Edexcel AS Paper 1 (AS Paper 1) Specimen

The line \(l\) passes through the points \(A (3, 1)\) and \(B (4, -2)\). Find an equation for \(l\). [3]

The curve \(C\) has equation $$y = 2x^2 - 12x + 16$$ Find the gradient of the curve at the point \(P (5, 6)\). (Solutions based entirely on graphical or numerical methods are not acceptable.) [4]

Given that the point \(A\) has position vector \(3\mathbf{i} - 7\mathbf{j}\) and the point \(B\) has position vector \(8\mathbf{i} + 3\mathbf{j}\).

1. find the vector \(\overrightarrow{AB}\) [2]
2. Find \(|\overrightarrow{AB}|\). Give your answer as a simplified surd. [2]

$$f(x) = 4x^3 - 12x^2 + 2x - 6$$

1. Use the factor theorem to show that \((x - 3)\) is a factor of \(f(x)\). [2]
2. Hence show that \(3\) is the only real root of the equation \(f(x) = 0\) [4]

Given that $$f(x) = 2x + 3 + \frac{12}{x^2}, \quad x > 0$$ show that \(\int_1^{2\sqrt{2}} f(x)\,dx = 16 + 3\sqrt{2}\) [5]

Prove, from first principles, that the derivative of \(3x^2\) is \(6x\). [4]

1. Find the first \(3\) terms, in ascending powers of \(x\), of the binomial expansion of $$\left(2 - \frac{x}{2}\right)^7$$ giving each term in its simplest form. [4]
2. Explain how you would use your expansion to give an estimate for the value of \(1.995^7\) [1]

\includegraphics{figure_1} A triangular lawn is modelled by the triangle \(ABC\), shown in Figure 1. The length \(AB\) is to be \(30\text{m}\) long. Given that angle \(BAC = 70°\) and angle \(ABC = 60°\),

1. calculate the area of the lawn to \(3\) significant figures. [4]
2. Why is your answer unlikely to be accurate to the nearest square metre? [1]

Solve, for \(360° \leqslant x < 540°\), $$12\sin^2 x + 7\cos x - 13 = 0$$ Give your answers to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.) [5]

The equation \(kx^2 + 4kx + 3 = 0\), where \(k\) is a constant, has no real roots. Prove that $$0 \leqslant k < \frac{3}{4}$$ [4]

1. Prove that for all positive values of \(x\) and \(y\) $$\sqrt{xy} \leqslant \frac{x + y}{2}$$ [2]
2. Prove by counter example that this is not true when \(x\) and \(y\) are both negative. [1]

A student was asked to give the exact solution to the equation $$2^{2x+4} - 9(2^x) = 0$$ The student's attempt is shown below: $$2^{2x+4} - 9(2^x) = 0$$ $$2^{2x} + 2^4 - 9(2^x) = 0$$ Let \(2^x = y\) $$y^2 - 9y + 8 = 0$$ $$(y - 8)(y - 1) = 0$$ $$y = 8 \text{ or } y = 1$$ $$\text{So } x = 3 \text{ or } x = 0$$

1. Identify the two errors made by the student. [2]
2. Find the exact solution to the equation. [2]

1. Factorise completely \(x^3 + 10x^2 + 25x\) [2]
2. Sketch the curve with equation $$y = x^3 + 10x^2 + 25x$$ showing the coordinates of the points at which the curve cuts or touches the \(x\)-axis. [2]

The point with coordinates \((-3, 0)\) lies on the curve with equation $$y = (x + a)^3 + 10(x + a)^2 + 25(x + a)$$ where \(a\) is a constant.

1. Find the two possible values of \(a\). [3]

\includegraphics{figure_2} A town's population, \(P\), is modelled by the equation \(P = ab^t\), where \(a\) and \(b\) are constants and \(t\) is the number of years since the population was first recorded. The line \(l\) shown in Figure 2 illustrates the linear relationship between \(t\) and \(\log_{10} P\) for the population over a period of 100 years. The line \(l\) meets the vertical axis at \((0, 5)\) as shown. The gradient of \(l\) is \(\frac{1}{200}\).

1. Write down an equation for \(l\). [2]
2. Find the value of \(a\) and the value of \(b\). [4]
3. With reference to the model interpret
   1. the value of the constant \(a\),
   2. the value of the constant \(b\).
   [2]
4. Find
   1. the population predicted by the model when \(t = 100\), giving your answer to the nearest hundred thousand,
   2. the number of years it takes the population to reach \(200\,000\), according to the model.
   [3]
5. State two reasons why this may not be a realistic population model. [2]

\includegraphics{figure_3} The curve \(C_1\), shown in Figure 3, has equation \(y = 4x^2 - 6x + 4\). The point \(P\left(\frac{1}{2}, 2\right)\) lies on \(C_1\) The curve \(C_2\), also shown in Figure 3, has equation \(y = \frac{1}{2}x + \ln(2x)\). The normal to \(C_1\) at the point \(P\) meets \(C_2\) at the point \(Q\). Find the exact coordinates of \(Q\). (Solutions based entirely on graphical or numerical methods are not acceptable.) [8]

\includegraphics{figure_4} Figure 4 shows the plan view of the design for a swimming pool. The shape of this pool \(ABCDEA\) consists of a rectangular section \(ABDE\) joined to a semicircular section \(BCD\) as shown in Figure 4. Given that \(AE = 2x\) metres, \(ED = y\) metres and the area of the pool is \(250\text{m}^2\),

1. show that the perimeter, \(P\) metres, of the pool is given by $$P = 2x + \frac{250}{x} + \frac{\pi x}{2}$$ [4]
2. Explain why \(0 < x < \sqrt{\frac{500}{\pi}}\) [2]
3. Find the minimum perimeter of the pool, giving your answer to \(3\) significant figures. [4]

A circle \(C\) with centre at \((-2, 6)\) passes through the point \((10, 11)\).

1. Show that the circle \(C\) also passes through the point \((10, 1)\). [3]

The tangent to the circle \(C\) at the point \((10, 11)\) meets the \(y\) axis at the point \(P\) and the tangent to the circle \(C\) at the point \((10, 1)\) meets the \(y\) axis at the point \(Q\).

1. Show that the distance \(PQ\) is \(58\) explaining your method clearly. [7]