Maths

A (α, 0), B (-α, 0)
⇒ D (α, α² − 1)
Area (ABCD) = (AB) (AD)
⇒ S = (2α) (1 − α²) = 2α – 2α³
dS/dα = 2 - 6α²
= 0 ⇒ α² = 1/3

⇒ α = 1/√3
Area = 2α – 2α³ = 2/√3 - 2/ (3√3)
= 4/ (3√3)

∫ [π/6 to π/3] (d/dx (tan? x) * sin³3x + tan? x * d/dx (sin³3x) dx
= 1/2 ∫ [π/6 to π/3] d/dx (tan? x * sin?3x) dx
= 1/2 [tan? x · sin?3x] from π/6 to π/3
= 1/2 [ (√3)? · 0 - 1/ (√3)? )]
= -1/18

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

Here,
| 1 |
| 2 4 -1 | = 0 ⇒ λ = 9/2
| 3 2 λ |

Also,
| 1 2 |
| 2 4 6 | = 0 ⇒ μ = 5
| 3 2 μ |

∴ Option B is correct.

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

Since, lim (x→0) f (x)/x exist ⇒ f (0) = 0
Now, f' (x) = lim (h→0) (f (x+h)-f (x)/h = lim (h→0) (f (h)+xh²+x²h)/h (take y = h)
= lim (h→0) f (h)/h + lim (h→0) (xh) + x²
⇒ f' (x) = 1 + 0 + x²
⇒ f' (3) = 10