A-Level Mathematics October/November 2024 Q9(b): It is given that y = 9x+k is a tangent to the curve. Find the value of the constant k.

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  The equation of the curve is $y = 3x^3 - 6x^2 + 13x + \frac{20}{9}$. The equation of the tangent is $y = 9x + k$.

  At the point of tangency, the gradient of the curve is equal to the gradient of the tangent line. The gradient of the tangent line $y = 9x + k$ is 9.

  First, find the derivative of the curve's equation to get an expression for the gradient: $\frac{dy}{dx} = \frac{d}{dx} (3x^3 - 6x^2 + 13x + \frac{20}{9})$ $\frac{dy}{dx} = 9x^2 - 12x + 13$

  Now, set the gradient of the curve equal to the gradient of the tangent: $9x^2 - 12x + 13 = 9$ $9x^2 - 12x + 4 = 0$

  Solve this quadratic equation for $x$. This is a perfect square: $(3x - 2)^2 = 0$ $3x - 2 = 0$ $x = \frac{2}{3}$

  Now find the y-coordinate of the point of tangency by substituting $x = \frac{2}{3}$ into the curve's equation: $y = 3(\frac{2}{3})^3 - 6(\frac{2}{3})^2 + 13(\frac{2}{3}) + \frac{20}{9}$ $y = 3(\frac{8}{27}) - 6(\frac{4}{9}) + \frac{26}{3} + \frac{20}{9}$ $y = \frac{8}{9} - \frac{24}{9} + \frac{78}{9} + \frac{20}{9}$ $y = \frac{8 - 24 + 78 + 20}{9} = \frac{82}{9}$

  The point of tangency is $(\frac{2}{3}, \frac{82}{9})$.

  This point must lie on the tangent line $y = 9x + k$. Substitute the coordinates into the line's equation to find $k$: $\frac{82}{9} = 9(\frac{2}{3}) + k$ $\frac{82}{9} = 6 + k$ $k = \frac{82}{9} - 6$ $k = \frac{82}{9} - \frac{54}{9}$ $k = \frac{28}{9}$

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  How the marks are awarded

  • *M1 — The first method mark (*M1) is awarded for finding the derivative of the curve, equating it to the gradient of the tangent (9), and simplifying the result into a 3-term quadratic equation, $9x^2 - 12x + 4 = 0$.
  • A1 — The first accuracy mark (A1) is for correctly solving the quadratic equation to find the x-coordinate of the point of tangency, $x = 2/3$.
  • DM1 — This dependent method mark (DM1) is for a complete method to find k. This involves substituting the found x-value into the curve's equation to find the y-coordinate, and then substituting both the x and y coordinates into the tangent line's equation, $y=9x+k$.
  • A1 — The final accuracy mark (A1) is awarded for obtaining the correct value of $k = 28/9$ or an equivalent exact form.

  Common mistakes

  • Equating the curve's equation directly to the tangent's equation ($3x^3 - ... = 9x+k$) and then attempting to solve without using the derivative, which is not a valid method unless using the discriminant on the resulting cubic.
  • Making an arithmetic error when substituting the fraction $x=2/3$ into the curve's equation to find the y-coordinate, for example, by miscalculating powers or finding a common denominator.
  • After finding $x=2/3$, substituting this value into $y=9x+k$ to find the y-coordinate. This is incorrect as it creates an equation with two unknowns ($y$ and $k$) and cannot be solved.
  • Incorrectly solving the quadratic equation $9x^2 - 12x + 4 = 0$, perhaps by making a sign error in the quadratic formula or by incorrect factorization.

  Examiner tip: To find the point where a line is tangent to a curve, remember that at that specific point, the gradient of the curve (its derivative) is equal to the gradient of the line.

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