1956 AMC 12 Problem 46

Question

For the equation

\dfrac{1+x}{1-x}=\dfrac{N+1}{N}

to be true where

N

is positive,

x

can have:

any positive value less than

1

any value less than

1

the value zero only

any non-negative value

any value

Cross-multiply and solve for

x

in terms of

N

The equation reduces to

x=\frac{1}{2N+1};

also solve this relation for

N

Cross-multiplication gives

N(1+x)=(N+1)(1-x),

so

x(2N+1)=1

and

x=\frac1{2N+1}.

Every positive

N

gives

0\lt x\lt1.

Conversely, for any

0\lt x\lt1,

N=\frac{1-x}{2x}>0,

so such an

N

exists.

Thus,

x

may be any positive value less than

1,

and the correct answer is A.

Problem 46 in other years