Class 11th

Since AM of two positive quantities ≥ their G.M.
(2^sinx + 2^cosx)/2 ≥ √ (2^sinx * 2^cosx)
= √ (2^ (sinx+cosx)
= √2^ (√2cos (x-π/4)
≥ √2^ (-√2) ⇒ 2^sinx + 2^cosx ≥ 2 · 2^ (-1/√2) = 2^ (1-1/√2)

By family of circle, passing through intersection of given circle will be member of S + λS? = 0 family (λ ≠ 1)
(x² + y² – 6x) + λ (x² + y² – 4y) = 0
(λ + 1)x² + (λ + 1)y² – 6x – 4λy = 0
x² + y² - 6/ (λ+1) x - 4λ/ (λ+1) y = 0
Centre (3/ (λ+1), 2λ/ (λ+1)
Centre lies on 2x – 3y + 12 = 0
2 (3/ (λ+1) - 3 (2λ/ (λ+1) + 12 = 0
6λ + 18 = 0
λ = -3
Circle x² + y² – 3x – 6y = 0

Given: 3α² – 10α + 27λ = 0
3α² – 3α + 6λ = 0
Subtract -7α + 21λ = 0
3λ = 0
By (ii) 9λ² – 3λ + 2λ = 0
⇒ λ = 0, 1/9
∴ α = 1/3, β = 2/3, α = 1/3, γ = 3
∴ (βγ)/λ = (2/3) * 3) / (1/9) = 18

N+5C_ {R-1}: N+5C_R: N+5C_ {R+1}
= 5:10:4
2 (N+5C_ {R-1}) = N+5C_R ⇒ 3R = N + 6
7 (N+5C_R) = 5 (N+5C_ {R+1}) ⇒ 12R = 18 + 5N
Solving: N = 6, R = 4
∴ the largest coefficient is N+5C_ {R+1} = 11C_5 = 462

Contrapositive of p ↔ q is ~ q →~ p.

= [√3e^ (iπ/3)]^4
= 9 (cos (2π/3) + isin (2π/3)
= -9/2 + 9√3i/2
⇒ 0 + 9 (-1 + i√3)/2)
∴ a = 0, b = 9

Dihydrogen of high degree of purity (>99.95%) is obtained by the electrolysis of warm aqueous barium hydroxide solution between nickel electrodes.

³? Cl ³? Cl
Molar ratio x 1 – x
M? 35 * x + 37 (1 − x) = 35.5
35x + 37 – 37x = 35.5
2x = 1.5
x = 3/4
So, ratio of ³? Cl: ³? Cl = 3: 1