WAEC 2011 · Paper 2 · Q10

(a)
The gradient of a curve is given by $2x - 3x^2$ . Find the equation of the curve if the point $(1, 2)$ lies on it.
(b)
(i) Find the equations of the normals to the curve $y = x^2 - 1$ at the points where it cuts the $x$ -axis. (ii) Find the coordinates of the point of intersection of the normals in (b)(i).
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Integrate the gradient: $y = x^2 - x^3 + c$ .
$(1, 2)$ lies on the curve: $2 = 1 - 1 + c$ , so $c = 2$ and $y = x^2 - x^3 + 2$ .

(b)(i)

The curve cuts the $x$ -axis where $x^2 - 1 = 0$ : $x = -1$ or $x = 1$ .
The gradient is $\dfrac{dy}{dx} = 2x$ .
At $(-1, 0)$ the tangent's gradient is $-2$ , so the normal's is $\frac12$ : $y = \frac12(x + 1)$ , that is $x - 2y + 1 = 0$ .
At $(1, 0)$ the tangent's gradient is $2$ , so the normal's is $-\frac12$ : $y = -\frac12(x - 1)$ , that is $x + 2y - 1 = 0$ .

(ii)

Add the two equations: $2x = 0$ , so $x = 0$ .
Then $2y = 1$ , so $y = \frac12$ .
They meet at $\left(0, \frac12\right)$ .

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