Class 12th

A line passes through (1,3). Its equation is y - 3 = m (x - 1) or y = mx + (3-m).
The angle θ between this line and the line y = 3√2x - 1 (with slope m? = 3√2) is given by tanθ = √2.

tanθ = | (m - m? )/ (1 + m*m? )|
√2 = | (m - 3√2) / (1 + m*3√2)|

This gives two cases:

Case 1 (+ve):
√2 = (m - 3√2) / (1 + 3√2m)
√2 (1 + 3√2m) = m - 3√2
√2 + 6m = m - 3√2
5m = -4√2
m = -4√2 / 5

Case 2 (-ve):
-√2 = (m - 3√2) / (1 + 3√2m)
-√2 (1 + 3√2m) = m - 3√2
-√2 - 6m = m - 3√2
7m = 2√2
m = 2√2 / 7

The reactions in the Solvay process are:

NaCl + H? O + NH? + CO? → NH? Cl + NaHCO?

2NaHCO? → Na? CO? + CO? + H? O

2NH? Cl + Ca (OH)? → CaCl? + 2NH? + H? O

CaCl? is a byproduct of the process.

o   Chlorophyll: Magnesium present in chlorophyll.

o   Vitamin B? : Cobalt (cyanocobalamin).

o   Anticancer drug: Platinum (Coordination compound of platinum).

o   Grubbs catalyst: Ruthenium.

Consider the series:
1/ (3²-1) + 1/ (5²-1) + 1/ (7²-1) + . + 1/ (201)²-1)

The general term T? can be written for the r-th term starting with r=1 for 3, r=2 for 5.
T? = 1/ (2r+1)² - 1) = 1/ (2r+1-1) (2r+1+1) = 1/ (2r * (2r+2) = 1/4 * 1/ (r (r+1)
T? = 1/4 * (1/r - 1/ (r+1)

The sum of the first n terms is:
S? = Σ T? = 1/4 * Σ (1/r - 1/ (r+1) from r=1 to n
S? = 1/4 * [ (1 - 1/2) + (1/2 - 1/3) + . + (1/n - 1/ (n+1) ]
S? = 1/4 * (1 - 1/ (n+1)

The last term is (201)²-1, so 2r+1 = 201, which gives r = 100. So, n=100.
S? = 1/4 * (1 - 1/101) = 1/4 * (100/101) = 25/101.

When we add cryolite (Na? AlF? ) in the extraction of aluminium, the melting point of alumina decreases

We need to evaluate the sum:
Σ (100 - r) (100 + r) for r from 0 to 99.

Σ (100² - r²) for r from 0 to 99
= Σ 100² - Σ r² for r from 0 to 99
= 100 * 100² - [99 (99+1) (2*99+1)]/6
= 100³ - [99 * 100 * 199]/6
= 100³ - (1650 * 199)

Comparing this with (100)³ – 199β, we get:
β = 1650
If we consider a comparison form α = 3β, this part of the solution seems to have a typo, but based on the final calculation, Slope = α/β = 1650 / 3 = 550 seems to be intended.

Given the function f (x) = cosec? ¹ (x) / √ {x - [x]} where [x] is the greatest integer function.

The domain of cosec? ¹ (x) is (-∞, -1] U [1, ∞).
For the denominator to be defined, x - [x] ≠ 0, which means {x} ≠ 0 (the fractional part of x is not zero). This implies that x cannot be an integer (x ∉ I).

Combining these conditions, the domain is all non-integer numbers except for those in the interval (-1, 1).

Functional group isomers

Given the equations:
t? (A + 2B) = -1 which expands to t? (A) + 2t? (B) = -1 . (I)
t? (2A - B) = 3 which expands to 2t? (A) - t? (B) = 3 . (II)

Solving equations (I) and (II) simultaneously, we get:
t? (A) = 1
t? (B) = -1

Therefore, t? (A) - t? (B) = 1 - (-1) = 2.