Maths

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.

The equation is for a hyperbola: x²/4 - y²/2 = 1.
The eccentricity e is given by e = √ (1 + b²/a²) = √ (1 + 2/4) = √6/2.
The focus F is at (ae, 0), which is (2 * √6/2, 0) = (√6, 0).
The equation of the tangent at a point P (x? , y? ) is xx? /a² - yy? /b² = 1.
The equation of the tangent at P is given as x - (√6 y)/2 = 1.
This tangent cuts the x-axis (y=0) at x=1, so Q is (1, 0).
The latus rectum is the line x = ae = √6.
To find the point R where the tangent intersects the latus rectum, we substitute x=√6 into the tangent equation:
√6 - (√6 y)/2 = 1 ⇒ √6 - 1 = (√6 y)/2 ⇒ y = 2 (√6 - 1)/√6.
The vertices of the triangle QFR are Q (1,0), F (√6,0), and R (√6, 2 (√6 - 1)/√6).
The base QF has length √6 - 1.
The height is the y-coordinate of R.
Area of ΔQFR = 1/2 * base * height = 1/2 * (√6 - 1) * [2 (√6 - 1)/√6].
Area = (√6 - 1)² / √6 = (6 - 2√6 + 1) / √6 = (7 - 2√6) / √6.

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.

Let x = m(a + λb).
Given m(a + λb) ⋅ (3i + 2j - k) = 0, which leads to λ = -3/8.

The projection of vector x on vector a is given by x ⋅ â, where â is the unit vector of a.
Projection = (x ⋅ a) / |a| = 17√6 / 2
x ⋅ a = (m(a + λb)) ⋅ a = m(a ⋅ a + λ(b ⋅ a)) = m(|a|^2 + λ(b ⋅ a))

The provided text simplifies this to:
m(6 - 3/8 * (-1)) = 17√6 / 2
m * (51/8) = 17 * 6 / 2 (The text seems to have a typo 17x6/2 instead of 17√6 / 2)
Assuming it is 17 * 6 / 2, m * 51/8 = 51, so m = 8.

x = 8(a + (-3/8)b) = 8a - 3b
x = 8( (13/8)i - (14/8)j + (11/8)k ) (The vectors a and b are not fully defined in the provided text)
The final vector is given as x = 13i - 14j + 11k.

|x|^2 = 13^2 + (-14)^2 + 11^2 = 169 + 196 + 121 = 486.

The problem provides an equation involving the coordinates (α, β, γ) of a point P:
((α + β + γ) / √3)^2 + ((α - nγ) / √(l^2 + n^2))^2 + ((α - 2β + γ) / √6)^2 = 9

The locus of P(α, β, γ) is given by replacing (α, β, γ) with (x, y, z):
((x + y + z) / √3)^2 + ((lx - nz) / √(l^2 + n^2))^2 + ((x - 2y + z) / √6)^2 = 9

This represents the equation of an ellipsoid. The text proceeds by comparing coefficients. By expanding the equation, the coefficients of x^2, y^2, z^2, and cross-product terms are collected. From the given conditions:
Coefficient of x^2: 1/3 + l^2 / (l^2 + n^2) + 1/6 = 1
Coefficient of y^2: 1/3 + 0 + 4/6 = 1
Coefficient of z^2: 1/3 + n^2 / (l^2 + n^2) + 1/6 = 1
Coefficient of xz: -2ln / (l^2 + n^2) + 2/6 = 0
(The document seems to simplify this differently).

The text provides a shortcut:
1/3 + (l^2 / (l^2 + n^2)) + 1/2 = 1 (Combining 1/3 and 1/6 to 1/2)
And also, (ln / (l^2 + n^2)) = 0. This implies ln = 0.

If ln = 0, then from l^2/(l^2+n^2) = 1/2, this means 2l^2 = l^2+n^2, so l^2 = n^2, which implies l = ±n.
The text states l^2 + n^2 - 2ln = 0, which simplifies to (l - n)^2 = 0, therefore l - n = 0 or l = n.

If the orthocenter and circumcenter of a triangle both lie on the y-axis, the centroid also lies on the y-axis.
The x-coordinate of the centroid is (x1 + x2 + x3) / 3. If the vertices are (cos α, sin α), (cos β, sin β), (cos γ, sin γ), then the x-coordinate of the centroid is (cos α + cos β + cos γ) / 3.
Since the centroid lies on the y-axis, its x-coordinate is 0.
cos α + cos β + cos γ = 0

Using the identity: If a + b + c = 0, then a^3 + b^3 + c^3 = 3abc.
Let a = cos α, b = cos β, c = cos γ.
Then, cos^3 α + cos^3 β + cos^3 γ = 3 * cos α * cos β * cos γ.

We need to evaluate the expression:
(cos 3α + cos 3β + cos 3γ) / (cos α * cos β * cos γ)

Using the identity cos 3θ = 4cos^3 θ - 3cos θ:
cos 3α + cos 3β + cos 3γ = (4cos^3 α - 3cos α) + (4cos^3 β - 3cos β) + (4cos^3 γ - 3cos γ)
= 4(cos^3 α + cos^3 β + cos^3 γ) - 3(cos α + cos β + cos γ)

Substitute the known values:
= 4(3 * cos α * cos β * cos γ) - 3(0)
= 12 * cos α * cos β * cos γ

So, the expression becomes:
(12 * cos α * cos β * cos γ) / (cos α * cos β * cos γ) = 12.
The value of the expression squared is 12^2 = 144.

The numbers 1, log10(4^x - 2), and log10(4^x + 18/5) are in an Arithmetic Progression (A.P.).
This means that the corresponding numbers 10^1, 10^(log10(4^x - 2)), and 10^(log10(4^x + 18/5)) are in a Geometric Progression (G.P.).
So, 10, 4^x - 2, and 4^x + 18/5 are in G.P.

For a G.P., the square of the middle term is equal to the product of the other two terms:
(4^x - 2)^2 = 10 * (4^x + 18/5)
Let y = 4^x.
(y - 2)^2 = 10y + 36
y^2 - 4y + 4 = 10y + 36
y^2 - 14y - 32 = 0
(y - 16)(y + 2) = 0
So, y = 16 or y = -2.

Since y = 4^x, y must be positive. Thus, 4^x = 16, which gives x = 2.

The determinant calculation that follows appears to be unrelated to the A.P./G.P. problem.
| [3, 1, 4], [1, 0, 2], [2, 1, 0] | = 3(0 - 2) - 1(0 - 4) + 4(1 - 0) = -6 + 4 + 4 = 2.