Question from MockCat, SimCat, Aim Cat,...... for CAT'06

[warrior]

Every now and then we come across some good questions in mock that we write. these questions can be
(a) the one which we are unable to solve
(b) the question seems easy on face but answer come out to be a different one ( a tricky question)
(c) a question which has been solved in some newer or shorter or different way in aimcat solutions
(d) we came across a different/newer sort of question

I have started this thread so that we can discuss questions from mocks by various institutes over here so that at the end we can have a good collection of question for revision.Also as different people join different test series we can have them all at one place.

A small contribution by all of us ( esp from those who join CL,CF which are less popular as compared to TIME,IMS) can have a big impact for all of us.

Note: Also when posting post your question in bold so that when browing through the page its easy to figure out and keep them numbered

[warrior]

Aimcat 717 was easy and it dosen't have may questions to include here but let start with this one

Problem #1: If the equation has x^4-4*x^3+a*x^2+b*x+1 = 0 has four positive roots, the ordered pair (a,b) is

(a) (-4,6) (b) (6,-4) (c) (-2,-3) (d) (-3,2)

[sidbadshah]

Originally Posted by warrior

Aimcat 717 was easy and it dosen't have may questions to include here but let start with this one

Problem #1: If the equation has x^4-4*x^3+a*x^2+b*x+1 = 0 has four positive roots, the ordered pair (a,b) is

(a) (-4,6) (b) (6,-4) (c) (-2,-3) (d) (-3,2)

if the roots are positive, the coefficient of x^2 must be greater than 0. which means a must be greater than 0, that leaves us withonly one choice and thats b.

[Quagmire]

Originally Posted by warrior

Aimcat 717 was easy and it dosen't have may questions to include here but let start with this one

Problem #1: If the equation has x^4-4*x^3+a*x^2+b*x+1 = 0 has four positive roots, the ordered pair (a,b) is

(a) (-4,6) (b) (6,-4) (c) (-2,-3) (d) (-3,2)

I would have taken my chances. What I did was on intuition take x=1, which turns the eqn to a+b=2. Here only option b satisfies a+b=2. Though the approach taken by sidbadshah seems very accurate and concise.

Also sidbadshah would like to ask u about this funda of the coefficient of x^2 being greater than zero when the eqns solutions r all positive. Is it true only for an eqn of degree 4 or for eqns with any degree of x.

[sidbadshah]

Originally Posted by Quagmire

I would have taken my chances. What I did was on intuition take x=1, which turns the eqn to a+b=2. Here only option b satisfies a+b=2. Though the approach taken by sidbadshah seems very accurate and concise.

Also sidbadshah would like to ask u about this funda of the coefficient of x^2 being greater than zero when the eqns solutions r all positive. Is it true only for an eqn of degree 4 or for eqns with any degree of x.

HI Quagmire

for an equation of degree n
sum of the roots = - coeff(x^n-1)/coeff(x^n)

sum of the roots taken two at a time = coeff(x^n-2) / coeff(x^n)

sum of roots taken three at a time = -coeff(x^n-3) / coeff(x^n)

In this case coeff of x^n = 1 which is positivem and sum of roots taken two at a time is also positive as all roots are positive, which means coeff of x^n-2 must be positive

Hope i am clear

Sid

[warrior]

Originally Posted by warrior

Aimcat 717 was easy and it dosen't have may questions to include here but let start with this one

Problem #1: If the equation has x^4-4*x^3+a*x^2+b*x+1 = 0 has four positive roots, the ordered pair (a,b) is

(a) (-4,6) (b) (6,-4) (c) (-2,-3) (d) (-3,2)

time has given the solu'n using descreates rule.
to have four roots the sign should change four times and that could happen only if we have (b) as choice.

[abhimax]

yup man it can be solved by descretes rule......

consider f(x)=x^2-4x+4
then,
f(1) will 2 sign change & f(-1) will give 0 sign change this means there is 2 +ve roots and 0 -ve roots

In this question 'a' shud be +ve and 'b' shud be -ve bcoz it has four roots and all +ve.
putting x=1 will give four sign change only when 'a' is ve and 'b' is -ve.
but still we cannot get the answer from the option coz there r two option with
'a' +ve and 'b' -ve.

here we get, product of root is 1 and sum of root is 4, so we get all four roots equal to 1.
so by putting the value of roots (by taking two at a time and three at a time ) we get the answer (6,-4)

[warrior]

Can anyone who has appeared for previous two Aimcats post the good questions from those papers here.

[Quagmire]

Originally Posted by warrior

time has given the solu'n using descreates rule.
to have four roots the sign should change four times and that could happen only if we have (b) as choice.

@warrior
Can u please state Descartes rule in explicit terms for me

@sidbadshah
The formulae that u have given above is for the sum of roots. Will the product of roots always be
c/a for a degree 2 eqn
- d/a for a degree 3 eqn
e/a for a degree 4 eqn and so on.....?

[warrior]

Originally Posted by Quagmire

@warrior
Can u please state Descartes rule in explicit terms for me

It says:For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)