Maths

Equation PQ
(x-1)/2 = (y+2)/3 = (z-3)/ (-6) = λ
Let Q = (2λ + 1, 3λ − 2, −6λ + 3)
Q lies on x - y + z = 5
⇒ (2λ + 1) − (3λ − 2) + (−6λ + 3) = 5
⇒ λ = -1/7
Q = (5/7, -17/7, 15/7)

∴ PQ = √ (2/7)² + (3/7)² + (6/7)²)

Applying L'Hôpital's Rule
Lim (t→x) [2tf² (x) – x² (2f (t)f' (t)] / 1
∴ 2xf² (x) – x² (2f (x)f' (x) = 0
⇒ f (x) – xf' (x) = 0
⇒ f' (x)/f (x) = 1/x ⇒ lnf (x) = lnx + C
At x=1, c=1
∴ lnf (x) = lnx + 1
when f (x) = 1
then lnx = -1
x = 1/e

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)

dy/dx = (y-3x)/ln (y-3x) - 3
dy/dx + 3 = (y+3x)/ln (y+3x)
d/dx (y+3x) = (y+3x)/ln (y+3x)
∫ (ln (y+3x)/ (y+3x) d (y+3x) = ∫ dx
Let ln (y+3x) = t
1/ (y+3x) d (y+3x) = dt
∫ tdt = ∫ dx
t²/2 = x+c
(ln (y+3x)²/2 = x+c

Let A =
| a? |
| b? |
| c? |

Ax? = B?
a? + a? + a? = 1
b? + b? + b? = 0
c? + c? + c? = 0
Similar 2a? + a? = 0 and a? = 0
2b? + b? = 2, b? = 0
2c? + c? = 0, c? = 2
∴ a? = 0, b? = 1, c? = -1,
a? = 1, b? = -1, c? = -1
A =
| 1 0 |
| -1 0 |
| -1 -1 2 |
∴ |A| = 2

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.