Maths

(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

 x²/a² + y²/b² = 1 (a > b); 2b²/a = 10 ⇒ b² = 5a
Now, φ (t) = -5/12 + t - t² = 8/12 - (t - 1/2)²
φ (t)max = 8/12 = 2/3 = e ⇒ e² = 1 - b²/a² = 4/9
⇒ a² = 81 (From (i) and (ii)
So, a² + b² = 81 + 45 = 126

f (2)=8, f' (2)=5, f' (x) ≥ 1, f' (x) ≥ 4, ∀x ∈ (1,6)
Using LMVT
f' (x) = (f' (5) - f' (2)/ (5-2) ≥ 4 ⇒ f' (5) ≥ 17
f' (x) = (f (5) - f (2)/ (5-2) ≥ 1 ⇒ f (5) ≥ 11
Therefore f' (5) + f (5) ≥ 28

x² - 3x + p = 0
α, β, γ, δ in G.P.
α + αr = 3
x² - 6x + q = 0
ar² + ar³ = 6
(2) ÷ (1) ⇒ r² = 2
So, 2q+p/2q-p = (2r? +r)/ (2r? -r) = (2r? +1)/ (2r? -1) = 9/7