Edexcel IAL P1 June 2019 question paper and mark scheme (WMA11/01)

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1. Algebra · 7 marks

   Let $f(x) = 5 - 4x - 2x^2$.
   
   (a) Express $f(x)$ in the form $A - B(x+C)^2$, where $A$, $B$, and $C$ are integers.
   (b) State the maximum value of $f(x)$ and the value of $x$ at which it occurs.
   (c) Solve $f(x) = 0$, giving your answers in exact form.

   Worked answer

   (a) M1 Factors out $-2$ from the $x$ terms: $5 - 2(x^2 + 2x)$.

   1. M1 Completes the square inside the bracket: $5 - 2((x+1)^2 - 1)$.
   2. A1 Expands and simplifies to obtain $7 - 2(x+1)^2$.

   (b) B1 States the maximum value is $7$.

   1. B1 States this occurs at $x = -1$.

   (c) M1 Sets $7 - 2(x+1)^2 = 0$ and rearranges to $(x+1)^2 = \frac{7}{2}$.

   1. A1 Solves for $x$ by square rooting, yielding $x = -1 \pm \sqrt{\frac{7}{2}}$ (or equivalent exact surd).

2. Differentiation · 6 marks

   The normal to the curve $y = \sqrt{x}$ at the point where $x = 4$ cuts the x-axis at $A$ and the y-axis at $B$. Calculate the exact area of the triangle $AOB$, where $O$ is the origin.

   Worked answer

   1. M1 for finding the y-coordinate at $x=4$: $y=2$.
   2. M1 for finding $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ and evaluating at $x=4$ to get gradient $m_t = \frac{1}{4}$.
   3. M1 for finding the normal gradient $m_n = -4$ and its equation: $y - 2 = -4(x - 4) \implies y = -4x + 18$.
   4. M1 for finding the axis intercepts of the normal: at $y=0, 4x = 18 \implies x = 4.5$ (point $A$) and at $x=0, y=18$ (point $B$).
   5. M1 for using the area of a triangle formula $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.5 \times 18$.
   6. A1 for the final area of $40.5$ square units.

3. Integration · 6 marks

   Given that $f''(x) = 12x - 4$, $f'(1) = 2$ and $f(-1) = 5$, find $f(x)$.

   Worked answer

   1. M1 for integrating $f''(x)$ to find $f'(x) = 6x^2 - 4x + c_1$.
   2. M1 for substituting $f'(1) = 2$ to find $c_1$: $2 = 6(1)^2 - 4(1) + c_1 \implies c_1 = 0$.
   3. M1 for substituting $c_1 = 0$ and integrating again to find $f(x) = 2x^3 - 2x^2 + c_2$.
   4. M1 for substituting $f(-1) = 5$: $5 = 2(-1)^3 - 2(-1)^2 + c_2$.
   5. M1 for solving for $c_2$: $5 = -2 - 2 + c_2 \implies c_2 = 9$.
   6. A1 for the final correct function $f(x) = 2x^3 - 2x^2 + 9$.

More on these topics: Algebra · Differentiation · Integration

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• Algebra
• Coordinate Geometry
• Trigonometry
• Differentiation
• Integration
• Graphs and Transformations

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Formula booklet

Mathematical Formulae and Statistical Tables (International AS/A Level Mathematics) · Pearson Edexcel · For the 2018 IAL specification (current)