Maths

  $\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$

Let sin x = t, t  $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$         

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$           

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$           

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$          

Minimum value of a for which solution exist = 9

T? = (1/3)T?
⇒ T? , T? , T? G.P.
T = 1/3
T? = 1/243
ar? = 1/243
a (1/3)? = 1/243
a/729 = 1/243 ⇒ a = 3

Sum of infinite series = T? + T? + T?
= (T? ) / (1-r²) = (3 * 1) / (1 - 1/9) = 3 / (8/9) = 27/8

y² = 4x
(x - 3)² + y² = 9
y = mx + 1/m
(3, 0), r = 3

|3m + 1/m| / √ (1+m²) = 3

9m² + 1/m² + 6 = 9 (1+m²)
9m² + 1/m² + 6 = 9 + 9m²
1/m² = 3 ⇒ m = ±1/√3
m = 1/√3 in first quadrant
y = x/√3 + √3 ⇒ √3y = x + 3

  $M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill c_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$

$M^{T}M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill \\ \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill \\ \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$     

$Tr\left(M^{T}M\right)=a_1^2+b_1^2+c_1^2+a_2^2+b_2^2+c_2^2+a_3^2+b_3^2+c_3^2=7$           

all  $a_i,b_i,c_i\in \left\{0,1,2\right\}for$  i = 1, 2, 3

Case 1 7 one's and two zeroes which can occur in ways

Case 2 One 2 three 1's five zeroes =

total such matrices = 504 + 36 = 540

(I + M)² = (I + M) (I + M) = I + 2M + M² = (I + 2M)
(I + M)³ = (I + 2M) (I + M) = I + 3M + 2M² = (I + 3M)
(I + M)? = I + 50M
det (I + M)? - 50M) = det (I) = 1

b? , b? , b?
a, ar, ar².
a + ar = 1 ⇒ a = 1 / (1+r)

Σ (from k=1 to ∞) b? = 2 ⇒ a / (1-r) = 2
(1 / (1+r) / (1-r) = 2
1 / (1-r²) = 2
1/2 = 1 - r²
r² = 1/2 ⇒ r = ±1/√2

b? < 0 r = -1/2
⇒ a = b? = 2 + √2

$z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

for $\alpha \in Ry+2=0\Rightarrow y=-2$

 $x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$

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

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$          

= 0

Δ ≠ 0
| 1 -λ -1 |
| λ -1 -1 | ≠ 0
| 1 -1 |

1 (1+λ) + λ (-λ+1) - 1 (λ+1) ≠ 0
1 + λ - λ² + λ - λ - 1 ≠ 0
-λ² + λ ≠ 0
λ² = 1 ⇒ λ = 1, -1

a² + b² = 2

p (10hr) = .1
p (7hr) = .2
p (4hr) = .7

p (s / 10hr) = .8
p (s / 7hr) = .6
p (s / 4hr) = .4

p (s) = .1 * .8 + .2 * .6 + .7 * .4
p (4hr / s) = (.7 * .4) / (.1 * .8 + .2 * .6 + .7 * .4) = 28 / (8 + 12 + 28) = 28 / 48 = 7 / 12

${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$  

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ ->value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$