Class 12th

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

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

$\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]a_1,b_1,c_{1}t\left\{-1,0,1\right\}$

$\underset{9times}{\underset{?}{a_1+a_2+a_3+....}}=5$

Total = 414

$\lambda =1+\mu$

$\begin{array}{l}\underset{\_}{2\lambda =3\lambda \mu }\\ \lambda =2\to S\left(2,4,6\right)\end{array}$

$\mu =1$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \end{array}\right|=\left(7,-2,-1\right)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

g (1) =?

$=f\left(f\left(f\left(1\right)\right)\right)+f\left(f\left(1\right)\right)$

$f\left(1\right)={\left(2\left(1-\frac{1}{2}\right)\left(2+1\right)\right)}^{\frac{1}{50}}=3^{\frac{1}{50}}$

$3^{\frac{1}{50}}+13^{\frac{1}{50}}>1^{\frac{1}{50}}$

3 > 1

$3^{\frac{1}{50}}>2$   $2>3^{\frac{1}{50}}>1$

$3<2^{50}$   $\left[3^{\frac{1}{50}}+1\right]=1+1=2$

${\left|2\left(\overrightarrow{a}*\overrightarrow{b}\right)\right|}^2+4{\left(\overrightarrow{a}.\overrightarrow{b}\right)}^2$

= $4{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2$

=4 * 16 * 9 = 576