Class 12th

|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 ${\stackrel{?}{B}}_2={\hat{c}}_2*{\stackrel{?}{E}}_2=-\hat{k}*\hat{j}={\stackrel{?}{B}}_2=\hat{i}$

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

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

f (x) = x? /20 - x? /12 + 5
f' (x) = x? /4 - x³/3 = x³ (x/4 - 1/3)
Local maxima at 0, Local minima at 4/3
f' (x) = x³ - x² = x² (x-1)
x = 1 point of inflection

dy/√ (1-y²) = dx/x²
sin? ¹ (y) = -1/x + c ⇒ c = π/2
sin? ¹ (y) = -π/3 + π/2 = π/6
y = 1/2