Class 12th

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

 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

In the gold foil scattering experiment by Rutherford and his team, almost all alpha particles passed straight through the foil. There were minor deflections mostly, and only a small amount of them showed deflections at really large angles. Only a few of them rebounded. This was the main observation that led Rutherford to conclude that the positive charge and mass of the atom concentrate at a very tiny section of the atom. That's the nucleus. The other observation was that the entire atom must be space where the alpha particles could go undeflected. 

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

Based on what classical electromagnetic theory says, we know that an accelerated charged particle must radiate energy or electromagnetic waves. That remains continuous. 

By looking at the Rutherford's atomic model we can assume that electrons would revolve around the nucleus, which in this logic, held in orbit by electrostatic attraction. The reason is that, their path is circular that's always accelerating. That would mean the electron would start to lose energy and finally let the atom to implode. But that doesn't happen. Atoms are stable and that's where classical physics that led to Rutherford model failed. 

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

(a + √2bcosx) (a - √2bcosy) = a² - b²
⇒ a² - √2abcosy + √2abcosx - 2b²cosxcosy = a² - b²
Differentiating both sides:
0 - √2ab (-siny dy/dx) + √2ab (-sinx) - 2b² [cosx (-siny dy/dx) + cosy (-sinx)] = 0
At (π/4, π/4):
ab dy/dx - ab - 2b² (-1/2 dy/dx + 1/2) = 0
⇒ dy/dx = (ab+b²)/ (ab-b²) = (a+b)/ (a-b); a, b > 0

f (x) = a? ⋅ (b? * c? ) = |x -2 3; -2 x -1; 7 -2 x|
= x³ - 27x + 26
f' (x) = 3x² - 27 = 0 ⇒ x = ±3 and f' (-3) < 0
⇒ local maxima at x = x? = -3
Thus, a? = -3i? - 2j? + 3k? , b? = 2i? - 3j? - k? , and c? = 7i? - 2j? - 3k?
⇒ a? ⋅ b? + b? ⋅ c? + c? ⋅ a? = 9 - 5 - 26 = -22