Maths

Since (3,3) lies on x²/a² - y²/b² = 1
9/a² - 9/b² = 1
Now, normal at (3,3) is y-3 = -a²/b² (x-3),

which passes through (9,0) ⇒ b² = 2a²

So, e² = 1 + b²/a² = 3
Also, a² = 9/2
(From (i) and (ii)
Thus, (a², e²) = (9/2, 3)

u = (2z+i)/ (z-ki)
= (2x² + (2y+1) (y-k)/ (x²+ (y-k)²) + I (x (2y+1) - 2x (y-k)/ (x²+ (y-k)²)
Since Re (u) + Im (u) = 1
⇒ 2x² + (2y+1) (y-k) + x (2y+1) - 2x (y-k) = x² + (y-k)²
P (0, y? )
Q (0, y? )
⇒ y² + y - k - k² = 0 {y? + y? = -1, y? = -k-k²}
∴ PQ = 5
⇒ |y? - y? | = 5 ⇒ k² + k - 6 = 0
⇒ k = -3, 2
So, k = 2 (k > 0)

Let TV (r) denotes truth value of a statement r.
Now, if TV (p) = TV (q) = T
⇒ TV (S? ) = F
Also, if TV (p) = T and TV (q) = F
⇒ TV (S? ) = T

1 + (1 - 2²⋅1) + (1 - 4²⋅3) + . + (1 - 20²⋅19)
= α - 220β
= 11 - (2²⋅1 + 4²⋅3 + . + 20²⋅19)
= 11 - 2² ⋅ Σ? (r=1) r² (2r-1) = 11 - 4 (110²/2) - 35 x 11)
= 11 - 220 (103)
⇒ α = 11, β = 103

(a + √2bcosx) (a - √2bcosy) = a² - b²
⇒ a² - √2abcosy + √2abcosx - 2b²cosxcosy = a² - b²
Differentiating both sides:
0 - √2ab (-siny dy/dx) + √2ab (-sinx) - 2b² [cosx (-siny dy/dx) + cosy (-sinx)] = 0
At (π/4, π/4):
ab dy/dx - ab - 2b² (-1/2 dy/dx + 1/2) = 0
⇒ dy/dx = (ab+b²)/ (ab-b²) = (a+b)/ (a-b); a, b > 0

f (x) = a? ⋅ (b? * c? ) = |x -2 3; -2 x -1; 7 -2 x|
= x³ - 27x + 26
f' (x) = 3x² - 27 = 0 ⇒ x = ±3 and f' (-3) < 0
⇒ local maxima at x = x? = -3
Thus, a? = -3i? - 2j? + 3k? , b? = 2i? - 3j? - k? , and c? = 7i? - 2j? - 3k?
⇒ a? ⋅ b? + b? ⋅ c? + c? ⋅ a? = 9 - 5 - 26 = -22

 x? = 10
⇒ x? = (63 + a + b)/8 = 10
⇒ a + b = 17
Since, variance is independent of origin.
So, we subtract 10 from each observation.
So, σ² = 13.5 = (79 + (a-10)² + (b-10)²)/8
⇒ a² + b² - 20 (a+b) = -171
⇒ a² + b² = 169
From (1) and (2) ; a = 12 and b = 5