Class 12th

Let X1, ., X2n be the first 2n observations and Y1, ., Yn be the last n observations.
Given:
ΣXi / 2n = 6 => ΣXi = 12n (i)
ΣYi / n = 3 => ΣYi = 3n (ii)

Combined mean: (ΣXi + ΣYi) / 3n = 5 => ΣXi + ΣYi = 15n. This is consistent with (i) and (ii).

Combined variance: (ΣXi^2 + ΣYi^2) / 3n - (mean)^2 = 4
(ΣXi^2 + ΣYi^2) / 3n - 5^2 = 4
ΣXi^2 + ΣYi^2 = (4 + 25) * 3n = 87n.

New observations are Xi + 1 and Yi - 1.
New mean = (Σ (Xi + 1) + Σ (Yi - 1) / 3n = (ΣXi + 2n + ΣYi - n) / 3n = (15n + n) / 3n = 16n / 3n = 16/3.

New variance k:
k = (Σ (Xi + 1)^2 + Σ (Yi - 1)^2) / 3n - (new mean)^2
k = (Σ (Xi^2 + 2Xi + 1) + Σ (Yi^2 - 2Yi + 1) / 3n - (16/3)^2
k = (ΣXi^2 + ΣYi^2 + 2ΣXi - 2ΣYi + 2n + n) / 3n - 256/9
k = (87n + 2 (12n) - 2 (3n) + 3n) / 3n - 256/9
k = (87n + 24n - 6n + 3n) / 3n - 256/9
k = 108n / 3n - 256/9 = 36 - 256/9 = (324 - 256)/9 = 68/9.

So, k = 68/9, which means 9k = 68.

Given vectors a? and b? such that |a? | = |b? | and a? ⋅ b? = 0 (they are orthogonal).
The problem implies |a? |=|b? |=1.
Let c? = a? + b? + a? x b?
To find the magnitude of c? , we calculate |c? |²:
|c? |² = c? ⋅ c? = (a? + b? + a? x b? ) ⋅ (a? + b? + a? x b? ).
This expands to |a? |² + |b? |² + |a? x b? |² because all other dot products are zero (e.g., a? ⋅ b? = 0, a? ⋅ (a? x b? ) = 0).
|a? x b? |² = (|a? |b? |sin (90°)² = |a? |²|b? |².
So, |c? |² = |a? |² + |b? |² + |a? |²|b? |² = 1² + 1² + 1²*1² = 3.
∴ |c? | = √3.
To find the angle θ between c? and a? , we compute their dot product:
c? ⋅ a? = (a? + b? + a? x b? ) ⋅ a? = a? ⋅ a? + b? ⋅ a? + (a? x b? ) ⋅ a?
c? ⋅ a? = |a? |² + 0 + 0 = 1.
Now, cos θ = (c? ⋅ a? ) / (|c? |a? |) = 1 / (√3 * 1) = 1/√3.
⇒ θ = cos? ¹ (1/√3).

Given functions f (x) = (x-2)/ (x-3) and g (x) = 2x-3.
First, find the inverse functions f? ¹ (x) and g? ¹ (x).
For f? ¹ (x): y = (x-2)/ (x-3) ⇒ y (x-3) = x-2 ⇒ xy - 3y = x-2 ⇒ xy-x = 3y-2 ⇒ x (y-1) = 3y-2 ⇒ x = (3y-2)/ (y-1). So, f? ¹ (y) = (3y-2)/ (y-1).
For g? ¹ (x): y = 2x-3 ⇒ y+3 = 2x ⇒ x = (y+3)/2. So, g? ¹ (y) = (y+3)/2.
We are given f? ¹ (x) + g? ¹ (x) = 13/2.
(3x-2)/ (x-1) + (x+3)/2 = 13/2.
Multiply by 2 (x-1): 2 (3x-2) + (x+3) (x-1) = 13 (x-1).
6x - 4 + x² + 2x - 3 = 13x - 13.
x² + 8x - 7 = 13x - 13.
x² - 5x + 6 = 0.
(x-2) (x-3) = 0.
The possible values of x are 2 and 3. Note that x=3 is not in the domain of the original f (x), and x=1 is not in the domain of f? ¹ (x). However, the equation is solved for x.
The sum of possible values is 2 + 3 = 5.

A relation R is defined as ARB if PAP? ¹ = B for a non-singular matrix P.

·       Reflexive: ARA requires PAP? ¹ = A. This holds if P is the identity matrix I, as IAI? ¹ = A. Assuming P can be I, the relation is reflexive.

·       Symmetric: We need to show that if ARB, then BRA.
ARB ⇒ PAP? ¹ = B.
To get the reverse, we need to express A in terms of B.
From PAP? ¹ = B, pre-multiply by P? ¹ and post-multiply by P:
P? ¹ (PAP? ¹)P = P? ¹BP ⇒ A = P? ¹BP. This shows BRA where the matrix is P? ¹. Thus, the relation is symmetric.

·       Transitive: We need to show that if ARB and BRC, then ARC.
ARB ⇒ PAP? ¹ = B — (I)
BRC ⇒ PBP? ¹ = C — (II)
Substitute B from (I) into (II):
P (PAP? ¹)P? ¹ = C
⇒ (P²)A (P? ¹)² = C
⇒ (P²)A (P²)? ¹ = C
This is in the form QAQ? ¹ = C with Q = P². This implies ARC. Thus, the relation is transitive.
Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.