161 - Mathematics Questions Answers

Please send the complete question

limn->∞ ∑_r=1ⁿ e^r/n/n

This question is related to definite integration, consider 1 and divide it into n parts, upto nth part total value = r/n is equivalent to x and 1/n is dx

so question will be

₀∫¹ e^xdx 

now solve it

first draw a cirle and triange inside it , consider angle B and angle C asα so angle A = 180-2α, let centre of the circle is O and D is the point in the line BC where altitude meets in the line BC, given that altitude = h and radius = r

so AB = AC = h cosecα and BC = 2 h cotα so 

p = 2 h (cosecα + cotα)

and φ = 1/2  2 h cotα h = h² cotαa

and in triangle OBD angle OBD = 2α - 90    so cos(2α - 90) = h cotα /r   implies that h = 2 r sin²α 

now i think limit will be based on h not n

 lim _{h−» 0}   φ/p³  = 1/128r

for getting solution put p and φ first then put h in terms of r, you will get an equation based on α and r,

convert lim_h->0  to  limα->0     because when h will be 0, α will also be 0 

given equation can be written as (y-2)²= 4(x+1)  so coordinate of focus (a, 0) will be x+1 = 1 and y - 2 = 0 so x = 0 and y = 2

let the centre of largest circle inside parabola is (k, 2) so equation of that circle will be (x-k)² + (y-2)² = k²

on solving parabola and circle we get x² + (4-2k) x + 4 = 0 

for largest circle roots of this quadratic equation should be equal so b²= 4ac

solve and get k = 4 so radius will be 4

f(x) is monotonically increasing so f '(x)>0 and f '' (x) >0 implies that increment of function increases rapidly with increase in x 

These informations provide the following informations about the nature of inverse of f(x) 

1) f ^-1 (x) will also be an increasing function but its rate of increase decreases with increasing x

2) for x₁ < x₂ < x₃ ,  å{f ^-1(x_i)/3} < f ^-1({x₁+x₂+x₃}/3), 

The same result for f(x) will be 

  å{f (x_i)/3} > f ({x₁+x₂+x₃}/3), 

let I = ₀ò¹1/{ (5+2x-2x²)(1+e^(2-4x)) } dx

on using the property ₀∫^a f(x) dx = ₀∫^af(a-x)dx

we get 2I = ₀ò¹1/(5+2x-2x²) dx

now solve this 

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 

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

so ₀ò^x [x] dx = ₀ò¹ 0 dx + ₁ò² 1 dx + ₂ò³ 2 dx + ..................... + _[x]-1∫^[x] [x]-1 dx + _[x]∫^x [x] dx

now solve it

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 = n² ₀ò¹ 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

 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