Level 2 Certificate Further Mathematics Paper 1 (2019)

Why we do this: We need to find the equation of a straight line in the form \( y = mx + c \). To do this, we first need the gradient (slope), \( m \), and then the y-intercept, \( c \).

We are given two points: \( (x_1, y_1) = (-2, 11) \) and \( (x_2, y_2) = (1, 2) \).

Why we do this: The gradient tells us how steep the line is. It is the “change in y” divided by the “change in x”.

Why we do this: Now that we know \( y = -3x + c \), we can substitute the coordinates of one of the points to find \( c \).

Why we do this: To add algebraic fractions, the denominators must be the same. We need the Lowest Common Multiple (LCM) of \( 6a \) and \( 4 \).

Multiples of \( 6 \): 6, 12, 18…

Multiples of \( 4 \): 4, 8, 12…

The LCM of 6 and 4 is 12. Since there is an ‘a’, the common denominator is \( 12a \).

Why we do this: Now that the denominators are the same, we can add the numerators into a single fraction.

Why we do this: We want to solve for \( x \), so we move the constant term to the other side first.

Why we do this: To get \( x \) on its own. Crucial Rule: When you divide or multiply an inequality by a negative number, you must FLIP the inequality sign.

Why we do this: The question asks for the “smallest integer” (whole number) that is strictly greater than -3.6.

Integers greater than -3.6 are: -3, -2, -1, 0…

-4 is smaller than -3.6, so it’s not included.

Why we do this: To compare the two sides of the identity, we need to expand everything to see the coefficients of \( x \) and the constant terms clearly.

Why we do this: Since this is an identity (\( \equiv \)), the coefficient of \( x \) on the left must equal the coefficient of \( x \) on the right, and the constant term on the left must equal the constant term on the right.

Why we do this: We have two equations. Solve the first one for \( p \), then substitute into the second to find \( k \).

Why we do this: To get to the \( x \) inside, we first need to remove the outermost operation, which is the cube root. The inverse of cube root is cubing.

Why we do this: Now we rearrange the equation to get the term with \( \sqrt{x} \) on its own.

Why we do this: Finally, to find \( x \), we square both sides (the inverse of square root).

Why we do this: A matrix transformation multiplies the matrix by the coordinate vector (written as a column vector) to get the new coordinate vector.

Why we do this: Multiplying the rows of the matrix by the column vector gives us two simultaneous equations.

Top row: \( (2a \times 3) + (b \times 4) = 8 \)

Bottom row: \( (-b \times 3) + (-a \times 4) = -7 \)

Why we do this: We have two equations with two unknowns. We can use elimination or substitution.

Why we do this: The function is \( y = -2x \) in this interval.

Calculate points:

• At \( x = -1 \): \( y = -2(-1) = 2 \). Plot point \( (-1, 2) \).
• At \( x = 0 \): \( y = -2(0) = 0 \). Plot point \( (0, 0) \).

This is a straight line segment.

Why we do this: The function is \( y = x(4-x) \) or \( y = 4x – x^2 \) in this interval. This is a quadratic (parabola).

Calculate points:

• At \( x = 0 \): \( y = 0(4) = 0 \). (Joins previous section).
• At \( x = 1 \): \( y = 1(3) = 3 \). Plot \( (1, 3) \).
• At \( x = 2 \): \( y = 2(2) = 4 \). Plot \( (2, 4) \). (This is the turning point).
• At \( x = 3 \): \( y = 3(1) = 3 \). Plot \( (3, 3) \).

Draw a smooth curve through these points.

Why we do this: The function is \( y = 2x – 3 \) in this interval.

Calculate points:

• At \( x = 3 \): \( y = 2(3) – 3 = 3 \). (Joins previous section).
• At \( x = 4 \): \( y = 2(4) – 3 = 5 \). Plot point \( (4, 5) \).

This is a straight line segment.

Why we do this: The ratio \( AB : BC = 5 : 3 \) means the total length from \( A \) to \( C \) is split into \( 5 + 3 = 8 \) parts.

Point \( B \) is located \( \frac{5}{8} \) of the way along the line from \( A \) to \( C \).

Why we do this: We need to find the total difference between the coordinates of \( A \) and \( C \) first.

Why we do this: We need \( \frac{5}{8} \) of these changes to get from \( A \) to \( B \).

Why we do this: Add the displacements to the coordinates of \( A \) to find \( B \).

Why we do this: Before differentiating, it is much easier to have the expression in the form \( ax^n + bx^m \).

Why we do this: To find \( \frac{dy}{dx} \), multiply by the power and decrease the power by 1.

Why we do this: Always look for a common factor first to simplify the expression. Here, all coefficients (6, 26, 20) are even.

Why we do this: We need to factorise \( 3x^2 + 13xy – 10y^2 \). We can use the “ac” method.

Multiply \( a \) (3) by \( c \) (-10) = -30.

Find factors of -30 that add to give \( b \) (13).

Factors: 15 and -2.

Why we do this: Don’t forget the 2 we took out at the start!

Why we do this: We are given the Curved Surface Area (CSA) is \( 60\pi \). The formula for the CSA of a cone is \( \pi r l \).

Why we do this: We are given \( l = 3r \). Substituting this into our equation allows us to find \( r \).

Why we do this: The height \( h \), radius \( r \), and slant height \( l \) form a right-angled triangle where \( l \) is the hypotenuse.

Why we do this: We know \( r^2 = 20 \), so we can find \( h^2 \) and then \( h \).

Why we do this: The gradient of a curve is given by its derivative, \( \frac{dy}{dx} \).

Why we do this: Calculate the expression for the gradient at \( x = 4 \) and \( x = -1 \).

Why we do this: The question states the gradient at \( x = 4 \) is twice the gradient at \( x = -1 \).

Why we do this: To simplify the algebraic expression, we first expand the brackets.

Why we do this: Collect like terms to see what we are left with.

Why we do this: To prove an expression is always non-negative (\( \geqslant 0 \)), we try to write it as a perfect square. Any real number squared is positive or zero.

Why we do this: We must explicitly state why this proves the inequality.

Why we do this: We transform the unit vectors \( I(1,0) \) and \( J(0,1) \).

Rotation 90° Clockwise:

• \( (1, 0) \) moves to \( (0, -1) \)
• \( (0, 1) \) moves to \( (1, 0) \)

Why we do this: Reflection in \( y = x \) swaps the x and y coordinates.

• \( (1, 0) \) moves to \( (0, 1) \)
• \( (0, 1) \) moves to \( (1, 0) \)

Why we do this: “A followed by B” means we apply A first, then B. In matrix algebra, this is written as \( BA \) (right to left).

Why we do this: We need to find where \( \cos x \) is negative. Looking at the graph or using the CAST diagram:

• Cosine is positive in Quadrants 1 and 4.
• Cosine is negative in Quadrants 2 and 3 (\( 90^\circ \) to \( 270^\circ \)).

Why we do this: The value \( 0.8090 \) corresponds to \( 36^\circ \). Since we want \( -0.8090 \), we look for angles related to \( 36^\circ \) in the 2nd and 3rd quadrants.

Why we do this: To remove the surd from the denominator, we multiply the numerator and denominator by the conjugate of the denominator. The conjugate of \( 3 – \sqrt{5} \) is \( 3 + \sqrt{5} \).

Why we do this: We expand \( (21 – 11\sqrt{5})(3 + \sqrt{5}) \) using FOIL (First, Outside, Inside, Last).

Why we do this: The denominator is in the form \( (a-b)(a+b) = a^2 – b^2 \).

Why we do this: Divide each term in the numerator by the denominator.

Why we do this: The angle between a tangent and a chord (\( \angle ACD \)) is equal to the angle in the alternate segment (\( \angle ABC \)).

Why we do this: \( \triangle AOB \) is isosceles because \( OA \) and \( OB \) are radii. However, looking at the whole angle \( \angle ABC \), we can split it into \( \angle ABO \) and \( \angle OBC \).

We know \( \angle ABC = 3x \) and \( \angle OBC = 37^\circ \).

Why we do this: Since \( \triangle AOB \) is isosceles (\( OA = OB \)), the base angles are equal: \( \angle BAO = \angle ABO = 3x – 37 \).

The angles in a triangle sum to \( 180^\circ \).

Why we do this: Focus on the right-angled triangle \( ABF \). We know height \( FB = \sqrt{6} \) (side of square) and angle \( A = 60^\circ \).

Why we do this: Focus on the right-angled triangle \( DCE \). We know height \( CE = \sqrt{6} \) and angle \( D = 30^\circ \). \( DE \) is the hypotenuse.

Why we do this: The perimeter is \( AD + DE + EF + FA \). We need \( CD \) (to find \( AD \)) and \( FA \).

We know: \( AB = \sqrt{2} \), \( BC = \sqrt{6} \) (square), \( EF = \sqrt{6} \) (square), \( DE = 2\sqrt{6} \).

Why we do this: Substitute \( (x+1) \) and \( (2x) \) into the function definition.

Why we do this: To subtract the algebraic fractions, we need a common denominator. We can use \( 4x(x+1) \).

Why we do this: Simplify the numerator and then solve for x.