Mathematics Test - 3

The total number of terms in the binomial expansion of (a + b)ⁿ is n + 1, i.e. one more than the exponent n.

a = 4, l = 31, n = 8
(l-a)/(n+1) = (31-4)/(8+1)
= 27/9  
= 3

(a ⁿ⁺¹+ b ⁿ⁺¹)/(a ⁿ+ b ⁿ) is mean between a &b
Mean of a &b =  ( a + b)/2
=>(a ⁿ⁺¹+ b ⁿ⁺¹)/(a ⁿ+ b ⁿ) = ( a + b)/2
=>2a ⁿ⁺¹+ 2b ⁿ⁺¹ = a ⁿ⁺¹ + b ⁿ⁺¹ + ba ⁿ+ b ⁿa
=>a ⁿ⁺¹+ b ⁿ⁺¹= ba ⁿ+ b ⁿa
=>a ⁿ(a - b) = b ⁿ(a - b)
=>a ⁿ= b ⁿ
=>(a/b)ⁿ= 1
=>n = 0   or a  = b

Let the digits at ones, tens and hundreds place be (a −d)a and (a+d) respectively. The, the number is
(a+d)×100+a ×10+(a −d) = 111a+99d
The number obtained by reversing the digits is
(a −d)×+a ×10+(a+d) = 111a −99d
It is given that the sum of the digits is 21.
(a −d)+a+(a+d) = 21             ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a −99d = 111a+99d −396      ...(ii)
⟹3a = 21 and 198d = 396
⟹a = 7 and d = -2
Original number = (a −d)×+a ×10+(a+d)
= 100(9) + 10(7) + 5
= 975

Let A(1,3), B(-2,9), and C(x,-1)
For to be points collinear,
x₁ (y₂ -y₃ ) + x₂ (y₃ -y₁ ) + x₃ (y₁ -y₂ )=0
⇒1(9-(-1)) + (-2)(-1-3) + x(3-9)=0
⇒1(10)+(-2)(-4)+x(-6)=0
⇒10+8-6x=0
⇒18 = 6x
⇒x = 3

P(2,3) Q(3,5) R(1,2)
R is at centre between P and Q, using section formula for internal division
Therefore, (1,2) = ((3 λ+2)/(λ+1), (5 λ+3)/(λ+1))
1 = (3 λ+2)/(λ+1)
(λ+1) = (3 λ+2)
λ= -1/2
- sign indicates the external division

Given circle = (2,3)
Given line (x+y-1) = 0
Distance between point to line is:
d = |ax₁ + by₁ + c|/√(a² + b² )
where a = 1, b = 1, c = -1 and x1 = 2, y1 = 3
d = |1(2) + 1(3) - 1|/√(1+1)
d = 4/(√2)
d = 2 √2

 Equation : 3x + 4y = 15
Points : (1,-3)
3(x-1) + 4(y-(-3)) = 15
3(x-1) + 4(y+3) = 15
3x - 3 + 4y + 12 = 15
3x + 4y = 6

Equation : 5x + 8y = 10
Points (2, -3)
(x-2, y+3)
⇒5(x-2) + 8(y+3) = 10
= 5x - 10 + 8y + 24 = 10
⇒5x + 8 = - 4