CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

Aarav

Well done all

most of you have taken different approaches to solving this problem and have arrived at the right answer.

If r is the root of f(x) = 0, then f(-1/x) = f(x) tells us that -1/r is also the root of
f(x) = 0 -> (1)

When we are checking that for any r, f(r) = 0 we want to check if f((1+r)/(1-r)) = 0 or not
then (1+x)/(1-x) = y and x = (y-1)/(y+1), on substitution in f(x) we get
y^4 + ay^3 - 6y^2 -ay + 1 -> (2)

The 3rd choice follows from (1) and (2).

To drive this point once more.
Suppose we were given that p, q, r, s as roots of f(x) = 0 and asked to find
(p+q+r)(q+r+s)(r+s+p)(s+p+q) we just need to substitute p+q+r+s - s = -a - s or -a - x = y i.e. x = -a-y and get a bi-quadratic in y => the product of the roots will give us the answer.

krsh.vik

I didn't know why, but seriously had a hunch that a can be equal to zero
Went on with the assumption & solved the equation
x^4-6x^2+1 = 0
let k=x^2
k = 3+2root2 or 3-2root2
x=+/- (root2+1) +/- (root2-1)
assuming x=root2+1
all the other values fit in with the options

shyamnaren

I did it by substitution....assumed 2 is a root of f(x), which made a= 7/6. Then I computed the value of all the answer options & saw they fitted the equation really well...crude method, but it worked!

All of these

krsh.vik

Quote:

Originally Posted by krsh.vik

I didn't know why, but seriously had a hunch that a can be equal to zero
Went on with the assumption & solved the equation
x^4-6x^2+1 = 0
let k=x^2
k = 3+2root2 or 3-2root2
x=+/- (root2+1) +/- (root2-1)
assuming x=root2+1
all the other values fit in with the options

Another small observation .... if sum of the roots is devoid of a, then the value of a shouldn't effect the equation ... hence, my assumption of a = 0 works

thebornattitude

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Quantitative Question # 032
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If r is the root of f(x) = x^4 + ax^3 - 6x^2 - ax + 1 = 0, then which among the following is also a root of f(x) = 0?

(1) -1/r (2) (1+r)/(1-r) (3) (r-1)/(r+1) (4) All of these (5) Exactly two of these

Unfortunately for me, by the time I solve the question after doing my job in office,,,,, most of you guys finish doing it ..............

Neways will post the solution .

put f(-1/r), we get the same equation as

f(-1/r)= r^4 + ar^3 -6r^2 -ar+1= f(r)=0,
hence -1/r is a root for sure....................................(i)

Now chcking for (r-1)/(r+1)
f(r-1/r+1)= same equation as got above= f(r)=0, hence r-1/r+1 is also a root...
as r-1/r+1 is a root, hence r+1/1-r will be a root as got from (i).....

hence answer- (4)- all of these

mehrozkhn

using rolle's theorem. We hav f(1)=-4 and f(-1)=-4 hence their exists 'p' such that -1<p<1 and f'(p)=0 which gives a=0 . Hence the function is f(x)=x^4-6x^2+1 = 0 which clearly means if f(r)=0 then f(-1/r) =0. And we can get f(r-1/r+1)=0=f(1+r/1-r) .

Aarav · 01:46 PM

Quote:

Originally Posted by mehrozkhn

using rolle's theorem. We hav f(1)=-4 and f(-1)=-4 hence their exists 'p' such that -1<p<1 and f'(p)=0 which gives a=0 . Hence the function is f(x)=x^4-6x^2+1 = 0 which clearly means if f(r)=0 then f(-1/r) =0. And we can get f(r-1/r+1)=0=f(1+r/1-r) .

Baap re baap

Rolle bhi aa gaya scene mein.

Ok -> A new problem.

x^3 + x^2 - 3x + 2 = 0 has roots p, q, r.
Compute
(pq)^2 + (qr)^2 + (rp)^2.
(2-p)^3 + (2-q)^3 + (2-r)^3.
p^4 + q^4 + r^4.

No choices as I haven't computed the answers yet.

sabsebadapaagal

Quote:

Originally Posted by Aarav

Baap re baap

Rolle bhi aa gaya scene mein.

Ok -> A new problem.

x^3 + x^2 - 3x + 1 = 0 has roots p, q, r.
Compute
(pq)^2 + (qr)^2 + (rs)^2.
(2-p)^3 + (2-q)^3 + (2-r)^3.

No choices as I haven't computed the answers yet.

What is s above? Is it a typing mistake for p? pls confirm .....