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

[fiction]

Originally Posted by 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)

least time taken by me to solve a QQAD problem , ever

[v-factor]

f(x+f(x)) = x----->1
and f(ax+bf(x)) = cx+df(x)----->2

Comapring the coefficients we get a=1, b=1, c=1 and d=0
So b=c and b=d+1

Putting x=x+f(x) in equation 1 we get
f(2x+f(x)) = x+f(x)
Comapring the coefficients with equation 2
a=2,b=1,c=1,d=1

So if we go on repeated substitution we see only b=c is true for all cases
Option 1)

[rohith_s_theone]

We have to recursively substitute and compare

f(x+f(x))=x ;f(ax+bf(x))=cx+df(x) which means a=1 b=1 c=1 and d= 0
This eliminates option 3 (a=d)

Now substitute x= x+f(x)
f(x+f(x)+f(x+f(x))=x+f(x)
f(2x+f(x))=x+f(x) ---[2]
Compare with f(ax+bf(x))=cx+df(x)
a=2 b=1 c=1 d= 1
which eliminates options (2) b=d+1 and (4) atleast 2

Now we are left with options (1) and (5).
By substituting x=x+f(x) again in [2] we get
f[3x+2f(x)]=2x+f(x)
which suggests that option (1) is correct

ANS: (1) b=c

[IIM maniac]

My attempt to this question..

Given : f(x+f(x)) = x --(1)
f(ax+bf(x)) = cx+df(x) --(2)
Comparing 1 & 2 we get a=1, b=1, c=1, d=0
Putting these values back to eq 2, we get f(x + f(x)) = x --(3)

Putting the value of x in eq 1,
f[(x+f(x)) + f(x+f(x))] = x+f(x)

But f(x+f(x)) = x
therefore eq reduces to f[x+f(x) + x] = x+f(x)
f(2x + f(x)) = x+f(x) --(4)

Comparing 4 with 2,
a=2, b=1, c=1, d= 1

therefore b=c, Option 1

Ameya

[prvineeth]

Originally Posted by 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)

How can you put x=c+f(c) ?

here x is a variable and c is a constant.........

As per your assumption variable= constant

[madhulikasambit]

please correct my approach if flaw....

given f(x+f(x))=x
now let f(0)=y
then f(0+f(0))=f(y)=x
now f(ax+bf(x))=cx+df(x)
so f(0+bf(0))=f(by)=bf(Y)=bx
hence b=c
hence option 1
could get just this....
please help me if my approach is wrong

[Fuzon]

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

Comparing the given equations viz: f(x + f(x)) = x and f(ax + bf(x)) = cx + df(x),
we know a=1 b=1 c=1 d=0
Thus option three can be shown the door.

Now for the above compared values 1) and 2) are correct. So lets check for another set of values

Lets put x=x+f(x) and try to get the equation in the form of ax , where a =2

f(x+f(x) + f(x+f(x))) = x+f(x)
thus
f(x+f(x)+x)=x+f(x)

f(2x+f(x))=x+f(x) compare with f(ax + bf(x)) = cx + df(x)

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

Now here b!=d+1 but b=c still persists

jitni chaadar hove hai utne pair pasaar

Answer option 1) seems correct to me ...

[madhulikasambit]

please correct my approach if wrong

[nhniesl]

well this is my first post...i hope the solution is right

SOLUTION :

since f(x + f(x)) = x i.e x + f(x) = f_inverse(x) --------- 1

and f(ax + bf(x)) = cx + df(x)
i.e ax + bf(x) = f_inverse( cx + df(x) )
so ax + bf(x) = f_inverse( cx ) + f_inverse( df(x) )
ax + bf(x) = c * f_inverse( x ) + d * f_inverse( f(x) )

putting the value of f_inverse(x) from 1 in the above

ax + bf(x) = c ( x + f(x) ) + d * x

simplifies to ax + bf(x) = ( c + d )x + c f(x)

from the above by equating the coefficient of the variables it can be said that

a = c + d and b = c.

thus from above OPTION 1 is right

what say u ?

[hismajesty143]

Originally Posted by 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

We have:
f(x + f(x)) = x ....................(i)
f(ax + bf(x)) = cx + df(x) .....................(ii)

Substituting the value of x from (i) in (ii) below we get:

f(ax + bf(x)) = c[f(x + f(x))] + df(x)

Since x is real the function is linear. Every linear function can be expanded in the manner below:

af(x) + bf(f(x)) = cf(x) + cf(f(x)) + df(x)

Now comparing the coefficient of the like terms on both sides we get the follwing two results:

• Comparing coefficients of f(x):
  a=c+d
• Comparing coefficients of f(f(x)):
  b=c

Hence option (1) is definitely true, which is the result obtained from above.
Option (2) is not true, as if b is to be expressed in terms of d we will get b=a-d
Option (3) is also not true as a=c+d
Hence, Option (4) is also not true.

MY ANSWER: OPTION (1)

PLEASE CORRECT ME IF MY EXPLANATION IS WRONG..