CAT 2008: QQAD Practice Test Discussions

[implex]

Originally Posted by karanverma86

Hi,
Shouldn't the answer this question be (B)

because g(-1)= 1- f(-1+1)
= 1- f(0)
similarly for g(1)

g(3) would render it out of domain of f(x) because f(4) is not defined.

yupp, an error on my part.
the key i published had two errors this one and question 19

[srikar2097]

Originally Posted by arun_juee

Hi,

My Take ( after 2 peg of Smirnoff )
Time 12:20 am to 1:10 am ( July 4th 2008 ).

6-D,11-B,14-B,15-E,17-A,20-A,21-B,22-B,23-D,24-E
Where are the options/solutions available?


Later.
Arun

Just travel back a couple of pages on the same thread. You'll have the official answers and detailed solutions for all problems...

[milindagrawal]

Originally Posted by v_v_n

is there any way to check whether 7^1/3 lies between what range.?
how did u come to know that 7^1/3 lie between 1.5 and 2? any method?

@v_v_n...
just thinking about the possible values led me there..
and i guess, in this question, this is the first nut to crack..
the rest just follows...

[nbangalorekar]

Originally Posted by implex

24) Let f(x) be a polynomial, and p and q be integers. Is f(p)/(p-q) an integer?

(X)Every coefficient in f(x) is an integer

(Y) f(p).f(q) = -(p-q)^2

X does not lead us anywhere here
y gives
f(p)/(p-q)=-(p-q)/f(q)

when we combine the two we know f(p),f(q) and (p-q) and (p-q) ^2 are all integers

equation is symmetric
so f(p)=-f(q)
f(p)=+-(p-q)
so f(p)/(p-q)=+-1 so integer

hence option d)

implex yaar when doing the paper,id solved this qn the same way... then one thing struk me:
suppose say (p-q)=6 & f(q)=4
then we get f(p)=-9
for this we wont get f(p)/(p-q) an integer
so i marked option as 'e'

Aarav please clarify this....

[implex]

Originally Posted by nbangalorekar

implex yaar when doing the paper,id solved this qn the same way... then one thing struk me:
suppose say (p-q)=6 & f(q)=4
then we get f(p)=-9
for this we wont get f(p)/(p-q) an integer
so i marked option as 'e'

Aarav please clarify this....

we can't just suppose, it should be possible in the polynomial man!

suppose the polynomial is 3x^2+1
q=1
f(q)=4
f(p)=128..

so we will need a polynomial such that
my solution is conclusive !

[nbangalorekar]

Originally Posted by implex

we can't just suppose, it should be possible in the polynomial man!

suppose the polynomial is 3x^2+1
q=1
f(q)=4
f(p)=128..

so we will need a polynomial such that
my solution is conclusive !

i just took one scenario... but wudnt such a case be possible???
will try to come up with an exception like this...
the proof u have given is not very rigorous...no doubt the answer probably is correct as it is the official answer to the qn..
Aarav ur inputs?

[Aarav]

Problem 24

For a natural number n, f(p) - f(q) is divisible by p-q since p^n-q^n is divisible by p - q (from X)

Thus, [f(p) - f(q)]/(p-q) is an integer. The quadratic equation that has 2 solutions f(p)/(p-q) and -f(q)/(p-q) is x^2 - x[f(p) - f(q)]/(p-q) + 1 = 0 (from Y).

If the roots of x^2 + ax + b = 0 are rational where a and b are integers then the roots are integers => f(p)/(p-q) is an integer

[nbangalorekar]

Originally Posted by Aarav

Problem 24

For a natural number n, f(p) - f(q) is divisible by p-q since p^n-q^n is divisible by p - q (from X)

Thus, [f(p) - f(q)]/(p-q) is an integer. The quadratic equation that has 2 solutions f(p)/(p-q) and -f(q)/(p-q) is x^2 - x[f(p) - f(q)]/(p-q) + 1 = 0 (from Y).

If the roots of x^2 + ax + b = 0 are rational where a and b are integers then the roots are integers => f(p)/(p-q) is an integer

wow mannn!!!! that was one of the mind blowing questions... n brilliantly explained!!!! thanks Aarav..

[implex]

Originally Posted by nbangalorekar

i just took one scenario... but wudnt such a case be possible???
will try to come up with an exception like this...
the proof u have given is not very rigorous...no doubt the answer probably is correct as it is the official answer to the qn..
Aarav ur inputs?

in general, in all such cases, looking for symmetry is a very good tool. and more often than not! it will give you the correct answer!

[nbangalorekar]

Originally Posted by Aarav

Problem 24

For a natural number n, f(p) - f(q) is divisible by p-q since p^n-q^n is divisible by p - q (from X)

Thus, [f(p) - f(q)]/(p-q) is an integer. The quadratic equation that has 2 solutions f(p)/(p-q) and -f(q)/(p-q) is x^2 - x[f(p) - f(q)]/(p-q) + 1 = 0 (from Y).

If the roots of x^2 + ax + b = 0 are rational where a and b are integers then the roots are integers => f(p)/(p-q) is an integer

Extremely sorry for spamming but i absolutely cant stop praising the beauty of this soln...
Kudos to you yaar Aarav....Awesum man!!!
but it wud have been impossible to spot this in an exam...