Maths

 max {n (A), n (B)} ≤ n (A U B) ≤ n (U)
⇒ 76 ≤ 76 + 63 - x ≤ 100
⇒ -63 ≤ -x ≤ -39
⇒ 63 ≥ x ≥ 39

 f (x) = ∫ (from 1 to 3) (√x dx)/ (1+x)² = ∫ (from 1 to √3) (t⋅2tdt)/ (1+t²)² (put √x = t)
= [ (-t/ (1+t²)] (from 1 to √3) + [tan? ¹t] (from 1 to √3) [Applying by parts]
= (-√3/4 + 1/2) + (π/3 - π/4)
= (-√3+2)/4 + π/12

  [x]² + 2 [x+2] - 7 = 0
⇒ [x]² + 2 [x] + 4 - 7 = 0
⇒ [x] = 1, -3
⇒ x ∈ [1,2) U [-3, -2)

Σ (r=30 to 50) 50-rC? =? C? +? C? +? C? + . + ³? C?
=? C? +? C? +? C? + . + (³? C? + ³? C? ) - ³? C?
=? C? +? C? +? C? + . + (³¹C? + ³¹C? ) - ³? C?
=? C? +? C? - ³? C?
=? ¹C? - ³? C?

A² = (cos2θ isin2θ cos2θ)
Similarly, A? = (cos5θ isin5θ cos5θ) = (a b; c d)
(1) a²+b² = cos²5θ - sin²5θ = cos10θ = cos75°
(2) a²-d² = cos²5θ - cos²5θ = 0
(3) a²-b² = cos²5θ + sin²5θ = 1
(4) a²-c² = cos²5θ + sin²5θ = 1

∫ (x/ (xsinx+cosx)²dx = ∫ (xcosx⋅xcosx)/ (xsinx+cosx)² dx
= x/cosx (-1/ (xsinx+cosx) + ∫ (cosx-xsinx)/cosx² (1/ (xsinx+cosx) dx
= -xsecx/ (xsinx+cosx) + ∫sec²xdx
= -xsecx/ (xsinx+cosx) + tanx + C

Since (3,3) lies on x²/a² - y²/b² = 1
9/a² - 9/b² = 1
Now, normal at (3,3) is y-3 = -a²/b² (x-3),

which passes through (9,0) ⇒ b² = 2a²

So, e² = 1 + b²/a² = 3
Also, a² = 9/2
(From (i) and (ii)
Thus, (a², e²) = (9/2, 3)

u = (2z+i)/ (z-ki)
= (2x² + (2y+1) (y-k)/ (x²+ (y-k)²) + I (x (2y+1) - 2x (y-k)/ (x²+ (y-k)²)
Since Re (u) + Im (u) = 1
⇒ 2x² + (2y+1) (y-k) + x (2y+1) - 2x (y-k) = x² + (y-k)²
P (0, y? )
Q (0, y? )
⇒ y² + y - k - k² = 0 {y? + y? = -1, y? = -k-k²}
∴ PQ = 5
⇒ |y? - y? | = 5 ⇒ k² + k - 6 = 0
⇒ k = -3, 2
So, k = 2 (k > 0)