Maths

Given P (X=3) = 5P (X=4) and n=7
=>? C? p³q? = 5? C? p? q³
=> q = 5p and also p + q = 1
=> p = 1/6 and q = 5/6
Mean = 7/6 and variance = 35/36
Mean + Variance
= 7/6 + 35/36 = 77/36

Total case = ¹? C?
Favourable cases
? C? (Select x? )
³C? (Select x? )
? C? (Select x? )
P = (6.3.7)/¹? C? = 1/68

a = I + j – k
c = 2i – 3j + 2k
Now,
b x c = a
=> (i+j–k) (2i–3j+2k) = 0
=> 2 – 3 – 2 = 0
=> –3 = 0 (Not possible)
=> No possible value of b is possible.

l + m – n = 0 => n = l + m
3l² + m² + cnl = 0
3l² + m² + cl (l+m) = 0
= (3+c) (l/m)² + c (l/m) + 1 = 0
? Lines are parallel
D = 0
c = 4 (as c > 0)

x²/a² + y²/b² = 1
(-4/√5)²)/a² + (2/√5)²)/b² = 1
=> 32/a² + 9/b² = 1
=> 32/ (5a²) + 9/b² = 1 . (i)
From (i)
6/b² + 9/b² = 1 => b²=15 & a²=16
a²+b² = 15+16=31

2x + y = 4
2x + 6y = 14
} y=2, x=3
B (1, 2)
Let C (k, 4–2k)
Now AB² = AC²
=> 5k² – 24k + 19 = 0
α = (6+1+10/5)/3 = 18/5
Now 15 (α+β)
15 (17/5) = 51

dy/dx + (2? (2? -1)/ (2? ¹ (2? -1) = 0
x, y > 0, y (1) = 1
dy/dx = – (2? (2? -1)/ (2? (2? -1)
∫ (2? -1)/2? dy = –∫ (2? -1)/2? dx
log? (2? -1)/log?2 = – log? (2? -1)/log?2 + log? c/log?2
Taking log of base 2.
∴ y = 2 – log?3

I = ∫? ² (x³+|x|)/ (e|x|+1) dx . (i)
I = ∫? ² (x³+|x|)/ (e? |x|+1) dx . (ii)
= ∫? ² |x| dx = 2∫? ² x dx
= [x²/2]? ² = (16/4 + 4/2) - 0
= 4+2=6

I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4

cos? ¹ (y/2) = log? (x/5)? , |y| < 2
Differentiating on both side
(1/√ (1- (y/2)²) * (-y'/2) = 5 * (5/x) * (1/5)
(-xy')/ (2√ (1- (y²/4) = 5
Square on both side
(x²y'²)/4 = 25 * (4-y²)/4
Diff on both side
xy' + y'x² + 25y = 0