Let X=[x] +{x} where [x] and {x} are respectively the integer and fractional parts of a real number x. Find the number of real number which exist such that they satisfy the following conditions:
a) 0<=x<=100
b) {x^2}={x}
A) 9000 B) 9900 C) 9990 D) 9901 E) 9991
here is the official solution to this problem
ReplyDeleteIf x and x^2 have the same fractional part then x^2 - x is in N. Consider the quadratic function f(x) = x^2 - x. Since it is monotonous and strictly increasing on [0, 100], all the integer values between f(0) and f(100) will be assumed by f exactly once. For each positive integer m, the equation x^2 - x = m has precisely one positive solution. Now f(0) = 0 and f(100) = 9900. So the 9901 possible values of m are 0, 1, 2, ..., 9900. The answer for 0<=x<=100 is 9901.
Didn't understand the last 2 lines of the solution. Please elaborate.
ReplyDeletef(x)=x^2-x
ReplyDeletenow {x^}={x}
so x^2-x is always a whole number in 0<=x<100
now
max f(x)=10000-100=9900 and min is 0
and all these values will be achieved for one x
so at max 9901 values