IB Math Analysis & Approaches Questionbank-Topic: SL 2.9 Exponential and Logarithmic functions-SL Paper 1

Question

Dominic jumps out of an airplane that is flying at constant altitude. Before opening his

parachute, he goes through a period of freefall.

Dominic’s vertical speed during the time of freefall, S , in m s1 , is modelled by the following function.

\(S(t)=K-6-(1.2^t),t\geqslant 0\)

where t , is the number of seconds after he jumps out of the airplane, and K is a constant. A sketch

of Dominic’s vertical speed against time is shown below.

Dominic’s initial vertical speed is \(0\:ms^{-1}\)

    1. Find the value of K . [2]

    2. In the context of the model, state what the horizontal asymptote represents. [1]

    3. Find Dominic’s vertical speed after 10 seconds. Give your answer in km h−1. [3]

       

Answer/Explanation

Ans: 

(a)

0 = K – 60(1.20)

K= 60

(b)

the (vertical) speed that Dominic is approaching (as t increases)

OR

the limit of the (vertical) speed of Dominic

(c)

S=60-60\((1.2^{-10})\)

\(S=50.3096..(ms^{-1})\)

\(181(kmh^{-1})(181.144..(kmh^{-1}))\)

Question

Let f (x)a log3 (x – 4) , for x > 4 , where a > 0 .

Point A(13 , 7) lies on the graph of f .

  1. Find the value of a . [3]

    The x-intercept of the graph of f is (5 , 0) .

  2. On the following grid, sketch the graph of f . [3]

Answer/Explanation

Ans:

(a)

attempt to substitute coordinates (in any order) into f

eg a \(log_{3} (13-4)=7, alog_{3}\)(7-4)=13 , alog9 = 7

finding \(log_{3}\) 9=2 (seen anywhere)

eg \(log_{3}9=2, 2a=7\)

\(a= \frac{7}{2}\)

(b)

Question

Let \(f(x) = {\log _p}(x + 3)\) for \(x >  – 3\) . Part of the graph of f is shown below.


The graph passes through A(6, 2) , has an x-intercept at (−2, 0) and has an asymptote at \(x =  – 3\) .

Find p .

[4]
a.

The graph of f is reflected in the line \(y = x\) to give the graph of g .

(i)     Write down the y-intercept of the graph of g .

(ii)    Sketch the graph of g , noting clearly any asymptotes and the image of A.

[5]
b.

The graph of \(f\) is reflected in the line \(y = x\) to give the graph of \(g\) .

Find \(g(x)\) .

[4]
c.
Answer/Explanation

Markscheme

evidence of substituting the point A     (M1)

e.g. \(2 = {\log _p}(6 + 3)\)

manipulating logs     A1

e.g. \({p^2} = 9\)

\(p = 3\)     A2     N2

[4 marks]

a.

(i) \(y = – 2\) (accept \((0{\text{, }} – 2))\)     A1     N1

(ii)


     A1A1A1A1     N4

Note: Award A1 for asymptote at \(y = – 3\) , A1 for an increasing function that is concave up, A1 for a positive x-intercept and a negative y-intercept, A1 for passing through the point \((2{\text{, }}6)\) .

[5 marks] 

b.

METHOD 1

recognizing that \(g = {f^{ – 1}}\)     (R1)

evidence of valid approach     (M1)

e.g. switching x and y (seen anywhere), solving for x

correct manipulation     (A1)

e.g. \({3^x} = y + 3\)

\(g(x) = {3^x} – 3\)     A1     N3

METHOD 2

recognizing that \(g(x) = {a^x} + b\)     (R1) 

identifying vertical translation     (A1)

e.g. graph shifted down 3 units, \(f(x) – 3\)

evidence of valid approach     (M1)

e.g. substituting point to identify the base

\(g(x) = {3^x} – 3\)     A1     N3

[4 marks]

c.

Question

Find the value of \({\log _2}40 – {\log _2}5\) .

[3]
a.

Find the value of \({8^{{{\log }_2}5}}\) .

[4]
b.
Answer/Explanation

Markscheme

evidence of correct formula    (M1)

eg   \(\log a – \log b = \log \frac{a}{b}\) , \(\log \left( {\frac{{40}}{5}} \right)\) , \(\log 8 + \log 5 – \log 5\)

Note: Ignore missing or incorrect base.

correct working     (A1)

eg   \({\log _2}8\) , \({2^3} = 8\)

\({\log _2}40 – {\log _2}5 = 3\)     A1     N2

[3 marks]

a.

attempt to write \(8\) as a power of \(2\) (seen anywhere)     (M1)

eg   \({({2^3})^{{{\log }_2}5}}\) , \({2^3} = 8\) , \({2^a}\)

multiplying powers     (M1)

eg   \({2^{3{{\log }_2}5}}\) , \(a{\log _2}5\)

correct working     (A1)

eg   \({2^{{{\log }_2}125}}\) , \({\log _2}{5^3}\) , \({\left( {{2^{{{\log }_2}5}}} \right)^3}\)

\({8^{{{\log }_2}5}} = 125\)     A1     N3

[4 marks]

b.

Question

Let \(f(x) = {{\rm{e}}^{x + 3}}\) .

(i)     Show that \({f^{ – 1}}(x) = \ln x – 3\) .

(ii)    Write down the domain of \({f^{ – 1}}\) .

[3]
a.

Solve the equation \({f^{ – 1}}(x) = \ln \frac{1}{x}\) .

[4]
b.
Answer/Explanation

Markscheme

(i) interchanging x and y (seen anywhere)     M1

e.g. \(x = {{\rm{e}}^{y + 3}}\)

correct manipulation     A1

e.g. \(\ln x = y + 3\) , \(\ln y = x + 3\)

\({f^{ – 1}}(x) = \ln x – 3\)     AG     N0

(ii) \(x > 0\)     A1     N1 

[3 marks]

a.

collecting like terms; using laws of logs     (A1)(A1)

e.g. \(\ln x – \ln \left( {\frac{1}{x}} \right) = 3\) , \(\ln x + \ln x = 3\) , \(\ln \left( {\frac{x}{{\frac{1}{x}}}} \right) = 3\) , \(\ln {x^2} = 3\)

simplify     (A1)

e.g. \(\ln x = \frac{3}{2}\) ,  \({x^2} = {{\rm{e}}^3}\)

\(x = {{\rm{e}}^{\frac{3}{2}}}\left( { = \sqrt {{{\rm{e}}^3}} } \right)\)     A1     N2

[4 marks]

b.

Examiners report

Many candidates interchanged the \(x\) and \(y\) to find the inverse function, but very few could write down the correct domain of the inverse, often giving \(x \ge 0\) , \(x > 3\) and “all real numbers” as responses.

a.

Where students attempted to solve the equation in (b), most treated \(\ln x – 3\) as \(\ln (x – 3)\) and created an incorrect equation from the outset. The few who applied laws of logarithms often carried the algebra through to completion.

b.

Question

Find the value of \({\log _2}40 – {\log _2}5\) .

[3]
a.

Find the value of \({8^{{{\log }_2}5}}\) .

[4]
b.
Answer/Explanation

Markscheme

evidence of correct formula    (M1)

eg   \(\log a – \log b = \log \frac{a}{b}\) , \(\log \left( {\frac{{40}}{5}} \right)\) , \(\log 8 + \log 5 – \log 5\)

Note: Ignore missing or incorrect base.

correct working     (A1)

eg   \({\log _2}8\) , \({2^3} = 8\)

\({\log _2}40 – {\log _2}5 = 3\)     A1     N2

[3 marks]

a.

attempt to write \(8\) as a power of \(2\) (seen anywhere)     (M1)

eg   \({({2^3})^{{{\log }_2}5}}\) , \({2^3} = 8\) , \({2^a}\)

multiplying powers     (M1)

eg   \({2^{3{{\log }_2}5}}\) , \(a{\log _2}5\)

correct working     (A1)

eg   \({2^{{{\log }_2}125}}\) , \({\log _2}{5^3}\) , \({\left( {{2^{{{\log }_2}5}}} \right)^3}\)

\({8^{{{\log }_2}5}} = 125\)     A1     N3

[4 marks]

b.

Examiners report

Many candidates readily earned marks in part (a). Some interpreted \({\log _2}40 – {\log _2}5\) to mean \(\frac{{{{\log }_2}40}}{{{{\log }_2}5}}\) , an error which led to no further marks. Others left the answer as \({\log _2}5\) where an integer answer is expected.

a.

Part (b) proved challenging for most candidates, with few recognizing that changing \(8\) to base \(2\) is a helpful move. Some made it as far as \({2^{3{{\log }_2}5}}\) yet could not make that final leap to an integer.

b.

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