Maths

D = |1 -2 3; 2 1; 1 -7 a| = 0 ⇒ a = 8
also, D? = |9 -2 3; b 1; 24 -7 8| = 0 ⇒ b = 3
hence, a-b = 8-3=5

Given (2x² + 3x + 4)¹? = Σ (r=0 to 20) a? x?
replace x by 2/x in above identity :-
2¹? (2x²+3x+4)¹? / x²? = Σ (r=0 to 20) a?2? /x?
⇒ 2¹? Σ (r=0 to 20) a? x? = Σ (r=0 to 20) a?2? x²? (from (i)
now, comparing coefficient of x? from both sides
(take r = 7 in L.H.S. and r = 13 in R.H.S.)
2¹? a? = a?2¹³ ⇒ a? /a? = 2³ = 8

x dy/dx - y = x² (xcosx + sinx), x > 0
dy/dx - y/x = x (xcosx+sinx) ⇒ dy/dx + Py = Q

So, I.F. = e^ (∫-1/x dx) = 1/|x| = 1/x (x > 0)
Thus, y/x = ∫ 1/x (x (xcosx+sinx)dx
⇒ y/x = xsinx + C
? y (π) = π ⇒ C = 1
So, y = x²sinx + x ⇒ (y)? /? = π²/4 + π/2
Also, dy/dx = x²cosx + 2xsinx + 1
⇒ d²y/dx² = -x²sinx + 4xcosx + 2sinx
⇒ [d²y/dx²] at π/2 = -π²/4 + 2

 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