Class 11th

|x|/2 + |y|/3 = 1
x²/4 + y²/9 = 1
Area of Ellipse = πab = 6π
Required area = π*2*3 - (Area of quadrilateral)
= 6π - 1/2*6*4
= 6π-12
= 6 (π-2)

For ideal gas
PM = dRT
d = [PM]/R * 1/T
So graph between dV ST is not straight line

In this acid base Titrating there is no use of Bunsen burner and measuring cylinder other laboratory equipments will be required for getting the end point of titration.

The value of (1+sin (2π/9)+icos (2π/9)/ (1+sin (2π/9)-icos (2π/9)³
= (1+cos (5π/18)+isin (5π/18)/ (1+cos (5π/18)-isin (5π/18)³
= (2cos² (5π/36)+2isin (5π/36)cos (5π/36)/ (2cos² (5π/36)-2isin (5π/36)cos (5π/36)³
= (cos (5π/36)+isin (5π/36)/ (cos (5π/36)-isin (5π/36)³
= (e^ (i5π/36)/e^ (-i5π/36)³ = (e^ (i5π/18)³ = e^ (i5π/6) = cos (5π/6)+isin (5π/6)
= -√3/2 + i/2

Let L be the common normal to parabola
y = x²+7x+2 and line y = 3x-3
⇒ slope of tangent of y=x²+7x+2 at P=3
⇒ dy/dx|for p = 3
⇒ 2x+7=3 ⇒ x=-2 ⇒ y=-8
So P (-2, -8)
Normal at P: x+3y+C=0
⇒ C=26 (satisfies the line)
Normal: x+3y+26=0

In presence of sunlight CFC's molecule divides & release chlorine free radical, which react with ozone give chlorine monoxide radical (CIO°) and oxygen.
CF? Cl? (g) → C (g) + C? Cl (g)
Cl° (g) + O? (g) → ClO° (g) + O? (g)
ClO° (g) + O (g) → Cl° (g) + O? (g)

Sol. Let t? denotes r+1th term of (αx? + βx? )¹?
t? = ¹? C? (αx? )¹? (βx? )? = ¹? C? α¹? β? x? ¹?
If t? is independent of x
90-15r=0 ⇒ r=6
This differs from the solution.
Let's follow the solution's powers.
(10-r)/9 - r/6 = 0 ⇒ r=4
maximum value of t? is 10K (given)
⇒ ¹? C? α? β? is maximum
By AM ≥ GM (for positive numbers)
(α³/2+α³/2+β²/2+β²/2)/4 ≥ (α? β? /16)¹/?
⇒ α? β? ≤ 16
So, 10K = ¹? C?16
⇒ K=336