Maths

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)

I = ∫ (e²? + 2e? - e? - 1)e^ (e? +e? ) dx
I = ∫ (e²? + e? - 1)e^ (e? +e? ) dx + ∫ (e? - e? )e^ (e? +e? ) dx
I = ∫ (e? + 1 - e? )e^ (e? +e? ) dx + e^ (e? +e? )
(e? - e? + 1)dx = du
I = e^ (e? +e? ) + e^ (e? +e? ) = e^ (e? +e? ) (e? + 1) then g (x) = e? + 1
g (0) = 2

For the parabola y = x², the tangent at (2,4) is given by (y+4)/2 = 2x, which simplifies to 4x - y - 4 = 0.
The equation of a circle touching the line 4x - y - 4 = 0 at the point (2,4) is
(x-2)² + (y-4)² + λ (4x-y-4) = 0.
It passes through (0,1).
∴ (0-2)² + (1-4)² + λ (4 (0) - 1 - 4) = 0
4 + 9 + λ (-5) = 0 ⇒ 13 = 5λ ⇒ λ = 13/5
∴ the circle is x² - 4x + 4 + y² - 8y + 16 + (13/5) (4x-y-4) = 0
x² + y² + (-4 + 52/5)x + (-8 - 13/5)y + (20 - 52/5) = 0
x² + y² + (32/5)x - (53/5)y + 48/5 = 0
∴ centre is (-g, -f) = (-16/5, 53/10).

Here D = | 2 -4 λ |
| 1 -6 1 | = (λ-3) (3λ+2)
| λ -10 4 |
D = 0 ⇒ λ = 3, -2/3

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