# 59 - Calculus Questions Answers

f(x+y) = f(x) + f(y) + 2xy - 1 " x,y. f is differentiable and f'(0) = cos a. Prove that f(x) > 0 " x Î R.

Joshi sir comment

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

Find area of region bounded by the curve y=[sinx+cosx] between x=0 to x=2p.

Joshi sir comment

[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

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

Joshi sir comment

[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

Solution by Joshi sir

[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,

Let g(x) be a continuous function such that    0ò1 g(t) dt = 2.   Let f(x) = 1/2 0òx (x-t)2 g(t) dt then find f '(x) and hence evaluate f "(x).

Joshi sir comment

f(x) = 1/2 0òx (x-t)2 g(t) dt

or f(x) = 1/2 0òx2 g(t) dt + 1/2 0òt2 g(t) dt - 0òx t g(t) dt

or f(x) = 1/2 x2 0ò g(t) dt + 1/2 0òt2 g(t) dt - x 0ò 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)

limit  n tends to ∞

then

(x^n)/(n!) equals

Joshi sir comment

let y = lim n->∞ (xn/n!)

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

or y = 0

limit n tends to ∞

x{ [tan‾¹ (x+1/x+4)] - (π/4)}

Joshi sir comment

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

Solution by Joshi sir

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

limit n tends  to ∞

then

[³√(n²-n³) + n ] equals

Joshi sir comment

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

if f(X)=xlxl then find f-1(x) can you please explain me the meaning and use of sgn

Joshi sir comment

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

which of the following is differntiable at x=0 ?

cos(lxl)+lxl

sin(lxl)-lxl