WAEC 2020 · Paper 2 · Q9✱

(a)
Differentiate $y = 3t^4 - 4t^3 + 2t^2 - \dfrac1t - 5$ .
(b)
The probability that a civil servant owns a car is $\frac16$ . If two civil servants are selected at random, find the probability that: (i) each owns a car; (ii) only one owns a car.
Separate values with commas, e.g. 3, −2
(c)
Solve: $\log_{10}(16x + 2) - \log_{10}(4x + 2) = \log_{10}(2x + 1)$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Write $\frac1t$ as $t^{-1}$ and differentiate term by term: $\frac{dy}{dt} = 12t^3 - 12t^2 + 4t - (-1)t^{-2}$
$= 12t^3 - 12t^2 + 4t + \frac{1}{t^2}$ .
The constant $-5$ gives 0.

(b)

$P(\text{owns a car}) = \frac16$ , so $P(\text{doesn't}) = \frac56$ .

(i)

Each owns one: $\frac16 \times \frac16 = \frac{1}{36}$ .

(ii)

Only one owns one: the first does and the second doesn't, $\frac16 \times \frac56$ .
Or the other way round, $\frac56 \times \frac16$ .
Together: $\frac{5}{36} + \frac{5}{36} = \frac{10}{36}$
$= \frac{5}{18}$ .

(c)

A difference of logs is the log of a quotient: $\log_{10}\frac{16x + 2}{4x + 2} = \log_{10}(2x + 1)$ , so $\frac{16x + 2}{4x + 2} = 2x + 1$ .
Multiply out: $16x + 2 = (2x + 1)(4x + 2)$
$= 8x^2 + 8x + 2$ , so $8x^2 - 8x = 0$ , $8x(x - 1) = 0$ , and $x = 0$ or $x = 1$ .
Check both in the original: at $x = 0$ , $\log 2 - \log 2 = 0 = \log 1$ ✓.
At $x = 1$ , $\log 18 - \log 6 = \log 3$ ✓.
Both are solutions.

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