If x is a real number such that 0 < x < 1, which of the following expressions is equivalen… — GRE General Test

If $x$ is a real number such that $0 < x < 1$ , which of the following expressions is equivalent to $\sqrt{\frac{x^{-2} - 1}{x^{-2} + 2 + x^2}}$ ?

The expression

\frac{\sqrt{1-x^2}}{1+x^2}

is equivalent to the given radical expression.

Expressing

x^{-2}

\frac{1}{x^2}

allows the numerator to be rewritten as

\frac{1-x^2}{x^2}

and the denominator as

\frac{(1+x^2)^2}{x^2}

. Dividing these fractions cancels out

x^2

, leaving

\frac{1-x^2}{(1+x^2)^2}

under the radical. Taking the square root of the numerator and denominator separately gives

\frac{\sqrt{1-x^2}}{1+x^2}

Rewrite negative exponents as fractions in both the numerator and denominator.

Numerator:

x^{-2} - 1 = \frac{1}{x^2} - 1 = \frac{1-x^2}{x^2}

. Denominator:

x^{-2} + 2 + x^2 = \frac{1}{x^2} + 2 + x^2 = \frac{1 + 2x^2 + x^4}{x^2}

Converting negative exponents into positive fractional exponents allows common denominators to be established.

Factor the perfect square trinomial in the denominator.

The numerator of the denominator expression is

1 + 2x^2 + x^4 = (1 + x^2)^2

. Thus, the entire denominator is

\frac{(1+x^2)^2}{x^2}

Recognizing

1 + 2x^2 + x^4

(1+x^2)^2

simplifies taking the square root.

Simplify the quotient inside the radical.

\frac{\frac{1-x^2}{x^2}}{\frac{(1+x^2)^2}{x^2}} = \frac{1-x^2}{(1+x^2)^2}.

Canceling the common factor of

x^2

in the denominators reduces the nested fraction.

Apply the square root rule

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

\sqrt{\frac{1-x^2}{(1+x^2)^2}} = \frac{\sqrt{1-x^2}}{\sqrt{(1+x^2)^2}} = \frac{\sqrt{1-x^2}}{1+x^2}.

Since

1+x^2 > 0

for all real

x

\sqrt{(1+x^2)^2} = 1+x^2

Simplifying radical expressions containing negative powers and algebraic fractions

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