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• Mathematics - General questions

$\cot h^{-1}(3)+ \tanh^{-1} \frac{1}{3}- cosec h^{-1} (-\sqrt{3})=$

Answer:

Explanation:

 We know that ,  $\cot h^{-1}x= \frac{1}{2}\log _{e}\left(\frac{x+1}{x-1}\right),|x|>1$

 $\tan h^{-1}x= \frac{1}{2}\log _{e}\left(\frac{1+x}{1-x}\right)$

and   $cosec h^{-1}x=\log _{e}\left(\frac{1-\sqrt{1+x^{2}}}{x}\right),x<0$

 $\therefore$      $\cot h^{-1}(3)=\frac{1}{2}\log _{e}\left(\frac{4}{2}\right)=\log_{e} \sqrt{2}$

 $\tan h^{-1}\left(\frac{1}{3}\right)=\frac{1}{2}\log _{e}\left(\frac{1+1/3}{1-1/3}\right)=\frac{1}{2}\log_{e}\left(\frac{4}{2}\right)=\log_{e} \sqrt{2}$

 and     $cosec h^{-1}(-\sqrt{3})=\log_{e}\left(\frac{1-\sqrt{1+3}}{-\sqrt{3}}\right)$

   =   $\log_{e}\left(\frac{1-2}{-\sqrt{3}}\right)=\log_{e}\left(\frac{1}{\sqrt{3}}\right)$

 Hence,   $\cot h^{-1}(3)+ \tanh^{-1} \frac{1}{3}- cosec h^{-1} (-\sqrt{3})=$

 = $\log _{e}\sqrt{2}+\log_{e}\sqrt{2}-\log_{e}\left(\frac{1}{\sqrt{3}}\right)$

 $=\log _{e}(\sqrt{2}.\sqrt{2})+\log_{e}(\sqrt{3})=\log_{e}(2\sqrt{3})$

$\frac{d}{dx}\left(\frac{x+5}{(x+1)^{2}(x+2)}\right)=$

Answer:

Explanation:

$\frac{x+5}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)^{}}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)^{}}$  ..........(i)

$\Rightarrow$     $x+5= A(x+1)(x+2)+B(x+2)+c(x+1)^{2}$

$\Rightarrow$   $x+5= x^{2}(A+C)+x(3A+B+2C)+2A+2B+C$

on comparing  the coefficients of like power of x , we get

 $A+C=0$,

 $3A+B+2C=1$  and $2A+2B+C=5$

 on substituting  A=-C  in last two equations, we get

    B-C=1  .....(ii)

    and  2B-C=5   .......(iii)

 On subtracting Eqs(ii) and (iii) , we get

              $B=4 \Rightarrow C=3$      [from Eq.(ii)]

and hence     A=-3

 Now,    $\frac{x+5}{(x+1)^{2}(x+2)}=\frac{-3}{(x+1)^{}}+\frac{4}{(x+1)^{2}}+\frac{3}{(x+2)^{}}$

On differentiating  both sides  w.r.t x, we get

$\frac{d}{dx}\left(\frac{x+5}{(x+1)^{2}(x+2)}\right)=$   $\frac{3}{(x+1)^{2}}-\frac{8}{(x+1)^{3}}-\frac{3}{(x+2)^{2}}$

The modulus -amplitude  form of  $\frac{(1-i)^{3}(2-i)}{(2+i)(1+i)}$  is 

Answer:

Explanation:

We have,

   $\frac{(1-i)^{3}(2-i)}{(2+i)(1+i)}=\frac{(1-i)^{2}(1-i)(2-i)}{(1+i)(2+i)}$

$=\frac{-2i(1-i)^{}(1-i)(2-i)(2-i)}{(1+i)(1-i)(2+i)(2-i)}$

$=\frac{-2i(-2i)(3-4i)}{2.5}=-2\left(\frac{3}{5}-\frac{4i}{5}\right)$

$=-2\left(\frac{-3}{5}+\frac{4}{5}i\right)$

$2cis\left( \pi-\tan^{-1}\frac{4}{3}\right)$

If x,y are any two  non-zero real numbers , $a_{ij}= xi+yj, A=(a_{ij})_{n xn}$ and P.Q are two n x n  matrices such that A= xP+ yQ,  then

Answer:

Explanation:

We have  , $a_{ij}=x_{j}+y_{i}$

$A= (a_{i})_{n xn}$

$A=\begin{bmatrix}x+y & 2x+y&3x+y&.....nx+y \\x+2y & 2x+2y&3x+2y &......nx+2y\\ x+3y&...&....&.....\\ x+ny &...&....&nx+ny \end{bmatrix}_{n\times n}$

 $A=\begin{bmatrix}1 & 2&3......&n \\1 & 2&3..... &n\\ 1&...&....&.....\\ 1 &2&3.....&n \end{bmatrix}+y\begin{bmatrix}1 & 1&1.....&1 \\2 & ..&.. &2\\ 3&...&....&3\\ n &..&..&n \end{bmatrix}$

A=xP+yQ

where

$P=\begin{bmatrix}1 & 2&3&n \\1 & 2&3 &n\\ 1&2&....&.....\\ 1 &2&3&n \end{bmatrix}and Q\begin{bmatrix}1 & 1&1....&1 \\2 &2 &2.... &2\\ 3&...&....&3\\ ....n &..&n...&n \end{bmatrix}$

 $\therefore$   (P+Q)   is symmetric and (P-Q) is skew symmetric

If A= $\begin{bmatrix}1 & 2&2 \\2 & 1&2\\2&2&1 \end{bmatrix}$ then $A^{-1}$=

Answer:

Explanation:

We have ,

A= $\begin{bmatrix}1 & 2&2 \\2 & 1&2\\2&2&1 \end{bmatrix}$

|A|= 1(1-4)-2(2-4)+2(4-2)=-3+4+4=5

Adj A= $\begin{bmatrix}-3 & 2&2 \\2 & -3&2\\2&2&-3 \end{bmatrix}$

$A^{-1}=\frac{1}{|A|} Adj A= \frac{1}{5} \begin{bmatrix}-3& 2&2 \\2 & -3&2\\2&2&-3 \end{bmatrix}$

 $A^{-1}= \frac{1}{5}  \begin{bmatrix}1 & 2&2 \\2 & 1&2\\2&2&1 \end{bmatrix}$-$\frac{1}{5} \begin{bmatrix}4 & 0&0 \\0 & 4&0\\0&0&4\end{bmatrix}$

 $A^{-1}= \frac{1}{5} [A-4l]$

• Mathematics - General questions