Class 12th

∫ (from 0 to 3) (x-1) (x-2) (x-3) dx = ∫ (from 0 to 3) (x³ - 6x² + 11x - 6) dx

A = ∫ (from 0 to 1) (x³ - 6x² + 11x - 6) dx - ∫ (from 1 to 2) (x³ - 6x² + 11x - 6) dx + ∫ (from 2 to 3) (x³ - 6x² + 11x - 6) dx
A = 11/4

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) =  $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

-> b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$        

y = (3/2)sin (2θ)
x = e^θ sinθ
dy/dθ = 3cos (2θ)
dx/dθ = e^θ (cosθ + sinθ)
dy/dx = (3cos (2θ) / (e^θ (cosθ + sinθ) = (3 (cosθ - sinθ) / e^θ

Nucleic acids are biomolecules which is made of nucleotides. They store and transmit the genetic information. Nucleic acids are of two types: DNA (Deoxyribonucleic acid) and RNA (Ribonucleic acid). The following are the functions of nucleic acids.

1. They store and transfer the genetic information.

2. Nucleic acids help in protein synthesis

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

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

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

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

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 = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

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