Maths

a? + a? = 4 ⇒ a? + a?r = 4
a? + a? = 16 ⇒ a?r² + a?r³ = 16
⇒ r = ±2
r = 2 ⇒ a? = 4/3
r = -2 ⇒ a? = -4
Σ(i=1 to 9) a? = (a(r?-1))/(r-1) = (-4)((-2)? - 1)/(-2-1)
= 4/3 (-513) = 4λ
⇒ λ = -171

y = (8²?-8?²?)/(8²?+8?²?) ⇒ (1+y)/(1-y) = 8? ⇒ 4x = log?((1+y)/(1-y))
x = (1/4)log?((1+y)/(1-y)), f?¹(x) = (1/4)log?((1+x)/(1-x))

A → 5Q; B → 5Q; C → 5QA
A?, A?, A?, A?, A?; B?, B?, B?, B?, B?; C?, C?, C?, C?, C?
A?A?A?B?C? → ?C? * ?C? * ?C? = 750
A?A?B?B?C? → ³C? * ?C? * ?C? = 1500
∴ Total = 2250

Slope AB = (k-α)/(h-2α) = -1
⇒ α = (k+h)/3
also (β+2α)/2 = h, (β+α)/2 = k
α = 2h - 2k
From (1) and (2)
(h+k)/3 = 2h - 2k
⇒ 5h = 7k
⇒ 5x = 7y

sinx = t, cosxdx = dt
I = ∫dt/(t³(1+t?)^(2/3)) = ∫dt/(t?(1+1/t?)^(2/3))
Put 1+1/t? = r³ ⇒ -6/t? dt = 3r²dr
I = (-1/2)∫dr = -1/2 r + c = (-1/2)(1+1/sin?x)^(1/3) + c
f(x) = (-1/2)cosec²x(1+sin?x)^(1/3) and λ=3
⇒ f(π/3) = -2

AB = r, AD = r/2
CD = rsin60° = √3r/2
|0+0-3|/√ (1²+2²) = √3r/2 ⇒ 3/√5 = √3r/2 ⇒ r = 2√3/√5 ⇒ r² = 12/5

α+β = 3/7, αβ = -2/7
α/ (1-α²) + β/ (1-β²) = (α+β) - αβ (α+β) / (1-α²) (1-β²)
= (α+β) - αβ (α+β) / (1 + (αβ)² - (α+β)² - 2αβ)
= (3/7) + (2/7) (3/7) / (1 + (-2/7)² - (3/7)² - 2 (-2/7) = 27/16

Note D = |3, 4, 5; 1, 2, 3; 4, 4| = 0 (R? → R? - 2R? + 3R? )
Now let P? = 4x+4y+4z-δ=0. If the system has solutions it will have infinite solution, so P? = αP? + βP?
Hence 3α+β=4 and 4α+2β=4 ⇒ α=2 and β=-2
So for infinite solution 2µ-2=δ ⇒ for 2µ ≠ δ+2 System inconsistent

I = ∫ (cosθ / (2sin²θ + 7sinθ + 3) dθ
sinθ = t ⇒ cosθdθ = dt
= (1/2) ∫ (1 / (t² + (7/2)t + 3/2) dt
= (1/2) ∫ (1 / (t+7/4)² - (5/4)²) dt
= (1/5) ln | (2t+1)/ (t+3)| + c
= (1/5) ln | (2sinθ+1)/ (sinθ+3)| + C
So A = 1/5
B (θ) = 5 (2sinθ+1)/ (sinθ+3)

y = |x| (x-1)
= { 0,   0 ≤ x < 1
{ x-1, 1 ≤ x < 2
Area = ∫? ² 2√x dx - ∫? ² (1) (1)
= [ (4x³/²)/3]? ² - 1/2 = (8√2)/3 - 1/2