Maths

The general term is T? = ¹? C? (K/x²)? (√x)¹?
= ¹? C? K? x? ²? x? /² = ¹? C? K? x? /²
For the constant term, the power of x is 0.
5 - 5r/2 = 0 ⇒ r = 2
The term is T? = ¹? C? · K² = 405
45 · K² = 405
⇒ K² = 9 ⇒ |K| = 3

S' = 2¹? + 2? ⋅ 3 + 2? ⋅ 3² + . + 2 ⋅ 3? + 3¹?
G.P. → a = 2¹? , r = 3/2, n = 11
S' = 2¹? ⋅ [ (3/2)¹¹ - 1)/ (3/2 - 1)] = 2¹¹ (3¹¹/2¹¹) - 1)
= 3¹¹ - 2¹¹

Negation of x ↔~ y
≡ ~ (x ↔ ~y)
≡ x ↔ ~ (~y)
≡ x ↔ y
≡ (x ∧ y) ∨ (~x ∧ ~y)

f (f (x) = (a-f (x)/ (a+f (x) = x
Let f (x) = y. (a-y)/ (a+y) = x ⇒ a-y = ax + xy ⇒ a (1-x) = y (1+x) ⇒ y = a (1-x)/ (1+x)
⇒ f (x) = a (1-x)/ (1+x)
From the given options, we infer that comparing the derived f (x) leads to a=1.
⇒ a = 1
So f (x) = (1-x)/ (1+x)
f (-1/2) = (1 - (-1/2)/ (1 + (-1/2) = (3/2)/ (1/2) = 3

Equation of normal to the ellipse x²/a² + y²/b² = 1 at (x? , y? ) is a²x/x? - b²y/y? = a² - b².
At the point (ae, b²/a):
a²x/ (ae) - b²y/ (b²/a) = a² - b²
It passes through (0, -b).
a² (0)/ (ae) - b² (-b)/ (b²/a) = a² - b²
ab = a² - b²
Since b² = a² (1-e²), a²-b² = a²e².
ab = a²e²
a²b² = a? e?
a² (a² (1-e²) = a? e?
1 - e² = e?
e? + e² - 1 = 0

Let the first A.P. be a? , a? + d, a? + 2d.
a? = a? + 39d = -159
a? = a? + 99d = -399
Subtracting the equations, 60d = -240 ⇒ d = -4.
Substituting d back, a? + 39 (-4) = -159 ⇒ a? - 156 = -159 ⇒ a? = -3.
Now, for the second A.P. with first term b? and common difference D = d+2 = -2.
b? = a?
⇒ b? + 99D = a? + 69d
⇒ b? + 99 (-2) = -3 + 69 (-4)
⇒ b? - 198 = -3 - 276
⇒ b? = -279 + 198 = -81

Let A (α, 0,0), B (0, β, 0), C (0,0, γ), then the centroid is G (α/3, β/3, γ/3) = (1,1,2).
α = 3, β = 3, γ = 6
∴ Equation of plane is x/α + y/β + z/γ = 1
⇒ x/3 + y/3 + z/6 = 1
⇒ 2x + 2y + z = 6
∴ Required line passing through G (1,1,2) and normal to the plane is (x-1)/2 = (y-1)/2 = (z-2)/1.

A = [cosθ, sinθ], [-sinθ, cosθ]
A² = [cos2θ, sin2θ], [-sin2θ, cos2θ]
⇒ A? = [cos4θ, sin4θ], [-sin4θ, cos4θ]
B = [cos4θ, sin4θ], [-sin4θ, cos4θ] + [cosθ, sinθ], [-sinθ, cosθ]
= [cos4θ + cosθ, sin4θ + sinθ], [- (sin4θ + sinθ), cos4θ + cosθ]
det (B) = (cos4θ + cosθ)² + (sin4θ + sinθ)²
= (cos²4θ + sin²4θ) + (cos²θ + sin²θ) + 2 (cos4θcosθ + sin4θsinθ)
= 1 + 1 + 2cos (4θ - θ)
= 2 + 2cos3θ
Given 3θ = 3π/5
|B| = 2 + 2cos (3π/5)
= 2 + 2 (- (√5-1)/4) = 2 - (√5-1)/2 = (4-√5+1)/2 = (5-√5)/2 ∈ (1,2)

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