Test Index

Indices

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Question : 24 of 86

Marks: +1, -0

\left[ \frac{x^{\frac{a}{a-b}}}{x^{\frac{a}{a+b}}} \times \frac{x^{\frac{b}{b-a}}}{x^{\frac{b}{b+a}}} \right]^{a+b},

then the value of

z

is

0
$x^{ab}$
$x^{a+b+c}$
1.

Solution:

Concept:

Use exponent laws to combine powers inside the bracket.

Explanation:

Rewrite the expression as a single power of

x

:

\left[ \frac{x^{\frac{a}{a-b}}}{x^{\frac{a}{a+b}}} \times \frac{x^{\frac{b}{b-a}}}{x^{\frac{b}{b+a}}} \right]^{a+b} = \left[ x^{\frac{a}{a-b} - \frac{a}{a+b}} \times x^{\frac{b}{b-a} - \frac{b}{b+a}} \right]^{a+b}

Note

b-a = -(a-b)

and

b+a = a+b

. So the second exponent becomes

-\frac{b}{a-b} - \frac{b}{a+b}

.

Add exponents:

\left(\frac{a}{a-b} - \frac{a}{a+b}\right) + \left(-\frac{b}{a-b} - \frac{b}{a+b}\right) = \frac{a-b}{a-b} - \frac{a+b}{a+b} = 1 - 1 = 0

Thus the bracket simplifies to

x^0 = 1

. Then

(1)^{a+b} = 1

.

Answer:

1 (Option D)

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