Edexcel A level Maths (9MA0) · Pure mathematics › Exponentials and logarithms
Practise Exponential functions and graphs. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
The curve \(C\) has equation \(y = 4 \times 2^{x}\).
(a) Sketch \(C\), showing the coordinates of the point where \(C\) meets the \(y\)-axis and the equation of the asymptote.[3]
(b) Describe fully a single transformation that maps the graph of \(y = 2^{x}\) onto \(C\).[2]
(c) Find the exact value of \(y\) on \(C\) when \(x = 3\).[1]
Show the answer and mark scheme
(a)Answer: An exponential curve increasing through \(\left(0, 4\right)\) with asymptote \(y = 0\).
B1 for the correct shape (increasing exponential curve)
B1 for \(\left(0, 4\right)\)
B1 for the asymptote \(y = 0\) shown
Worked solution: When \(x = 0\), \(y = 4\). As \(x \to -\infty\), \(y \to 0\), so the asymptote is \(y = 0\).
(b)Answer: A stretch parallel to the \(y\)-axis with scale factor 4.
B1 for stretch parallel to the \(y\)-axis
B1 for scale factor 4
Worked solution: \(y = 4 \times 2^{x}\) is \(y = 2^{x}\) after a stretch parallel to the \(y\)-axis with scale factor 4.
(c)Answer: \(32\)
B1 for \(32\)
Worked solution: \(y = 4 \times 2^{(3)} = 32\)
Question 2Medium7 marks
The diagram shows part of the curve with equation \(y = pq^{x}\), where \(p\) and \(q\) are positive constants. The curve passes through the points \(A\left(1, 6\right)\) and \(B\left(3, 54\right)\).
[object Object]
(a) Find the value of \(p\) and the value of \(q\).[4]
(b) Find the value of \(x\) for which \(y = \frac{2}{9}\).[2]
(c) State the range of possible values of \(y\) on the curve.[1]
Show the answer and mark scheme
(a)Answer: \(p = 2, \ q = 3\)
M1 for forming two equations, e.g. \(pq = 6\) and \(pq^{3} = 54\)
M1 for dividing to obtain \(q^2 = 9\)
A1 for \(q = 3\) (positive root)
A1 for \(p = 2\)
Worked solution: \(\frac{pq^{3}}{pq} = q^2 = \frac{54}{6} = 9\), so \(q = 3\) (as \(q \gt 0\)). Then \(p = \frac{6}{3} = 2\).
(b)Answer: \(x = -2\)
M1 for \(3^{x} = \frac{1}{9}\) oe
A1 for \(x = -2\)
Worked solution: \(2 \times 3^{x} = \frac{2}{9} \Rightarrow 3^{x} = \frac{1}{9} = 3^{-2}\), so \(x = -2\).
(c)Answer: \(y \gt 0\)
B1 for \(y \gt 0\)
Worked solution: Since \(q^{x} \gt 0\) for all \(x\) and \(p \gt 0\), \(y \gt 0\).
Question 3Hard7 marks
The diagram shows the curves with equations \(y = 9^{x}\) (solid) and \(y = 12 \times 3^{x} - 27\) (dashed).
[object Object]
(a) Show that the \(x\)-coordinates of the points where the curves meet satisfy \[u^2 - 12u + 27 = 0,\] where \(u = 3^{x}\).[2]
(b) Hence find the coordinates of the points where the curves meet, giving exact values.[5]
Show the answer and mark scheme
(a)Answer: Shown
M1 for \(9^{x} = (3^{2})^{x} = (3^{x})^2 = u^2\)
A1* for \(u^2 - 12u + 27 = 0\)
Worked solution: \(9^{x} = 3^{2x} = u^2\), so \(u^2 = 12u - 27\), i.e. \(u^2 - 12u + 27 = 0\).