NECO 2023 · Paper 2 · Q8

(a)
Evaluate without using tables $5\log2 + \log40 - \log12.8$ .
(b)
Find the equation of the curve which passes through $(-2, 5)$ and has gradient $6x^2 + 8x - 3$ at any point.
(c)
Differentiate $y = 3x^2 + 4x - 5$ with respect to $x$ .

Worked solution (try it first)

(a)

Move the 5 inside as a power: $5\log 2 = \log 2^5 = \log 32$ .
Then combine: $\log 32 + \log 40 - \log 12.8 = \log\frac{32 \times 40}{12.8}$
$= \log\frac{1280}{12.8}$
$= \log 100$
$= 2$ .

(b)

The gradient is $\frac{dy}{dx}$ , so integrate it: $y = \frac{6x^3}{3} + \frac{8x^2}{2} - 3x + c$
$= 2x^3 + 4x^2 - 3x + c$ .
The curve passes through $(-2, 5)$ : $2(-8) + 4(4) - 3(-2) + c = 5$ , so $-16 + 16 + 6 + c = 5$ and $c = -1$ .
The curve is $y = 2x^3 + 4x^2 - 3x - 1$ .

(c)

$\frac{dy}{dx} = 6x + 4$ (the constant $-5$ gives 0).

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