Mathematics 9709/31 — October/November 2023

Walkthrough

We need the gradient of the tangent, so we differentiate the given function. The function is a quotient, so we use the quotient rule.

Let the numerator be $u = x^2$ and the denominator be $v = 1 - 3x$. Then $u' = 2x$ and $v' = -3$. The quotient rule gives:

$$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$$

Substituting the expressions gives:

$$\frac{dy}{dx} = \frac{(1-3x)(2x) - x^2(-3)}{(1-3x)^2}$$

Expanding the numerator, we get $2x - 6x^2 + 3x^2 = 2x - 3x^2$. So the derivative is:

$$\frac{dy}{dx} = \frac{2x - 3x^2}{(1-3x)^2}$$

Now set this derivative equal to the given gradient, $8$:

$$\frac{2x - 3x^2}{(1-3x)^2} = 8$$

Multiply both sides by $(1-3x)^2$ to clear the denominator:

$$2x - 3x^2 = 8(1-3x)^2$$

Expand the right-hand side:

$$8(1 - 6x + 9x^2) = 8 - 48x + 72x^2$$

So:

$$2x - 3x^2 = 8 - 48x + 72x^2$$

Bring all terms to one side:

$$0 = 8 - 50x + 75x^2$$

or equivalently:

$$75x^2 - 50x + 8 = 0$$

Factorise this quadratic:

$$75x^2 - 50x + 8 = (15x - 4)(5x - 2)$$

Thus:

$$15x - 4 = 0 \quad \text{or} \quad 5x - 2 = 0$$

So:

$$x = \frac{4}{15} \quad \text{or} \quad x = \frac{2}{5}$$

Finally, substitute each $x$-value back into the original equation to find the corresponding $y$-coordinate.

For $x = \frac{2}{5}$:

$$y = \frac{\left(\frac{2}{5}\right)^2}{1 - 3\left(\frac{2}{5}\right)} = \frac{\frac{4}{25}}{1 - \frac{6}{5}} = \frac{\frac{4}{25}}{-\frac{1}{5}} = -\frac{4}{5}$$

For $x = \frac{4}{15}$:

$$y = \frac{\left(\frac{4}{15}\right)^2}{1 - 3\left(\frac{4}{15}\right)} = \frac{\frac{16}{225}}{1 - \frac{12}{15}} = \frac{\frac{16}{225}}{\frac{1}{5}} = \frac{16}{45}$$

Therefore the exact coordinates are $\left(\frac{2}{5}, -\frac{4}{5}\right)$ and $\left(\frac{4}{15}, \frac{16}{45}\right)$.

Key Takeaways

• The quotient rule is essential when differentiating a fraction of two functions.
• Setting the derivative equal to a given gradient turns the problem into solving an equation.
• After finding $x$, always substitute back into the original curve equation to find the corresponding $y$-coordinate.
• Exact fractional answers are required, not decimal approximations.

Common Mistakes

• Forgetting the minus sign in the derivative of $1-3x$, which is $-3$.
• Misapplying the quotient rule, especially the order of the terms in the numerator.
• Expanding $(1-3x)^2$ incorrectly.
• Solving the quadratic incorrectly or failing to factorise it.
• Stopping after finding $x$ and not computing the corresponding $y$-values.
• Giving decimal answers instead of exact fractions.

Things to Be Careful About

• The denominator $1-3x$ cannot be zero, so $x \neq \frac{1}{3}$. Neither solution equals $\frac{1}{3}$, so both are valid.
• When multiplying both sides by $(1-3x)^2$, this quantity is positive for all valid $x$, so no sign changes occur.
• The mark scheme requires exact values, so leave answers as fractions.
• Show the quotient rule clearly to earn the method mark; an unsupported derivative may not receive full credit.