TargetIIT

MULTIPLE ANSWER QUESTION

If f(x) = 0 is a polynomial whose coefficents are all ± 1 and whose roots are all real, then degree of f(x) can be equal to

(A)  1          (B) 2         (C) 3        (D) 4

a,c ??

ya i also feel the answer shud be A and C...
logic : number of terms shud be even so that they can be cancelled out...

x[p]2[/p]+x-1=0 also has real roots

so why cant it be 2?

Ya Rahul It can also be 2

In fact I was getting  A B C D

But ans given is A B C

I say it can also be 4 ( Considering repeated roots)

oops i misread the question ..i thought both roots and coefficients have to be ± 1 [13]

Uttara can u show  how four??

Acc. to me, A,B,C should be right.........

(x[p]2[/p] + x)(x[p]2[/p] - 1) = 0

the eq[p]n[/p] that u have posted can be written as x[p]4[/p] - 2x[p]2[/p] + x = 0...which do not satisfy the condition given abv...

EDITTED

but u have taken the constant term in the above eqn as 0,
but ques says that every term should be +1 or -1

is there any mathematical proof for this??

yeah ans should be ABC

i thot we shud also include repeated roots

y=x[p]2[/p](x[p]2[/p]-1)=x[p]4[/p]-x[p]2[/p].

@Rahul (#12 )

The qs mentions that the coefficients shud be ± 1

So it is not necessary that the constant term is also ±1 na ??

@msp :  I too thought the same way

@ Asish : Ya u're ans is correct

but my doubt is shud there be a constant term at all ??

ya..in a general polynomial, there should be a constant term ......................which u have taken as '0'..
but qs says it should be ±1

I think there need not b a constant term becz the qs said coefficients shud be +- 1  & nothing is mentioned  abt constant term

its like this

if v say an eqn has all positive coeffs then it need not be that constant term shud also be positive no???

constant is nothing but co-efficient   of  '[im]http://codecogs.izyba.com/gif.latex?x%5E%7B0%7D[/im]' in the polynomial....which i think u hav to take  ±1
[hide](Hope i m not talkink any blunder [3])[/hide]

ya right Sorry .....[3]  [hide]So stupid of me[/hide]