Edexcel A level Maths (9MA0) · Pure mathematics › Exponentials and logarithms
Practise The function eˣ and its gradient. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Worked solution: The gradient of \(\mathrm{e}^{kx}\) is \(k\mathrm{e}^{kx}\), so \(\frac{\mathrm{d}y}{\mathrm{d}x} = 8\mathrm{e}^{4x}\).
(b)Answer: \(8\mathrm{e}^{-4}\)
B1 for \(8\mathrm{e}^{-4}\) oe
Worked solution: When \(x = -1\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 8\mathrm{e}^{-4} = 8\mathrm{e}^{-4}\).
(c)Answer: \(\left(\frac{1}{4}\ln 4, 13\right)\)
M1 for \(8\mathrm{e}^{4x} = 32\) leading to \(\mathrm{e}^{4x} = 4\)
A1 for \(x = \frac{1}{4}\ln 4\)
A1 for \(y = 13\)
Worked solution: \(8\mathrm{e}^{4x} = 32 \Rightarrow \mathrm{e}^{4x} = 4 \Rightarrow x = \frac{1}{4}\ln 4\). Then \(y = 2 \times 4 + 5 = 13\).
Question 2Medium7 marks
The curve \(C\) has equation \(y = 2\mathrm{e}^{2x} + 4\). The point \(P\) on \(C\) has \(x\)-coordinate \(\ln 2\).
(a) Find the exact \(y\)-coordinate of \(P\) and the exact gradient of \(C\) at \(P\).[3]
(b) Find an equation of the tangent to \(C\) at \(P\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are exact constants.[2]
(c) The tangent meets the \(x\)-axis at the point \(Q\). Find the exact \(x\)-coordinate of \(Q\).[2]
Show the answer and mark scheme
(a)Answer: \(y = 12\), gradient \(16\)
B1 for \(y = 12\) (using \(\mathrm{e}^{2\ln 2} = 4\))
M1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4\mathrm{e}^{2x}\)
A1 for gradient \(16\)
Worked solution: \(\mathrm{e}^{2\ln 2} = 2^{2} = 4\), so \(y = 2 \times 4 + 4 = 12\). \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4\mathrm{e}^{2x}\), which at \(P\) is \(4 \times 4 = 16\).
(b)Answer: \(y = 16x + 12 - 16\ln 2\)
M1 for \(y - 12 = 16(x - \ln 2)\)
A1 for \(y = 16x + 12 - 16\ln 2\)
Worked solution: \(y - 12 = 16(x - \ln 2) \Rightarrow y = 16x + 12 - 16\ln 2\)
(c)Answer: \(x = \ln 2 - \frac{3}{4}\)
M1 for setting \(y = 0\) in their tangent
A1 for \(x = \ln 2 - \frac{3}{4}\) oe
Worked solution: \(0 = 16(x - \ln 2) + 12 \Rightarrow x - \ln 2 = -\frac{3}{4} \Rightarrow x = \ln 2 - \frac{3}{4}\)
Question 3Hard7 marks
The curve \(C\) has equation \(y = 3\mathrm{e}^{x}\). The point \(P\) on \(C\) has \(x\)-coordinate \(t\).
(a) Show that the tangent to \(C\) at \(P\) meets the \(x\)-axis at the point where \(x = t - 1\).[3]
(b) Given that the tangent at \(P\) passes through the point \((2, 0)\), find the exact coordinates of \(P\).[2]
(c) The tangent at \(P\), the \(x\)-axis and the line \(x = 3\) enclose a triangle. Find the exact area of this triangle.[2]
Show the answer and mark scheme
(a)Answer: Shown
M1 for the gradient \(3\mathrm{e}^{t}\)
M1 for the tangent \(y - 3\mathrm{e}^{t} = 3\mathrm{e}^{t}(x - t)\) and setting \(y = 0\)
A1* for \(x = t - 1\)
Worked solution: At \(P\), \(y = 3\mathrm{e}^{t}\) and the gradient is \(3\mathrm{e}^{t}\). Setting \(y = 0\) in the tangent: \(-3\mathrm{e}^{t} = 3\mathrm{e}^{t}(x - t) \Rightarrow x - t = -1\), so \(x = t - 1\).
(b)Answer: \(\left(3, 3\mathrm{e}^{3}\right)\)
M1 for \(t - 1 = 2\)
A1 for \(P\left(3, 3\mathrm{e}^{3}\right)\)
Worked solution: \(t = 2 + 1 = 3\), so \(y = 3\mathrm{e}^{ \times 3} = 3\mathrm{e}^{3}\).
Worked solution: The triangle has base from \(x = 2\) to \(x = 3\), of length \(1\), and height \(3\mathrm{e}^{3}\). Area \(= \frac{1}{2} \times 1 \times 3\mathrm{e}^{3} = \frac{3\mathrm{e}^{3}}{2}\) (\(\approx 30.1\)).