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WAEC Further Mathematics 2010 Paper

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Question : 25

Total: 40

Find the least value of n for which

\binom{3n}{2} > 0, n \in \mathbb{R}

.

$\frac{1}{3}$
$\frac{1}{6}$
$\frac{2}{3}$
1

Solution:

Concept:

The binomial expression

\binom{3n}{2}

expands to

\frac{3n(3n-1)}{2}

, so its sign is determined by the product

3n(3n-1)

.

Explanation:

Write:

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

Since the denominator is positive, we need:

3n(3n-1) > 0

The critical points are given by

3n = 0

and

3n - 1 = 0

.

Hence,

n = 0 \quad \text{or} \quad n = \frac{1}{3}

The product is positive when

n < 0

or

n > \frac{1}{3}

.

Check the given options:

n = \frac{1}{3}

gives

\binom{1}{2} = 0

, not greater than zero.

n = \frac{1}{6}

makes

\binom{0.5}{2} < 0

.

The smallest option satisfying

n > \frac{1}{3}

is

n = \frac{2}{3}

.

Answer:

C.

\frac{2}{3}

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