Fractional Part
{ x 1 } = { x }
Find the number of solution(s) of x in the range [ 1 , 6 ] such that the equation above is satisfied.
Note that { x } denotes the fractional part of x .
The answer is 6.
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1 solution
Moderator note:
When solving a quadratic, we have 2 roots. You should explain why one of them has to be rejected.
When a = 0 , we have x = 1 . However this is not one of the answers.
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Yeah answer should be 7.
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Why should it be 7?
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@Jesse Nieminen – You haven't included x=1 I think.
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@Harsh Shrivastava – If a = 0 , x = 1 since substituting a = 0 to x = 2 − a + a 2 + 4 gives x = 1
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@Jesse Nieminen – @Jesse Nieminen he means that there will be number solutions =7 and not the value of x as 7. And yeah, if you can explain how did one get -5≤a≤0? thanks.
I was too lazy to write an explanation because it is obvious. x = 2 − a − a 2 + 4 ∣ a ∣ < a 2 + 4 Therefore x is always negative, which is not possible.
I have a question when we take 1 / x = f r a c ( x ) if x is an integer, then 1/x returns a fraction..(x is between 1 and 6), but then frac x returns 0.. how can it be equal? Shouldn't the answer be just 1? The only solution is one.
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Why are you assuming that x is an integer?
I agree that if x is an integer, then there are no solutions (other than x = 1 ). However, in this problem, x is a real number.
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@Calvin Lin – Thank you sir... i finally got it! Sorry for the misconception
Hi, can someone please explain how did one get -5≤a≤0? Thanks.
The answer is 5, because for a=6 the value of x is not in the range. (In fact x can't be integer, because {1/x} is never zero.)
IMHO there are five solutions: 1.618033989, 2.414213562, 3.302775638, 4.236067977, 5.192582404. Eg.
1/1.618033989=0.618033989
{1/1.618033989}={1.618033989}
First of all, we know that {1/x} = 1/x - floor(1/x). We are working in the domain [1,6]. For any x > 1, we know that 0 < 1/x < 1 and therefore floor(1/x) = 0. So by this being the case, we can say {1/x} = 1/x for x ϵ (1,6]. But what about 1? Well substituting 1 into x we get LHS = {1} and RHS = {1} so LHS=RHS which means we already have one solution: x=1. So now the problem has been reduced to the domain (1,6] which is useful because in this scenario the problem simplifies to: 1/x = {x}. And by definition: 1/x = x - floor(x) x - 1/x = floor(x). We recall that we're looking for solutions of x within [1,6] and therefore we know that floor(x) must be either: 1 if 1 ≤ x < 2; 2 if 2 ≤ x < 3; 3 if 3 ≤ x < 4; 4 if 4 ≤ x < 5; or 5 if 5 ≤ x < 6 Meaning that to solve for x, we must solve for each of these cases. We can substitute floor(x)=a to be able to generalise for each of those cases. We end up getting: x - 1/x = a, x² - 1 = ax, x² - ax - 1 = 0. Using the quadratic formula: xₐ = a + √(a² + 4) / 2 (there's no ± since x > 0 in the problem we're solving). Therefore, we can solve for 5 unique solutions by substituting a = 1, 2, 3, 4, 5. To this set of solutions we can add the first solution we've found: x = 1. Finally giving us a total of 6 solutions. ☐
(This is just how I understood it so here you go, I hope this is helpful for someone).
There are 7 solutions.x=1 is also a solution as it is given [1,6]
{ x 1 } = { x } , 1 ≤ x ≤ 6 ⇒ x 1 = x + a , a ∈ Z ⇒ x 2 + a x − 1 = 0 ⇒ x = 2 − a + a 2 + 4 1 ≤ x ≤ 6 ⇔ − 5 ≤ a ≤ 0 There are 6 possible values for a in total, which implies that there are 6 possible values for x.
Therefore the answer is 6 .