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

[Aarav]

Error at my end - please find f(4) - everything else remains the same.
It was hell hectic in office last day - hence made slight error in asking for f(3) when I meant f(4).

Please solve for f(4).

[pavanpadekal]

f(x).f(y)-f(xy)=3(x+y+2)
Putting y=1
f(x).f(1)-f(x)=3(x+3)
f(1).f(1)-f(1)=12
i.e.,a^2-a-12=0 where a=f(1)
a=[1+-sqrt(1+4]/2
a=[1+-7]/2
a=4 or -3
3f(x)=3(x+3)
-4f(x)=3(x+3)
f(3)=18/3 = 6
or f(3)=-18/4=-9/2
Hence f(3)= 6 or -4.5
Hence (5) none of these


Now for f(4)
f(x).f(1)-f(x)=3(x+3)
f(4).f(1)-f(4)=3(7)
so 4f(4)-f(4)=3f(4)=21
i.e.,f(4)=7
OR
-3f(4)-f(4)=21
-4f(4)=21
f(4)=-21/4
Hence (2) 7

[Aarav]

Originally Posted by pavanpadekal

f(x).f(y)-f(xy)=3(x+y+2)
Putting y=1
f(x).f(1)-f(x)=3(x+3)
f(1).f(1)-f(1)=12
i.e.,a^2-a-12=0 where a=f(1)
a=[1+-sqrt(1+4]/2
a=[1+-7]/2
a=4 or -3
3f(x)=3(x+3)
-4f(x)=3(x+3)
f(3)=18/3 = 6
or f(3)=-18/4=-9/2
Hence f(3)= 6 or -4.5
Hence (5) none of these

Please find f(4) with the given options. I have already seen the flawed approach of all of you - perhaps this was why the question was asked.

[comradeharsh]

Well, I am posting my question again, for visibility:

Originally Posted by comradeharsh

A silly question:
Can we take x=y=0, solve for f(0), and then take x=3 and y=0, to get f(3)?
I got f(3)=4 or f(3)=-7/2.
Please bear in mind this comes from a quant weakling

[nsitaditya]

acc. to me option 5
f(0)[f(2)-1]=0.....(1)
f(2)=1 or f(0)=0
f(2)[f(1)-1]=3........(2)
f(0)[f(1)-1]=-3.......(3)
(2) divided by (3) gives
f(2)/f(0)=-1
so f(2)=1 (as f(0) can't be 0 from 1)
from (2) f(1)=4

4f(3)-f(3)=3(1+3-2)=6;
f(3)=2

hence option 5

[Aarav]

Originally Posted by comradeharsh

Well, I am posting my question again, for visibility:

You can but requires extra check - that's why the question was put - f(3) has been calculated wrongly by all - jha16june made it look too simplistic by ignoring some value of f(0).

[Aarav]

Originally Posted by nsitaditya

acc. to me option 5
f(0)[f(2)-1]=0.....(1)
f(2)=1 or f(0)=0
f(2)[f(1)-1]=3........(2)
f(0)[f(1)-1]=-3.......(3)
(2) divided by (3) gives
f(2)/f(0)=-1
so f(2)=1 (as f(0) can't be 0 from 1)
from (2) f(1)=4

4f(3)-f(3)=3(1+3-2)=6;
f(3)=2

hence option 5

Bad mistake from my end - everyone will arrive at the right option (5
) with some screwed logic. Please find f(4) in the question.

[nsitaditya]

continuing with my previous post

f(0)=-1
f(4).f(0)-f(0)=3(4+0-2)
f(4)=-5

so the answer still remains none of these ie (5)

[implex]

Originally Posted by Aarav

Bad mistake from my end - everyone will arrive at the right option (5
) with some screwed logic. Please find f(4) in the question.

f(x).f(y)-f(xy)=3(x+y+2)...(1)

x=y=0 gives
f(0)=3 or -2
and x=y=1 gives
f(1)=4 or -3
x=1 y=0 gives
f(1)f(0)-f(0)=9...(2)


so f(1)=4 and f(0)=3 satisfy eqn (2)

f(x).f(0)-f(0)=3(x+2)
f(x)=x+3
putting x=4
we get f(4)=7 hence option (2)

[nsitaditya]

sir is my logic correct?