Edexcel A Level Pure Mathematics 1 (June 2023)

💡 What are we doing? We cannot integrate a product or quotient directly like this. We need to expand the brackets and divide each term to write the expression as a sum of individual powers of \(x\) (in the form \(ax^n\)).

The constant factor of \(\frac{1}{3}\) can be kept outside the terms to keep things tidy.

💡 Strategy: Use the power rule for integration: \(\int x^n \mathrm{d}x = \frac{x^{n+1}}{n+1}\). Increase the power by 1 and divide by the new power.

💡 Tip: Dividing by a fraction is the same as multiplying by its reciprocal. For example, dividing by \(\frac{5}{2}\) is multiplying by \(\frac{2}{5}\).

💡 Strategy: The Factor Theorem states that if \((x-a)\) is a factor of \(f(x)\), then \(f(a) = 0\). We substitute \(x = a\) into the function and set it to zero.

💡 Reasoning: We have the equation \(a(4a^2 + 5a – 6) = 0\). The question states \(a\) is a positive constant. This allows us to reject \(a=0\) and any negative solutions.

💡 Strategy: First, write out the full expression for \(f(x)\) using \(a = \frac{3}{4}\). Then set it equal to 3 and solve.

💡 Formula: The magnitude of a vector \(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) is \(\sqrt{x^2 + y^2 + z^2}\).

💡 Strategy: Find the magnitude of \(\vec{OB}\) in terms of \(a\), then set up the inequality \(|\vec{OB}| > \sqrt{38}\). Since both sides are positive, we can square both sides to remove the square roots.

💡 Knowledge: A stationary point occurs where the derivative \(f'(x) = 0\). The small angle approximation for \(\cos x\) (in radians) is \(\cos x \approx 1 – \frac{x^2}{2}\).

💡 Strategy: We need the gradient of the curve at \(P(0,3)\). The gradient is found by evaluating \(f'(x)\) at \(x=0\). Then use \(y – y_1 = m(x – x_1)\).

💡 Formula: Area \(\approx \frac{h}{2} [y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})]\).

First, identify \(h\) (the strip width) and separate the end ordinates from the middle ones.

\(h = 3.2 – 3 = 0.2\).

💡 Strategy: We have one equation from part (a): \(a + 2b = 51\). The second piece of information gives us the sum of all \(y\) values. This gives a second linear equation to solve simultaneously.

💡 Law of Logarithms: Power rule \(\log_n (x^k) = k \log_n x\).

Recall that \(\sqrt{x} = x^{0.5}\).

💡 Strategy: Factorise the expression inside the logarithm first. Then use the product rule: \(\log(xy) = \log x + \log y\).

💡 Strategy: Simplify the fraction inside the log to create a single term. Look for expressions that match our \(a\) and \(b\) definitions.

💡 Reasoning: Since \(x > 2\), the term \(\sqrt{x-2}\) takes all positive values greater than 0. Therefore, \(3 + \text{positive}\) must be greater than 3.

💡 Strategy: Let \(y = f(x)\), rearrange to make \(x\) the subject, then swap \(x\) and \(y\).

💡 Strategy: Work from the inside out. First calculate \(f(6)\), then put that result into \(g(x)\).

💡 Strategy: Substitute \(a^2+2\) into \(f(x)\) and equate it to \(g(a)\). Be careful with the square root simplification: \(\sqrt{a^2} = |a|\), but we usually check if \(a\) must be positive or negative based on the domains.

💡 Formula: Arc length \(l = r\theta\).

💡 Reasoning: AOG is a straight line, so the angles on it sum to \(\pi\) radians (180°). The diagram is symmetrical with congruent triangles, so angle \(AOB\) = angle \(FOG\).

\( \angle AOB + \angle BOC + \angle COE + \angle EOF + \angle FOG \) … wait. The diagram shows B, C, O are collinear? Let’s check.

Actually, the sector is “joined” to the triangles. Usually, this implies the boundaries match. Specifically, the line segment \(OC\) is part of the line segment \(OB\).

Angles on the line AOG: \(\angle AOB + \angle COE + \angle FOG = \pi\)? No, the sector is in the middle. The “gaps” are filled by the triangles.

The total angle \(\pi\) is made up of: \(\angle AOB + \angle COE + \angle FOG\). (Since C lies on OB and E lies on OF).

💡 Strategy: 1. Calculate length \(BC\). We know total front length and arc length. 2. Find total length \(OB\). 3. Calculate area of sector. 4. Calculate area of triangles using \(\frac{1}{2}ab\sin C\). 5. Sum them up.

💡 Strategy: For a geometric sequence, the common ratio \(r\) is constant. This means \(\frac{u_2}{u_1} = \frac{u_3}{u_2}\), or more simply \(u_2^2 = u_1 u_3\).

💡 Condition: A geometric series converges if and only if \(|r| < 1\). We need to solve for \(k\), find the ratio \(r\) for each solution, and choose the valid one.

💡 Strategy: Complete the square for both \(x\) and \(y\) to convert to \((x-a)^2 + (y-b)^2 = r^2\) form.

💡 Strategy: To find intersection points, solve the equations simultaneously. Substitute \(y = 2x – 1\) into the circle equation. Since there are “2 distinct points”, the resulting quadratic discriminant must be positive (\(b^2 – 4ac > 0\)).

💡 Strategy: Take logs of the model equation \(V = ab^t\) to match the graph axes.

This is in the form \(Y = c + mX\), where \(Y = \log_{10} V\) and \(X = t\).

The y-intercept is \(\log_{10} a\).

💡 Strategy: Find the gradient of the line to find \(b\).

Gradient \(m = \log_{10} b\).

💡 Strategy: Calculate the predicted value at \(t=24\) (2 years = 24 months) and compare it to the actual value (£320).

💡 Formula: First principles formula:

💡 Formula: \(\sin(A+B) = \sin A \cos B + \cos A \sin B\).

💡 Strategy: Group terms with \(\sin x\) and terms with \(\cos x\). Separate the fraction to match the standard limits given.

💡 Reasoning: The equation is a quadratic with a negative \(t^2\) term (inverted parabola). The maximum value occurs when the squared term is zero, i.e., at \(t=20\). The maximum height is \(a\).

💡 Strategy: Find \(\alpha\) and \(\beta\) using the max/min values.

Max value of cosine is 1, min is -1.

Max \(H = 29(1) + \beta = 29 + \beta\).

Min \(H = 29(-1) + \beta = -29 + \beta\).

Wait, we just need to fit it to the initial conditions. Max is 60, min is usually implied by amplitude or symmetry. Center line \(\beta\).

Max = 60. Amplitude = 29. So \(60 = 29 + \beta \implies \beta = 31\).

Now use \(t=0, H=2\) to find \(\alpha\).

💡 Reasoning: A roller coaster does loops. A quadratic model goes up and comes down once, then goes to negative infinity (crash!). A cosine model repeats (periodic).

💡 Strategy: Expand \((n+1)^3\) using the binomial expansion or simply multiplying out brackets.

💡 Reasoning: To prove a number is odd, we can write it in the form \(2k + 1\) (an even number plus 1). Alternatively, factorise terms to show they are even.

💡 Strategy: Use the quotient rule: \(y = \frac{u}{v} \implies y’ = \frac{v u’ – u v’}{v^2}\).

\(u = 7xe^x\) (use product rule for this).

\(v = (e^{3x} – 2)^{1/2}\) (use chain rule for this).

💡 Strategy: To show \(\alpha = 0.432\), check the function \(g(x) = x – \frac{2e^{3x}-4}{e^{3x}+4}\) or the numerator \(h(x) = e^{3x}(2-x) – 4x – 4\) at the bounds \(0.4315\) and \(0.4325\).