Maths

Any point (x, y) on perpendicular bisector equidistant from p and q
∴ (x − 1)² + (y − 4)² = (x − k)² + (y − 3)²
At x = 0, y = -4
∴ 1 + 64 = k² + 49
k² = 16

Let number of elements in T is R.
∴ 20R = 500 ⇒ R = 25
and 6R = 5N ⇒ N = 30

Given : a/e = 4 and 1/4 = 1 - b²/a²
Solving : a = 2, b = √3
Parametric co - ordinates are
(2cosθ, √3sinθ) = (1, β)
∴ θ = 60°
∴ Equation of normal is axsecθ − bycosecθ = a² − b²
⇒ 4x - 2y = 1

tan 30° = x/y ⇒ y = √3x
and tan 60° = (x+400)/y ⇒ √3y = x+400
= x + 400
Solving (i) and (ii), we get
2x = 400, x = 200
sin 30° = x/PC = 200/PC ⇒ PC = 400

sum 6 → (1,5), (5,1), (3,3), (2,4), (4,2)
sum 7 → (1,6), (6,1), (5,2), (2,5), (3,4), (4,3)
= P (A) + P (? ) · P (B) · P (A) + P (? )P (B)P (? ) · P (B) · P (A) + …
This is infinite G.P. with common ratio P (? ) * P (B)
Probability of A wins = P (A) / (1 - P (? )P (B? )
= (5/36) / (1 - (31/36)* (30/36) = 30/61

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