Tangents, normals and stationary points - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 19 exam-style questions on Tangents, normals and stationary points for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

19 questions
3–11 marks each
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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

4 marks

A curve has equation \(y = 3\sqrt{x} - \dfrac{8}{x}\), \(x > 0\).
Find an equation of the tangent to the curve at the point where \(x = 4\). Write it in the form \(y = mx + c\), where \(m\) and \(c\) are constants to be found.

Model answer

When \(x = 4\): \(y = 3 \times 2 - \dfrac{8}{4} = 4\)
\(y = 3x^{\frac{1}{2}} - 8x^{-1}\)
\(\dfrac{dy}{dx} = \dfrac{3}{2}x^{-\frac{1}{2}} + 8x^{-2}\)
At \(x = 4\): \(\dfrac{dy}{dx} = \dfrac{3}{2} \times \dfrac{1}{2} + \dfrac{8}{16} = \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{5}{4}\)
Tangent: \(y - 4 = \dfrac{5}{4}(x - 4)\)
\(y = \dfrac{5}{4}x - 1\)

Explanation

Find the \(y\)-coordinate of the point, differentiate and substitute \(x = 4\) to get the gradient of the tangent, then use \(y - y_1 = m(x - x_1)\). The final mark needs the full equation in the form \(y = mx + c\), not just the values of \(m\) and \(c\). Common errors are the sign when differentiating \(-8x^{-1}\) and substituting \(x = 4\) into \(\dfrac{dy}{dx}\) to find the \(y\)-coordinate.

Mark scheme

1 mark
M1 Differentiates: \(x^{\frac{1}{2}} \to x^{-\frac{1}{2}}\) or \(x^{-1} \to x^{-2}\); or a correct final answer of \(y = \dfrac{5}{4}x - 1\) with no working
An answer of \(y = \dfrac{5}{4}x - 1\) with no working scores 4 marks.
1 mark
A1 \(\dfrac{dy}{dx} = \dfrac{3}{2}x^{-\frac{1}{2}} + 8x^{-2}\), giving a gradient of \(\dfrac{5}{4}\) at \(x = 4\)
Accept 1.25. A common error is \(-\dfrac{8}{x} \to -\dfrac{8}{x^2}\) (sign error).
1 mark
M1 (dM1) Finds \(y = 4\) at \(x = 4\) and uses their gradient with \((4, 4)\) in a straight-line equation: \(y - 4 = m(x - 4)\), or \(y = mx + c\) with \(c\) found
Depends on the first M mark. The gradient must come from substituting \(x = 4\) into their derivative; using the gradient of the normal is M0.
1 mark
A1 \(y = \dfrac{5}{4}x - 1\)
Accept \(y = 1.25x - 1\). Stating just \(m = \dfrac{5}{4}\) and \(c = -1\) without the equation is A0. \(5x - 4y - 4 = 0\) on its own is A0 (not in the form asked for) unless rearranged correctly.

Question 2

5 marks

The curve \(C\) has equation
\[y = 8\sqrt{x} - x, \quad x > 0\]
The tangent to \(C\) at the point \(P(4, 12)\) meets the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).

Find the area of the triangle \(OAB\), where \(O\) is the origin.

Model answer

\(\dfrac{dy}{dx} = 4x^{-\frac{1}{2}} - 1\)
At \(x = 4\): \(\dfrac{dy}{dx} = \dfrac{4}{2} - 1 = 1\)
Tangent: \(y - 12 = x - 4 \Rightarrow y = x + 8\)
\(y = 0\): \(x = -8\), so \(A(-8, 0)\); \(x = 0\): \(y = 8\), so \(B(0, 8)\)
Area of \(OAB = \dfrac{1}{2} \times 8 \times 8 = 32\)

Explanation

Find the gradient at \(P\), write the equation of the tangent, then put \(y = 0\) and \(x = 0\) to find \(A\) and \(B\). The triangle \(OAB\) is right-angled at \(O\), so its area is half the product of the two intercept lengths. Common errors are differentiating \(8\sqrt{x}\) as \(4\sqrt{x}\) and using a negative length for \(OA\).

Mark scheme

1 mark
M1 Differentiates: \(x^{\frac{1}{2}} \to x^{-\frac{1}{2}}\). A correct final answer of 32 with no working scores 1 mark only (SC)
An answer of 32 with no working scores 1 mark (SC: the final A1). This is a multi-step problem: the gradient, the tangent and its intercepts must be seen for the other marks. An answer of 32 following the correct tangent \(y = x + 8\) scores full marks.
1 mark
A1 \(\dfrac{dy}{dx} = 4x^{-\frac{1}{2}} - 1\), giving a gradient of 1 at \(P\)
1 mark
M1 (dM1) Forms the equation of the tangent at \(P\) using their gradient: \(y - 12 = 1(x - 4)\)
Depends on the first M mark.
1 mark
M1 Finds where their tangent meets both axes
Their tangent must be a straight line with non-zero gradient. \(y = x + 8\) gives \(A(-8, 0)\) and \(B(0, 8)\).
1 mark
A1 Area \(= 32\)
Units not needed. A negative area (\(-32\)) is A0 unless corrected.

Question 3

7 marks

The curve \(C\) has equation
\[y = x^2 - \dfrac{k}{x}, \quad x ≠ 0\]
where \(k\) is a constant.
Given that \(C\) has a stationary point at \(x = -2\), find the value of \(k\), and find the coordinates of this stationary point and determine its nature.

Model answer

\(y = x^2 - kx^{-1}\)
\(\dfrac{dy}{dx} = 2x + kx^{-2}\)
At \(x = -2\): \(-4 + \dfrac{k}{4} = 0 \Rightarrow k = 16\)
\(y = (-2)^2 - \dfrac{16}{-2} = 4 + 8 = 12\), so the stationary point is \((-2, 12)\)
\(\dfrac{d^2y}{dx^2} = 2 - 2kx^{-3} = 2 - \dfrac{32}{x^3}\)
At \(x = -2\): \(\dfrac{d^2y}{dx^2} = 2 - \dfrac{32}{-8} = 6 > 0\), so \((-2, 12)\) is a minimum.

Explanation

At a stationary point \(\dfrac{dy}{dx} = 0\), so substituting \(x = -2\) into the derivative gives an equation for \(k\); then find \(y\) and use the second derivative for the nature. The marks are for each derivative, the value of \(k\), the coordinates and the conclusion with a reason. Common errors are the sign when differentiating \(-kx^{-1}\) and the sign of \(x^{-3}\) at \(x = -2\).

Mark scheme

1 mark
M1 Differentiates: \(x^2 \to 2x\) or \(x^{-1} \to x^{-2}\). Correct final answers of \(k = 16\) and a minimum at \((-2, 12)\) with no working score 2 marks only (SC)
Correct answers of \(k = 16\) and a minimum at \((-2, 12)\) with no working score 2 marks (SC: the A marks for \(k = 16\) and for \((-2, 12)\)). This is a multi-step problem: the derivative, the equation for \(k\) and a reason for the nature must be seen for the other marks.
1 mark
A1 \(\dfrac{dy}{dx} = 2x + kx^{-2}\)
A common error is \(-\dfrac{k}{x} \to -\dfrac{k}{x^2}\).
1 mark
M1 Substitutes \(x = -2\) into their \(\dfrac{dy}{dx}\) and sets it equal to 0
Depends on the first M mark.
1 mark
A1 \(k = 16\)
1 mark
A1 Coordinates \((-2, 12)\)
Follows from \(k = 16\): \(y = 4 - \dfrac{16}{-2} = 12\).
1 mark
M1 Finds \(\dfrac{d^2y}{dx^2} = 2 - 2kx^{-3}\) (ft their \(\dfrac{dy}{dx}\)) and evaluates it at \(x = -2\) with their \(k\), or checks the sign of \(\dfrac{dy}{dx}\) either side
1 mark
A1 \(\dfrac{d^2y}{dx^2} = 6 > 0\), so \((-2, 12)\) is a minimum
Needs the correct value (or sign) of the second derivative and the conclusion 'minimum'.

Question 4

10 marks

A builders' merchant sells 18 000 bags of cement each year, at a steady rate. The merchant orders the bags in batches of \(x\) bags. Each delivery of a batch costs £50, whatever the size of the batch. Keeping one bag in stock for a year costs £5, and the number of bags in stock falls steadily from \(x\), when a batch arrives, to 0, when the next batch arrives.

The total annual cost, £\(C\), of the deliveries and of keeping the stock is modelled by
\[C = \dfrac{900\,000}{x} + 2.5x, \quad x > 0\]

Use calculus to find the value of \(x\) for which \(C\) is a minimum, justifying that it gives a minimum. Find also the minimum annual cost.

Explain how each of the two terms of the model arises.

The supplier will deliver only batches whose size is a multiple of 250 bags. Find the batch size that the merchant should order, giving a reason for your answer.

Model answer

\(C = 900\,000x^{-1} + 2.5x\)
\(\dfrac{dC}{dx} = -900\,000x^{-2} + 2.5\)
\(\dfrac{dC}{dx} = 0 \Rightarrow x^2 = \dfrac{900\,000}{2.5} = 360\,000 \Rightarrow x = 600\) (since \(x > 0\))
\(\dfrac{d^2C}{dx^2} = 1\,800\,000x^{-3} = \dfrac{1\,800\,000}{600^3} = \dfrac{1}{120} > 0\), so this gives a minimum.
Minimum annual cost \(= \dfrac{900\,000}{600} + 2.5(600) = 1500 + 1500 =\) £3000
\(\dfrac{900\,000}{x}\): there are \(\dfrac{18\,000}{x}\) deliveries each year at £50 each, costing \(\dfrac{50 \times 18\,000}{x} = \dfrac{900\,000}{x}\).
\(2.5x\): the stock falls steadily from \(x\) to 0, so on average there are \(\dfrac{x}{2}\) bags in stock, costing \(5 \times \dfrac{x}{2} = 2.5x\) a year.
\(C(500) = 1800 + 1250 = 3050\) and \(C(750) = 1200 + 1875 = 3075\)
So the merchant should order batches of 500 bags, which is cheaper.

Explanation

Differentiating \(900\,000x^{-1} + 2.5x\) and setting the derivative to zero gives \(x^2 = 360\,000\), so \(x = 600\), and the positive second derivative confirms a minimum cost of £3000. The first term is the number of deliveries, \(\dfrac{18\,000}{x}\), times £50, and the second is the average stock \(\dfrac{x}{2}\) times £5. When only multiples of 250 are allowed, the costs at 500 and 750 must both be calculated, because the cost curve is not symmetrical about \(x = 600\); a common error is to choose the nearer value without checking.

Mark scheme

1 mark
M1 Differentiates: \(\dfrac{dC}{dx} = \pm 900\,000x^{-2} + 2.5\) (at least one power reduced by 1). An answer of \(x = 600\) and £3000 with no working scores 0 marks for the calculus part (see comments).
An answer of \(x = 600\) and £3000 with no working scores 0 marks, because the question says use calculus; the two explanation marks and the batch-size mark are marked on their own merits.
1 mark
A1 \(\dfrac{dC}{dx} = 2.5 - \dfrac{900\,000}{x^2}\)
1 mark
M1 Sets their \(\dfrac{dC}{dx} = 0\) and solves for \(x\): \(x^2 = 360\,000\)
1 mark
A1 \(x = 600\)
The negative root must not be given as an answer (or must be rejected, since \(x > 0\)).
1 mark
M1 Finds \(\dfrac{d^2C}{dx^2} = \dfrac{1\,800\,000}{x^3}\) and considers its sign at their \(x = 600\) (or considers the sign of \(\dfrac{dC}{dx}\) on either side of \(x = 600\))
1 mark
A1 \(\dfrac{d^2C}{dx^2} = \dfrac{1\,800\,000}{600^3} = \dfrac{1}{120} > 0\), so \(C\) is a minimum, cso
Accept any correct value with its sign, e.g. awrt 0.0083, i.e. in [0.00825, 0.00835), "> 0". A conclusion is needed.
1 mark
A1 Minimum annual cost \(= \dfrac{900\,000}{600} + 2.5 \times 600 = 1500 + 1500 =\) £3000
Depends on the second M mark (not on the point just before). cao.
1 mark
B1 Explains \(\dfrac{900\,000}{x}\): there are \(\dfrac{18\,000}{x}\) deliveries a year, each costing £50, so the delivery cost is \(50 \times \dfrac{18\,000}{x}\)
The number of deliveries must be referred to.
1 mark
B1 Explains \(2.5x\): the stock falls steadily from \(x\) to 0, so the average stock is \(\dfrac{x}{2}\) bags, each costing £5 a year: \(5 \times \dfrac{x}{2}\)
The average stock of \(\dfrac{x}{2}\) must be referred to.
1 mark
B1 500 bags, with \(C(500) =\) £3050 and \(C(750) =\) £3075 both found and compared
Both costs are needed: the multiples of 250 either side of 600 are 500 and 750. "500 because it is nearer to 600" is B0, since nearness to 600 does not decide which is cheaper.

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