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

[Aarav]

Attached is the cumulative points earned by 73 students who participated in discussions from problem 5th to 16th.

@Implex -> the divisor found by you isn't correct. Please check again.

[implex]

Originally Posted by Aarav

Attached is the cumulative points earned by 73 students who participated in discussions from problem 5th to 16th.

@Implex -> the divisor found by you isn't correct. Please check again.

sorry !!
D(x)=x^2+(1-sqrt(3))x+2
R(x)-D(x)=x^2+(1+sqrt(3))x
clealry for all non positive x
Rhs is not positve

option 5)!!

[Aarav]

Originally Posted by implex

sorry !!
D(x)=x^2+(1-sqrt(3))x+2
R(x)-D(x)=x^2-(1+sqrt(3))x
clealry for any non positive x
rhs is positive
so is lhs
R(x)>D(x)

option 3)!!

Ok -> here is how we factorize x^6 + 4x^3 + 8 -> we all know the factorization of a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca). In our question, a = x^2, b = -2x, c = 2.

A small reconfirmation on the concept in the choices -> non-positive or non-negative includes 0 and hence choice (5) is our answer.

Sorry for revealing this a bit early -> but somehow I didn't want to spend more time in this dull environment where the participation is almost zilch.

[rajatkapoor1986]

by dividing given polynomial by x^3-1 we get,R(x)=2x^2+2x+2
(or split given polynomial in powers of 3 put x^3=1 n remaining gives remainder)
now for D(X),
x^6 + 4x^3 + 8=O FOR x^3 =-2+i*2 & x^3 =-2- i*2
so,x^6 + 4x^3 + 8= [x^3-(-2+i*2)][x^3-(-2-i*2)]
D(X)=[x^3-(-2+i*2)] or [x^3-(-2-i*2)]
consider D(X)=[x^3-(-2+i*2)]
option 1 & 2 certainly don't apply
for option 3,
f(x)={R(x)=2x^2+2x+2}- {D(X)=[x^3-(-2+i*2)]}
=2x^2+2x+2-x^3-(-2+i*2)
f(x) can be =+ve or -ve depending upon x
so we are left with option 5.

[implex]

Originally Posted by Aarav

Ok -> here is how we factorize x^6 + 4x^3 + 8 -> we all know the factorization of a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca). In our question, a = x^2, b = -2x, c = 2.

A small reconfirmation on the concept in the choices -> non-positive or non-negative includes 0 and hence choice (5) is our answer.

Sorry for revealing this a bit early -> but somehow I didn't want to spend more time in this dull environment where the participation is almost zilch.

@Aarav if we put the values of a,b and c as suggested lhs will becomes
x^6-20x^3+8

[Aarav]

Originally Posted by rajatkapoor1986

by dividing given polynomial by x^3-1 we get,R(x)=2x^2+2x+2
(or split given polynomial in powers of 3 put x^3=1 n remaining gives remainder)
now for D(X),
x^6 + 4x^3 + 8=O FOR x^3 =-2+i*2 & x^3 =-2- i*2
so,x^6 + 4x^3 + 8= [x^3-(-2+i*2)][x^3-(-2-i*2)]
D(X)=[x^3-(-2+i*2)] or [x^3-(-2-i*2)]
consider D(X)=[x^3-(-2+i*2)]
option 1 & 2 certainly don't apply
for option 3,
f(x)={R(x)=2x^2+2x+2}- {D(X)=[x^3-(-2+i*2)]}
=2x^2+2x+2-x^3-(-2+i*2)
f(x) can be =+ve or -ve depending upon x
so we are left with option 5.

Answers like these should fetch negative marks. You have made mockery of quant up there.

[Aarav]

Originally Posted by implex

@Aarav if we put the values of a,b and c as suggested lhs will becomes
x^6-20x^3+8

Come on yaar, don't spoil my already spoilt mood.
-3abc yields +12x^3.

[implex]

Originally Posted by Aarav

Come on yaar, don't spoil my already spoilt mood.
-3abc yields +12x^3.

oops, my apologies! i took it as +3abc

[implex]

Originally Posted by rajatkapoor1986

by dividing given polynomial by x^3-1 we get,R(x)=2x^2+2x+2
(or split given polynomial in powers of 3 put x^3=1 n remaining gives remainder)
now for D(X),
x^6 + 4x^3 + 8=O FOR x^3 =-2+i*2 & x^3 =-2- i*2
so,x^6 + 4x^3 + 8= [x^3-(-2+i*2)][x^3-(-2-i*2)]
D(X)=[x^3-(-2+i*2)] or [x^3-(-2-i*2)]
consider D(X)=[x^3-(-2+i*2)]
option 1 & 2 certainly don't apply
for option 3,
f(x)={R(x)=2x^2+2x+2}- {D(X)=[x^3-(-2+i*2)]}
=2x^2+2x+2-x^3-(-2+i*2)
f(x) can be =+ve or -ve depending upon x
so we are left with option 5.

complex numbers can't be compared
so you can't really say R(x)-D(x) is greater than 0 or not

There is no notion of value of complex numbers so their comparison is absurd

[schinnar]

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Quantitative Question # 017
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Let R(x) be the remainder when x^16 + x^8 + x^6 + x^4 + x^2 + 1 is divided by x^3 - 1. Let D(x) be the divisor (less than degree 4) of x^6 + 4x^3 + 8. Then which among the following is true?

(1) The sum of the coefficients of R(x) and D(x) is equal

(2) The sum of the absolute value of coefficients of R(x) and D(x) is equal
(3) R(x) > D(x) for all non-positive x

(4) at least 2 of the above

(5) none of the above

Ans:

The remainder when x^16 + x^8 + x^6 + x^4 + x^2 + 1 is divided by X^3-1 is 2x^2+2 +1 ---> 1

R(x)=2x^2+2+1

x^2+2 divides x^6+4X^3+8 which is degree < 4

D(x)=x^2+2

option 1 is not true

option 2 I am bit confused i assume absolute value of coefficient is constant term then this is false

option 3 is clearly false
take the value of -1 R(x) = 1 D(x) = 3
if my assumption is correct then ans is 5