A-Level Mathematics May/June 2024 Q11(b): The function g is defined by g(x) = 4x+k for x ∈ R where k is a constant. It is given t…

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  The function $f(x)$ is given as $f(x) = 10 + 6x - x^2$ from part (a). The function $g(x)$ is given by $g(x) = 4x+k$.

  First, find the inverse function $g^{-1}(x)$. Let $y = g(x)$, so $y = 4x+k$. To find the inverse, swap $x$ and $y$ and make $y$ the subject: $x = 4y+k$ $4y = x-k$ $y = \frac{1}{4}(x-k)$ So, $g^{-1}(x) = \frac{1}{4}(x-k)$.

  The graphs of $y = g^{-1}f(x)$ and $y = g(x)$ meet at a single point $P$. This means the equation $g^{-1}f(x) = g(x)$ has exactly one solution.

  First, find the composite function $g^{-1}f(x)$: $g^{-1}f(x) = g^{-1}(10+6x-x^2) = \frac{1}{4}(10+6x-x^2 - k)$.

  Now, set $g^{-1}f(x) = g(x)$: $\frac{1}{4}(10+6x-x^2 - k) = 4x+k$

  Multiply by 4 to clear the fraction: $10+6x-x^2 - k = 4(4x+k)$ $10+6x-x^2 - k = 16x+4k$

  Rearrange into the standard quadratic form $ax^2+bx+c=0$: $x^2 + 16x - 6x + 4k + k - 10 = 0$ $x^2 + 10x + (5k-10) = 0$

  For the equation to have exactly one solution (a single point of intersection), the discriminant must be zero: $b^2-4ac=0$. Here, $a=1$, $b=10$, and $c=5k-10$.

  $(10)^2 - 4(1)(5k-10) = 0$ $100 - 4(5k-10) = 0$ $100 - 20k + 40 = 0$ $140 - 20k = 0$ $20k = 140$ $k = 7$

  Now substitute $k=7$ back into the quadratic equation to find the x-coordinate of $P$: $x^2 + 10x + (5(7)-10) = 0$ $x^2 + 10x + (35-10) = 0$ $x^2 + 10x + 25 = 0$ $(x+5)^2 = 0$ $x = -5$

  To find the y-coordinate of $P$, substitute $x=-5$ into the equation for $y=g(x)$ (since $P$ lies on this line). $y = g(-5) = 4(-5) + k$ Using our found value $k=7$: $y = 4(-5) + 7$ $y = -20 + 7$ $y = -13$

  Therefore, the coordinates of $P$ are $(-5, -13)$.

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  How the marks are awarded

  • B1 — Correctly finding the inverse function $g^{-1}(x) = \frac{1}{4}(x-k)$.
  • M1 — Setting up the correct equation by equating the two functions: $g^{-1}f(x) = g(x)$, leading to $\frac{1}{4}(10+6x-x^2 - k) = 4x+k$.
  • M1 — Rearranging the equation into a quadratic in $x$ ($x^2 + 10x + 5k - 10 = 0$) and applying the discriminant condition $b^2-4ac=0$.
  • A1 — Correctly forming the equation for $k$ from the discriminant, $100-4(5k-10)=0$, and solving it to find $k=7$.
  • DM1 — Substituting the calculated value of $k$ back into the quadratic equation and solving to find the value of $x$. This mark is dependent on the previous M1 mark.
  • A1 — Using the value of $x$ to find the corresponding $y$-coordinate and stating the final coordinates of $P$ as $(-5, -13)$ only.

  Common mistakes

  • Incorrectly finding the inverse function, for example getting $g^{-1}(x) = \frac{x+k}{4}$ or $g^{-1}(x) = \frac{1}{4x+k}$.
  • Making an algebraic error when expanding brackets or rearranging the equation, such as forgetting to multiply the $k$ on the right side by 4, leading to an incorrect quadratic.
  • Not knowing that 'meets at a single point' implies that the discriminant of the resulting quadratic equation must be zero ($b^2-4ac=0$).
  • Stopping after finding the value of $k=7$ and not proceeding to find the coordinates of the point $P$ as required by the question.

  Examiner tip: Recognise that the number of intersection points between graphs corresponds to the number of real roots of the equation formed by setting them equal, which can be determined using the discriminant.

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