Maths

1st observation: n? =10, mean x? =2, variance σ? ²=2.
Σx? = n? x? = 20.
σ? ² = (Σx? ² / n? ) - x? ² => 2 = (Σx? ²/10) - 2² => 6 = Σx? ²/10 => Σx? ² = 60.
2nd observation: n? , mean y? =3, variance σ? ²=1. Let n? =n.
Σy? = ny? = 3n.
σ? ² = (Σy? ² / n) - y? ² => 1 = (Σy? ²/n) - 3² => 10 = Σy? ²/n => Σy? ² = 10n.
Combined variance σ² = 17/9. n_total = 10+n.
Combined mean = (Σx? +Σy? )/ (10+n) = (20+3n)/ (10+n).
Combined Σ (squares) = 60+10n.
σ² = (Combined Σsq / n_total) - (Combined mean)²
17/9 = (60+10n)/ (10+n) - [ (20+3n)/ (10+n)]²
Multiply by 9 (10+n)²:
17 (10+n)² = 9 (60+10n) (10+n) - 9 (20+3n)²
17 (100+20n+n²) = 9 (600+160n+10n²) - 9 (400+120n+9n²)
1700+340n+17n² = 5400+1440n+90n² - 3600-1080n-81n²
1700+340n+17n² = 1800+360n+9n²
8n² - 20n - 100 = 0
2n² - 5n - 25 = 0
(2n+5) (n-5) = 0.
Since n>0, n=5.

f (x) = {x+a, if x<0; |x-1|, if x≥0}
g (x) = {x+1, if x<0; (x-1)²+b, if x≥0}
g (f (x) must be continuous. The potential points of discontinuity are where the definitions of f (x) and g (f (x) change. This is at x=0 and where f (x)=0.
f (x)=0 when x=-a (if a>0) or when x=1.
Continuity at x = 0:
lim (x→0? ) g (f (x) = lim (x→0? ) g (x+a). Since a could be anything, let's analyze f (0? )=a. So, lim is g (a).
lim (x→0? ) g (f (x) = g (f (0? ) = g (|0-1|) = g (1). Since 1≥0, g (1) = (1-1)²+b = b.
g (f (0) = g (|0-1|) = g (1) = b.
So we need g (a) = b.
Case 1: a < 0. g (a) = a+1. So a+1=b.
Case 2: a ≥ 0. g (a) = (a-1)²+b. So (a-1)²+b=b => (a-1)²=0 => a=1.
Now consider continuity at x=-a (assuming a>0).
lim (x→ (-a)? ) g (f (x) = g (f (-a)? ) = g (-a+a) = g (0? ) = 0+1=1.
lim (x→ (-a)? ) g (f (x) = g (f (-a)? ) = g (-a+a) = g (0? ) = (0-1)²+b = 1+b.
For continuity, 1 = 1+b => b=0.
If b=0, then we look at the condition g (a)=b from x=0 continuity.
If a<0, a+1=0 => a=-1.
If a≥0, a=1.
The solution gets a=1, b=0. So a+b=1.