Maths

$f\left(x\right)=\left|x^2-3x-2\right|-x$

$=\left|\left(x-\frac{3-\sqrt[]{17}}{2}\right)\left(x-\frac{3+\sqrt[]{17}}{2}\right)\right|-x$

$\Rightarrow f\left(x\right)=\left[\begin{array}{cc}\hfill x^2-4x-2;-1\le x\le \frac{3-\sqrt[]{17}}{2}\hfill & \hfill -x^2+2x+2;\frac{3-\sqrt[]{17}}{2}<x\le 2\hfill \end{array}\right]$

absolute minimum $f\left(\frac{3-\sqrt[]{17}}{2}\right)=\frac{-3+\sqrt[]{17}}{2}$  

 absolute maximum = 3

$\therefore sum3+\frac{-3+\sqrt[]{17}}{2}=\frac{3+\sqrt[]{17}}{2}$  

Kindly consider the following figure

B = (I – adjA)⁵

Kindly consider the following figure

B = (I – adjA)⁵

$?$ gof is differentiable at x = 0

 So R.H.D = L.H.D.

$\frac{d}{dx}\left(4e^x+k_2\right)=\frac{d}{dx}\left({\left(-\left|x+3\right|\right)}^2-k_1\left|x+3\right|\right)$  

⇒ 4 = 6 – k₁ Þ k₁ = 2

Now g (f (-4) + g (f (4)

=2 (2e⁴ – 1)

$\left|z\right|=3\Rightarrow$ $$ circle with radius = 3

arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4},$ $\frac{}{}\frac{}{}$ part of a circle (with radius $\text{exception 25:}$ $\sqrt[]{2}$ ). no common points

$\underset{x\to \frac{1}{\sqrt[]{2}}}{lim}\frac{sin\left(co{s}^{-1}x\right)-x}{1-tan\left(co{s}^{-1}x\right)}$

Let $co{s}^{-1}x=\frac{\pi }{4}+\theta$

= $\underset{\theta \to \infty }{lim}\frac{\sqrt[]{2}sin\theta }{-2tan\theta }\left(1-tan\theta \right)=-\frac{1}{\sqrt[]{2}}$

$2021\equiv -2$ mod (7)

$\Rightarrow {\left(2021\right)}^{2023}\equiv {\left(-2\right)}^{2023}mod\left(7\right)$   …… (i)

Now,   ${\left(-2\right)}^3\equiv -1mod\left(7\right)$  

$\Rightarrow {\left(-2\right)}^{2023}\equiv \left(-2\right)mod\left(7\right)\equiv 5mod\left(7\right)$       ……. (ii)

(i) & (ii)

$\Rightarrow {\left(2021\right)}^{2023}\equiv 5mod\left(7\right)$  

$\therefore$  Remainder = 5

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

  $\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

$x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -8\hfill & \hfill 9\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 3\\ -1\end{array}\right)$

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 15\hfill & \hfill -24\hfill & \hfill 27\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 3a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 9\\ -3\end{array}\right)$

$R_3-2R_1,R_2-5R_1$

for no solution

3a + 9 = 0 but $\frac{3}{2}-\frac{9b}{2}\ne 0$

$\Rightarrow a=-3\Rightarrow b\ne \frac{1}{3}$

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$