Maths

First we arrange 5 red cubes in a row and assume x? , x? , x? , x? , x? and x? number of blue cubes between them
Here, x? + x? + x? + x? + x? + x? = 11
and x? , x? , x? , x? ≥ 2
so x? + x? + x? + x? + x? + x? = 3
No. of solutions =? C? = 56

|adj (adj (A)| = |A|? ¹² = |A|?
|A| = | (14, 28, -14), (-14, 28), (25, -14, 14) |
= (14)² | (1, 2, -1), (-1, 2), (25/14, -1, 1) |
= (14)² (3–2 (–5)–1 (–1)
|A|? = (14)? => |A| = 14²

Let e? = t then equation reduces to
t³ – 11t – 45/t + 81/t² = 0
=> 2t? – 22t³ + 81t – 45 = 0 . (i)
If roots of
e³? – 11e²? – 45e? + 81/2 = 0
=> α? + α? + α? = ln45 => p = 45

f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99

| 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.