Maths

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

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)