# 135 - Mathematics Questions Answers

Maths >> Calculus >> Functions IIT JEE

Let f(x) be a real valued function not identically equal to zero such that f(x+yn)=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).

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

Maths >> Calculus >> Integrations IIT JEE

Evaluate 0òx [x] dx .

Joshi sir comment

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

so 0òx [x] dx = 0ò1 0 dx + 1ò2 1 dx + 2ò3 2 dx + ..................... + [x]-1[x] [x]-1 dx + [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 I1 + I2 + I3 + I+ I5 = 450, where In = n.0òn f(x) dx. Find f(x).

Joshi sir comment

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 = n0ò1 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

Show that 0òp q3 ln sin q dq = 3p/2 0òp q2 ln [Ö2 sin q] dq.

Joshi sir comment

let I = 0òp q ln sin q dq

on aplying property of definite integral

I =  0òp (π-q) ln sin q dq

so 2 I =  0òp π ln sin q dq

or I = π/2  0ò ln sin q dq

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

similarly solve  0òp q2 ln sin q dq  and   then 0òp q3 ln sin q d

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.

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)

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

Joshi sir comment

Angle between two lines is given by the formula

tanθ = (m1-m2)/(1+m1m2)

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 ?

Joshi sir comment

(1+λ)n = C0 + C1x + ..................................Cnxn

similarly for (1+μ)n and

(λ+μ)n = C0λn + C1λn-1μ + ........................ + Cnμn

On multiplying these 3, we will get the coefficient of  λnμ= ∑Cn3      (Here you should remember that C0 = Cn, C1 = Cn-1 and etc)

For getting the value of  ∑Cn:

multiply the expansions of (1+x)n, (1-x)n and [1-(1/x2)]n and calculate the coefficient of x0, we get ∑Cn3

and if we multiply (1+x)n, (1-x)n and [1-(1/x2)]n, then we get (-1)n(1-x2)2n/x2n so coefficient of xin this expression will be (-1)n 2nC(-1)2nCn