Maths

R { (x, y):x, y∈z, x²+3y²≤8}
For domain of R? ¹
Collection of all integral of 's
For x=0, 3y²≤8
⇒ y∈ {-1,0,1}

α and β are roots of 5x²+6x-2=0
⇒ 5α²+6α-2=0
⇒ 5α? ²+6α? ¹-2α? =0
(By multiplying α? )
Similarly 5β? ²+6β? ¹-2β? =0
By adding (1) and (2)
5S? +6S? -2S? =0
For n=4
5S? +6S? =2S?

f (x)=sin (|x|+5)/ (x²+1)
For domain:
-1 ≤ (|x|+5)/ (x²+1) ≤ 1
Since |x|+5 and x²+1 is always positive
So (|x|+5)/ (x²+1) ≥ 0 ∀x∈R
So for domain:
(|x|+5)/ (x²+1) ≤ 1
⇒ |x|+5 ≤ x²+1
⇒ 0 ≤ x²-|x|-4
⇒ 0 ≤ (|x|- (1+√17)/2) (|x|- (1-√17)/2)
⇒ |x| ≥ (1+√17)/2 or |x|≤ (1-√17)/2 (Rejected)
⇒ x∈ (-∞, - (1+√17)/2] ∪ [ (1+√17)/2, ∞)
So, a = (1+√17)/2

Let p denotes statement
p: I reach the station in time.
q: I will catch the train.
Contrapositive of p→q is q→p
q→p: I will not catch the train, then I do not reach the station in time.

(2+sin x)/ (y+1) dy/dx = -cosx, y>0
⇒ dy/ (y+1) = -cosx/ (2+sinx) dx
By integrating both sides:
ln|y+1| = -ln|2+sinx|+lnK
⇒ y+1 = K/ (2+sinx) (y+1>0)
⇒ y (x) = K/ (2+sinx) - 1
Given y (0)=1 ⇒ 1=K/2-1 ⇒ K=4
So, y (x)=4/ (2+sinx)-1
a=y (π)=1
b=dy/dx|x=π = -cosx/ (y (x)+1)|x=π = 1
So, (a, b)= (1,1)

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