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WAEC Further Mathematics 2012 Paper

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Question : 40

Total: 40

Simplify

\frac{\; x^{3n\;+\;1}}{x^{2n\;+\;\;\frac{5}{2}}(x^{2n-3})^{\;\frac{1}{2}}}

$0$
$-\;\frac{1}{2}$
$1$
$10$

Solution:

Concept:

The expression is simplified using exponent laws:

(x^a)^b = x^{ab}

and

x^a \cdot x^b = x^{a+b}

.

Explanation:

Simplify the power inside the denominator:

(x^{2n-3})^{\frac{1}{2}} = x^{\frac{2n-3}{2}}

.

So the denominator becomes

x^{2n+\frac{5}{2}} \cdot x^{\frac{2n-3}{2}}

.

Add the exponents:

2n + \frac{5}{2} + \frac{2n-3}{2} = 2n + \frac{2n+2}{2}

.

This simplifies to

2n + n + 1 = 3n+1

.

Thus the denominator is

x^{3n+1}

.

The original expression becomes

\frac{x^{3n+1}}{x^{3n+1}} = 1

.

Answer:

C.

1

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