Mathematics

1. 1.

   Solve |2x - 3| < 7.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A modulus inequality |A| < k means A lies between -k and k.
   2. Solve the double inequality in one go, performing each operation on all three parts.

   Marking points

   • Writes -7 < 2x - 3 < 7.
   • Adds 3 to all parts: -4 < 2x < 10.
   • Divides by 2: -2 < x < 5.

   Examiner tip: |A| < k gives one bounded interval; |A| > k gives two separate regions. Do not mix them up.

2. 2.

   Given f(x) = 3x + 2 and g(x) = x^2 - 1, find fg(x) and gf(x), and show that fg and gf are not the same function.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. fg(x) means apply g first, then f. gf(x) means apply f first, then g.
   2. Substitute carefully, then compare at one value to show they are different.

   Marking points

   • fg(x) = 3(x^2 - 1) + 2 = 3x^2 - 1.
   • gf(x) = (3x + 2)^2 - 1 = 9x^2 + 12x + 3.
   • They differ, for example at x = 1: fg(1) = 2 but gf(1) = 24.

   Examiner tip: The order matters: fg(x) = f(g(x)). Reversing the order is the most common error.

3. 3.

   Express (5x + 1) / ((x - 1)(x + 2)) in partial fractions.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Set up the form with one constant over each linear factor.
   2. Multiply through by the denominator and substitute the roots x = 1 and x = -2 to isolate each constant.
   3. Check by recombining: 2(x + 2) + 3(x - 1) = 5x + 1.

   Marking points

   • Writes A/(x - 1) + B/(x + 2).
   • Obtains 5x + 1 = A(x + 2) + B(x - 1).
   • Finds A = 2 (putting x = 1) and B = 3 (putting x = -2).
   • States 2/(x - 1) + 3/(x + 2).

   Examiner tip: Always check by recombining the fractions. Substituting the roots is much faster than equating coefficients.

4. 4.

   The function f is defined by f(x) = (2x + 1)/(x - 3), x != 3. Find f^-1(x) and state its domain.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. To invert, swap the roles of x and y by solving for x in terms of y.
   2. The value 2 is excluded because f(x) approaches 2 as x grows large but never reaches it.

   Marking points

   • Sets y = (2x + 1)/(x - 3) and rearranges to yx - 3y = 2x + 1.
   • Collects x terms: x(y - 2) = 3y + 1.
   • States f^-1(x) = (3x + 1)/(x - 2).
   • The domain of f^-1 is the range of f, so x != 2.

   Examiner tip: The domain of an inverse equals the range of the original function, not its domain.

5. 5.

   The curve y = f(x) has a maximum point at (2, 5). Write down the coordinates of the corresponding turning point on each of: (a) y = 2f(x - 3), (b) y = f(2x) + 4, (c) y = -f(x + 1). State the nature of the turning point in (c).

   [5 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Transformations inside the bracket act on x and work 'backwards': f(x - 3) moves right by 3; f(2x) halves x-values.
   2. Transformations outside act on y and work as written: 2f scales y by 2, +4 moves up, the minus sign reflects in the x-axis.

   Marking points

   • (a) The point is (5, 10): translation 3 right and stretch 2 in y.
   • (b) The x-coordinate is 1 (stretch by 1/2 in x).
   • (b) The y-coordinate is 9 (translation 4 up).
   • (c) The point is (1, -5): translation 1 left and reflection in the x-axis.
   • (c) It is a minimum because the reflection turns the maximum upside down.

   Examiner tip: Do the x-changes and y-changes separately. Inside the bracket, the effect on x is the opposite of what the sign suggests.

6. 6.

   Solve 2^(2x+1) - 9(2^x) + 4 = 0, giving exact values of x.

   [5 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Spot the hidden quadratic: 2^(2x+1) = 2 * (2^x)^2, so the equation is quadratic in u = 2^x.
   2. Solve for u, then convert back. Both u values are positive, so both give valid x.

   Marking points

   • Writes 2^(2x+1) = 2(2^x)^2.
   • With u = 2^x obtains 2u^2 - 9u + 4 = 0.
   • Factorises (2u - 1)(u - 4) = 0 so u = 1/2 or u = 4.
   • 2^x = 4 gives x = 2.
   • 2^x = 1/2 gives x = -1.

   Examiner tip: After substituting u = 2^x, reject any u <= 0 because 2^x is always positive. Here both roots survive.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.