Maths

lim (x→1) (x+x²+.+x? -n)/ (x-1) = 820
⇒ lim (x→1) (x-1)/ (x-1)+ (x²-1)/ (x-1)+.+ (x? -1)/ (x-1) = 820
⇒ 1+2+.+n = 820
⇒ n (n+1)=2*820
⇒ n (n+1)=40*41
Since n∈N, so n=40

|a|=|b|=|c|=1
|a-b|²+|a-c|²=8
⇒|a|²+|b|²-2a.b+|a|²+|c|²-2a.c=8
⇒4-2 (a.b+a.c)=8
⇒a.b+a.c=-2
|a+2b|²+|a+2c|²
=|a|²+4|b|²+4a.b+|a|²+4|c|²+4a.c
=10+4 (a.b+a.c)
=10-8=2

MOTHER
1? E
2? H
3? M
4? O
5? R
6? T
So position of word MOTHER in dictionary
2*5!+2*4!+3*3!+2!+1
=240+48+18+2+1
=309

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