# 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.

**Asked By: KAMAL**

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**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) = x^{2} + cosα x + 1

its descreminant is negative and coefficient of x^{2} is positive so f(x) > 0

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

**Asked By: KAMAL**

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**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.

**Asked By: KAMAL**

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**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 _{0}ò^{1} g(t) dt = 2. Let f(x) = 1/2 _{0}ò^{x} (x-t)^{2} g(t) dt then find f '(x) and hence evaluate f "(x).

**Asked By: KAMAL**

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**Joshi sir comment**

f(x) = 1/2 _{0}ò^{x} (x-t)^{2} g(t) dt

or f(x) = 1/2 _{0}ò^{x }x^{2 }g(t) dt + 1/2 _{0}ò^{x }t^{2 }g(t) dt - _{0}ò^{x }x t g(t) dt

or f(x) = 1/2 x^{2} _{0}ò^{x }^{ }g(t) dt + 1/2 _{0}ò^{x }t^{2 }g(t) dt - x _{0}ò^{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)

limit n tends to ∞

then

(x^n)/(n!) equals

**Asked By: AMIT DAS**

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**Joshi sir comment**

let y = lim n->∞ (x^{n}/n!)

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

or y = 0

limit n tends to ∞

x{ [tan‾¹ (x+1/x+4)] - (π/4)}

**Asked By: AMIT DAS**

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**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

**Asked By: AMIT DAS**

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**Joshi sir comment**

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

Use the formula a^{3}+b^{3} = (a+b)(a^{2}+b^{2}-ab) in the format

(a+b) = (a^{3}+b^{3})/(a^{2}+b^{2}-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

**Asked By: NIKHIL VARSHNEY**

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**Joshi sir comment**

f(x) = x|x| => f(x) = -x^{2} for negative real values of x and f(x) = x^{2} 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

**Asked By: NIKHIL VARSHNEY**

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**Joshi sir comment**

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 = (1-2sin^{2}h+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