Maths

| p | q | p^q | ¬ (p^q) | (¬ (p^q)vq | pvq | p? q | p? (pvq) |
|-|-|-|-|-|-|-|-|
| T | T | T | F | T | T | T | T |
| T | F | F | T | T | T | F | T |
| F | T | F | T | T | T | T | T |
| F | F | F | T | T | F | T | T |
Tautology, Tautology

sin? ¹ (√3/2) + cos? ¹ (-√3/2) + tan? ¹ (-1)
= π/3 + 5π/6 – π/4
= (4π+10π–3π)/12 = 11π/12

(cos (2π/7) + cos (4π/7) + cos (6π/7)/ (sin (3π/7)cos (π/7) = (2sin (π/7)cos (2π/7) + 2sin (π/7)cos (4π/7) + 2sin (π/7)cos (6π/7)/ (2sin (π/7)
= (sin (3π/7)-sin (π/7) + sin (5π/7)-sin (3π/7) + sin (π)-sin (5π/7)/ (2sin (π/7) = (-sin (π/7)/ (2sin (π/7) = -1/2

Given P (X=3) = 5P (X=4) and n=7
=>? C? p³q? = 5? C? p? q³
=> q = 5p and also p + q = 1
=> p = 1/6 and q = 5/6
Mean = 7/6 and variance = 35/36
Mean + Variance
= 7/6 + 35/36 = 77/36

Total case = ¹? C?
Favourable cases
? C? (Select x? )
³C? (Select x? )
? C? (Select x? )
P = (6.3.7)/¹? C? = 1/68

a = I + j – k
c = 2i – 3j + 2k
Now,
b x c = a
=> (i+j–k) (2i–3j+2k) = 0
=> 2 – 3 – 2 = 0
=> –3 = 0 (Not possible)
=> No possible value of b is possible.

l + m – n = 0 => n = l + m
3l² + m² + cnl = 0
3l² + m² + cl (l+m) = 0
= (3+c) (l/m)² + c (l/m) + 1 = 0
? Lines are parallel
D = 0
c = 4 (as c > 0)

x²/a² + y²/b² = 1
(-4/√5)²)/a² + (2/√5)²)/b² = 1
=> 32/a² + 9/b² = 1
=> 32/ (5a²) + 9/b² = 1 . (i)
From (i)
6/b² + 9/b² = 1 => b²=15 & a²=16
a²+b² = 15+16=31

2x + y = 4
2x + 6y = 14
} y=2, x=3
B (1, 2)
Let C (k, 4–2k)
Now AB² = AC²
=> 5k² – 24k + 19 = 0
α = (6+1+10/5)/3 = 18/5
Now 15 (α+β)
15 (17/5) = 51

dy/dx + (2? (2? -1)/ (2? ¹ (2? -1) = 0
x, y > 0, y (1) = 1
dy/dx = – (2? (2? -1)/ (2? (2? -1)
∫ (2? -1)/2? dy = –∫ (2? -1)/2? dx
log? (2? -1)/log?2 = – log? (2? -1)/log?2 + log? c/log?2
Taking log of base 2.
∴ y = 2 – log?3