Mathematics

1. 1.

   Differentiate y = x^2 e^(3x) with respect to x.

   [3 marks] · no calculator

   Answer explanation

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

   1. The function is a product of x^2 and e^(3x), so use d(uv)/dx = u'v + uv'.
   2. The derivative of e^(3x) needs the chain rule, giving the factor 3.

   Marking points

   • Uses the product rule with u = x^2 and v = e^(3x).
   • States u' = 2x and v' = 3e^(3x).
   • Obtains dy/dx = e^(3x)(2x + 3x^2) = x e^(3x)(2 + 3x).

   Examiner tip: The derivative of e^(kx) is k e^(kx). Forgetting the factor k is the usual slip.

2. 2.

   Find the gradient of the curve y = (3x^2 + 1)^4 at the point where x = 1.

   [3 marks] · no calculator

   Answer explanation

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

   1. Differentiate the outer power first, then multiply by the derivative of the inner function 3x^2 + 1.
   2. Substitute x = 1 only after differentiating.

   Marking points

   • Uses the chain rule: dy/dx = 4(3x^2 + 1)^3 * 6x.
   • Simplifies to 24x(3x^2 + 1)^3.
   • At x = 1 the gradient is 24 * 4^3 = 1536.

   Examiner tip: Substitute the value of x after differentiating, not before.

3. 3.

   Find the exact x-coordinates of the stationary points of y = (2x + 1)/(x^2 + 1).

   [4 marks] · no calculator

   Answer explanation

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

   1. Stationary points occur where dy/dx = 0, which for a fraction means the numerator is zero (denominator never zero here).
   2. The numerator quadratic does not factorise, so use the quadratic formula to keep exact surds.

   Marking points

   • Uses the quotient rule: dy/dx = [2(x^2 + 1) - (2x + 1)(2x)]/(x^2 + 1)^2.
   • Simplifies the numerator to -2x^2 - 2x + 2.
   • Sets the numerator to zero: x^2 + x - 1 = 0.
   • Solves: x = (-1 + sqrt(5))/2 or x = (-1 - sqrt(5))/2.

   Examiner tip: Only the numerator needs to be zero. Do not expand the squared denominator.

4. 4.

   The curve C has equation x^2 + xy + y^2 = 7 and passes through P(1, 2). Find the equation of the tangent to C at P in the form ax + by = c.

   [5 marks] · no calculator

   Answer explanation

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

   1. Differentiate every term with respect to x; terms in y get a factor dy/dx and xy needs the product rule.
   2. Collect the dy/dx terms, make dy/dx the subject, substitute (1, 2), then write the tangent in point-slope form and rearrange.

   Marking points

   • Checks P lies on C: 1 + 2 + 4 = 7.
   • Differentiates implicitly: 2x + y + x(dy/dx) + 2y(dy/dx) = 0.
   • Obtains dy/dx = -(2x + y)/(x + 2y).
   • At P the gradient is -4/5.
   • Tangent: y - 2 = -(4/5)(x - 1), so 4x + 5y = 14.

   Examiner tip: Remember the product rule for xy. A quick check: the point (1, 2) must satisfy your tangent equation.

5. 5.

   A closed cylinder has volume 500 cm^3. Find the radius that minimises its total surface area, and the minimum surface area, giving answers to 3 significant figures. Justify that it is a minimum.

   [6 marks]

   Answer explanation

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

   1. Eliminate h using the volume constraint so the surface area is a function of r only.
   2. Set dS/dr = 0 to find the stationary point and use the second derivative to confirm it is a minimum.
   3. At the optimum, h = 500/(pi r^2) is about 8.60 cm, which is twice r: a useful sanity check.

   Marking points

   • Uses V = pi r^2 h = 500 to write h = 500/(pi r^2).
   • Forms S = 2 pi r^2 + 2 pi r h = 2 pi r^2 + 1000/r.
   • Differentiates: dS/dr = 4 pi r - 1000/r^2.
   • Sets to zero: r^3 = 250/pi so r = 4.30 cm.
   • Shows d^2S/dr^2 = 4 pi + 2000/r^3 > 0, so it is a minimum.
   • Minimum surface area is 349 cm^2.

   Examiner tip: Always justify a minimum with the second derivative or a sign change. State both r and the minimum value if asked.

6. 6.

   Air is pumped into a spherical balloon at 100 cm^3 per second. Find (a) the rate at which the radius increases when r = 5 cm, (b) the rate at which the surface area increases at that instant. [V = 4/3 pi r^3, S = 4 pi r^2]

   [5 marks] · no calculator

   Answer explanation

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

   1. Link the known rate dV/dt to the wanted rate dr/dt with the chain rule: dV/dt = (dV/dr)(dr/dt).
   2. Reuse dr/dt in the second chain rule dS/dt = (dS/dr)(dr/dt).

   Marking points

   • Differentiates V: dV/dr = 4 pi r^2.
   • Uses the chain rule: dV/dt = 4 pi r^2 (dr/dt).
   • (a) dr/dt = 100/(4 pi * 25) = 1/pi, about 0.318 cm/s.
   • Differentiates S: dS/dt = 8 pi r (dr/dt).
   • (b) dS/dt = 8 pi * 5 * (1/pi) = 40 cm^2/s.

   Examiner tip: Differentiate with respect to t, not r, in related-rate problems. Substitute the instant's values only at the end.

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