Differentiation Test 44

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ 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 $${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$$
$$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 }-{ e }^{ 1 } }=-{1} $$
$$A+B+C=-1-1-1=-3$$

$$f(x)=\sqrt {x+2\sqrt {2x-4}}+\sqrt {x-2\sqrt {2x-4}}$$
$$f(x)=\sqrt { { \left( \sqrt { x-2 } +\sqrt { 2 }  \right)  }^{ 2 } } +\sqrt { { \left( \sqrt { x-2 } -\sqrt { 2 }  \right)  }^{ 2 } } $$
$$f(x)=\left| \sqrt { x-2 } +\sqrt { 2 }  \right| +\left| \sqrt { x-2 } -\sqrt { 2 }  \right| \\ $$
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 $$4\le x<\infty $$

Since total number of digits $$=6$$
So to get a number of a number of the form $$\dfrac { p }{ q } $$ we will have to choose two numbers out of the 6 numbers.
There is only one way to arrange them since the number $$x$$ has to satisfy 0 Hence no.  $$x = \left( \begin{matrix} 6 \\ 2 \end{matrix} \right) \times 1=15$$,  but some numbers have to be removed because they are repeated.
They are :
$$\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } \\ \dfrac { 4 }{ 6 } =\dfrac { 2 }{ 3 } \\ \dfrac { 2 }{ 6 } =\dfrac { 1 }{ 3 } \\ \dfrac { 3 }{ 6 } =\dfrac { 1 }{ 2 } $$
Thus the number of removed numbers$$=4$$
Therefore, total no. of distinct $$x = 15-4=11$$
Hence, option 'D' is correct.

In order to solve this question, we must observe the number of ways in which we can select the first term of the required A.P. for different values of the common difference($$r$$) starting from $$r=1$$ given that there are only $$3$$ terms before $$n$$.
For $$r=1$$, the number of ways in which we can select the first term of the A.P. $$=n-2$$
For $$r=2$$, the number of ways to select the first term $$=n-4$$
For $$r=3$$, the number of ways to select the first term $$=n-6$$
Now we see a pattern emerging, we also realize from this that $$r<=\dfrac{n-1}{2}$$ for an A.P. with 3 terms to exist in the given interval.
$$\therefore$$ The final answer $$=n-2+n-4+n-6+...+5+3+1$$
Now we use the formula to find the sum of an A.P. which is $$S_n=\dfrac{n}{2}[a_1+a_n]$$ 
$$\therefore$$ Answer $$=\dfrac{n-1}{4}[1+n-2]=\dfrac{(n-1)^2}{4}$$

We have, 

Total number of wrong answers $$= 2^{n-1}+2^{n-2}+...+2^{2}+2+1$$
$$=2^{n-1}+2^{n-2}+...+2^{4}+2^{3}+2^{2}+2+1$$
$$\Rightarrow 2^{n}-1=2043$$  [Using formula for sum of G.P]
$$ \Rightarrow 2^{n}=2048$$
$$ \Rightarrow 2^{n}=2^{11}$$
$$\Rightarrow n = 11$$
Hence, option 'B' is correct.

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