let y = x^(1/x)
ln y = ln x/x
x ln y = ln x
differentiate both sides
x/y dy/dx + ln y = 1/x
dy/dx = (1/x - ln y)/(x/y)
= (1/x-1/x ln x)/(x/y) = y/x^2(1-ln x)
dy/dx = 0 when 1- ln x = 0
this is maximum or minimum
or x = e
for x < e dy/dx >0 or y increases
x > e dy/dx < 0 or y decreases
so x^(1/x) is maximum at x = e
some short and selected math problems of different levels in random order I try to keep the ans simple
Saturday, June 23, 2007
Saturday, June 9, 2007
2007/ 001) Slightly hard cubic eqution
Find constants a, b, c, and d such that:
4x³ - 3x + √3/2=a * (x-b) * (x-c) * (x-d)
Solution
let p(x) be given polynomial
we reallize that sqrt(3)/2 = cos pi/6
so 4x^3-3x + cos pi/ 6= 0
or - cos pi/6 =-3x + 4x^3 = cos(3t) if x= cos t
so cos 3t = - cos pi/6 = cos 5pi/6
so 3t = 5pi/6 or 2pi-5pi/6 = 7pi/6 or 4pi-7pi/6 = 17pi/6
so t = 5pi/18 or 7pi/18 or 17pi/18
so b = cos 5pi/18,
c = cos 7pi/18
d= cos 17pi/18
as these are zeroes of polynomials
further comparing coefficient of x^3 we get a = 4
4x³ - 3x + √3/2=a * (x-b) * (x-c) * (x-d)
Solution
let p(x) be given polynomial
we reallize that sqrt(3)/2 = cos pi/6
so 4x^3-3x + cos pi/ 6= 0
or - cos pi/6 =-3x + 4x^3 = cos(3t) if x= cos t
so cos 3t = - cos pi/6 = cos 5pi/6
so 3t = 5pi/6 or 2pi-5pi/6 = 7pi/6 or 4pi-7pi/6 = 17pi/6
so t = 5pi/18 or 7pi/18 or 17pi/18
so b = cos 5pi/18,
c = cos 7pi/18
d= cos 17pi/18
as these are zeroes of polynomials
further comparing coefficient of x^3 we get a = 4
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