CAT 2008: Quantitative Questions a Day 101-133 - The discussions

[rahulsar]

Ans QQAD 101 :-

F<G<B<P

Since 2 single parents (known) , thus F=2, P=2 is known

Also logically 5<6<7<8, where for F=5, P=2+2*3=8

Rgds.

[gripened]

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Quantitative Question # 102
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Let f(x + f(x)) = x for all real x, and if f(ax + bf(x)) = cx + df(x), then which
among the following is necessarily true?

(1) b = c (2) b = d+1 (3) a = d (4) at least 2 of the foregoing (5) none

[b.s.swaminathan]

102)

f(x+f(x))= x
put x=x+f(x)

f(x+f(x)+f(x+f(x)))=x+f(x)
f(2x+f(x))=x+f(x)-----A

again putting x=x+f(x)
f(2x+2f(x)+f(x+f(x)))=x+f(x)+f(x+f(x))
f(3x+2f(x))=2x+f(x)------ B

comparing with f(ax+bf(x))=cx+df(x);

in equation A
a=2 b=1 c=1 d=1...so b=c;

in equation B
a=3 b=2 c=2 d=1 ...so b=c && b=d+1..

so b=c is true in each case...
option 1)

[gripened]

Originally Posted by gripened

------------------------------------------------------
Quantitative Question # 102
------------------------------------------------------


Let f(x + f(x)) = x for all real x, and if f(ax + bf(x)) = cx + df(x), then which
among the following is necessarily true?

(1) b = c (2) b = d+1 (3) a = d (4) at least 2 of the foregoing (5) none

f(x+f(x)) = x
==> f( (x+f(x)) + f(x+f(x)) ) = (x+f(x))
==> f( ax + bf(x)) = cx + df(x), where d < (b==c) < a are consecutive fibonacci terms...

Hence, we have , b = c && a = c + d = b + d.
Answer Option (1) b = c always...

[Varun Khullar]

Originally Posted by gripened

------------------------------------------------------
Quantitative Question # 102
------------------------------------------------------


Let f(x + f(x)) = x for all real x, and if f(ax + bf(x)) = cx + df(x), then which
among the following is necessarily true?

(1) b = c (2) b = d+1 (3) a = d (4) at least 2 of the foregoing (5) none

f(x+ f(x)) = x
then x = k+f(k)
so
f(2k+f(k)) = k+ f(k)
so b =c
(1) option one

[Varun Khullar]

Originally Posted by gripened

f(x+f(x)) = x
==> f( (x+f(x)) + f(x+f(x)) ) = (x+f(x))
==> f( ax + bf(x)) = cx + df(x), where d < (b==c) < a are consecutive fibonacci terms...

Hence, we have , b = c && a = c + d = b + d.
Answer Option (1) b = c always...

can u explain the part in bold

[gripened]

Originally Posted by Varun Khullar

can u explain the part in bold

since we have (x +f(x)) + f(x+f(x)), and and f(x+f(x)) gives us x,
the coefficient of X part grows by an amount equal to the coefficient of f(x)..., and we keep replacing x by x + f(x), hence the same thing happens at the RHS side...but the RHS is behind by one term, hence the bold part...

[shashank3012]

From the given expression,we can see that,
a,b,c=1 and d=0.
So,we are left with only 2 options,b=c or b=d+1.
So,now,put x=X+f(X)
Hence,
f(X+f(X)+f(X+f(X)))=X+f(X)
But,f(X+f(X))=X,from given condition.
So,
f(2X+f(X))=X+f(X)
Putting X=y+f(y)
f(2y+2f(y)+f(y+f(y)))=2y+f(y)
f(3y+2f(y))=2y+f(y)
In all these cases,
We can see that,b=c
So,option (1)

[fiction]

f(x+f(x)) = x

put x = c + f(c)

f(c+f(c) + f(c+f(c))) = c +f (c)

f(c + f(c) + c) = c + f(c) ...............(since f(c+f(c)) = c)

f(2c + f(c)) = c + f(c)

compare with given. f(ax + b f(x)) = cx + d f(x)

a = 2 b = 1 c = 1 d = 1

b = c

answer option (1)

[**genius_yogi**]

102)
given f(x +f(x)) = x
considering x = X + f(X)
we ll get f(X+f(X) + f(X + f(X)) = X + f(X)

now
f(X + f(X) = X

therefor f(2X+f(X)) = X + f(X)
now comparing it with f(ax + bf(x)) = cx + df(x)

we ll get a = 2, b = 1, c = 1 & d = 1
thus b = c & hence option (1)