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

[Aarav]

Originally Posted by prade

x^4-ax^3 +3a^2x^2 -a^3x + (a^4-b)/2 = 0 = (x^2+px+q)(x^2+mx+n)
p+m = -a
n+q+pm = 3a^2
qm+pn = -a^3
qn = (a^4-b)/2

for real roots p^2-4pq >= 0
and m^2-4mn > = 0

aarav help needed...

Work with options Prade. Getting at the exact condition would be tough here.

[kekambus]

A:

x^2+y^2+2xy = a^2
Now x^2+y^2>=2xy => 2(x^2+y^2)>=a^2

Similarily working on 2(x^2+y^2)>=a^2 we get a^4/8<=(x^4 + y^4)

=> a^4/8<=b
=> option 5

B:

x=10(11-n)/(n-1)
y = 20(6-n)/(n-1)
z = 10(13-3n)/(n-1)

=> 2<=n<=4 else x,y,z will become negative (From here we can just put in the values 2,3,4 and check)
also n-1 cld be a factor of 10
=> n-1=5 or n-1=2 or n-1=11
frm here we get n= 3 as the possible solution
now just see tht n = 2 is also satisfied but n = 4 is not

=> 2 distinct values possible => option 1

[Marshall]

Ans #134 a) option 5
Condition is b <= a^4
b) option 1 for n=2,3 satisfies the given relation.

[mejogi]

For Part A
option 2 & 3 are certainly wrong for b=0 and a=1
option 1 is wrong for a=1 and b=1
for option 4... not able to proof till now but got the same as Rohit did a^4 +b >=0.
So option 5 seems correct.....doin a general solution...

[mejogi]

Part A
a^4>= -b
for both " b" a +ve or -ve value makes option 4 invalid....doin Part B

[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

PART B

Only n=2,3 satisfy...Else z gets negative

[mejogi]

For Part B
(n-1)=100/(10+x)=100/(20+y)=100/(30+z).......1

0r m =(n-1)=100/(10+x)=100/(20+y)=100/(30+z) .....where m=n-1,& a +ve integer including 0.

for x=40 and 90, m has integral value (where 10+y=x,y=10+z,x=20+z =>x >y>z(though, not required))

Hence option 1 is true

[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

Descrate's theorem on change of signs doesn't get applied here. Please read kekambus's solution.

[Superstar]

Originally Posted by Aarav

Descrate's theorem on change of signs doesn't get applied here. Please read kekambus's solution.

Why??

[kondapalli]

Originally Posted by Superstar

Why??

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 .....