The value of {a¹/2}+a^{-1/2}}{1-a}+{1-a^{-1/2}}{1+{a}}

Correct Answer is : 21−a2/1-a1−a2

Correct Answer is : 21−a2/1-a1−a2

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Question : 58 of 86

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The value of

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

Solution:

Concept:

Simplify the given expression using substitution

x = \sqrt{a}

Explanation:

Let

x = a^{1/2} = \sqrt{a}

, so

a = x^2

and

a^{-1/2} = 1/x

Then

1 - a = 1 - x^2 = (1 - x)(1 + x)

Rewrite the first term:

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

Rewrite the second term:

\frac{1 - 1/x}{1 + x} = \frac{(x-1)/x}{1+x} = \frac{x-1}{x(1+x)}

Note

x-1 = -(1-x)

, so the second term becomes

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

Combine with common denominator

x(1-x)(1+x)

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

Simplify numerator:

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

Thus the expression becomes

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

Since

x^2 = a

, we get

\frac{2}{1-a}

Answer:

\frac{2}{1-a}

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