Maths

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)

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