WAEC 2009 · Paper 2 · Q10

(a)(i)
An exponential sequence is given by $18, 2, \frac29, \ldots$ Find an expression for the $n$ th term;
(a)(ii)
the sum of the first $n$ terms.
(b)(i)
Find the equation of the tangent to the curve $y = -x^2 + x + 1$ at the point $(2, -1)$ .
(b)(ii)
Find the intercepts of the tangent in (b)(i) with the axes.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

The first term is $a = 18$ and the common ratio is $r = \frac{2}{18} = \frac19$ .
$T_n = ar^{n-1}$
$= 18\left(\frac19\right)^{n-1}$ , which is also $162\left(\frac19\right)^n$ .

(ii)

$r < 1$ , so use $S_n = \dfrac{a(1 - r^n)}{1 - r}$
$= \dfrac{18\left(1 - (\frac19)^n\right)}{\frac89}$ .
$\dfrac{18}{\frac89} = \dfrac{18 \times 9}{8}$
$= \dfrac{81}{4}$ , so $S_n = \frac{81}{4}\left(1 - \left(\frac19\right)^n\right)$ .

(b)(i)

Differentiate: $\dfrac{dy}{dx} = -2x + 1$ .
At $x = 2$ the gradient is $-4 + 1 = -3$ .
Tangent through $(2, -1)$ : $y + 1 = -3(x - 2)$ , which gives $y + 3x - 5 = 0$ .

(ii)

On the $x$ -axis, $y = 0$ : $3x = 5$ , so $x = \frac53$ .
The intercept is $\left(\frac53, 0\right)$ .
On the $y$ -axis, $x = 0$ : $y = 5$ .
The intercept is $(0, 5)$ .

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