TS EAMCET 2018 :: Mathematics :: General questions

• Mathematics - General questions

$\sin h^{-1}2+\cos h^{-1}2-\tan h^{-1} \frac{2}{3}+\cot h^{-1}(-2)=$

Answer:

Explanation:

 We have 

$\sin h^{-1}2+\cos h^{-1}2-\tan h^{-1}\frac{ 2}{3}+\cot h^{-1}(-2)$

 = $ln(2+\sqrt{2^{2}+1})+ln(2+\sqrt{2^{2}-1})-\frac{1}{2}ln\left(\frac{1+2/3}{1-2/3}\right)+\frac{1}{2}ln\left(\frac{-2+1}{-2-1}\right)$

=$ln(2+\sqrt{5})+ln(2+\sqrt{3})-\frac{1}{2}ln 5+\frac{1}{2} ln \frac{1}{3}$

=$ln\left[\frac{(2+\sqrt{5})(2+\sqrt{3})\sqrt{3}}{\sqrt{5}}\right]$

                    $\because \sin h^{-1}x=ln(x+\sqrt{x^{2}+1)}, \cos h^{-1}x=ln(x+\sqrt{x^{2}-1)}$,

             $\tan h^{-1}x=\frac{1}{2}ln\left(\frac{1+x}{1-x}\right), \cot h^{-1}x=\frac{1}{2}ln \left(\frac{x+1}{x-1}\right)$

If n is a positive integer greater than 1, then  $3(^nC_{0})-8(^nC_{1})+13(^nC_{2})-18(^nC_{3})+... $    $upto (n+1)$ terms =

Answer:

Explanation:

 The general  term of the  given series  is 

$T_{r}=(-1)^{r}(3+5r)^{n}C_{r},r=0,1,2,.....,n$

 $\therefore$    $S_{n}=\sum_{r=0}^{n}(-1)^{r}(3+5r)^{n} C_{r}$

 $=3\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}+5\sum_{r=0}^{n}(-1)^{r}.{r^{n}}C_{r}$

    $=3\left[\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}\right]+5\left[\sum_{r=1}^{n}(-1)^{r}.{r^{}\frac{n}{r}}.{^{n-1}}C_{r-1}\right]$

$=3\left[\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}\right]+5n\left[\sum_{r=1}^{n}(-1)^{r}.{^{n-1}}C_{r-1}\right]$

 =$3(0)+5n(0)=0+0=0$

If $z=x+iy$ is a complex number satisfying  $|\frac{z-2i}{z+2i}|=2$ and the locus of z is a circle , then its radius is 

Answer:

Explanation:

 We have  

$|\frac{z-2i}{z+2i}|=2 \Rightarrow $ $|\frac{x+iy-2i}{x+iy+2i}|=2$

 $\Rightarrow$     $\mid\frac{x+(y-2)i}{x+(y+2)i}|=2\Rightarrow\frac{|x+(y-2)i|}{|x+(y+2)i|}=2$

$\Rightarrow$    $\frac{\sqrt{x^{2}+(y-2)^{2}}}{\sqrt{x^{2}+(y+2)^{2}}}=2$

$\Rightarrow$          $x^{2}+(y-2)^{2}=4[x^{2}+(y+2)^{2}]$

$\Rightarrow$            $x^{2}+y^{2}-4y+4=4x^{2}+4y^{2}+16y+16$

 $\Rightarrow$       $x^{2}+y^{2}+\frac{20}{3}y+4=0$

$\Rightarrow$      $ x^{2}+\left(y+\frac{10}{3}\right)^{2}+4-\frac{100}{9}=0$

 $\Rightarrow$    $  x^{2}+\left(y+\frac{10}{3}\right)^{2}=\left(\frac{8}{3}\right)^{2}$

 Which is equation of circle with centre $(0, \frac{-10}{3})$ and radius $\frac{8}{3}$

The rank of the matrix  $\begin{bmatrix}3 & 2&1&-4 \\2 & 3&0&-1\\1&-6&3&-8 \end{bmatrix}$ is 

Answer:

Explanation:

$\begin{bmatrix}3 & 2&1&-4 \\2 & 3&0&-1\\1&-6&3&-8 \end{bmatrix}$  

On applying  $R_{2} \rightarrow R_{2}-\frac{2}{3} R_{1}$  and $R_{3} \rightarrow  R_{3}-\frac{1}{3} R_{1}$

 we get

 A=   $\begin{bmatrix}3 & 2&1&-4 \\0 &5/3&-2/3&5/3\\0&-20/3&8/3&-20/3 \end{bmatrix} $

 On applying $R_{3} \rightarrow R_{3}+4R_{2}$, we get

$A=\begin{bmatrix}3 & 2&1&-4 \\0 &5/3&-2/3&5/3\\0&0&0&0 \end{bmatrix} $

$\therefore$   There are two linear  independent row, i.e.

$R_{1} and R_{2}$ 

$\therefore$    Rank of matrix A=2

 If  $A=\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}$ and   $ I=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$  then for all  $n\in N$

Answer:

Explanation:

 We have

$A=\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}$ and   $ I=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$

 Now,     $A^{2}=A.A=$  $\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 2 \\0 & 1 \end{bmatrix}$

 =$2\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}-\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}=2A-(2-1)I$

Again,     $A^{3}=A^{2}.A=$  $\begin{bmatrix}1 & 2 \\0 & 1 \end{bmatrix}\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 3 \\0 & 1 \end{bmatrix}$

 $=3\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}-2\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=3A-(3-1)I$

 $\therefore$     $A''= nA-(n-1)I$

• Mathematics - General questions