166 - Questions Answers

let I = ₀ò¹1/{ (5+2x-2x²)(1+e^(2-4x)) } dx

on using the property ₀∫^a f(x) dx = ₀∫^af(a-x)dx

we get 2I = ₀ò¹1/(5+2x-2x²) dx

now solve this 

take x = 0 and y = 0 and n = 3

we get f(0) = f(0) + f(0)ⁿ or f(0) = 0

now take x = 0,  y = 1 and n = 3

so f(1) = f(0) + f(1)³

so get f(1) = 1, other values are not excetable because f(x) be a real valued function not identically equal to zero and f'(0) ³ 0

similarly get the other values 

you will get f(5) = 5

so f(x) = x generally then find f ^' (x) , i think it will be 1 

integer nearst to x and less than x will be [x]

so ₀ò^x [x] dx = ₀ò¹ 0 dx + ₁ò² 1 dx + ₂ò³ 2 dx + ..................... + _[x]-1∫^[x] [x]-1 dx + _[x]∫^x [x] dx

now solve it

according to the given condition f(nx) = n f(x)

₀òⁿ f(x) dx

consider x = ny so this integration will become ₀ò¹ f(ny) ndy = n² ₀ò¹ f(y) dy = n² ₀ò¹ f(x) dx

now by using these conditions I₁ = 1³ ₀ò¹ f(x) dx similarly others 

put these values and get the answer

 let I = ₀ò^p q ln sin q dq

on aplying property of definite integral

I =  ₀ò^p (π-q) ln sin q dq

so 2 I =  ₀ò^p π ln sin q dq 

or I = π/2  ₀ò^p  ln sin q dq

or I = 2π/2  ₀ò^p/2 ln sin q dq    this is due to property

similarly solve  ₀ò^p q² ln sin q dq  and   then ₀ò^p q³ ln sin q dq 

finally solve the right hand side of the equation to prove LHS = RHS

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