Question 4. [Maximum mark: 18]
In a small village there are two doctors’ clinics, one owned by Doctor Black and the other owned by Doctor Green. It was noted after each year that 3.5 % of Doctor Black’s patients moved to Doctor Green’s clinic and 5 % of Doctor Green’s patients moved to Doctor Black’s clinic. All additional losses and gains of patients by the clinics may be ignored.
At the start of a particular year, it was noted that Doctor Black had 2100 patients on their register, compared to Doctor Green’s 3500 patients.
a. Write down a transition matrix T indicating the annual population movement between clinics. [2]
b. Find a prediction for the ratio of the number of patients Doctor Black will have, compared to Doctor Green, after two years. [2]
c. Find a matrix P , with integer elements, such that T = PDP–1 , where D is a diagonal matrix.
d. Hence, show that the long-term transition matrix T x is given by T x = 
e. Hence, or otherwise, determine the expected ratio of the number of patients Doctor Black would have compared to Doctor Green in the long term. [2]
▶️Answer/Explanation
(a) 
so ratio is \(2294 : 3306 = (1147 :1653, 0.693889…)\)
(c) to solve Ax = λx : 
\( (0.965 – λ) (0.95 – λ) -0.05 × 0.035 = 0\)
\(λ = 0.915 OR λ =1\)
attempt to find eigenvectors for at least one eigenvalue when \(λ = 0.915\) , \(x=(_{-1}^{1})\)
(or any real multiple) when λ =1, \(x=(_{7}^{10})\) therefore 
(accept integer valued multiples of their eigenvectors and columns in either order) 
(e) 
so ratio is \(3294 : 2306\) \((1647 :1153, 1.42844…, 0.700060…)\)
Question
The following table shows values of ln x and ln y.
The relationship between ln x and ln y can be modelled by the regression equation ln y = a ln x + b.
a.Find the value of a and of b.[3]
b.Use the regression equation to estimate the value of y when x = 3.57.[3]
c.The relationship between x and y can be modelled using the formula y = kxn, where k ≠ 0 , n ≠ 0 , n ≠ 1.
By expressing ln y in terms of ln x, find the value of n and of k.[7]
▶️Answer/Explanation
Markscheme
valid approach (M1)
eg one correct value
−0.453620, 6.14210
a = −0.454, b = 6.14 A1A1 N3
[3 marks]
correct substitution (A1)
eg −0.454 ln 3.57 + 6.14
correct working (A1)
eg ln y = 5.56484
261.083 (260.409 from 3 sf)
y = 261, (y = 260 from 3sf) A1 N3
Note: If no working shown, award N1 for 5.56484.
If no working shown, award N2 for ln y = 5.56484.
[3 marks]
METHOD 1
valid approach for expressing ln y in terms of ln x (M1)
eg \({\text{ln}}\,y = {\text{ln}}\,\left( {k{x^n}} \right),\,\,{\text{ln}}\,\left( {k{x^n}} \right) = a\,{\text{ln}}\,x + b\)
correct application of addition rule for logs (A1)
eg \({\text{ln}}\,k + {\text{ln}}\,\left( {{x^n}} \right)\)
correct application of exponent rule for logs A1
eg \({\text{ln}}\,k + n\,{\text{ln}}\,x\)
comparing one term with regression equation (check FT) (M1)
eg \(n = a,\,\,b = {\text{ln}}\,k\)
correct working for k (A1)
eg \({\text{ln}}\,k = 6.14210,\,\,\,k = {e^{6.14210}}\)
465.030
\(n = – 0.454,\,\,k = 465\) (464 from 3sf) A1A1 N2N2
METHOD 2
valid approach (M1)
eg \({e^{{\text{ln}}\,y}} = {e^{a\,{\text{ln}}\,x + b}}\)
correct use of exponent laws for \({e^{a\,{\text{ln}}\,x + b}}\) (A1)
eg \({e^{a\,{\text{ln}}\,x}} \times {e^b}\)
correct application of exponent rule for \(a\,{\text{ln}}\,x\) (A1)
eg \({\text{ln}}\,{x^a}\)
correct equation in y A1
eg \(y = {x^a} \times {e^b}\)
comparing one term with equation of model (check FT) (M1)
eg \(k = {e^b},\,\,n = a\)
465.030
\(n = – 0.454,\,\,k = 465\) (464 from 3sf) A1A1 N2N2
METHOD 3
valid approach for expressing ln y in terms of ln x (seen anywhere) (M1)
eg \({\text{ln}}\,y = {\text{ln}}\,\left( {k{x^n}} \right),\,\,{\text{ln}}\,\left( {k{x^n}} \right) = a\,{\text{ln}}\,x + b\)
correct application of exponent rule for logs (seen anywhere) (A1)
eg \({\text{ln}}\,\left( {{x^a}} \right) + b\)
correct working for b (seen anywhere) (A1)
eg \(b = {\text{ln}}\,\left( {{e^b}} \right)\)
correct application of addition rule for logs A1
eg \({\text{ln}}\,\left( {{e^b}{x^a}} \right)\)
comparing one term with equation of model (check FT) (M1)
eg \(k = {e^b},\,\,n = a\)
465.030
\(n = – 0.454,\,\,k = 465\) (464 from 3sf) A1A1 N2N2
[7 marks]
Question
a.Expand \({(x – 2)^4}\) and simplify your result.[3]
b.Find the term in \({x^3}\) in \((3x + 4){(x – 2)^4}\) .[3]
▶️Answer/Explanation
Markscheme
evidence of expanding M1
e.g. \({(x – 2)^4} = {x^4} + 4{x^3}( – 2) + 6{x^2}{( – 2)^2} + 4x{( – 2)^3} + {( – 2)^4}\) A2 N2
\({(x – 2)^4} = {x^4} – 8{x^3} + 24{x^2} – 32x + 16\)
[3 marks]
finding coefficients, \(3 \times 24( = 72)\) , \(4 \times( – 8)( = – 32)\) (A1)(A1)
term is \(40{x^3}\) A1 N3
[3 marks]