Maths

According to the question,  
  (7 * 4) + (n * 6) = 52
6n = 24 ⇒ n = 4

The system of equations has no solution if the determinant of the coefficient matrix is zero.
Δ = |k 1|
|1 k 1|
|1 k|
Δ = k (k² - 1) - 1 (k - 1) + 1 (1 - k) = 0
Δ = k³ - k - k + 1 + 1 - k = 0
⇒ k³ - 3k + 2 = 0 ⇒ (k - 1)² (k + 2) = 0
∴ k = -2, 1
If k = 1 then all the equations are identical (infinite solutions). Hence k = -2 for no solution.

dy/dx = xy - 1 + x - y = x (y + 1) - (y + 1) = (x - 1) (y + 1)
⇒ ∫ dy / (1 + y) = ∫ (x - 1) dx ⇒ ln (1 + y) = x²/2 - x + C
For y (0) = 0, c = 0. ∴ ln (1 + y) = x²/2 - x
1 + y = e^ (x²/2 - x)
y (1) = e^ (1/2 - 1) - 1 = e^ (-1/2) - 1

n (s) = 6² = 36
E = { (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), (2, 1), (2, 2), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 5), (5, 1), (5, 2), (5, 3), (7, 1)}
∴ n (E) = 17
Required prob. = n (E) / n (S) = 17 / 36

S-2, L-2, A, B, Y, U
Required = ²C? ⋅? C? ⋅ 4!/2! = 2 ⋅ 10 ⋅ 24/2 = 240

P (at least 2 show 3 or 5) =? C? (2/6)² (4/6)² +? C? (2/6)³ (4/6)¹ +? C? (2/6)?
= (384+128+16)/6? = 11/27
n=27
∴ expectation of number of times = np
= 27 ⋅ (11/27) = 11

2x-y+3=0
4x-2y+α=0 ⇒ 2x-y+α/2=0
6x-3y+β=0 ⇒ 2x-y+β/3=0
d? = |α/2-3|/√ (2²+1²) = 1/√5 ⇒ |α-6|=2 ⇒ α-6=2, -2 ⇒ α=8,4
d? = |β/3-3|/√ (2²+1²) = 2/√5 ⇒ |β-9|=6 ⇒ β-9=6, -6 ⇒ β=15,3
Sum of all value of α and β = 30.

We have, 1 - (probability of all shots result in failure) > 1/4
? 1 - (9/10)? > 1/4
? 3/4 > (9/10)? ? n? 3 > (9/10)? ? n? 3

-5 ≤ [x/2] < 5
I ⇒ [x/2] = -5, -4, -3, -2, -1,0,1,2,3,4
Hence, function is discontinues at = -4, -3, -2, -1,1,2,3,4 Number of values is 8.

-5? [x/2] < 5
I? [x/2] = -5, -4, -3, -2, -1,0,1,2,3,4
Hence, function is discontinues at = -4, -3, -2, -1,1,2,3,4 Number of values is 8.