A-Level Mathematics October/November 2024 Q2: The curve y = x² - a/x has a stationary point at (-3, b). Find the values of the consta…

1 answer

• ▲0
  accepted ✓

  To find the values of the constants $a$ and $b$, we need to use the information that the curve has a stationary point at $(-3, b)$.

  The equation of the curve is $y = x^2 - \frac{a}{x}$. First, we rewrite the equation in a form that is easier to differentiate: $y = x^2 - ax^{-1}$

  At a stationary point, the gradient of the curve, $\frac{dy}{dx}$, is equal to zero. We differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 2x - a(-1)x^{-2}$ $\frac{dy}{dx} = 2x + ax^{-2}$

  We are given that the stationary point occurs at $x = -3$. We set $\frac{dy}{dx} = 0$ and substitute $x = -3$: $2(-3) + a(-3)^{-2} = 0$ $-6 + \frac{a}{(-3)^2} = 0$ $-6 + \frac{a}{9} = 0$

  Now, we solve for $a$: $\frac{a}{9} = 6$ $a = 6 \times 9$ $a = 54$

  Now that we have the value of $a$, the equation of the curve is $y = x^2 - \frac{54}{x}$. The point $(-3, b)$ lies on this curve. Therefore, we can substitute $x = -3$ and $y = b$ into the equation to find $b$: $b = (-3)^2 - \frac{54}{-3}$ $b = 9 - (-18)$ $b = 9 + 18$ $b = 27$

  Thus, the values of the constants are $a = 54$ and $b = 27$.

  ---

  How the marks are awarded

  • B1 — The first mark is awarded for correctly differentiating the expression for $y$. Rewriting $y = x^2 - a/x$ as $y = x^2 - ax^{-1}$ is a helpful first step, leading to the derivative $\frac{dy}{dx} = 2x + ax^{-2}$.
  • M1 — The method mark is for using the information about the stationary point. This involves setting the derivative $\frac{dy}{dx}$ equal to zero and substituting the given x-coordinate, $x = -3$, to form an equation in terms of $a$, as shown in the step $2(-3) + a(-3)^{-2} = 0$.
  • A1 — The third mark is an accuracy mark for correctly solving the equation from the previous step to find the value of $a$. The calculation $-6 + a/9 = 0$ leads to $a = 54$.
  • A1 — The final accuracy mark is for finding the value of $b$. This is achieved by substituting the coordinates of the stationary point $(x=-3, y=b)$ and the value of $a=54$ into the original equation for the curve, leading to $b = 27$.

  Common mistakes

  • Incorrectly differentiating the $-a/x$ term. A common error is to get a sign wrong, resulting in $\frac{dy}{dx} = 2x - ax^{-2}$, which leads to an incorrect value for $a$.
  • Making an error when substituting $x=-3$ into the derivative, for example calculating $(-3)^2$ as $-9$ instead of $9$. This would lead to $-6 - a/9 = 0$ and $a=-54$.
  • Finding the value of $a$ correctly but then making a sign error when calculating $b$. For instance, calculating $b = (-3)^2 - \frac{54}{-3}$ as $9 - 18 = -9$ instead of the correct $9 - (-18) = 27$.
  • Confusing the conditions for the two unknowns, for example, by substituting $y=b$ into the derivative equation, or setting the original equation for $y$ to zero.

  Examiner tip: Remember that a given stationary point $(x, y)$ provides two key pieces of information: the coordinates satisfy the original equation $y=f(x)$, and the x-coordinate makes the derivative $f'(x)$ equal to zero.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.