Mathematics

1. 1.

   A curve has x = t^2 + 1 and y = 3t for t > 0. Find dy/dx and the tangent equation at t = 2.

   [3 marks] · no calculator

   Answer explanation

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

   1. Differentiate each coordinate with respect to the parameter and divide the rates.
   2. The tangent uses x and y coordinates computed from t; the parameter value 2 is not the x-coordinate.

   Marking points

   • dy/dx = (dy/dt)/(dx/dt) = 3/(2t).
   • At t = 2 the point is (5,6) and the gradient is 3/4.
   • The tangent is y - 6 = (3/4)(x - 5).

   Examiner tip: Do not divide x by y; divide their derivatives with respect to t.

2. 2.

   Differentiate xe^(2x). Then find its indefinite integral, showing integration by parts.

   [3 marks] · no calculator

   Answer explanation

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

   1. The chain rule supplies the factor 2 in differentiation, while reversing it supplies a factor 1/2 in integration.
   2. Apply uv - integral v du. Differentiating the resulting expression returns xe^(2x), which checks both coefficients.

   Marking points

   • Derivative = e^(2x)(1 + 2x).
   • Choose u = x and dv = e^(2x) dx, giving du = dx and v = e^(2x)/2.
   • Integral = e^(2x)(x/2 - 1/4) + C.

   Examiner tip: Integration by parts leaves a second integral; do not stop at uv.

3. 3.

   For x = t^2 and y = t^3 - 3t, with t > 0, find the horizontal tangent point and use d^2y/dx^2 to classify it.

   [4 marks] · no calculator

   Answer explanation

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

   1. Rewrite the first derivative as (3/2)(t - 1/t) before differentiating with respect to t.
   2. A derivative with respect to t is not yet a derivative with respect to x: divide again by 2t. The restriction t > 0 removes the other root.

   Marking points

   • dy/dx = 3(t^2 - 1)/(2t).
   • Horizontal tangent occurs at t = 1, giving (x,y) = (1,-2).
   • d^2y/dx^2 = [d/dt(dy/dx)]/(dx/dt) = (3/4)(1/t + 1/t^3).
   • At t = 1 the second derivative is 3/2 > 0, so the point is a local minimum.

   Examiner tip: For the second derivative, the extra division by dx/dt is essential.

4. 4.

   Evaluate the exact integral from 0 to 1 of 2x/(1 + x^2) dx by substitution.

   [4 marks] · no calculator

   Answer explanation

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

   1. The numerator is the derivative of the denominator, which suggests a logarithmic antiderivative.
   2. Changing limits with the variable avoids mixing x-limits with a u-expression. Since u stays positive, no sign ambiguity arises.

   Marking points

   • Use u = 1 + x^2 and du = 2x dx.
   • The transformed limits are u = 1 and u = 2.
   • Integral = integral from 1 to 2 of 1/u du = [ln u] from 1 to 2.
   • The exact value is ln 2.

   Examiner tip: Do not use the old limits 0 and 1 after changing the integration variable.

5. 5.

   Solve dy/dx = 2x(1 + y) with y(0) = 2. State the solution's real domain and find y(1) exactly.

   [5 marks] · no calculator

   Answer explanation

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

   1. The initial value has 1 + y = 3 > 0. The resulting exponential remains positive, so division by 1 + y is valid along this solution.
   2. Differentiating gives 6xe^(x^2), equal to 2x(1 + y). The constant solution y = -1 also satisfies the equation but not the initial condition.

   Marking points

   • Separate: dy/(1 + y) = 2x dx.
   • Integrate to ln|1 + y| = x^2 + C.
   • The initial condition fixes the positive branch and constant: 1 + y = 3e^(x^2).
   • y = 3e^(x^2) - 1, defined for all real x.
   • y(1) = 3e - 1.

   Examiner tip: Determine the constant from the initial condition before evaluating y(1).

6. 6.

   Use the trapezium rule with two equal strips to estimate the integral of 1/x from 1 to 3. Show that it overestimates the exact integral, using curvature, and give the error exactly.

   [5 marks] · no calculator

   Answer explanation

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

   1. Evaluate the function at 1, 2 and 3. The shared internal ordinate is counted twice in the rule.
   2. Convexity explains the direction of the error geometrically, while the antiderivative gives its exact amount (about 0.0681).

   Marking points

   • Strip width h = (3 - 1)/2 = 1.
   • Estimate = (1/2)[1 + 2(1/2) + 1/3] = 7/6.
   • The exact integral is [ln x] from 1 to 3 = ln 3.
   • f''(x) = 2/x^3 > 0 on [1,3]; the chords lie above the curve, so the estimate is too large.
   • Error = estimate minus exact value = 7/6 - ln 3.

   Examiner tip: A decreasing function need not give an overestimate; curvature controls this trapezium-rule error.

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