Maths

Variance of a, b, c & a+2, b+2, c+2, are same.
Given: b = a + c (i)
d² = (1/3) (a² + b² + c²) - [ (a+b+c)/3]²
as a + c = b
d² = (1/3) (a² + b² + c²) - (2b/3)²
9d² = 3 (a² + b² + c²) - 4b²
⇒ b² = 3 (a² + c²) - 9d²

log? /? [ (|z|+11)/ (|z|-1)²] < 2
(|z|+11)/ (|z|-1)² > (1/2)²
(|z|+11)/ (|z|-1)² > 1/4
⇒ 4|z| + 44 > |z|² - 2|z| + 1
⇒ |z|² - 6|z| - 43 < 0

⇒ |z| - 7 ≤ 0
∴ |z|max = 7

P (x) = x² + bx + c.
Given ∫? ¹ P (x) dx = 1.
∫? ¹ (x² + bx + c) dx = [x³/3 + bx²/2 + cx] from 0 to 1 = 1/3 + b/2 + c = 1.
2 + 3b + 6c = 6 => 3b + 6c = 4 - (i)
When P (x) is divided by (x-2), the remainder is 5. So, P (2) = 5.
(2)² + b (2) + c = 5 => 4 + 2b + c = 5 => 2b + c = 1 - (ii)
From (ii), c = 1 - 2b. Substitute into (i):
3b + 6 (1 - 2b) = 4
3b + 6 - 12b = 4
-9b = -2 => b = 2/9.
c = 1 - 2 (2/9) = 1 - 4/9 = 5/9.
We need to find 9 (b+c).
9 (2/9 + 5/9) = 9 (7/9) = 7.

dy/dx + (tanx)y = sinx. This is a linear differential equation.
Integrating Factor (I.F.) = e^ (∫tanx dx) = e^ (ln|secx|) = secx.
The solution is y * I.F. = ∫ (sinx * I.F.) dx + C.
y * secx = ∫ (sinx * secx) dx = ∫tanx dx = ln|secx| + C.
Given y (0) = 0.
0 * sec (0) = ln|sec (0)| + C => 0 = ln (1) + C => C = 0.
So, y * secx = ln (secx).
y = cosx * ln (secx).
At x = π/4:
y = cos (π/4) * ln (sec (π/4) = (1/√2) * ln (√2) = (1/√2) * (1/2)ln (2) = ln (2) / (2√2).

The inequality is experience ( (|z|+3) (|z|-1) / (|z|+1) * log?2 ) ≥ log√? 16.
The right side is log? (1/2) 16 = log? (2? ¹) 2? = (4/-1)log?2 = -4. This seems incorrect.
Let's assume the base of the log on the right is √2. log√? 16 = log? (1/2) 2? = 2 * log?2? = 8.
The inequality becomes: 2^ (|z|+3) (|z|-1) / (|z|+1) ≥ 8 = 2³.
So, (|z|+3) (|z|-1) / (|z|+1) ≥ 3.
Let |z| = t. (t+3) (t-1) / (t+1) ≥ 3
t² + 2t - 3 ≥ 3t + 3
t² - t - 6 ≥ 0
(t-3) (t+2) ≥ 0
Since t = |z| ≥ 0, we must have t-3 ≥ 0.
So, t ≥ 3, which means |z| ≥ 3.
The minimum value of |z| is 3.