The Official CAT 2007 Quant Thread

[vineet.nitd]

Originally Posted by CAT 2007

5)G(x) is a function such that G(x + 1) + G(x – 1) = G(x). For what minimum value of p does the equation G(x + p) = -G(x) necessary hold true if x is a positive real number?

a) 2 b) 3 c) 5 d) 6

put x=0 and x=1 , to get g(2)=-g(-1)
hence ans is 3

[the.bombardier]

Originally Posted by vineet.nitd

you have just equated the slopes .wot of the opposite side length being the same .It wont be holding for (a,b) = (2,2) .Check .

Oly option c suffices ..

oops..thanks for pointing that out..

[vineet.nitd]

Originally Posted by the.bombardier

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ans 4
a + b + c + d + e = 10
P = ac + bc + ad + bd + ae + be
=a(c+d+e)+b(c+d+e)
=(a+b)(c+d+e)=(a+b)(10-(a+b))
the maximum value of (a+b) is tending to 10- .so (10-(a+b)) will be tending to 0+.
so the value of P>0
the value of P will be maximum when both the factors are maximum and equal.
this will happen when (a+b)=5
so P<=25
0<P<=25
so option b is the answer.
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ur method is good

my method is

a/3=b/3=c/2=d/2=e/2 =a+b+c+d+e/3+3+2+2+2 = 10/12=5/6

hence Smax=25/6 *6=25

[the.bombardier]

Originally Posted by vineet.nitd

ur method is good

my method is

a/3=b/3=c/2=d/2=e/2 =a+b+c+d+e/3+3+2+2+2 = 10/12=5/6

hence Smax=25/6 *6=25

hey thats a really fast method..nice

[krsh.vik]

Originally Posted by vineet.nitd

ur method is good

my method is

a/3=b/3=c/2=d/2=e/2 =a+b+c+d+e/3+3+2+2+2 = 10/12=5/6

hence Smax=25/6 *6=25

well the max value of the expression is 25 hence option a also satisfies ...

[geturdream]

Originally Posted by vineet.nitd

ur method is good

my method is

a/3=b/3=c/2=d/2=e/2 =a+b+c+d+e/3+3+2+2+2 = 10/12=5/6

hence Smax=25/6 *6=25

can u please expalin the metod ...I am not getting it ....

[geturdream]

Originally Posted by vineet.nitd

put x=0 and x=1 , to get g(2)=-g(-1)
hence ans is 3

u cant have x=0 as x is a real positive number ...

it is coming g(3)=-g(0)
hence the ans is 3 (that is matching with u)

[the.bombardier]

<----snipped--->

[vineet.nitd]

Originally Posted by geturdream

can u please expalin the metod ...I am not getting it ....

ok ..lemmee explain the concept behind this ..
ac+bc+ad+bd+ae+be/6 >= (a^3b^3c^2d^2e^2)^1/6

So basically u ve got to maximise Z=a^3b^3c^2d^2e^2.
Z=3^3.3^3.2^2.2^2.2^2(a/3)^3(b/3)^3(c/2)^2(d/2)^2(e/2)^2


(a/3)^3(b/3)^3(c/2)^2(d/2)^2(e/2)^2 will be maximum when all the factors are equal ...

Hope u got the funda

[vineet.nitd]

Originally Posted by geturdream

u cant have x=0 as x is a real positive number ...

it is coming g(3)=-g(0)
hence the ans is 3 (that is matching with u)

ohh wo jaldbazee me dekhna bhool gaya tha.anyways hardly matters

Point noted sir