NECO 2022 · Paper 2 · Q7

(a)(i)
Given the arithmetic progression (A.P.) $6, 9, 12, 15, 18, \ldots$ , find its 99th term.
(a)(ii)
Find its $n$ th term. $T_n =$
(b)
Differentiate $y = \sqrt{1 - x^3}$ with respect to $x$ . $\frac{dy}{dx} =$
(c)
A curve passes through the point $(-3, 5)$ and its gradient at any point is given by $6x^2 + 10x - 7$ . Find the equation of the curve. $y =$

Worked solution (try it first)

(a)(i)

The first term is $a = 6$ and the common difference is $d = 9 - 6 = 3$ .
The $n$ th term of an A.P. is $T_n = a + (n - 1)d$ , so $T_{99} = 6 + 98 \times 3$ .
Work it out: $T_{99} = 6 + 294 = 300$ .

(ii)

Substitute $a$ and $d$ : $T_n = 6 + (n - 1) \times 3$ .
Expand and simplify: $T_n = 3n + 3$ .

(b)

Write $y = (1 - x^3)^{1/2}$ and let $u = 1 - x^3$ , so $\frac{du}{dx} = -3x^2$ .
By the chain rule, $\frac{dy}{dx} = \frac12 u^{-1/2} \times \frac{du}{dx}$
$= \frac12(1 - x^3)^{-1/2} \times (-3x^2)$ .
Simplify: $\frac{dy}{dx} = \frac{-3x^2}{2\sqrt{1 - x^3}}$ .

(c)

Integrate the gradient: $y = \int (6x^2 + 10x - 7)\,dx$
$= 2x^3 + 5x^2 - 7x + c$ .
Put in the point $(-3, 5)$ : $5 = 2(-27) + 5(9) - 7(-3) + c$ .
Simplify: $5 = -54 + 45 + 21 + c = 12 + c$ , so $c = -7$ .
The curve is $y = 2x^3 + 5x^2 - 7x - 7$ .

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