Class 12th

Biomolecules in class 12 Chemistry include the following important topics: classification and structure of carbohydrates, amino acids, and proteins, enzymes and their characteristics, nucleic acids (DNA and RNA structure and components), and vitamins.

You must also prepare the following topics that are frequently asked in exams: zwitterion structure in amino acids, enzyme action mechanisms, and differences between DNA and RNA.

∫ (from 0 to π/2) (x/2)|cos (2x)|dx
∫ (from 0 to π/4) (x/2)cos (2x)dx - ∫ (from π/4 to π/2) (x/2)cos (2x)dx
= [x sin (2x)/4 - ∫sin (2x)/4 dx] (from 0 to π/4) - [x sin (2x)/4 - ∫sin (2x)/4 dx] (from π/4 to π/2)
= [x sin (2x)/4 + cos (2x)/8] (from 0 to π/4) - [x sin (2x)/4 + cos (2x)/8] (from π/4 to π/2)
= (π/16 - 1/8) - (-1/8 - π/16) = π/8

|3a+4b|² = 9|a|² + 16|b|² + 24a·b
But a·b = 0, |a|=|b|=k
|3a+4b| = 5k
|4a-3b|
10k = 20? k = 2 = |a| = |b|

Biomolecules are organic compounds essential for life; they are found in all living organisms. In chemistry class 12, we talked about the following biomolecules: carbohydrates, proteins, lipids, nucleic acids, enzymes, and vitamins. 

Biomolecules chemistry class 12 connects biology and chemistry and helps us understand metabolism, heredity, and biochemical reactions. 

Differentiable ⇒ continuous
Continuous at x = 2 ⇒ 2a + b = 0
Continuous at x = 3
0 = 9p + 3q + 1
Differentiable at x=2
a = 2*2 - 5 ⇒ a = -1
Differentiable at x=3
2*3 - 5 = 2p*3 + q ⇒ 6p + q = 1
a = -1, b = 2, p = 4/9, q = -5/3

For point of intersection
1 + 3λ = 3 + μ
2 + λ = 1 + 2μ
5λ = 5 ⇒ λ = 1, μ = 1
Point of intersection (4, 3, 5)
For the greatest distance from origin perpendicular from meet plane at point of intersection
Hence equation r . (4i + 3j + 5k) = 50

Foods that contains nucleic acids are animal products such as meat, eggs, fish and few plant-based sources are legumes, spinach, mushrooms, yeast and cereals. Adding these foods in the diet will support the body's nucleic acid pool, which is important for cellular function.

f' (x) = x (1+y)
dy/ (y+1) = xdx
ln (y+1) = x²/2 + c
(0,0)
c = 0
y+1 = e^ (x²/2) ⇒ e^ (x²/2) = 2
x²/2 = k = (ln2 > 0)
x = ±√2k

${\stackrel{?}{B}}_1=\frac{E_1}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t){\stackrel{?}{B}}_1$    where   ${\stackrel{?}{B}}_1={\hat{c}}_1*{\stackrel{?}{E}}_1=\hat{k}*\hat{i}={\stackrel{?}{B}}_1=\hat{j}$

${\stackrel{?}{B}}_2=\frac{E_2}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t){\stackrel{?}{B}}_2$ where ${\stackrel{?}{B}}_2={\hat{c}}_2*{\stackrel{?}{E}}_2=-\hat{k}*\hat{j}={\stackrel{?}{B}}_2=\hat{i}$

$\therefore \mathrm{}\stackrel{?}{B}={\stackrel{?}{B}}_1+{\stackrel{?}{B}}_2$

$\stackrel{?}{B}=\frac{E_1}{c}\hat{j}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t)+\frac{E_2}{c}\hat{i}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t)$