Determinants

f (x) = |sin²x, -2+cos²x, cos2x; 2+sin²x, cos²x, cos2x; sin²x, cos²x, 1+cos2x|.
R? →R? -R? , R? →R? -R?
f (x) = |sin²x, -2+cos²x, cos2x; 2, 2-2cos²x, 0; 0, 2-2cos²x, 1|.
f (x) = sin²x (2-2cos²x) - (-2+cos²x) (2) + cos2x (2 (2-2cos²x).
This seems tedious. From the solution, f (x)=4+2cos2x.
Max value when cos2x=1, f (x)=6.

R? → R? -R? , R? → R? -R?
|sinx-cosx, cosx-sinx, 0; 0, sinx-cosx, cosx-sinx; cosx, sinx| = 0
(sinx-cosx)² |1, -1, 0; 0, 1, -1; cosx, sinx| = 0
(sinx-cosx)² (1 (sinx+cosx) + 1 (cosx) = 0
(sinx-cosx)² (sinx + 2cosx) = 0
sin x = cos x
tan x = 1 ⇒ x = π/4
or
sin x = -2cos x
tan x = -2
Not within given range.

$A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\Rightarrow \left|A\right|=4\Rightarrow \left|3adj\left(2A^{-1}\right)\right|=\left|3.2^{2}adj\left(A^{-1}\right)\right|$

$\Rightarrow 12^3\left|adj\left(A^{-1}\right)\right|=12^3{\left|A^{-1}\right|}^2=\frac{12^3}{{\left|A\right|}^2}=\frac{12*12*12}{16}=108$

$\left|2A\right|=2^3\left|A\right|$

replace A by adj 2A

  $\left|2adj2A\right|=2^3\left|adj2A\right|$           

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

$=2^9{\left|A\right|}^2$              

Again replace A by (adj A)

|2adj 2 (adj (adj 2A)| = 2⁹ |adj 2A|⁴

= 2⁹ (|2A|²)⁴

->|A²| = 4

  $\Delta =\left|\begin{array}{ccc}\hfill 3sin3\theta \hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 3cos2\theta \hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 6\hfill & \hfill 7\hfill & \hfill 7\hfill \end{array}\right|$

= 3 sin3θ (28 – 21) + (21 cos 2θ - 18) + 1 (21 cos 2θ - 24)

$\Delta =21sin3\theta +42cos2\theta -42$          

for no solution

sin 3θ + 2 cos 2θ = 2

$\theta =\pi ,2\pi ,3\pi ,\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}$

$\left(\overrightarrow{a}*\overrightarrow{b}\right)*\widehat{i}=\left(\overrightarrow{a}.\widehat{i}\right)\overrightarrow{b}-\left(\overrightarrow{b}.\widehat{i}\right)\overrightarrow{a}$

$\left(\left(\overrightarrow{a}*\overrightarrow{b}\right)*\widehat{i}\right).\widehat{k}=\left(\overrightarrow{a}.\widehat{i}\right)\left(\overrightarrow{b}.\widehat{k}\right)-\left(\overrightarrow{b}.\widehat{i}\right)\left(\overrightarrow{a}.\widehat{k}\right)$

$co{s}^{-1}\left(\frac{y}{2}\right)=lo{g}_e{\left(\frac{x}{5}\right)}^5,\left|y\right|<2$  

 Differentiating on both side

$-\frac{1}{\sqrt[]{1-{\left(\frac{y}{2}\right)}^2}}*\frac{y\text{'}}{2}=\frac{5}{\frac{x}{5}}*\frac{1}{5}$  

$\frac{-xy\text{'}}{2}=5\sqrt[]{1-{\left(\frac{y}{2}\right)}^2}$

Square on both side

$\frac{x^{2}y{\text{'}}^2}{4}=25\left(\frac{4-y^2}{4}\right)$

Diff on both side

$xy\text{'}+y\text{'}\text{'}{x}^2+25y=0$  

$\Delta =0$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 5\hfill & \hfill \alpha \hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right|=0$

15 - 2a + a - 6 – 1 = 0

a = 8

For a = 8, equations are

x + y + 3 = 6

2x + 5y + 8z = b

x + 2y + 3z = 14

$\left(2,5,8\right)=\mathcal{l}\left(1,1,1\right)+m\left(1,2,3\right)$

$\begin{array}{cc}\hfill 2=\mathcal{l}+m\hfill & \hfill 5=\mathcal{l}+2m\hfill \end{array}]\to 3=m,\mathcal{l}=-1$              

8 =  $\mathcal{l}+3m$

$\beta =6\mathcal{l}+14m$

= -6 + 42 = 36

a + b = 8 + 36 = 44

Let f(x) = $\frac{x^2-9}{x-5}$  

  $f\text{'}\left(x\right)=\frac{2x\left(x-5\right)-\left(x^2-9\right)}{{\left(x-5\right)}^2}$

$=\frac{x^2-10x+9}{{\left(x-5\right)}^2}=\frac{\left(x-1\right)\left(x-9\right)}{{\left(x-5\right)}^2}$

$\alpha =f\left(1\right)=2,\beta =\left\{f\left(0\right),f\left(2\right)\right\}=\frac{5}{3}$

$\therefore {\displaystyle \underset{-1}{\overset{3}{\int }}max}\left\{\frac{x^2-9}{x-5},x\right\}dx+{\left[\frac{x^2}{2}\right]}_{9/5}^3$           

a₁ = 18, a₂ = 16

a₁ + a₂ = 34