$$20=5\left( k+1 \right)$$
$$4=k+1$$
$$k=3$$

We have,
$$f(x) + f(y) + f(z) + f(x)f(y)f(z) = 14$$ for all $$x,\space y,\space z \in R \quad\quad ...(i)$$
Putting $$x = y = z = 0$$ , we get,
$$3f(0) + \left\{f(0)\right\}^3 = 14$$
$$\left\{f(0)\right\}^3 + 3f(0) - 14 = 0$$
$$f(0) = 2.$$
Now, putting $$y = z = x$$ in $$(i)$$ , we get
$$3f'(x) + 3\left\{f(x)\right\}^2f'(x) = 0 \quad$$ for all $$x \in R$$
$$\left\{\left\{f(x)\right\}^2 + 1\right\}f'(x) = 0 \quad$$ for all $$x \in R$$
$$f'(x) = 0 \quad$$ for all $$x \in R$$

Let $$g'\left( 1 \right) =a$$ and $$g'\left( 2 \right) =b$$ --------(1)
Then 
$$f\left( x \right) ={ x }^{ 2 }+ax+b,f\left( 1 \right) =1+a+b\\ f'\left( x \right) =2x+a,f''\left( x \right) =2\\ \therefore g\left( x \right) =\left( 1+a+b \right) { x }^{ 2 }+\left( 2x+a \right) x+2\\ ={ x }^{ 2 }\left( 3+a+b \right) +ax+2\\ \Rightarrow g'\left( x \right) =2x\left( 3+a+b \right) +a$$

$$\sqrt {y+x}+\sqrt {y-x}=c$$
Squaring both sides, we get
$${ \left( \sqrt { y+x } +\sqrt { y-x }  \right)  }^{ 2 }={ c }^{ 2 }$$
$$2y+2\sqrt { { y }^{ 2 }-{ x }^{ 2 } } ={ c }^{ 2 }$$
$$y+\sqrt { { y }^{ 2 }-{ x }^{ 2 } } =\displaystyle \frac { { c }^{ 2 } }{ 2 }$$ ...........(1)
Differentiating bothe the sides, we get
$$\displaystyle \frac { dy }{ dx } +\displaystyle \frac { 1 }{ 2\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  } \left( 2y\displaystyle \frac { dy }{ dx } -2x \right) =0$$
$$\displaystyle \frac { dy }{ dx } +\displaystyle \frac { y }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  } \displaystyle \frac { dy }{ dx } -\displaystyle \frac { x }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  } =0$$
$$\displaystyle \frac { dy }{ dx } \left( 1+\displaystyle \frac { y }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }  \right) =\displaystyle \frac { x }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }$$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { x }{ y+\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }$$ ..................Option (b)
On rationalizing we get,
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { x\left( y-\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  \right)  }{ { x }^{ 2 } }$$
$$=\displaystyle \frac { y-\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }{ x }$$ ........Option (c)
Putting $$y+\sqrt { { y }^{ 2 }-{ x }^{ 2 } } ={ c }^{ 2 }$$ . we get
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { x }{ \displaystyle \frac { { c }^{ 2 } }{ 2 }  } =\displaystyle \frac { 2x }{ { c }^{ 2 } }$$ ......... Option (a)

Equation $$x^n-1=0, n > 1, n\epsilon N$$ , has roots $$1, a_1, a_2, \dots a_n$$
We have to find value of  $$\displaystyle \sum_{r=2}^{n}\frac {1}{1-a_r}$$
$${ x }^{ n }-1=\left( x-1 \right) \left( x-a_{ 2 } \right)\dots\left( x-{ a }_{ n } \right)\rightarrow$$ {i}
Taking log on both the sides
$$\log { \left( { x }^{ n }-1 \right) =\log { \left( x-1 \right) + } \log { \left( x-{ a }_{ 2 } \right) +\dots+ } \log { \left( x-{ a }_{ n } \right)  }  }$$
Differentiating both the sides, we get
$$\displaystyle \frac { n{ x }^{ n-1 } }{ { x }^{ n }-1 } =\displaystyle \frac { 1 }{ x-1 } +\displaystyle \frac { 1 }{ x-{ a }_{ 2 } } +\cdots+\displaystyle \frac { 1 }{ x-{ a }_{ n } }\rightarrow$$ {ii}

$$\displaystyle \frac { n{ x }^{ n-1 } }{ { x }^{ n }-1 } -\displaystyle \frac { 1 }{ x-1 } =\displaystyle \frac { 1 }{ x-{ a }_{ 2 } } +\dots\displaystyle \frac { 1 }{ x-{ a }_{ n } }$$

$$\because (x^n-1)=(x-1)(1+x+x^2+x^3+x^4+\dots+x^{n-1})$$
$$\therefore \displaystyle \frac { n{ x }^{ n-1 }-1\left( 1+x+{ x }^{ 2 }+\dots+{ x }^{ n-1 } \right)  }{ { x }^{ n}-1 } =\displaystyle \frac { 1 }{ x-{ a }_{ 2 } } +\dots\displaystyle \frac { 1 }{ x-{ a }_{ n } }$$

$$\displaystyle\lim _{ x\rightarrow 1 }{ \left( \displaystyle \frac { n{ x }^{ n-1 }-1\left( 1+x+{ x }^{ 2 }+\dots+{ x }^{ n-1 } \right)  }{ { x }^{ n}-1  }  \right)  } =\lim _{ x\rightarrow 1 }{ \left(\displaystyle \frac { 1 }{ x-{ a }_{ 2 } } +\dots\displaystyle \frac { 1 }{ x-{ a }_{ n } }  \right)  }$$

Applying L'Hospital's rule on L.H.S
$$\displaystyle\lim _{ x\rightarrow 1 }{\displaystyle  \frac { n\left( n-1 \right) { x }^{ n-2 }-\left\{ 1+2x+\cdots+\left( n-1 \right) { x }^{ n-2 } \right\}  }{ n{ x }^{ n-1 } } =\displaystyle \frac { 1 }{ 1-{ a }_{ 2 } } +\dots\displaystyle \frac { 1 }{ 1-{ a }_{ n } }  }$$  

$$\Rightarrow \displaystyle \frac { n\left( n-1 \right) -\left[ 1+2+\dots+\left( n-1 \right)  \right]  }{ n } =\displaystyle \frac { 1 }{ 1-{ a }_{ 2 } } +\dots\displaystyle \frac { 1 }{ 1-{ a }_{ n } }$$

$$\therefore\displaystyle \frac { 1 }{ 1-{ a }_{ 2 } } +\dots\displaystyle \frac { 1 }{ 1-{ a }_{ n } } =\displaystyle \frac { n-1 }{ 2 }$$

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ 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 $$(\pi, \pi)$$
$$\cos { y\displaystyle \frac { dy }{ dx }  } +\cos { x } =\sin { x } \cos { y\displaystyle \frac { dy }{ dx } +\sin { y } \cos { x }  }$$
$$\displaystyle \frac { dy }{ dx } \left( \cos { y } -\sin { x } \cos { y }  \right) =\sin { y\cos { x-\cos { x }  }  }$$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x\left( \sin { y-1 }  \right)  }  }{ \cos { y\left( 1-\sin { x }  \right)  }  }$$
$$B=\displaystyle \frac { -1\left( 0-1 \right)  }{ -1\left( 1-0 \right)  } =-1$$
$$C: \displaystyle \frac {d}{dx}(2e^{xy}+e^xe^y-e^x=e^{xy+1})$$ at $$(1, 1)$$
$$\left[ 2{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right)  \right] +{ e }^{ x }{ e }^{ y }\displaystyle \frac { dy }{ dx } +{ e }^{ y }{ e }^{ x }-{ e }^{ x }={ e }^{ xy+1 }\left( x\displaystyle \frac { dy }{ dx } +y \right)$$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { y{ e }^{ xy+1 }-2y{ e }^{ xy }-{ e }^{ x+y }+{ e }^{ x } }{ 2x{ e }^{ xy }+{ e }^{ x+y }-x{ e }^{ xy+1 } }$$ .......... since $${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$$
$$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 } }$$
$$=\displaystyle \frac { -{ e }^{ 1 } }{ 2{ e }^{ 1 } } =-\displaystyle \frac { 1 }{ 2 }$$
Therefore $$A-B-C=\displaystyle \frac { 1 }{ 2 }$$

$$f(x)=\sqrt {x+2\sqrt {2x-4}}+\sqrt {x-2\sqrt {2x-4}}$$
$$f(x)=\sqrt { { \left( \sqrt { x-2 } +\sqrt { 2 }  \right)  }^{ 2 } } +\sqrt { { \left( \sqrt { x-2 } -\sqrt { 2 }  \right)  }^{ 2 } }$$
$$f(x)=\left| \sqrt { x-2 } +\sqrt { 2 }  \right| +\left| \sqrt { x-2 } -\sqrt { 2 }  \right| \\ $$
For $$\sqrt { x-2 }$$   to exist, $$x\ge 2$$
Also $$\sqrt { x-1 } +\sqrt { 2 } >0$$ .............(always true)
But $$\sqrt { x-1 } -\sqrt { 2 } \ge 0\quad$$ only if $$x\ge 4$$
                                         $$<0$$ only if $$x<4$$
Now $$f(x)$$ becomes
$$f(x)=\sqrt { x-2 } +\sqrt { 2 } -\sqrt { x-2 } +\sqrt { 2 }$$ for $$2\le x<4$$
$$f(x)=\sqrt { x-2 } +\sqrt { 2 } +\sqrt { x-2 } -\sqrt { 2 }$$ for $$x\ge 4$$
$$f(x)=2\sqrt { 2, }$$ for $$2\le x<4$$
$$f(x)=2\sqrt { x-2 }$$ for $$4\le x<\infty$$

Let $$f\left( x \right) ={ px }^{ 2 }+qx+r$$
Differentiatin g w.r. to $$x$$ , we get
$$f^{ ' }\left( x \right) =2px+q$$
Now,
$$f\left( -1 \right) =f\left( 1 \right)$$
$$\therefore p-q+r=p+q+r$$
$$\therefore q=0$$
$$\therefore f^{ ' }\left( x \right) =2px$$
$$f^{ ' }\left( a \right) =2pa.f^{ ' }\left( b \right) =2pb,f^{ ' }\left( c \right) =2pc$$
Since $$a,b,c$$ are in $$A.P.$$ , therefore
$$2b=a+c$$
$$4pb=2pa+2pc$$
$$2f^{ ' }\left( b \right) =f^{ ' }\left( a \right) +f^{ ' }\left( c \right)$$
$$\therefore f^{ ' }\left( a \right) ,f^{ ' }\left( b \right) ,f^{ ' }\left( c \right)$$ are in A.P

$$\displaystyle y=\frac{log(3^{x}-1)}{log2}$$
$$\displaystyle y'=\frac{3^{x}log3}{(3^{x}-1)log2}$$
At $$x\rightarrow \infty, \displaystyle\frac{3^{x}}{3^{x}-1} \rightarrow 1$$

So, $$\displaystyle y'=\frac{log3}{log2}$$
Hence, option C is the correct option.

$$\displaystyle x^{2}+y^{2}=R^{2}$$
$$\displaystyle x=-yy' => y'=\frac{-x}{y}$$
$$\displaystyle y''=\frac{-y+xy'}{y^{2}}=\frac{-y^{2}-x^{2}}{y^{3}}=-\frac{R^{2}}{y^{3}}$$
$$\displaystyle k=\frac{y''}{\sqrt{(1+y'^{2})^{3}}}=\frac{-\frac{R^{2}}{y^{3}}}{\sqrt{(1+\frac{x^{2}}{y^{2}})^{3}}}$$
$$\displaystyle =\frac{-\frac{R^{2}}{y^{3}}}{\sqrt{(\frac{R^{2}}{y^{2}})^{3}}}$$
$$\displaystyle =-\frac{R^{2}}{R^{3}}=\frac{-1}{R}$$

$$\Rightarrow \displaystyle f(0)=\frac{1}{2}, g(0)=4, f'(0)=16, g'(0)=8$$