Level 2 Certificate FURTHER MATHEMATICS (8365/1) – Paper 1 Non-Calculator

💡 What are we being asked to find?

We need to find the value of \(x\) by setting up an equation that represents the given information: increasing \((x+1)\) by 20% makes it equal to \((x+6)\).

🌱 Why we do this:

An increase of 20% means multiplying the original amount by a multiplier of \(1 + 0.20 = 1.2\). This lets us translate the word problem into a single algebraic equation.

🧠 What this tells us:

We now have a simple linear equation that we can solve by collecting the \(x\) terms on one side and the constant terms on the other.

🌱 Why we do this:

We isolate \(x\) to find its value.

🧠 What this tells us:

The value of \(x\) is 24. We can check this by substituting \(x=24\) into the original statement. \((x+1) = 25\). Increased by 20%: \(25 \times 1.2 = 30\). \((x+6) = 24 + 6 = 30\). Since \(30 = 30\), the answer is correct.

💡 What are we being asked to find?

The line crosses the \(y\)-axis at the \(y\)-intercept. The equation of a straight line is \(y = mx + c\), where \(m\) is the gradient and \(c\) is the \(y\)-intercept. We need to find \(c\).

🌱 Why we do this:

By substituting the coordinates \((x, y) = (-6, -4)\) and the gradient \(m = \frac{3}{2}\) into the equation \(y = mx + c\), we create a simple equation with only \(c\) as the unknown, which we can then solve.

🧠 What this tells us:

The \(y\)-intercept, \(c\), is 5. Since the line crosses the \(y\)-axis when \(x=0\), the coordinates of the point are \((0, 5)\).

🌱 Why we do this:

The function is defined by three different rules over three domains. We must calculate the endpoints and key features (like the vertex for the quadratic) for each segment.

🧠 What this tells us:

The three segments connect smoothly at \((1, 3)\) and \((4, 0)\), forming a continuous graph.

🌱 Why we do this:

To find the inverse function, we first write \(g(x)\) as \(y\), then swap \(x\) and \(y\), and finally rearrange the equation to make \(y\) the subject.

🧠 What this tells us:

The inverse function is \(g^{-1}(x) = \frac{6-x}{3}\).

🌱 Why we do this:

We must use the known exact value of \(\tan 30^\circ\) and then square the result.

🧠 What this tells us:

The correct value is \(\frac{1}{3}\).

🌱 Why we do this:

The cosine graph starts at its maximum value at \(0^\circ\), reaches zero at \(90^\circ\), its minimum value at \(180^\circ\), zero again at \(270^\circ\), and returns to its maximum at \(360^\circ\). We must accurately plot these five points.

🧠 What this tells us:

The resulting curve must be smooth, passing through the correct points and showing the correct shape of a cosine wave.

💡 What are we being asked to find?

We are told that the two expressions are identical (an identity). Our strategy is to expand the brackets on the LHS and then compare the coefficients of \(x\) and the constant terms to find \(a\) and \(b\).

🌱 Why we do this:

We expand \((3x + a)(5x – 4)\) using the FOIL method to write it as a standard quadratic expression, ready for comparison.

🧠 What this tells us:

The expanded form is \(15x^2 + (5a – 12)x – 4a\). This must equal \(15x^2 – 2x + b\). The \(x^2\) term matches, so we can now compare the other terms.

🌱 Why we do this:

The coefficient of \(x\) on the LHS is \((-12 + 5a)\), and on the RHS it is \(-2\). Setting them equal allows us to solve for \(a\).

🌱 Why we do this:

The constant term on the LHS is \(-4a\), and on the RHS it is \(b\). We substitute the value \(a=2\) to find \(b\).

🧠 What this tells us:

The values are \(a=2\) and \(b=-8\). The identity is \((3x+2)(5x-4) = 15x^2 – 2x – 8\), which is correct.

🌱 Why we do this:

The easiest way to differentiate this expression is to expand the brackets and rewrite the term \(\frac{3}{x}\) using a negative power, so the expression is in the form \(y = Ax^n + Bx^m + \dots\). We can then use the rule \(\frac{d}{dx}(ax^n) = anx^{n-1}\).

🧠 What this tells us:

The simplified expression \(y = 2x^7 + 4x^4 – 6x^3\) is now ready for term-by-term differentiation.

🌱 Why we do this:

We apply the power rule, \(\frac{d}{dx}(ax^n) = anx^{n-1}\), to each term to find the derivative \(\frac{dy}{dx}\).

🧠 What this tells us:

The resulting expression is the gradient function of the original curve \(y\).

🌱 Why we do this:

We need to find the distance between the points \(B(-2, 5)\) and \(C\left(-1, 5\frac{3}{4}\right)\). We use the distance formula: \(d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\).

🧠 What this tells us:

We need to calculate the difference in the \(x\) and \(y\) coordinates before squaring them.

🌱 Why we do this:

We calculate the changes in \(x\) and \(y\): \(\Delta x = x_2 – x_1\) and \(\Delta y = y_2 – y_1\).

🧠 What this tells us:

The square of the length \(BC\) is \(\frac{25}{16}\). We now take the square root to find the final length.

🌱 Why we do this:

We take the square root of \(BC^2\) to find the length \(BC\).

🧠 What this tells us:

The length of \(BC\) is \(\frac{5}{4}\) units, or 1.25 units.

🌱 Why we do this:

To use matrix multiplication, we must first find the correct \(2 \times 2\) matrix for each transformation. We must also identify the final transformation’s matrix for comparison.

🧠 What this tells us:

The combined transformation must be \(\mathbf{F} \times \mathbf{R}\), as the rotation is “followed by” the reflection, meaning the rotation is done first (closest to the object/coordinate vector). The matrix product must equal \(\mathbf{X}\).

🌱 Why we do this:

We multiply the matrices in the correct order to find the single matrix that represents the combined transformation.

🧠 What this tells us:

The resulting matrix is \(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\).

🌱 Why we do this:

We compare the matrix obtained from the combined transformations with the matrix for a reflection in the \(x\)-axis to complete the proof.

🧠 What this tells us:

The process of matrix multiplication successfully showed that the two sequential transformations yield the same result as a single reflection in the \(x\)-axis.

🌱 Why we do this:

For a quadratic sequence, the \(n\)th term is \(an^2 + bn + c\). The second difference is always \(2a\).

🌱 Why we do this:

We know the \(n\)th term is \(T_n = 2n^2 + bn + c\). We can use simultaneous equations or compare the sequence to \(2n^2\).

🧠 What this tells us:

The expression \(T_n = 2n^2 – 5n + 1\) generates the given sequence. We can quickly check the first term: \(2(1)^2 – 5(1) + 1 = 2 – 5 + 1 = -2\). Correct.

🌱 Why we do this:

We need to find the number of terms (\(n\)) for which the value is less than 2000. We set up an inequality and solve it algebraically, as required by the question.

🌱 Why we do this:

The roots of \(n^2 + 10n – 2000 = 0\) tell us the boundaries for \(n\). We can solve this by factoring or using the quadratic formula. Since \(10\) is the difference between \(-40\) and \(50\), and their product is \(-2000\), factoring is the fastest method.

🧠 What this tells us:

The graph of \(y = n^2 + 10n – 2000\) is a U-shaped parabola crossing the \(n\)-axis at \(-50\) and \(40\). The expression \(n^2 + 10n – 2000\) is less than 0 (i.e., the term is less than 2000) when \(-50 < n < 40\).

🌱 Why we do this:

Since \(n\) must be a positive integer (term number), we use the condition \(-50 < n < 40\) to find the valid values of \(n\).

🌱 Why we do this:

To rationalise a denominator of the form \(a + \sqrt{b}\), we multiply the numerator and the denominator by the conjugate, which is \(a – \sqrt{b}\). The conjugate of \(3 + \sqrt{3}\) is \(3 – \sqrt{3}\).

🧠 What this tells us:

The next step is to perform the multiplication for both the numerator and the denominator.

🌱 Why we do this:

The multiplication \((a+b)(a-b) = a^2 – b^2\) simplifies the denominator to remove the surd.

🧠 What this tells us:

The denominator is now rational (the surd is removed). We can now simplify the fraction by dividing all terms by the common factor of 3.

🌱 Why we do this:

We divide every term in the numerator and the denominator by the common factor of 3 to simplify the expression fully.

🧠 What this tells us:

The final, fully simplified and rationalised expression is \(\frac{\sqrt{3} – 1}{2}\).

🌱 Why we do this:

We use the binomial expansion \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\) with \(n=5\), \(a=3\), and \(b=2x\). We first need the coefficients for \(n=5\), which are the numbers in the 5th row of Pascal’s triangle.

🧠 What this tells us:

We will have 6 terms in the expansion, and we must apply the correct powers of \(a=3\) and \(b=2x\) to each coefficient.

🌱 Why we do this:

We systematically write out the 6 terms, ensuring the powers of \(3\) decrease from 5 to 0 and the powers of \((2x)\) increase from 0 to 5. Crucially, we treat \(2x\) as a single term \((2x)\) when applying the power.

🧠 What this tells us:

We now calculate the value of each term by applying the powers and multiplying the coefficients.

🌱 Why we do this:

We perform the arithmetic, noting that \((2x)^k = 2^k x^k\).

🧠 What this tells us:

We have now fully expanded and simplified the expression.

🌱 Why we do this:

We set the \(n\)th term expression equal to the given term value and solve for \(n\). Cross-multiplication is the algebraic method to clear the denominators.

🧠 What this tells us:

The result is a simple quadratic equation in terms of \(n^2\).

🌱 Why we do this:

We rearrange the equation to isolate \(n^2\) and then take the square root to find \(n\).

🌱 Why we do this:

As \(n \to \infty\), the constant term \(+2\) in the denominator becomes insignificant compared to \(n^2\). A formal way to find the limit of an algebraic fraction is to divide the numerator and the denominator by the highest power of \(n\), which is \(n^2\).

🧠 What this tells us:

For large \(n\), the terms of the sequence will get closer and closer to 3.

🌱 Why we do this:

The expression involves two terms being divided by a single term. We can simplify this by splitting the fraction into two separate terms, each divided by \(3x^2y^{-3}\).

🧠 What this tells us:

We now have two simpler fractions to simplify using the rules of indices: \(a^m \div a^n = a^{m-n}\) and constant division.

🌱 Why we do this:

We simplify the coefficients, and subtract the powers for the \(x\) and \(y\) terms.

🌱 Why we do this:

We repeat the process for the second term.

🧠 What this tells us:

The full simplified expression is the sum of the two simplified terms.

🌱 Why we do this:

To begin rearranging for \(e\), we must first clear the fraction by multiplying both sides of the equation by 3.

🧠 What this tells us:

The equation is now linear, but the subject variable \(e\) appears in two separate terms on opposite sides.

🌱 Why we do this:

We move all terms containing \(e\) to one side (the left side) and all other terms (constants and terms not containing \(e\)) to the other side.

🧠 What this tells us:

The \(e\) terms are now collected. We can now factor out \(e\) using the reverse of the distributive property.

🌱 Why we do this:

Factorising \(e\) allows us to have a single instance of \(e\) multiplied by a bracket, which we can then divide by to make \(e\) the subject.

🧠 What this tells us:

The variable \(e\) is now the subject of the formula.

🌱 Why we do this:

The radius to the point of tangency is perpendicular to the tangent. This establishes two right angles in the quadrilateral \(ABPC\). This is a key geometric relationship for this proof.

🌱 Why we do this:

The sum of interior angles in the quadrilateral \(ABPC\) is \(360^\circ\). We can use this to express the angle \(\angle BPC\) (let’s call it the minor angle for now) in terms of \(x\).

🌱 Why we do this:

The angle at the centre is twice the angle at the circumference subtended by the same arc. The reflex angle \(\angle BPC\) subtends the major arc \(BDC\), and \(\angle BDC\) (angle \(y\)) is the angle at the circumference.

🌱 Why we do this:

We simplify the fraction to match the required form \(y = 90 + \frac{x}{2}\).

🌱 Why we do this:

A full proof requires justification of the geometric steps. The main theorems used are the Tangent-Radius theorem, Angles in a Quadrilateral, and Angle at Centre vs. Angle at Circumference.

🌱 Why we do this:

We rearrange the linear equation (\(x – y = \frac{19}{4}\)) to express one variable in terms of the other, which allows us to substitute it into the quadratic equation (\(xy = -3\)). It’s easiest to make \(y\) the subject of the linear equation.

🧠 What this tells us:

The system is now reduced to a single quadratic equation in terms of \(x\), which we need to solve.

🌱 Why we do this:

We expand the brackets and clear the fraction to get a quadratic equation in the form \(ax^2 + bx + c = 0\) with integer coefficients, which is easier to solve.

🧠 What this tells us:

We have a quadratic equation we can now solve by factoring or using the quadratic formula.

🌱 Why we do this:

We find the two possible values of \(x\).

🌱 Why we do this:

We use the simpler quadratic equation \(xy = -3\) (or the rearranged linear equation) to find the \(y\) value for each \(x\) value.

🧠 What this tells us:

The solutions are the pairs \((4, -\frac{3}{4})\) and \((\frac{3}{4}, -4)\).

🌱 Why we do this:

The equation \(x^2 + y^2 = 16\) tells us the radius \(r\) is \(\sqrt{16} = 4\). The line \(OP\) forms a right-angled triangle with the axes, so we can find the coordinates \((x, y)\) using \(x = r \cos \theta\) and \(y = r \sin \theta\).

🧠 What this tells us:

The exact coordinates of \(P\) are confirmed to be \((2, 2\sqrt{3})\).

🌱 Why we do this:

The tangent is perpendicular to the radius at the point of contact, \(P\). We first find the gradient of the radius \(OP\) using the formula \(m = \frac{y_2 – y_1}{x_2 – x_1}\).

🌱 Why we do this:

The gradient of the tangent \(m_T\) is the negative reciprocal of the radius gradient \(m_{OP}\), so \(m_T = -\frac{1}{m_{OP}}\).

🌱 Why we do this:

We use the point-gradient form, \(y – y_1 = m(x – x_1)\), with \(m = -\frac{1}{\sqrt{3}}\) and \((x_1, y_1) = (2, 2\sqrt{3})\), and then rearrange it into the required form \(x + ay = b\).

🧠 What this tells us:

The equation of the tangent is \(x + \sqrt{3}y = 8\), which is in the required form with \(a = \sqrt{3}\) and \(b = 8\).

🌱 Why we do this:

We are given a side ratio (\(RS:ST\)) and two opposite angles (\(\angle T = \theta\) and \(\angle R = 135^\circ\)). The Sine Rule relates the ratio of a side length to the sine of its opposite angle: \(\frac{a}{\sin A} = \frac{b}{\sin B}\).

🧠 What this tells us:

We can rearrange this equation to solve for \(\sin \theta\): \(\sin \theta = \sin 135^\circ \times \frac{RS}{ST}\).

🌱 Why we do this:

We use the given ratio \(RS:ST = 1:4\), which means \(\frac{RS}{ST} = \frac{1}{4}\). We also recall the exact value for \(\sin 135^\circ\).

🧠 What this tells us:

The final calculation is simple multiplication.

🌱 Why we do this:

We multiply the two fractions to get the exact value of \(\sin \theta\).

🌱 Why we do this:

To put a quadratic expression into the completed square form \(a(x + b)^2 + c\), we first factor out the coefficient of \(x^2\), which is \(a=6\), from the terms involving \(x\).

🧠 What this tells us:

We now complete the square on the simpler quadratic expression inside the bracket, \(x^2 – 4x\).

🌱 Why we do this:

We write \(x^2 – 4x\) as a perfect square, \((x – 2)^2\), and adjust for the constant term introduced by squaring.

🌱 Why we do this:

We expand the outer bracket to combine the constant terms and achieve the final required form \(a(x + b)^2 + c\).

🌱 Why we do this:

Stationary points occur where the gradient is zero. We use differentiation to find the gradient function \(\frac{dy}{dx}\) and set it equal to 0 to find the \(x\)-coordinates.

🧠 What this tells us:

We have a cubic equation to solve for \(x\), which should yield three roots (the \(x\)-coordinates of the stationary points).

🌱 Why we do this:

We factorise the expression to find the roots, and then substitute each \(x\) value back into the original equation, \(y = x^4 – 18x^2\), to find the corresponding \(y\)-coordinate.

🧠 What this tells us:

The three stationary points are \((0, 0)\), \((3, -81)\), and \((-3, -81)\).

🌱 Why we do this:

The second derivative test determines the nature of each stationary point: if \(\frac{d^2y}{dx^2} > 0\) it’s a minimum, and if \(\frac{d^2y}{dx^2} < 0\) it's a maximum.

🌱 Why we do this:

We substitute each \(x\)-coordinate into \(\frac{d^2y}{dx^2}\) to find the nature.

🌱 Why we do this:

We start with the Left-Hand Side (LHS) and aim to simplify it to the Right-Hand Side (RHS), which is \(-\tan^2 x\). The core Pythagorean identity, \(\sin^2 x + \cos^2 x = 1\), allows us to rewrite the numerator in a simpler form, ideally in terms of a single trigonometric function.

🧠 What this tells us:

The numerator is now in a form where the terms can be combined.

🌱 Why we do this:

We expand the brackets in the numerator and simplify by collecting \(\cos^2 x\) and \(\sin^2 x\) terms.

🧠 What this tells us:

The expression is now in terms of a familiar quotient identity.

🌱 Why we do this:

We use the quotient identity \(\tan x = \frac{\sin x}{\cos x}\), which means \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\), to match the RHS.