IBDP Maths AA: Topic : SL 1.5: Laws of exponents: IB style Questions SL Paper 2

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.

Find the value of a and of b.[3]

a.

Use the regression equation to estimate the value of y when x = 3.57.[3]

b.

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]

c.
Answer/Explanation

Markscheme

valid approach (M1)

eg one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14 A1A1 N3

[3 marks]

a.

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]

b.

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]

c.

Question

Expand \({(x – 2)^4}\) and simplify your result.

[3]
a.

Find the term in \({x^3}\) in \((3x + 4){(x – 2)^4}\) .

[3]
b.
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]

a.

finding coefficients, \(3 \times 24( = 72)\) , \(4 \times( – 8)( = – 32)\)     (A1)(A1)

term is \(40{x^3}\)     A1     N3

[3 marks]

b.

Question

Let \(f(x) =log_ax,x>0\).
(a) Write down the value of (i) f(a)    (ii) f(1)     (iii) \(f(a^4)\)
(b) The diagram below shows part of the graph of f.
On the same diagram, sketch the graph of \(f^{-1}\).

Answer/Explanation

Ans
(a) (i) f(a)=1     (ii) f(1)=0     (iii) \(f(a^4)=4\)
(b)

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