166 - Questions Answers

In |z-3|=|z-4i|, z represents all points lying on the perpendicular bisector of the line joining 3 and 4i in x-y plane. In this perpendicular bisector, if we draw a perpendicular then it will be the minimum magnitude of z. Construct this as a diagram we will get

sinθ = |z| / (7/8)

or |z| = (7/8)(4/5) = 7/10   

let f(x) = x^3-3x+k

so f^'(x) = 3 x^2 - 3

on comparing with 0, x = +1, -1

on double differentiating we get that function f(x) is minimum at +1 and maximum at -1

so for getting the only +ve real root plot a graph such that it could intersect positive x axis at any point and nowhere else. From the graph it is clear that k will be negative.

at x = +1,  f(x) = k-2

at x = -1, f(x) = k+2

so on the basis of given conditions k-2<0 or k<2  and  k+2<0 or k<-2 

finally k<-2 so [|k|] has minimum value 2

x^7 - 3x^4+2x^3-k=0

according to sign change rule there are 3 pair of sign + -, - +, + - ,       see the sign before coefficient of different powers

so 3 positive real roots are confirmed

now take  x = -x

we get -x^7 - 3x^4 - 2x^3 - k = 0, there is no sign change so no negative root

total real roots = 3

so imaginary roots = 4

(1 + 0.0001) ¹⁰⁰⁰⁰ = 1 + 10000*0.0001 + other terms = greater then 2

besides it lim _x->0 (1+x)^1/x = e = less then 3

it means the given expression is more then 2 but less then 3 

so 2 will be the answer

1)   (10/12)(9/11)(8/10)(7/9)

2)   4(10/12)(9/11)(8/10)(2/9)

3)   (4*3/2)(10/12)(9/11)(2/10)(1/9)

4)   1 - (10/12)(9/11)(8/10)(7/9)

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),