Maths

Slope of tangent to the curve y=x+siny
at (a, b) is (a-2)/ (b-3)=1
⇒ dy/dx|x=a = 1
dy/dx = 1+cos y dy/dx (from equation of curve)
⇒ dy/dx|x=a and y=b = 1+cosb=1
⇒ cosb=0
⇒ sinb=±1
Now, from curve y=x+siny
b=a+sinb
⇒ |b-a|=|sinb|=1

σ²=variance
µ=mean
σ² = Σ (x? -µ)²/n
µ=17
⇒ Σ (ax+b)/17 = 17
⇒ 9a+b=17
σ²=216
⇒ Σ (ax+b-17)²/17 = 216
⇒ Σa² (x-9)²/17 = 216
⇒ a² (81-18*9+3*35) = 216
⇒ a² (24) = 216 ⇒ a²=9 ⇒ a=3 (a>0)
⇒ From (1), b=-10
So, a+b=-7

|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)

|A|≠0
For (P): A≠I?
So, A = [1 0; 0 1] or [1; 0 1] or [1 0; 1]
or [1; 1 0]
So (P) is false.
A = [1 0; 1 0] or [1; 0 1] or [1 0; 1]
⇒ tr (A)=2
⇒ Q is true

Two points on the line (say) x/3 = y/2, z=1 are (0,0,1) and (3,2,1)
So dr's of the line is (3,2,0)
Line passing through (1,2,1), parallel to L and coplanar with given plane is r = i+2j+k + t (3i+2j), t∈R (-2,0,1) satisfies the line (for t=-1)
⇒ (-2,0,1) lies on given plane.
Answer of the question is (B)
We can check other options by finding equation of plane
Equation plane: |x-1, y-2, z-1; 1+2, 2-0, 1-1; 2+2, 1-0, 2-1| = 0
⇒ 2 (x-1)-3 (y-2)-5 (z-1)=0
⇒ 2x-3y-5z+9=0

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

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