# 135 - Mathematics Questions Answers

Let f(x) be a real valued function not identically equal to zero such that f(x+y^{n})=f(x)+(f(y))^{n}; y is real, n is odd and n >1 and f'(0) ³ 0. Find out the value of f '(10) and f(5).

**Asked By: KAMAL**

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**Joshi sir comment**

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

Evaluate _{0}ò^{x} [x] dx .

**Asked By: KAMAL**

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**Joshi sir comment**

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

so _{0}ò^{x} [x] dx = _{0}ò^{1} 0 dx + _{1}ò^{2} 1 dx + _{2}ò^{3} 2 dx + ..................... + _{[x]-1}∫^{[x] }[x]-1 dx + _{[x]}∫^{x }[x] dx

now solve it

A function f : R® R satisfies f(x+y) = f(x) + f(y) for all x,y Î R and is continuous throughout the domain. If I_{1} + I_{2} + I_{3} + I_{4 }+ I_{5} = 450, where I_{n} = n._{0}ò^{n} f(x) dx. Find f(x).

**Asked By: KAMAL**

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**Joshi sir comment**

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

_{0}ò^{n} f(x) dx

consider x = ny so this integration will become _{0}ò^{1} f(ny) ndy = n^{2 }_{0}ò^{1} f(y) dy = n^{2} _{0}ò^{1} f(x) dx

now by using these conditions I_{1} = 1^{3} _{0}ò^{1} f(x) dx similarly others

put these values and get the answer

Show that _{0}ò^{p} q^{3} ln sin q dq = 3p/2 _{0}ò^{p} q^{2} ln [Ö2 sin q] dq.

**Asked By: KAMAL**

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**Joshi sir comment**

let I = _{0}ò^{p} q ln sin q dq

on aplying property of definite integral

I = _{0}ò^{p} (π-q) ln sin q dq

so 2 I = _{0}ò^{p} π ln sin q dq

or I = π/2 _{0}ò^{p } ln sin q dq

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

similarly solve _{0}ò^{p} q^{2} ln sin q dq and then _{0}ò^{p} q^{3} 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 " 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)

show that line perpendicular to x-axis and a line perpendecular to y-axis having slope multipication -1

**Asked By: HARISH**

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**Joshi sir comment**

Angle between two lines is given by the formula

tanθ = (m_{1}-m_{2})/(1+m_{1}m_{2})

solve this by taking θ = 90

The coefficient of λ^n μ^n in the expansion of [(1+λ)(1+μ)(λ+μ)]^n is ?

**Asked By: ASHISH RANA**

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**Joshi sir comment**

(1+λ)^{n} = C_{0} + C_{1}x + ..................................C_{n}x^{n}

similarly for (1+μ)^{n} and

(λ+μ)^{n} = C_{0}λ^{n} + C_{1}λ^{n-1}μ + ........................ + C_{n}μ^{n}

On multiplying these 3, we will get the coefficient of λ^{n}μ^{n }= ∑C_{n}^{3 }(Here you should remember that C_{0} = C_{n, }C_{1} = C_{n-1} and etc)

For getting the value of ∑C_{n}^{3 }:

multiply the expansions of (1+x)^{n}, (1-x)^{n} and [1-(1/x^{2})]^{n} and calculate the coefficient of x^{0}, we get ∑C_{n}^{3}

and if we multiply (1+x)^{n}, (1-x)^{n} and [1-(1/x^{2})]^{n}, then we get (-1)^{n}(1-x^{2})^{2}^{n}/x^{2n} so coefficient of x^{0 }in this expression will be (-1)^{n} ^{2n}C_{n }(-1)^{n } = ^{2n}C_{n}