Maths

Given f (1) = a = 3, and assuming the function form is f (x) = a?
So f (x) = 3?
∑? f (i) = 363
⇒ 3 + 3² + . + 3? = 363
This is a geometric progression. The sum is S? = a (r? -1)/ (r-1).
3 (3? -1)/ (3-1) = 363
3 (3? -1)/2 = 363
3? - 1 = 242
3? = 243
3? = 3? ⇒ n = 5

S = tan? ¹ (1/3) + tan? ¹ (1/7) + tan? ¹ (1/13) + . upto 10 term
S = tan? ¹ (2-1)/ (1+1⋅2) + tan? ¹ (3-2)/ (1+2⋅3) + tan? ¹ (4-3)/ (1+3⋅4) + . + tan? ¹ (11-10)/ (1+11⋅10)
S = (tan? ¹2 - tan? ¹1) + (tan? ¹3 - tan? ¹2) + . + (tan? ¹11 - tan? ¹10)
S = tan? ¹11 - tan? ¹1
S = tan? ¹ (11) - π/4
tan (S) = 5/6

P (A∪B∪C) = P (A) + P (B) + P (C) – P (A∩B) – P (B∩C) – P (C∩A) + P (A∩B∩C)
Given relations lead to: α = 1.4 – P (A∩B) – β ⇒ α + β = 1.4 - P (A∩B)
Again, from P (A∪B) = P (A) + P (B) – P (A∩B), and given values, it is found that P (A∩B) = 0.2.
From (1) and (2), α = 1.2 – β.
Now given 0.85 ≤ α ≤ 0.95
⇒ 0.85 ≤ 1.2 – β ≤ 0.95
⇒ -0.35 ≤ -β ≤ -0.25
⇒ 0.25 ≤ β ≤ 0.35, so β ∈ [0.25, 0.35]

I = ∫ from -π/2 to π/2 (1 / (1+e^ (sin x) dx
I = ∫ from -π/2 to π/2 (e^ (sin x) / (1+e^ (sin x) dx
2I = ∫ from -π/2 to π/2 1dx ⇒ I = 1/2 ∫ from -π/2 to π/2 dx
I = 1/2 [x] from -π/2 to π/2 ⇒ I = π/2

∴ x² = |x|² = t let
9t² - 18t + 5 = 0
(3t - 1) (3t - 5) = 0
|x| = 1/3, 5/3
Product of roots = (1/3) (-1/3) (5/3) (-5/3) = 25/81

P: n³ - 1 is even, q: n is odd.
The contrapositive of p → q is ~q → ~p.
~q: n is not odd, i.e., n is even.
~p: n³ - 1 is not even, i.e., n³ - 1 is odd.
⇒ "If n is not odd then n³ - 1 is not even"
⇒ "For an integer n, if n is even, then n³ – 1 is odd."

y² = 4x and x² = 4y
any tangent of y² = 4x is y = mx + 1/m
it also tangent for x² = 4y
1/m = -m² ⇒ m = -1
∴ common tangent is y = -x - 1, it also touches x² + y² = c²
∴ 1 = c² ⋅ (1+1) ⇒ c² = 1/2

n (C) = 73, n (T) = 65, n (C ∩ T) = x
n (C ∪ T) ≤ 100
⇒ n (C) + n (T) - n (C ∩ T) ≤ 100
⇒ x ≥ 38
n (C ∩ T) ≤ min (n (C), n (T) ⇒ x ≤ 65
⇒ 38 ≤ x ≤ 65

Assuming the equation is (x + iy)² = I (x² + y²)
x² - y² + 2ixy = I (x² + y²)
Compare real and imaginary parts
x² - y² = 0 ⇒ x = y or x = -y
2xy = x² + y²
If x=y, then 2x² = x² + x², which is true for all x.
If x=-y, then -2y² = y² + y² = 2y², which implies 4y²=0, so y=0 and x=0.
The non-trivial solution is x = y.

a, b, c are in Andhra Pradesh then
2b = a + c
28 = 3^ (2sin2θ-1) + 3^ (4-2sin2θ)
Put 3^ (2sin2θ) = x
28 = x/3 + 81/x ⇒ x² - 84x + 243 = 0
(x-3) (x-81) = 0
3^ (2sin2θ) = 3 or 3?
2sin2θ = 1 or 4
sin2θ = 1/2
terms are 1, 14, 27,
then T? = 1 + 5 (13)