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WAEC Further Mathematics 2022 Paper

Question : 19

Total: 45

^{2}

\left(x + \frac{2}{x^2}\right)^5

Solution:

Concept:

Use the general term of a binomial expansion to extract the coefficient of a required power of

x

Explanation:

For the expansion

\left(x + \frac{2}{x^2}\right)^5

, take

p = x

q = \frac{2}{x^2}

, and

n = 5

The general term is:

T_{r+1} = {}^5C_r \, x^{5-r} \left(\frac{2}{x^2}\right)^r

Rewrite it by combining powers of

x

T_{r+1} = {}^5C_r \, 2^r \, x^{5-r-2r} = {}^5C_r \, 2^r \, x^{5-3r}

We want the term that contains

x^2

, so set

5 - 3r = 2

Solving gives

r = 1

Substitute

r = 1

into the general term:

T_2 = {}^5C_1 \cdot 2^1 \cdot x^2 = 5 \times 2 \times x^2 = 10x^2

Therefore, the coefficient of

x^2

10

Answer:

Option A —

10

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