Maths

p →~ (p ∧ ~ q)
=~ p ∨ ~ (p ∧ ~ q)
=~ p ∨ ~ p ∨ q
=~ p ∨ q

Δ = | x-2 2x-3 3x-4 |
| 2x-3 3x-4 4x-5 |
| 3x-5 5x-8 10x-17|
= Ax³ + Bx² + Cx + D.
R? → R? - R? , R? → R? - R?
Δ = | x-2 2x-3 3x-4 |
| x-1 |
| x-2 (x-2) 6 (x-2) |
= (x-1) (x-2) | 1 2x-3 3x-4 |
| 1 |
| 1 2 6 |
= -3 (x - 1)² (x - 2) = -3x³ + 12x² - 15x + 6
∴ B + C = 12 - 15 = -3

∴ σ² ≤ 1/4 (M - m)²
Where M and m are upper and lower bounds of values of any random variable.
∴ σ² < 1/4 (10 - 0)
⇒ 0 < < 5
∴ σ ≠ 6

(1 + e? ) (1 + y²) dy/dx = y²
⇒ (1 + y? ²)dy = ( e? / (1 + e? ) ) dx
⇒ (y - 1/y) = ln (1 + e? ) + c
∴ It passes through (0,1) ⇒ c = -ln2
⇒ y² = 1 + yln ( (1+e? )/2 )

Equation of
AB = r = (î + j) + λ (3j - 3k)
Let coordinates of M
= (1, (1 + 3λ), -3λ).
PM = -3î + (3λ - 1)j - 3 (λ + 1)k
AB = 3j - 3k
? PM ⊥ AB ⇒ PM · AB = 0
⇒ 3 (3λ - 1) + 9 (λ + 1) = 0
⇒ λ = -1/3

∴ M = (1,0,1)
Clearly M lies on 2x + y - z = 1.

r = î (1 + 12l) + j (-1) + k (l)
r = î (2 + m) + j (m - 1) + k (-m)
For intersection
1 + 2l = 2 + m
-1 = m - 1
l = -m
from (ii) m = 0
from (iii) l = 0
These values of m and l do not satisfy equation (1).
Hence the two lines do not intersect for any values of l and m.

Let P = (3t², 6t); N = (3t²,0)
M = (3t², 3t)
Equation of MQ: y = 3t
∴ Q = (3/4 t², 3t)
Equation of NQ
y = ( 3t / (3/4 t² - 3t²) ) (x - 3t²)
y - intercept of NQ = 4t = 4/3 ⇒ t = 1/3
∴ MQ = 9/4 t² = 1/4
PN = 6t = 2

y² + ln (cos² x) = y x ∈ (-π/2, π/2)
for x = 0 y = 0 or 1
Differentiating wrt x
⇒ 2y' - 2tan x = y'
At (0,0)y' = 0
At (0,1)y' = 0
Differentiating wrt x
2yy' + 2 (y')² - 2sec² x = y'
At (0,0)y' = -2
At (0,1)y' = 2
∴ |y' (0)| = 2

A: D ≥ 0
⇒ (m + 1)² - 4 (m + 4) ≥ 0
⇒ m² + 2m + 1 - 4m - 16 ≥ 0
⇒ m² - 2m - 15 ≥ 0
⇒ (m - 5) (m + 3) ≥ 0
⇒ m ∈ (-∞, -3] U [5, ∞)
∴ A = (-∞, -3] U [5, ∞)
B = [-3,5)
A − B = (-∞, −3) U [5, ∞)
A ∩ B = {-3}
B - A = (-3,5)
A U B = R

Ellipse: x²/4 + y²/3 = 1
eccentricity = √ (1 - 3/4) = 1/2
∴ foci = (±1,0)
For hyperbola, given 2a = √2 ⇒ a = 1/√2
∴ hyperbola will be x²/ (1/2) - y²/b² = 1
eccentricity = √ (1 + 2b²)
∴ foci = (±√ ( (1+2b²)/2 ), 0)
∴ Ellipse and hyperbola have same foci
√ ( (1+2b²)/2 ) = 1
⇒ b² = 1/2
∴ Equation of hyperbola: x²/ (1/2) - y²/ (1/2) = 1
⇒ x² - y² = 1/2
Clearly, (√3/2, 1/2) does not lie on it.