Maths

$e^{2x}-4=0or6e^{2x}-5e^x+1=0$

$\Rightarrow x=\mathcal{l}n2x=\mathcal{l}n\left(\frac{1}{3}\right),\mathcal{l}n\left(\frac{1}{2}\right)$

= $-\mathcal{l}n3,-\mathcal{l}n2$

$\left(x^2+1\right)*1=x*2$

$\Rightarrow {\left(x^2+1\right)}^2+1=x^2+8\Rightarrow x^2=2$

$2si{n}^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)=2si{n}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

Apart from upper and lower triangular matrices, we have the unit triangular matrix, strictly-triangular matrix and an atomic-triangular matrix.

A few types of matrices are row and column matrices, singleton, null matrix, square, diagonal, scalar, identity, equal, triangular (both upper and lower), singular and non-singular matrices, symmetric, skew-symmetric, hermitian, and orthogonal matrices. NCERT excercise on matrices chapter covers questions related to this topic

An upper triangular matrix is a square matrix where all the elements above the diagonal are non-zero, and below it is zero. A lower triangular matrix is a square matrix where all the elements above the diagonal are zero.

The elements of a matrix may be real or complex numbers. If all the elements of a matrix are real, then the matrix is called a real matrix.

  $\left(1,1\right)\left(1,4\right)\left(4,1\right)\left(2,4\right)\left(4,2\right)\left(3,4\right)\left(4,3\right)\left(4,4\right)$  all have only one image.

(2, 1) (1, 2), (2, 2) each element has 3 choice.

(3, 2) (2, 3) (3, 1) (1, 3) (3, 3) each element has two choices.

$\therefore$ total function = 3 * 3 * 2 * 2 * 2 = 72

Case I

None of the pre image have 3 as image, total functions = 2 * 2 * 1 * 1 * 1 = 4

Case II

None of the pre images have 2 as image then number of function = 2⁵ = 32

Case III

None of the pre image have either 3 or 2 as image

Total function = 1⁵ = 1

$\therefore$ Total number of onto function

= 72 – 4 – 32 + 1 = 37

$\tilde{z}=i{z}^2+z^2-z$

$z+\overline{Z}=z^2\left(i+1\right)$

$z+\overline{Z}=z^2\left(i+1\right)$ Let z be equal to (x + iy)

(x + iy) + (x – iy) = (x + iy)² (i + 1)

$2x=\left(x^2-y^2+2ixy\right)\left(i+1\right)$               

Equating the real & in eg part.

$\left(x^2-y^2+2ixy\right)=0.........\left(i\right)$

$\left(x^2-y^2-2xy\right)=\left(2x\right)..............\left(ii\right)$               

(i) & (ii)

 4xy = -2x Þ x = 0 or y = $\left(\frac{-1}{2}\right)$  

(for x = 0, y = 0)

For y = $\frac{-1}{2}$  

x²   $-\frac{1}{4}+2\left(\frac{-1}{2}\right)x=0$

x =   $\frac{4\pm \sqrt[]{16+16}}{2.4}$

=  $\left(\frac{1+\sqrt[]{2}}{2}\right)or\left(\frac{1-\sqrt[]{2}}{2}\right)$  

$\therefore sum$ of   ${\left|z\right|}^2={\left(\frac{1+\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+{\left(\frac{1-\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+0^2+O^2$

=   $\frac{3}{4}+\frac{\sqrt[]{2}}{2}+\frac{1}{4}+\frac{3}{4}-\frac{\sqrt[]{2}}{2}+\frac{1}{4}=\frac{3}{2}+\frac{1}{2}=2$

$\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)$

$A^2=\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)$

$A^4=\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$

$\therefore A^5=A$

A⁹ = A

$\therefore forn=1,5,9,.....,97$

total possible values of n = 25