[amar_kashyap]

To begin with this is not my idea, and has been borrowed from some other math site (i am not putting the name, will be against the rules of PG, but i am sure our quant studs must be aware). So here are the rules.Ths basics: i start off with a problem, someone solves it(has to provide the full solution, and the solution can be discussed) and posts the new problem.
Few things worth noting:

1. Please keep the problems numbered.
2. The level of problems can be anyhting provided that it helps us in understanding the basics required to crack CAT(especially after trend in cat quant which focusses more on concepts rather than speed).
3. Please keep the thread moving by posting a new problem everytime u have solved a problem, if u have none with u..u can request somebody else to post a new problem.

So here we go:

PROB No.1: If 'a' is a real number and the equation (a-2)(x-[x])^2 +2(x-[x]) +a^2 =0 (where [x] denotes the greatest integer <=x) has no integral solution and has exactly one solution in (2,3), then 'a' lies in the interval

a. (-1,2) b. (0,1) c. (-1,0) d. (2,3)

Happy solving..

[amar_kashyap]

Where are quant gods?...c,mmon folks at least give a try here...If nobody comes out with the solution i will have to post the solution which will be so boring .....

[Aarav]

Originally Posted by amar_kashyap

To begin with this is not my idea, and has been borrowed from some other math site (i am not putting the name, will be against the rules of PG, but i am sure our quant studs must be aware). So here are the rules.Ths basics: i start off with a problem, someone solves it(has to provide the full solution, and the solution can be discussed) and posts the new problem.
Few things worth noting:

1. Please keep the problems numbered.
2. The level of problems can be anyhting provided that it helps us in understanding the basics required to crack CAT(especially after trend in cat quant which focusses more on concepts rather than speed).
3. Please keep the thread moving by posting a new problem everytime u have solved a problem, if u have none with u..u can request somebody else to post a new problem.

So here we go:

PROB No.1: If 'a' is a real number and the equation (a-2)(x-[x])^2 +2(x-[x]) +a^2 =0 (where [x] denotes the greatest integer <=x) has no integral solution and has exactly one solution in (2,3), then 'a' lies in the interval

a. (-1,2) b. (0,1) c. (-1,0) d. (2,3)

Happy solving..

artofproblemsolving.com
Intermediate section on that forum is something that is useful for CAT preparation.
That site runs marathon on many topics. I was interested in starting this, but decided against finally and instead went for "quant question a day" which gives you exactly what you need.
We are filtering out the best, and among many sources that site is one of our sources.

I see some competition to "quant question a day" from this thread.
And that's a good thing happening :-)

Happy solving.

f(2) = 4 + t
f(3) = 9 + 2t + t^2
where t = (x-[x])

=> f(2) < 0 as f(3) > 0 => t < -4, not possible so this eq. is in x and not in "a" as I considered initially.

(a-2)(x-2)^2 + 2(x-2) + a^2 = 0
=> 4 -4(a-2)a^2 >= 0
=> 1 - a^3 - 2a^2 >= 0
=> (1+a) (a^2 + a -1) <= 0
=> a lies in (-1, (5^1/2 -1)/2)

Hence, (a)

changing later...
F(0)F(1) < 0, obviously as the question says exactly one.
I missed this basic point. Hence, (c) is the solution.

[chautauqua]

Originally Posted by amar_kashyap

PROB No.1: If 'a' is a real number and the equation (a-2)(x-[x])^2 +2(x-[x]) +a^2 =0 (where [x] denotes the greatest integer <=x) has no integral solution and has exactly one solution in (2,3), then 'a' lies in the interval

a. (-1,2) b. (0,1) c. (-1,0) d. (2,3)

Happy solving..

the problem has only soln. in x belonging to (2,3) => (x-[x]) belongs to (0,1) ....... (the key thing is the open interval and not the numbers 2 and 3)


let z = (x-[x])

=> (a-2)*z^2 + 2*z + a^2 = 0

=> z = -1 + sqrt(1 - (a-2)*a^2) .... (the other solution is

ignored as z cant be negative)

we have 0 < -1 + sqrt(1 - (a-2)*a^2) < 1

=> 1 < sqrt(1 - (a-2)*a^2) < 2

=> 1 < (1 - (a-2)*a^2) < 4

=> 0 < - (a-2)*a^2 < 3

=> i) (a-2)*a^2 < 0

ii) (a-2)*a^2 > -3


i) => a < 2

ii) => a^3 - 2*a^2 + 3 > 0

=> (a + 1)*(a^2 - 3*a + 3) > 0

=> (a + 1)*( (a - 1.5)^2 + 0.75 ) > 0

=> a > -1

i), ii) => -1 < a < 2 .... hence option a) is the answer

[amar_kashyap]

Originally Posted by Aarav

artofproblemsolving.com
Intermediate section on that forum is something that is useful for CAT preparation.
That site runs marathon on many topics. I was interested in starting this, but decided against finally and instead went for "quant question a day" which gives you exactly what you need.
We are filtering out the best, and among many sources that site is one of our sources.

I see some competition to "quant question a day" from this thread.
And that's a good thing happening :-)

Happy solving.

f(2) = 4 + t
f(3) = 9 + 2t + t^2
where t = (x-[x])

=> f(2) < 0 as f(3) > 0 => t < -4, not possible so this eq. is in x and not in "a" as I considered initially.

(a-2)(x-2)^2 + 2(x-2) + a^2 = 0
=> 4 -4(a-2)a^2 >= 0
=> 1 - a^3 - 2a^2 >= 0
=> (1+a) (a^2 + a -1) <= 0
=> a lies in (-1, (5^1/2 -1)/2)

Hence, (a)

The interval in option a) includes the interval in option c) also...How do we check for that..
I think there is more to it. We can proceed by taking into account the fact that if 2<x<3 (according to the question, there is one solution between 2 and 3), then the quadratic equation in x-[x] will have a solution in the interval 0<x-[x] <1.That means F(x-[x]) will have one solution in the interval (0, 1). That implies F(0).F(1)<0 , tht implies a^2.(a^2+2+a-2)<0, tht implies a(a+1)<0. Hence -1<a<0. Hope i make sense .

anyways where is next problem, Aarav bhai and Chautauqa bhai. Any one of you please post the new problem

[Aarav]

Since, the norm is to give a question after solving it, here is one from me.
I solved this, but looking for an alternate approach.

Problem #2
Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

Originally Posted by amar_kashyap

The interval in option a) includes the interval in option c) also...How do we check for that..
I think there is more to it. We can proceed by taking into account the fact that if 2<x<3 (according to the question, there is one solution between 2 and 3), then the quadratic equation in x-[x] will have a solution in the interval 0<x-[x] <1.That means F(x-[x]) will have one solution in the interval (0, 1). That implies F(0).F(1)<0 , tht implies a^2.(a^2+2+a-2)<0, tht implies a(a+1)<0. Hence -1<a<0. Hope i make sense .

Amazing solution. I'm impressed.
And I also realise my error.

[amar_kashyap]

Please give no to the problem also that you have posted...It keeps things easier ...I hope you don't mind doing that . Regarding the options, i think i have tried to explain the previous post

[amar_kashyap]

For problem no.2, ie m/n + (n+1)/m=4..I am just giving a try...

We can write the given equation as m/n+n/m+1/m=4

Using A.M>=G.M , we can write 4>= 2+1/m

ie 1/m <=2.

I dont know wht to do next

[Aarav]

Originally Posted by amar_kashyap

For problem no.2, ie m/n + (n+1)/m=4..I am just giving a try...

We can write the given equation as m/n+n/m+1/m=4

Using A.M>=G.M , we can write 4>= 2+1/m

ie 1/m <=2.

I dont know wht to do next

AM-GM doesn't apply for 2 numbers when m/n < 0.

Requires a bit of number theory unless you come up with some other approach :-)

[siddhesh_j]

[quote=Aarav]Since, the norm is to give a question after solving it, here is one from me.
I solved this, but looking for an alternate approach.

Problem #2
Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

lhs= m/n+n/m+1/m=4

= (root(m/n)-root(n/m))^2+1/m=2

clearly , m, n have to be greater than zero , which limits the set of integers to 0<m,n, for real m,n.

now for integers greater than one 1/m is a fraction.

case 1:m=1 then square term =1 , and no integral n exists to satisfy such an eqn.

case 2:m>1 then either of root(m/n) or root(n/m) will be irrational and square of irrational = irrational. to that add 1/m , answer again irrational , while2 is rational.

so lhs= irrational and rhs = rational , hence the proof.

i realise this isnt a gr8 solution ... so shud I still post a problem or not??

anyways here is problem #3

This one a bit easier than above 2:

At what time are the hands of clock together between 2 and 3?
1. 10 10/11 mins past 2
2. 10 10/11 mins to 3
3. 9 1/11 mins past 2
4. 11 1/11 mins past 2