Maths

Kindly consider the following figure

|x|<1, |y|<1, x? y
(x+y) + (x²+xy+y²) + (x³+x²y+xy²+y³) + .
By multiplying and dividing by x-y:
(x²-y²)+ (x³-y³)+ (x? -y? )+.)/ (x-y)
= (x²+x³+x? +.)- (y²+y³+y? +.)/ (x-y)
= (x²/ (1-x)- (y²/ (1-y)/ (x-y)
= (x²-y²-xy (x-y)/ (1-x) (1-y) (x-y)
= (x+y-xy)/ (1-x) (1-y)                                         

Let B? be the event where Box-I is selected and B? →where box-II selected
P (B? )=P (B? )=1/2
Let E be the event where selected card is non prime.
For B? : Prime numbers: {2,3,5,7,11,13,17,19,23,29}
For B? : Prime numbers: {31,37,41,43,47}
P (E)=P (B? )*P (E/B? )+P (B? )P (E/B? )
= 1/2*20/30+1/2*15/20
Required probability:
P (B? /E) = (P (E/B? )P (B? )/P (E) = (1/2*20/30)/ (1/2*20/30+1/2*15/20) = (2/3)/ (2/3+3/4) = 8/17

Slope of tangent is 2, Tangent of hyperbola
x²/4-y²/2=1 at the point (x? , y? ) is xx? /4-yy? /2=1 (T=0)
Slope: x? /2y? =2 ⇒ x? =4y?
(x? , y? ) lies on hyperbola
⇒ x? ²/4-y? ²/2=1
From (1) and (2)
(4y? )²/4-y? ²/2=1 ⇒ 4y? ²-y? ²/2=1
⇒ 7y? ²=2 ⇒ y? ²=2/7
Now x? ²+5y? ² = (4y? )²+5y? ² = 21y? ² = 21*2/7=6

2x-y+2z=2
x-2y+λz=-4
x+λy+z=4
For no solution:
D=|2, -1,2; 1, -2, λ 1,1,1|=0
⇒ 2 (-2-λ²)+1 (1-λ)+2 (λ+2)=0
⇒ -2λ²+λ+1=0
⇒ λ=1, -1/2
D? =|2, -1,2; -4, -2, λ 4,1,1|
=2 (-2-λ)+1 (-4-4λ)+2 (-4+8)
=2 (1+λ) which is not equal to zero for λ=1, -1/2

Let three terms of G.P. are a/r, a, ar product=27
⇒ a³=27 ⇒ a=3
S=3/r+3r+3
For r>0
(3/r+3r)/2 ≥ √9 = 3
⇒ 3/r+3r≥6
For r<0, 3/r+3r-6
From (1) and (2)
S∈ (-∞, -3]∪ [9, ∞)

f (x) = {ae? +be? , -1For continuity at x=1
Lim (x→1? )f (x) = Lim (x→1? )f (x)
⇒ ae+be? ¹=c ⇒ b=ce-ae²
For continuity at x=3
Lim (x→3? )f (x) = Lim (x→3? )f (x)
⇒ 9c=9a+6c ⇒ c=3a
f' (0)+f' (2)=e
(ae? -be? ) at x=0 + (2cx) at x=2 = e
⇒ a-b+4c=e
From (1), (2) and (3)
a-3ae+ae²+12a=e
⇒ a (e²+13-3e)=e
⇒ a=e/ (e²-3e+13)

Since p (x) has relative extreme at x=1 and 2
so p' (x)=0 at x=1 and 2
⇒ p' (x)=A (x-1) (x-2)
⇒ p (x)=∫A (x²-3x+2)dx
p (x)=A (x³/3 - 3x²/2 + 2x)+C
P (1)=8
From (1)
8=A (1/3-3/2+2)+C
⇒ 8=5A/6+C ⇒ 48=5A+6C
P (2)=4
⇒ 4=A (8/3-6+4)+C
⇒ 4=-2A/3+C ⇒ 12=-2A+3C
From 3 and 4, C=-12
So P (0)=C=-12

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