161 - Mathematics Questions Answers

f(x+y) = f(x) + f(y) + 2xy - 1 

so f(0) = 1      obtain this by putting x = 0 and y = 0

now f ^' (x+y) = f ^' (x)  + 2y         on differentiating with respect to x

take x = 0, we get f ^' (y) = f ^'(0) + 2y

or f ' (x) = cosα + 2x

now integrate this equation within the limits 0 to x

we get f(x) = x² + cosα x + 1

its descreminant is negative and coefficient of x² is positive so f(x) > 0

[sinx + cosx] = [√2 sin{x+(π/4)}]    here [ ] is greatest integer function

its value in different interval are

1       for 0 to π/2

0       for π/2 to 3π/4

-1        for 3π/4 to π

-2     for π to 3π/2

-1     for 3π/2 to 7π/4

0      for 7π/4 to 2π

now solve

[3Öxy + 4y - 7Öy]dx + [4x - 7Öxy + 5Öx]dy = 0

=> dy/dx = [3Öxy + 4y - 7Öy] / [-4x + 7Öxy - 5Öx]

=> dy/dx = √y/√x [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]

=> (dy/√y)/(dx/√x) = [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]

now take √x = X + h and √y = Y + k

you will get a homogeneous equation dY/dX = (3X+4Y+3h+4k-7)/(7Y-4X+7k-4h-5)

take 3h+4k-7 = 0 and 7k-4h-5 = 0 and solve the equation

[3Öxy + 4y - 7Öy]dx + [4x - 7Öxy + 5Öx]dy = 0

=> dy/dx = [3Öxy + 4y - 7Öy] / [-4x + 7Öxy - 5Öx]

=> dy/dx = √y/√x [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]

=> (dy/√y)/(dx/√x) = [(3√x + 4√y - 7) / (-4√x + 7√y - 5)]

now take √x = X + h and √y = Y + k

you will get a homogeneous equation dY/dX = (3X+4Y+3h+4k-7)/(7Y-4X+7k-4h-5)

take 3h+4k-7 = 0 and 7k-4h-5 = 0 and solve the equation

dY/dX = (3X+4Y)/(7Y-4X)

take Y = uX then dY/dX = u + Xdu/dX

after solving put the values of X and Y in terms of x and y, 

f(x) = 1/2 ₀ò^x (x-t)² g(t) dt

or f(x) = 1/2 ₀ò^x x² g(t) dt + 1/2 ₀ò^x t² g(t) dt - ₀ò^x x t g(t) dt 

or f(x) = 1/2 x² ₀ò^x  g(t) dt + 1/2 ₀ò^x t² g(t) dt - x ₀ò^x  t g(t) dt

now use newton leibniz formula for middle function and product rule for first and last functions for finding f^' (x)

similarly get f^'' (x)

Angle between two lines is given by the formula

tanθ = (m₁-m₂)/(1+m₁m₂)

solve this by taking θ = 90

(1+λ)ⁿ = C₀ + C₁x + ..................................C_nxⁿ

similarly for (1+μ)ⁿ and 

(λ+μ)ⁿ = C₀λⁿ + C₁λ^n-1μ + ........................ + C_nμⁿ

On multiplying these 3, we will get the coefficient of  λⁿμⁿ = ∑C_n³      (Here you should remember that C₀ = C_n, C₁ = C_n-1 and etc)

For getting the value of  ∑C_n³ :

multiply the expansions of (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ and calculate the coefficient of x⁰, we get ∑C_n³

and if we multiply (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ, then we get (-1)ⁿ(1-x²)²ⁿ/x²ⁿ so coefficient of x⁰ in this expression will be (-1)^n 2nC_n (-1)ⁿ  = ²ⁿC_n

Let x = (5√5 + 11)²ⁿ⁺¹ and y = (5√5 - 11)²ⁿ⁺¹

so x*y = 4²ⁿ⁺¹ = even number

similarly x - y = integer

if we consider the value of y , it will be (5*2.3-5)²ⁿ⁺¹ = (0.5)²ⁿ⁺¹  = fraction number less than 1   (approx)

so definitely y will be the fractional part of x, and difference of x and y is even so integer part of x will be even

let z = r(cosθ+i sinθ) and it is given that z¹⁵ = 1  =>  r¹⁵(cosθ+i sinθ)¹⁵ = 1

or r¹⁵(cos15θ+i sin15θ) = 1+0i                   (Here we use Demoivre's theorem)

on comparison we get sin 15θ = 0 => 15θ = nπ          here n is an integer

for n = -7 to +7 we will get 15 answers for which θ will lie between -π/2 to +π/2

max of sin and cos are 1 so max of left is 2

but x² + 1/x² ≥ 2x(1/x) = 2

so only solution will be obtained for 2cos²(x/2)sin²(x) = 2

so (1+cosx)(1-cos²x) = 2 which is impossible,so i think no solution

I think the last term would be cos(14π/15) 

We know that cos(14π/15) = -cos(π/15)

so     16cos(2π/15)cos(4π/15)cos(8π/15)cos(14π/15)

   = -16cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

   = -[8/sin(π/15)]2sin(π/15)cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

  =  -[8/sin(π/15)]sin(2π/15)cos(2π/15)cos(4π/15)cos(8π/15)   

take the similar steps