TS EAMCET 2018 :: Mathematics :: General questions

• Mathematics - General questions

 if z and w are complex  numbers such that 

$\overline{z}-i\overline{w}=0 $  and Arg (zw) = $\frac{3 \pi}{4}$, then  Arg z=

Answer:

Explanation:

 We have,     $\overline{z}-i\overline{w}=0 $

$\Rightarrow$   $i\overline{w}=\overline{z}\Rightarrow w=\frac{1}{i}\overline{z}$

$\Rightarrow$     $w=-\frac{1}{i}z\Rightarrow w=iz$

Now , we have

             arg(zw)= $\frac{3\pi}{4}$

$\Rightarrow$    arg(z(iz))= $\frac{3\pi}{4}$

$\Rightarrow$      $arg(iz^{2})$= $\frac{3\pi}{4}$

$\Rightarrow$   $arg(i)+arg(z^{2})$= $\frac{3 \pi}{4}$

                                              $[ \because arg(z_{1} z_{2})= arg(z_{1})+arg(z_{2})]$

$\Rightarrow$     $arg(i)+2arg(z)= \frac{3\pi}{4}$             [ $\because arg (z^{n})=n arg(z)]$

$\Rightarrow$       $\frac{\pi}{2}+2 arg(z)= \frac{3 \pi}{4}$

$\Rightarrow$       arg(z)= $\frac{\pi}{8}$

The set of real values of $\alpha$ for which the system of linear equations

 $x+(\sin \alpha)y+(\cos \alpha)z=0$

$x+(\cos \alpha)y+(\sin \alpha )z=0$

$-x+(\sin  \alpha )y-(\cos \alpha  )z=0$

has a non-trivial solution is 

Answer:

Explanation:

 $x+\sin \alpha y+\cos \alpha z=0, x+ \cos \alpha y+ \sin \alpha z=0$

  $-x+\sin  \alpha y-\cos \alpha z=0$

 Non-trivial sol, so  

$\begin{bmatrix}1 & \sin \alpha&  \cos \alpha \\1 & \cos \alpha& \sin \alpha\\ -1 & \sin \alpha & -\cos \alpha \end{bmatrix}=0$

  $\Rightarrow$    $| 1(-\sin^{2}  \alpha -\cos^{2} \alpha)-\sin \alpha(-\cos \alpha+\sin \alpha)+ \cos \alpha( \sin \alpha+\cos \alpha)|$=0

 $ -1- \sin  \alpha (- \cos \alpha+ \sin \alpha] +\cos \alpha [ \sin \alpha+\cos  \alpha]=0$

$-1+ \sin \alpha \cos \alpha- \sin^{2} \alpha+ \sin \alpha \cos \alpha + \cos ^{2} \alpha=0$

    $-1+ \sin 2 \alpha+\cos 2 \alpha=0$

$\sin 2 \alpha+\cos 2 \alpha=1$

  $\Rightarrow$       $\frac{1}{\sqrt{2}}\sin\alpha+\frac{1}{\sqrt{2}} \cos\alpha=\frac{1}{\sqrt{2}}$

  $\Rightarrow$   $\sin\left(2\alpha+\frac{\pi}{4}\right)=\sin\frac{\pi}{4}$

  $\Rightarrow$  $2\alpha+\frac{\pi}{4}=n\pi+(-1)^{n}\frac{\pi}{4}$ 

               $2\alpha=n\pi+(-1)^{n}\frac{\pi}{4}-\frac{\pi}{4}$

  $\Rightarrow$  $\alpha$= $\frac{n \pi}{2}+(-1)^{n} \frac{\pi}{8}-\frac{\pi}{8}$ 

$\begin{bmatrix}1 & bc+ad&b^{2}c^{2}+a^{2}{d}^{2} \\1 & ca+bd & c^{2}a^{2}+b^{2}d^{2} \\ 1 & ab+cd & a^{2}b^{2}+c^{2}d^{2} \end{bmatrix}$=

Answer:

Explanation:

$\begin{vmatrix}1 & 1&1 \\bc+ad & ca+bd &ab+cd \\ b^{2}c^{2}+a^{2}{d}^{2} & c^{2}a^{2}+b^{2}d^{2} & a^{2}b^{2}+c^{2}d^{2} \end{vmatrix}$

 Applying $c_{1} \rightarrow  c_{1}-c_{2}  $  and  $c_{2}= c_{2}-c_{3}$

$\begin{vmatrix}0 & 0 &1\\ c(b-a)+d(a-b) & c(a-d)+b(d-a)&ab+cd \\ c^{2}(b^{2}-a^{2})+d^{2}(a^{2}-b^{2}) & c^{2}(a^{2}-d^{2})+b^{2}(d^{2}-a^{2}) & a^{2}b^{2}+c^{2}d^{2} \end{vmatrix}$

$\Rightarrow 1[\left\{c(b-a)+d(a-b)\right\}\left\{c^{2}(a^{2}-d^{2})+b^{2}(d^{2}-a^{2})\right\}$

$\Rightarrow-[\left\{c(a-d)+b(d-a)\right\}\left\{c^{2}(b^{2}-a^{2})+d^{2}(a^{2}-b^{2})\right\}]$

 often after simplify we are getting

  (a-b)(a-c)(b-c)(b-d)(a-d)(c-d)

$x^{n}+y^{n}$  is divisible by 

Answer:

Explanation:

Given , $x^{n}+y^{n}$

 at n=1,x+y , which is divisible by x+y

 n=2, $x^{2}+y^{2}$, which is  not divisble by x+y

n=3, $x^{3}+y^{3}$ ,  which is divisble  by x+y

 Hence, clearly $x^{n}+y^{n}$ is divisble by n= odd numbers  as n=2m-1 , where   $M \in  N$.

A function  $f:R-\left\{0\right\}\rightarrow R$ is defined as  

$f(x)=\begin{cases}x^{2}+3x-7, & x > 0\\h(x) ,& x < 0\end{cases}$

 If f(x)  is an odd function, then h(x)=

Answer:

Explanation:

$f(x)=\begin{cases}x^{2}+3x-7, & x > 0\\h(x) ,& x < 0\end{cases}$

 f(x)  is odd function, then h(x)=?

 forr odd function

 $h(-x)= -h(x) \Rightarrow   h(-x)= (-x)^{2}-3x-7$

$\Rightarrow$  $ h(-x)=x^{2}-3x-7$

$\Rightarrow$  $-h(x)=-[x^{2}-3x-7] \Rightarrow $   $-x^{2}+3x+7$

• Mathematics - General questions