CAT 2007: Quantitative Questions a Day 134 till the end - The Discussions

[KOOL030]

can someone please explain me what is descartes theorem?

[Marshall]

Hi Aarav,

Even I applied the same logic as explained by SuperStar, Although i didnot posted the whole solution earlier.....Just wanted to know why we cant use the Descrat's theorem. Is there something which i am missing.
Thanks in advance

[Aarav]

Originally Posted by Marshall

Hi Aarav,

Even I applied the same logic as explained by SuperStar, Although i didnot posted the whole solution earlier.....Just wanted to know why we cant use the Descrat's theorem. Is there something which i am missing.
Thanks in advance

I'm not sure what your understanding of descrates theorem on sign changes is, so let's be on common ground here Descartes’ Rule of Signs

Just try for b = 1 and a = 2 for necessary and sufficient condition in the question.

[Superstar]

Originally Posted by kondapalli

There are a total of 4 sign changes in f(x). Hence there could be 0, 2 or 4 +ve roots. But we can't be sure about 0 or 2 or 4. It can be any of those.

f(-x) has 0 sign changes. Hence no -ve root. So we have either +ve roots only or +ve and complex.

Someone can correct me if I am wrong .....

But in case of 0 or 2 +ve roots, the other roots will be imaginary (complex) which goes against the assumption of real roots.....

What does Aarav Bhaiyya feel??

[vineet.nitd]

Originally Posted by Aarav

------------------------------------------------------
Quantitative Question # 134
------------------------------------------------------

Part (A)

The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is

(1) b >= a^4 (2) a >= 4b^4 (3) a >= b^4 (4) b >= 4a^4 (5) none of these

To solve take x=0,y=2 => options (2) ,(3) and (4) are eliminated.

Now take x=1,y=2 =>option (1) gone and hence the ans

[Aarav]

Originally Posted by Marshall

Hi Aarav,

Even I applied the same logic as explained by SuperStar, Although i didnot posted the whole solution earlier.....Just wanted to know why we cant use the Descrat's theorem. Is there something which i am missing.
Thanks in advance

Perhaps, this can be answered using Descrates rule of signs had the question been just sufficient condition and not necessary.

For, x^4 + (x-a)^4 = b, we have 2x^4 - 4ax^3 + 6a^2.x^2 - 4a^3.x + (a^4-b) = 0 => b >= a^4. The change of signs is 3 in this case and hence we have atleast one positive real solution.

[Aarav]

Originally Posted by Superstar

PART A

Option 5 none of these is the answer.

a^4 > b .

I did it by forming the equation which is of 4 degree which has a constant term as a^4 - b

Apply the logic ( There is some theorem) which helps to find out the maximum nember of positive and negative real roots. And One finds that the above condition has to hold true.

Aarav Bhaiyya , I hope the logic is fine

Descrates is sufficient in this case but not necessary. Sorry, for last minute hiccups ... just got terribly confused in the midst of work here.

[Superstar]

Aarav I am writing the full solution to this problem as I took it...

Please let me know where I am wrong (if I am )

So, Finally

2x^4 - 4ax^3 + 6a^2.x^2 - 4a^3.x + (a^4-b) = 0

Now if assume a^4 > b then effectively there will be 4 sign changes. ( I included the sign for 2x^4 also ) ..........Am I correct

So, there are maximum 4 positive real roots to this equation.

For f (-x), I have no sign changes So there will be no negative roots to this equation.

So, all roots will have to be positive and they have to be 4 in number only as any number of roots less than 4 would introduce imaginary roots into the picture which is not possible.....So a^4 > b is a condition for roots to be real and satisy both equations

Aarav, Please finally enlighten.... So, that this freezes for everyone now.

Thanks for support..

Acchcha when to expect the next Practice test....

[Aarav]

Originally Posted by Superstar

Aarav I am writing the full solution to this problem as I took it...

Please let me know where I am wrong (if I am )

So, Finally

2x^4 - 4ax^3 + 6a^2.x^2 - 4a^3.x + (a^4-b) = 0

Now if assume a^4 > b then effectively there will be 4 sign changes. ( I included the sign for 2x^4 also ) ..........Am I correct

So, there are maximum 4 positive real roots to this equation.

For f (-x), I have no sign changes So there will be no negative roots to this equation.

So, all roots will have to be positive and they have to be 4 in number only as any number of roots less than 4 would introduce imaginary roots into the picture which is not possible.....So a^4 > b is a condition for roots to be real and satisy both equations

Aarav, Please finally enlighten.... So, that this freezes for everyone now.

Thanks for support..

Acchcha when to expect the next Practice test....

See, if you read the link pasted on Descrates rule of sign, you will see that the number of positive roots are either 4 or 2 or 0. Hence, you are not sure if this equation will have any real roots.
One example that explains this is (x-1)^4 = -1 as this has 0 real roots.

Next practice test is on 03/10.

[aravindva]

------------------------------------------------------
Quantitative Question # 135 (The answer to both Part A and Part B are in the same set of options)
------------------------------------------------------
Part (A)

Let X = {1, 2, 3, 4}. How many functions f: X->X are there satisfying (f.f)(x) = x for all x in X?



Part (B)

In a hockey match India beat Pakistan 5-4. India scored first and kept the lead until the end. In how many different orders could the goals been scored?


(1) 13 (2) 15 (3) 14 (4) 16 (5) none of these
-------------------------------------------------------------------