60 - Calculus Questions Answers

 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)

let y = lim n->∞ (xⁿ/n!)

or y = x lim n->∞ (1/n) lim n->∞ x^n-1/(n-1)!

or y = 0

I think n is printed by mistake, so considering it as x

the question will be limit x tends to ∞

x{ [tan^-1 (x+1/x+4)] - (π/4)}

consider [tan^-1 (x+1/x+4)] - (π/4) = θ

so [tan^-1 (x+1/x+4)] = (π/4)+θ

so (x+1/x+4) = tan(π/4+θ)

or x = (-3-5tanθ)/2tanθ

so converted question will be limθ->0  (-3-5tanθ)θ/2tanθ

solve

I think n is printed by mistake, so considering it as x

the question will be limit x tends to ∞

x{ [tan^-1 (x+1/x+4)] - (π/4)}

consider [tan^-1 (x+1/x+4)] - (π/4) = θ

so [tan^-1 (x+1/x+4)] = (π/4)+θ

so (x+1/x+4) = tan(π/4+θ)

or x = (-3-5tanθ)/2tanθ

so converted question will be limθ->0  (-3-5tanθ)θ/2tanθ

now we know that limθ->0 tanθ/θ = limθ->0 θ/tanθ = 1

so next line will be limθ->0 (-3-5tanθ)/2 = -3/2 

We have to calculate limit n -> ∞ [{³√(n²-n³) }+ n ]

Use the formula a³+b³ = (a+b)(a²+b²-ab) in the format 

(a+b) = (a³+b³)/(a²+b²-ab)

here a = ³√(n²-n³)  and b = n

on solving we get 1/(1+1+1) = 1/3

f(x) = x|x|  => f(x) = -x²  for negative real values of x  and f(x) = x² for positive real values of x

so f^-1(x) = -√|x|  for negative real values and = +√|x| for positive real values of x

sin(|x|)+|x| is differentiable at x = 0

LHD of first = {cos|0+h|+|0+h|-cos|0|-|0|}/0+h-0  = (cosh+h-1)/h = (1-2sin²h+h-1)/h = 1 (on taking limits)

now check RHD, its value will be -1

similarly in second both values are 0 so it is differentiable