Edexcel A level Maths (9MA0) · Pure mathematics › Differentiation
Practise Tangents, normals and stationary points. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
M1 for \(x^{n} \to x^{n - 1}\) for at least one term
A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = -3x^{2} - 4x + 2\)
Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = -3x^{2} - 4x + 2\)
(b)Answer: \(y = 3x + 3\)
B1 for \(y = 0\) at \(P\)
M1 for substituting \(x = -1\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient of \(C\) at \(P\)
dM1 for a correct straight-line method using \((-1, 0)\) and their gradient
A1 for \(y = 3x + 3\)
Worked solution: At \(x = -1\): \(y = 0\) and \(\frac{\mathrm{d}y}{\mathrm{d}x} = -3(-1)^{2} - 4(-1) + 2 = 3\). \(y = 3(x + 1)\) \(y = 3x + 3\)
Question 2Medium10 marks
The curve \(C\) has equation \(y = x + 4\sqrt{x} - \frac{4}{\sqrt{x}}\), \(x \gt 0\). The point \(P(4, 10)\) lies on \(C\).
(a) Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\), giving each term in its simplest form.[3]
(b) Find an equation of the normal to \(C\) at \(P\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.[4]
(c) The normal to \(C\) at \(P\) meets the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\). Find the area of triangle \(OAB\), where \(O\) is the origin.[3]
Worked solution: Writing each term as a power of \(x\): \(y = x + 4x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}\), so \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + 2x^{-\frac{1}{2}} + 2x^{-\frac{3}{2}}\).
(b)Answer: \(4x + 9y - 106 = 0\)
M1 for substituting \(x = 4\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\), giving the gradient \(\frac{9}{4}\)
M1 for the gradient of the normal, \(-1 \div \left(\frac{9}{4}\right) = -\frac{4}{9}\)
dM1 for a correct straight-line method using \((4, 10)\)
A1 for \(4x + 9y - 106 = 0\) or any integer multiple
Worked solution: At \(x = 4\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \frac{2}{\sqrt{4}} + \frac{2}{4\sqrt{4}} = \frac{9}{4}\). Gradient of the normal \(= -1 \div \left(\frac{9}{4}\right) = -\frac{4}{9}\). \(y - 10 = -\frac{4}{9}(x - 4)\) \(4x + 9y - 106 = 0\)
(c)Answer: \(\frac{2809}{18}\)
M1 for using their equation to find the coordinates of \(A\) and \(B\)
A1 for \(A(\frac{53}{2}, 0)\) and \(B(0, \frac{106}{9})\)
A1 for area \(= \frac{2809}{18}\)
Worked solution: \(A(\frac{53}{2}, 0)\), \(B(0, \frac{106}{9})\). Area of \(OAB = \frac{1}{2} \times \frac{53}{2} \times \frac{106}{9} = \frac{2809}{18}\).
Question 3Hard8 marks
The curve \(C\) has equation \(y = x^{3} - 2x^{2} + 5x + 2\). The point \(P(-2, -24)\) lies on \(C\).
(a) Find an equation of the tangent to \(C\) at \(P\), giving your answer in the form \(y = mx + c\).[4]
(b) The tangent to \(C\) at \(P\) meets \(C\) again at the point \(Q\). Find the coordinates of \(Q\).[4]
Show the answer and mark scheme
(a)Answer: \(y = 25x + 26\)
M1 for \(x^{n} \to x^{n - 1}\) for at least one term
A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - 4x + 5\)
dM1 for finding the gradient at \(P\) and using a correct straight-line method with \((-2, -24)\)
A1 for \(y = 25x + 26\)
Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - 4x + 5\); at \(x = -2\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 25\). \(y + 24 = 25(x + 2)\), so \(y = 25x + 26\).
(b)Answer: \(Q(6, 176)\)
M1 for equating the equations of \(C\) and the tangent and rearranging to a cubic \(= 0\)
M1 for using \((x + 2)^{2}\) as a factor (the line touches \(C\) at \(P\)) to find the remaining linear factor
A1 for \(x = 6\)
A1 for \(Q(6, 176)\)
Worked solution: \(x^{3} - 2x^{2} + 5x + 2 = 25x + 26\) gives \(x^{3} - 2x^{2} - 20x - 24 = 0\). The tangent touches \(C\) at \(P\), so \((x + 2)^{2}\) is a factor: \((x + 2)^{2}(x - 6) = 0\). So at \(Q\), \(x = 6\) and \(y = 176\).