IB Math Analysis & Approaches Questionbank-Topic: SL 2.8-The reciprocal and rational function SL Paper 1

Question

Let \(f(x) = m – \frac{1}{x}\), for \(x \ne 0\). The line \(y = x – m\) intersects the graph of \(f\) in two distinct points. Find the possible values of \(m\).

Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f = y,{\text{ }}m – \frac{1}{x} = x – m\)

correct working to eliminate denominator     (A1)

eg\(\,\,\,\,\,\)\(mx – 1 = x(x – m),{\text{ }}mx – 1 = {x^2} – mx\)

correct quadratic equal to zero     A1

eg\(\,\,\,\,\,\)\({x^2} – 2mx + 1 = 0\)

correct reasoning     R1

eg\(\,\,\,\,\,\)for two solutions, \({b^2} – 4ac > 0\)

correct substitution into the discriminant formula     (A1)

eg\(\,\,\,\,\,\)\({( – 2m)^2} – 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(4{m^2} > 4,{\text{ }}{m^2} = 1\), sketch of positive parabola on the \(x\)-axis

correct interval     A1     N4

eg\(\,\,\,\,\,\)\(\left| m \right| > 1,{\text{ }}m <  – 1\) or \(m > 1\)

[7 marks]

Question

Let \(f(x) = 3x – 2\) and \(g(x) = \frac{5}{{3x}}\), for \(x \ne 0\).

Let \(h(x) = \frac{5}{{x + 2}}\), for \(x \geqslant 0\). The graph of h has a horizontal asymptote at \(y = 0\).

Find \({f^{ – 1}}(x)\).

[2]
a.

Show that \(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\).

[2]
b.

Find the \(y\)-intercept of the graph of \(h\).

[2]
c(i).

Hence, sketch the graph of \(h\).

[3]
c(ii).

For the graph of \({h^{ – 1}}\), write down the \(x\)-intercept;

[1]
d(i).

For the graph of \({h^{ – 1}}\), write down the equation of the vertical asymptote.

[1]
d(ii).

Given that \({h^{ – 1}}(a) = 3\), find the value of \(a\).

[3]
e.
Answer/Explanation

Markscheme

interchanging \(x\) and \(y\)     (M1)

eg     \(x = 3y – 2\)

\({f^{ – 1}}(x) = \frac{{x + 2}}{3}{\text{   }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)\)     A1     N2

[2 marks]

a.

attempt to form composite (in any order)     (M1)

eg     \(g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}\)

correct substitution     A1

eg     \(\frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}\)

\(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\)     AG     N0

[2 marks]

b.

valid approach     (M1)

eg     \(h(0),{\text{ }}\frac{5}{{0 + 2}}\)

\(y = \frac{5}{2}{\text{   }}\left( {{\text{accept (0, 2.5)}}} \right)\)     A1     N2

[2 marks]

c(i).

     A1A2     N3

Notes:     Award A1 for approximately correct shape (reciprocal, decreasing, concave up).

     Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at \((0, 2.5)\), asymptotic to x-axis, correct domain \(x \geqslant 0\).

     If only two of these features are correct, award A1.

[3 marks]

c(ii).

\(x = \frac{5}{2}{\text{   }}\left( {{\text{accept (2.5, 0)}}} \right)\)     A1     N1

[1 mark]

d(i).

\(x = 0\)   (must be an equation)     A1     N1

[1 mark]

d(ii).

METHOD 1

attempt to substitute \(3\) into \(h\) (seen anywhere)     (M1)

eg     \(h(3),{\text{ }}\frac{5}{{3 + 2}}\)

correct equation     (A1)

eg     \(a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a\)

\(a = 1\)     A1     N2

[3 marks]

METHOD 2

attempt to find inverse (may be seen in (d))     (M1)

eg     \(x = \frac{5}{{y + 2}},{\text{ }}{h^{ – 1}} = \frac{5}{x} – 2,{\text{ }}\frac{5}{x} + 2\)

correct equation, \(\frac{5}{x} – 2 = 3\)     (A1)

\(a = 1\)     A1     N2

[3 marks]

e.

Question

Let \(f(x) = p + \frac{9}{{x – q}}\), for \(x \ne q\). The line \(x = 3\) is a vertical asymptote to the graph of \(f\).

Write down the value of \(q\).

[1]
a.

The graph of \(f\) has a \(y\)-intercept at \((0,{\text{ }}4)\).

Find the value of \(p\).

[4]
b.

The graph of \(f\) has a \(y\)-intercept at \((0,{\text{ }}4)\).

Write down the equation of the horizontal asymptote of the graph of \(f\).

[1]
c.
Answer/Explanation

Markscheme

\(q = 3\)     A1     N1

[1 mark]

a.

correct expression for \(f(0)\)     (A1)

eg\(\;\;\;p + \frac{9}{{0 – 3}},{\text{ }}4 = p + \frac{9}{{ – q}}\)

recognizing that \(f(0) = 4\;\;\;\)(may be seen in equation)     (M1)

correct working     (A1)

eg\(\;\;\;4 = p – 3\)

\(p = 7\)     A1     N3

[3 marks]

b.

\(y = 7\;\;\;\)(must be an equation, do not accept \(p = 7\)     A1     N1

[1 mark]

Total [6 marks]

c.

Question

Let \(f(x) = p + \frac{9}{{x – q}}\), for \(x \ne q\). The line \(x = 3\) is a vertical asymptote to the graph of \(f\).

Write down the value of \(q\).

[1]
a.

The graph of \(f\) has a \(y\)-intercept at \((0,{\text{ }}4)\).

Find the value of \(p\).

[4]
b.

The graph of \(f\) has a \(y\)-intercept at \((0,{\text{ }}4)\).

Write down the equation of the horizontal asymptote of the graph of \(f\).

[1]
c.
Answer/Explanation

Markscheme

\(q = 3\)     A1     N1

[1 mark]

a.

correct expression for \(f(0)\)     (A1)

eg\(\;\;\;p + \frac{9}{{0 – 3}},{\text{ }}4 = p + \frac{9}{{ – q}}\)

recognizing that \(f(0) = 4\;\;\;\)(may be seen in equation)     (M1)

correct working     (A1)

eg\(\;\;\;4 = p – 3\)

\(p = 7\)     A1     N3

[3 marks]

b.

\(y = 7\;\;\;\)(must be an equation, do not accept \(p = 7\)     A1     N1

[1 mark]

Total [6 marks]

c.

MAA SL 2.8-2.10 EXPONENTS AND LOGARITHMS [concise]-ashok

Question

Solve the equations
(a) \(2^{2x}=8^{1-x}\)
(b) \(4^{2x}=8^{1-x}\)
(c) \(2^{x^2+3}=16^x\)

Answer/Explanation

Ans
(a) x = 3/5      (b) x = 3/7     (c) x = 1 or x = 3

Question

Solve the equations
(a) \(25^{x+1}=\frac{1}{5^x}\)     (b) \(25^{x+1}=\sqrt{5^x}\)
(c) \(25^{x^2}=125^x\)          (d) \(7^{x^2-5x} = 1\)

Answer/Explanation

Ans
(a) x = -2/3   (b) x = -4/3    (c) x = 0 or x = 3/2    (d) x = 0 or x = 5

Question

Let \(f(x)=e^x+2\). The points A(1,e+2) and B(2,b) of the graph are shown in the diagram below.

(a) Write down the value of b.
(b) On the diagram above, sketch the graph of f. Indicate the y-intercept and the horizontal asymptote of the graph.
(c) Write down the domain and the range of f.

Answer/Explanation

Ans
(a) \(b=e^2+2\).
(b) y-intercept: (0,3),  horizontal asymptote: y = 2
(c) Domain: x ∈ R, Range: y > 2

Question

Sketch the graph of the function \(f(x)=2^x+2\). Indicate the y-intercept, the horizontal asymptote and points A(1,a), B(2,b) and C(3,c) of the graph.

Answer/Explanation

Ans
y-intercept: (0,3), horizontal asymptote: y = 2
Graph for 3 and 4 are shown below

Question

For each of the following functions sketch the graph and complete the table
(a) \(f(x)=2e^x+3\)     (the point A lies on the curve).


(b) \(f(x)=2e^x-3\)     (the point A lies on the curve).

Answer/Explanation

Ans

Question

Write down the following values

Answer/Explanation

Ans

Question

(a) Find the value of x in the following equations
(i) \(x = log_28\)     (ii) \(log_2x=3\)     (iii) \(log_x8=3\)
(b) Find the value of x in the following equations
(i) x = log100     (ii) log x = 2      (iii) x = In e4       (iv) In x = 4

Answer/Explanation

Ans
(a) (i) x = 3     (ii) x = 8     (iii) x = 2
(b) (i) x = 2     (ii) x = 100   (iii) x = 4     (iv) \(x = e^e\)

Question

Solve the following equations
(i) \(log_3(x+1)=2\)      (ii) log(x+1)=2         (iii) In(x+1)=2

Answer/Explanation

Ans
(a) x = 8    (b) x = 99    (c) \(x=e^2-1\)

Question

Find the following integers

Answer/Explanation

Ans
the answers for any set of 4 questions are 5,5,25,125

Question

Confirm the following properties for x = 8 and y = 4

Answer/Explanation

Ans

Question

Let \(log_5x=a,log_5y=b\) and \(log_5z=c\). Express the following in terms of a,b,c.

Answer/Explanation

Ans

Question

Solve the equations
(a) \(log_2x+log_2(x+1)=log_26\)
(b) \(log_2x+log_2(x+1)=1\)
(c) \(log_2(x+5)-log_2x=1\)

Answer/Explanation

Ans
(a) x = 2       (b) x = 1       (c) x = 5

Question

Solve the equations
(a) \(log_2(x+14)-2log_2x=2\)
(b) \(log_4(x+14)-log_2x=1\)         by using change of base on \(log_4(x+14)\)
(c) \(log_2(x+14)=2+log_{\sqrt{2}}x\)      by using change of base on \(log_{\sqrt{2}}x\)

Answer/Explanation

Ans
(a) x = 2     (b) x = 2     (c) x = 2    ( it is the same equation in all 3 cases )

Question

Solve the following equations by using the natural logarithm In().
(a) \(2^{x+3}=5\)
(b) \(2^{x+3}=5^x\)
(c) \(e^{x+3}=5\)
(d) \(2e^{x+3}=5\)

Answer/Explanation

Ans
(a) \(x=\frac{In5}{In2}-3\)   (b) \(x=\frac{In8}{In5-In2} or \frac{In8}{In5/2}\)    (c) x = In 5 – 3    (d) \(x=In\frac{5}{2}-3\)

Question

(a) Solve the quadratic equation \(y^2-2y+1=0\)
(b) Hence solve the following equations
(i) \(2^x+\frac{1}{2^x}=2\)          by letting \(y=2^x\)
(ii) \(e^x+\frac{1}{e^x}=2\)          by letting \(y=e^x\)
(iii) \(4^x-2(2^x)+1=0\)          by using an appropriate substitution,
(iv) \(e^{2x}-2e^x+1=0\)          by using an appropriate substitution.

Answer/Explanation

Ans
(a) x = 0       (b) x = 0      (c) x = 0       (d) x = 0

Question

Solve the following equations
(a) \((Inx)^2-Inx^2+1=0\)          by letting y = In x
(b) \(Inx+\frac{1}{ln x}=2\)          by using an appropriate substitution
(c) \(logx+\frac{1}{logx}=2\)          by using an appropriate substitution,

Answer/Explanation

Ans
(a) x = e     (b) x = e     (c) x = 10

Question

Find the exact solution of the equation \(9^{2x}=27^{1-x}\)

Answer/Explanation

Ans
\(9^{2x}=27^{1-x} \Leftrightarrow 3^{4x} = 3^{3-3x} \Leftrightarrow 7x = 3 \Leftrightarrow x=\frac{3}{7}\)

Question

Solve the equation \(9^{x-1}=(\frac{1}{3})^{2x}\)

Answer/Explanation

Ans
\(9^{x-1}=(\frac{1}{3})^{2x} \Rightarrow 3^{2x-2}=3^{-2x}\)
\(2x-2=-2x \Rightarrow x = \frac{1}{2}\)

Question

(a) Find \(log_2 32\).
(b) Given that \(log_2(\frac{32^x}{8^y})\) can be written as px+qy, find the value of p and of q.

Answer/Explanation

Ans
(a) 5
(b) METHOD 1
\(log_2(\frac{32^x}{8^y})=log)_232^x-log_28^y=xlog_232-ylog_28=5x-3y\)
p = 5, q = -3
    MEHTOD 2
\(\frac{32^x}{8^y}=\frac{(2^5)^x}{(2^3)^y}=\frac{2^{5x}}{2^{3y}}=2^{5x-3y}\)
\(log_2(2^{5x-3y})=5x-3y\)
p=5, q=-3

Question

Let \(log_10P=x, log_{10}Q=y\) and \(log_{10}R=z\). Express \(log_{10}(\frac{P}{QR^3})^2\) in terms of x, y, and z.

Answer/Explanation

Ans
\(log_{10}(\frac{P}{QR^3})=2log_{10}(\frac{P}{QR^3})=2(log_{10}P-log_{10}Q-3log_{10}R)=2(x-y-3z)=2x-2y-6z\)

Question

Let a = log x, b = log y, and c = log z. Write log\((\frac{x^2\sqrt{y}}{z^3})\) in terms of a, b, c.

Answer/Explanation

Ans
METHOD 1
log\((\frac{x^2\sqrt{y}}{z^3})\)=\(2 logx+\frac{1}{2} log y-3log z=2a+\frac{1}{2}b-3c\)
METHOD 2
\(x^2=10^{2a}, \sqrt{y}=10^{\frac{b}{2}},z^3=10^{3c}\)
\(log_{10}(\frac{x^2sqrt{y}}{z^3})=log_{10}\)\((\frac{10^{2a} \times 10^{\frac{b}{2}}{10^{3c}})=log_{10}\)\((10^{2a+\frac{b}{2}-3c)=2a+\frac{1}{2}b-3c\)

Question

Let \(p = log_{10}x,q=log_{10}y\) and \(r=log_{10}z\). Write \(log_{10}\((\frac{x}{y^2\sqrt{z}})\) in terms of p, q and r.

Answer/Explanation

Ans
METHOD 1

\(log_{10}(\frac{x}{y^2\sqrt{z}})=log_{10}x-2logy-\frac{1}{2}logz=p-2q-\frac{1}{2}r\)
METHOD 2
\(x=10, y^2=10^{2p}, \sqrt{z}=10^{\frac{r}{2}}\)
\(log_{10}(\frac{x}{y^2\sqrt{z}})=log_{10}(\frac{10^p}{10^{2a}10^{\frac{r}{2}}})=log_{10}(10^{p-2q-\frac{r}{2}})\)

Question

Let In a = p, In b = q. Write the following expressions in terms of p and q.
(a) \(In a^3b         (b) \(In(\frac{\sqrt{a}}{b})\)

Answer/Explanation

Ans
(a) \(Ina^3b=3Ina+Inb=3p+q\)
(b) \(In\frac{\sqrt{a}}{b}=\frac{1}{2}\) In a – In b = 1/2 p-q

Question

Given that \(p=log_a5,q=loga2\), express the following in terms of p and/or q.
(a) \(log_e10\)      (b) \(log_e8\)       (c) \(log_a2.5\)

Answer/Explanation

Ans
(a) \(log_a10=log_a(5\times 2)=log_a5+log_a2=p+q\)
(b) \(log_a8=log_a2^3=3log_a2=3q\)
(c) \(log_a2.5=log_a\frac{5}{2}=log_a5-log_a2=p+q\)

Question

Let \(log_c3=p\) and \(log_c5=q\). Find an expression in terms of p and q for
(i) \(log_5x^2\)    (b) \(log_5(\frac{1}{x})\)     (c) \(log_{25}x\)

Answer/Explanation

Ans
(a) (i) \(log_c15=log_c3+log_c5=p+q\)
(ii) \(log_c25=2 log_c5=2q\)
(b) \(a^{\frac{1}{2}}=6 \Leftrightarrow d = 36\)

Question

Given that \(log_5 x=y\), express each of the following in terms of y.
(a) \(log_5x^2\)     (b) \(log_5(\frac{1}{x})\)    (c) \(log_{25}x\)

Answer/Explanation

Ans
(a) \(log_5x^2=2log_5x=2y\)
(b) \(log_5\frac{1}{x}=-log_5x=-y\)
(c) \(log_25x=\frac{log_5x}{log_525}=\frac{1}{2}y\)

Question

If \(log_a2=x\) and \(log_e5=y\), find in terms of x and y, expressions for
(a) \(log_25;\)   (b) \(log_a20\).

Answer/Explanation

Ans
(a) \(log_25=\frac{log_a5}{log_a2}=\frac{y}{x}\)
(b) \(log_a20=log_a4+log_a5=2log_a2+log_a5=2x+y\)

Question

Solve the equation \(log_981+log_9\frac{1}{9}+log_93=log_9x\)

Answer/Explanation

Ans
METHOD 1

\(log81+log_9(\frac{1}{9})+log_93=log_9x \Rightarrow log_9[81(\frac{1}{9})3]=log_9x\)
\(\Rightarrow log_927=log_9x \Rightarrow x=27\)
METHOD 2
\(log_981+log_9(\frac{1}{9})+log_93=2-1+\frac{1}{2}\)
\(\Rightarrow \frac{3}{2} = log_9x \Rightarrow x = 9^{\frac{3}{2}} \Rightarrow x = 27\)

Question

Solve \(log_2x+log_2(x-2)=3\), for x>2.

Answer/Explanation

Ans
\(\log_2(x(x-2))=3 \Leftrightarrow x(x-2)=2^3 \Leftrightarrow x^2 – 2x – 8 = 0 \Leftrightarrow (x-4)(x+2)\)
x = 4

Question

Solve the equation \(log_27 x=1-log_27(x-0.4)\).

Answer/Explanation

Ans
\(log_27(x(x-0.4))=1\)
\(x^2-0.4x=27\)
x = 5.4 or x = -5
x = 5.

Question

(a) Given that \(log_3x-log_3(x-5)=log_3A\), express A in terms of x.
(b) Hence or otherwise, solve the equation \(log_3x-log_3(x-5)=1\)

Answer/Explanation

Ans
(a) \(log_3x-log_3(x-5)=log_3(\frac{x}{x-5})\)        \(A=\frac{x}{x-5}\)
(b) \(log_3(\frac{x}{x-5})=1 \Leftrightarrow \frac{x}{x-5}=3^1(=3) \Leftrightarrow x =3x-15 \Leftrightarrow x=\frac{15}{2}\)

Question

Find the exact value of x in each of the following equations.
(a) \(5^{x+1}=625\)     (b) \(log_a(3x+5)=2\)

Answer/Explanation

Ans
(a) \(5^{x+1}=5^4 \Leftrightarrow x+1 = 4 \Leftrightarrow x=3\)
(b) \(3x+5=a^2 \Leftrightarrow x=\frac{a^2-5}{3}\)

Question

Solve the following equations.
(a) In(x+2) = 3.    (b) \(10^{2x} = 500\).

Answer/Explanation

Ans
\(x+2=e^3 \Leftrightarrow x = e^3-2\)
(b) \(log_{10}(10^{2x})=log_{10}500\)
\(2x=log_{10}500 \Leftrightarrow x=\frac{1}{2} log_{10} 500\)       \((=\frac{log500}{log100}=1.35)\)

Question

Let \(f(x) = e^x\).
(a) Find an expression for \(A(x)=9f(x) \times f(3x) – f(x)^2\)
(b) Find the value of x for which \(A(x) = 0\). Express your answer in the form a Inb, where a,b \(\epsilon \mathbb{Z}\)

Answer/Explanation

Ans
(a) \( A=9e^x \times e^{3x} -e^x \times e^x = 9c^{4x} – e^{2x}\)
(b) \(e^{2x}=\frac{1}{9}  \Leftrightarrow 2x = In\frac{1}{9} \Leftrightarrow x =-In3\)      (a=-1, b=3)

Question

The diagram shows three graphs.

A is part of the graph of y=x, B of the graph of \(y=2^x\), C is the reflection of graph B in line A. Write down.
(a) the equation of C in the form y = f(x).
(b) the coordinates of the point where C cut the x-axis.

Answer/Explanation

Ans
(a) C has equation \(y = log_2x\)
(b) Cuts x-axis \(\Rightarrow log_2x=0 \Rightarrow x = 2^0 \Rightarrow x=1\)
Point is (1,0)

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)

Question

Let \(f(x) = log_2x\).
(a) Given that \(f^{-1}(1)=8, find the value of k.
(b) Find \(f^{-1}(\frac{2}{3})\).

Answer/Explanation

Ans
(a) METHOD 1
\(f(8)=1 \Leftrightarrow k log_28 \Leftrightarrow k = \frac{1}{3}\)
     MEHTOD 2
find the inverse of \(f(x) =log_2x\)
\(y=2^{\frac{x}{k}}\)
substituting 1 and 8
\(2^{1}{k}=8 \Leftrightarrow k=\frac{1}{3}\)
(b) METHOD 1
\(f(x) =\frac{2}{3} \Leftrightarrow \frac{2}{3}=\frac{1}{3} log_2x \Leftrightarrow  log_2x=2 \Leftrightarrow x=4 \)     \(f^{-1}(\frac{2}{3})=4\)
     METHOD 2
inverse of \(f(x)=\frac{1}{3} log_2x\) is \(f^{-1}(x)=2^{3x}\)      \(f^1(\frac{2}{3})=4\)

Question

Let f(x) = 3 In x and \(g(x) = In 5x^3\).
(a) Express g(x) in the form f(x) + In a, where \(a\epsilon \mathbb{Z}^+\)
(b) The graph of g is a transformation of the graph of f. Give a full geometric description of this transformation.

Answer/Explanation

Ans
(a) \(In5x^3=In 5+ In x^3 = In 5+3 In x\)
\(g(x)=f(x) + In5\)
(b) translation by \(\begin{pmatrix}
0\\In 5\end{pmatrix}\) OR shift up by In 5 OR vertical translation of In 5

Question

Let \(f(x) = log_3\sqrt{x}\) for x>0.
(a) Show that \(f^{-1}(x)=3^{2x}\).
(b) Write down the range of \(f^{-1}\).
Let \(f(x)=log_3x\), for x>0
(c) Find the value of \((f^{-1}0g)(2)\), giving your answer as an integer.

Answer/Explanation

Ans
(a) \(log_3\sqrt{x}=y \Leftrightarrow \sqrt{x}=3^y \Leftrightarrow x=3^{2y}\)
\(f^{-1}(x)=3^{2x}\(
(b) y>0
(c) METHOD 1
     \(g(2)=log_32\)
\((f^{-1}0g)(2)=f(x)=3^{2log_32^2}=4\).
    METHOD 2
\((f^{-1}0g)(x) = 3^{log_3x^2}=x^2\)
\((f^{-1}0g)(2)=4\)

Question

Let f(x) = In(x + 2), x>-2 and \(g(x)=e^{(x-4)} g(x)=e^{(x-4)},x>0\).
(a) Write down the x-intercept of the graph of f.
(b)    (i) Write down f(-1.999).
(ii) Write down g(4)
(c) Find the coordinates of the point of intersection of the graphs of f and g.

Answer/Explanation

Ans
(a) x = -1
(b)   (i)   f(-1.999) = In (0.001) = -6.91     (ii) g(4) = 1
(c) (4.64, 1.89)

Question

The functions f(x) and g(x) are defined by \(f(x) = e^2\) and g(x) = In( 1 + 2x )
(a) Write down \(f^{-1}(x)\).
(b) (i) Find (fog)(x)        (ii) Find \((fog)^{-1}(x)\)
Extra question
Find (gof)(x) and \(g^{-1}(x)\)

Answer/Explanation

Ans
(a) \(f^{-1}(x)= In x\)
(b) (i) \((fog)(x)=e^{In(1+2x)}=1+2x\)
(ii) \(y = 1+2x \Leftrightarrow x = (y-1)/2\)
\((fog)^{-1}(x)=\frac{x-1}{2}\)
Extra question
\((gof)(x)=In(1+2e^x)\) and \(g^{-1}(x)=\frac{e^x-1}{2}\)

Question

A group of ten leopards is introduced into a game park. After t years the number of leopards, N, is modelled by \(N = 10e^{0.4t}\)
(a) How many leopards are there after 2 years?
(b) How long will it take for the number of leopards to reach 100?
Give your answers to an appropriate degree of accuracy.
Extra question
1. Find the initial number of leopards
2. How long will it take for the number of leopards to double?

Answer/Explanation

Ans
(a) At \(t= 2,N=10e^{0.4(2)}=22.3(3 sf)\)
Number of leopards = 22
(b) If N = 100, then solve 100 = 10e^{0.4t}\)
\(t=\frac{In 10}{0.4}∼5.76\) years (3 sf)
Extra question
N = 0   \(t=\frac{In2}{0.4}\)(=1.73 years)

Question

The mass m kg of a radio-active substance at time t hours is given by \(m = 4e^{-0.2t}\)
(a) Write down the initial mass.
(b) The mass is reduced to 1.5 kg. How long does this take?
Extra question
Find the half-life time of m.

Answer/Explanation

Ans
(a) Initial mass \(\Rightarrow t=0\), mass = 4
(b) \(1.5 = 4e^{0.2t} \Leftrightarrow In 0.375 = – 0.2t \Leftrightarrow t=4.90\) hours
Extra question
When m = 2 (half of the initial value) \(t = \frac{In 0.5}{-0.2}\) (-3.47  years)

Question

The population p of bacteria at time t is given by \(p = 100e^{0.05t}\). Calculate
(a) the value of p when t = 0.
(b) How long will it take for the size of the population to double?

Answer/Explanation

Ans
(a) \(p=100e^0=100\)
(b) \(200=100e^{0.05t} \Leftrightarrow In2 = 0.05r\)
t = In2/0.05

Question

A machine was purchased for \($ 10000\). Its value V after t years is given by \(V=10000e^{-0.3r}\). The machine must be replaced at the end of the year in which its value drops below \($1500\). Determine in how many years the machine will need to be replaced.

Answer/Explanation

Ans
\(10000e^{-0.3r}=1500\)
\((-0.3tIne=In0.15 \Leftrightarrow t=\frac{In 0.15}{-0.3})\)
t = 6.32       7 (years)

Question

The area A km2 affected by a forest fire at time t hours is given by \(A=A_oe^{kt}\).
When t=5, the area affected is 1 km2. Given that \(A_o=\frac{1}{e}\),
(a) Show that k = 0.2;
(b) Find the value of t when 100 km2 are affected.

Answer/Explanation

Ans
(a) \(1=(1/e)e^{5k}\), so \(e=e^{5k}\) so 5k = l
k = 0.2
(b) \(100=\frac{1}{e}e^{0.2t}\)
\(t=\frac{In100+1}{0.2}(=28.0)\)

Question

The population of a city at the end of 1972 was 250 000. The population increases by 1.3% per year.
(a) Write down the population at the end of 1973.
(b) Find the population at the end of 2002.
Extra question:
How long will it take for the size of the population to exceed 400 000?

Answer/Explanation

Ans
(a) 253250
(b) 1972 → 2002 is 30 years, increase of 1.3% → 1.013
Extra question:
(c) \(250000 \times 1.013^n > 400000 \Rightarrow 1.013^n > 40/25 \Rightarrow 1.013^n > 1.6\)
\(\Rightarrow n log 1.013^n > log 1.6 \Rightarrow n>log1.6/ log1.013 \Rightarrow n>36.4\)
OR directly by GDC n>364

Question

A population of bacteria is growing at the rate of 2.3% per minute. How long will it take for the size of the population to double? Give your answer to the nearest minute.

Answer/Explanation

Ans
\(1.023^t=2\)
\(\Rightarrow t=\frac{In2}{In1.023}=30.48…\)
30 minutes (nearest minute)

Question

Each year for the past five years the population of a certain country has increased at a steady rate of 2.7% per annum. The present population is 15.2 million.
(a) What was the population one year ago?
(b) What was the population five years ago?

Answer/Explanation

Ans
(a) \(\frac{15.2}{1.027}=14.8\) million
(b) \(\frac{15.2}{(1.027)^5}=13.3\) million    (OR 15.2(1.027)-5)

Question

$1000 is invested at 15% per annum interest, compounded monthly, Calculate the minimum number of months required for the value of the investment to exceed $3000.

Answer/Explanation

Ans
15% per annum \(=\frac{15}{12}%=1.25% \) per month
Total value of investment after n months, 1000(1.0125)n>3000
\(\Rightarrow (1.0125)n>3\)
n log (1.0125)> log (3) \(\Rightarrow n>\frac{log(3)}{log(1.0125)}\)
Whole number of month required so n = 89 months.

Question

Solve the following equation.
(a) \(log_x49=2\).
(b) \(log_28=x\)
(c) \(log_25x=-\frac{1}{2}\)
(d) \(log_2x+log_2(x-7)=3\)

Answer/Explanation

Ans
(a) \(x^2=49\)    x=±7   x=7
(b) \(2^x=8\)    x=3
(c) \(x=25^{\frac{1}{2}},x=\frac{1}{\sqrt{25}}, x=\frac{1}{5}\)
(d) \(log_2(x(x-7))=3\)
\(log_2(x^2-7x)=3\)
\(2^3=x^2-7x\)
\(x^2-7x-8=0\)
\((x-8)(x+1)=0(x=8,x=-1)\)
x=8

Question

Michele invested 1500 francs at an annual rate of interest of 5.25 percent, compound annually.
(a) Find the value of Michele’s investment after 3 years (to the nearest franc).
(b) How many complete years will it take for Michele’s initial investment to double?
(c) What should the interest rate be if Michele’s initial investment were to double in value in 10 years?

Answer/Explanation

Ans
(a) Value = \(1500(1.0525)^3=1748.87=1749\)   (nearest franc)
(b) \(3000=1500(1.0525)^t \Rightarrow 2 = 1.0525^t \Rightarrow t=\frac{log2}{log1.0525}=13.546\)
It takes 14 years
(c) \(3000=1500(1+r)^{10}\)
r=0.0718[or7.18%]

Question

Let \(f(x)=log_2\frac{x}{2}+log_316-log_34\), for x>0.
(a) Show that \(f(x) = log_32x\).
(b) Find the value of f(0.5) and of f(4.5).
The function f can also be written in the form \(f(x)=\frac{In ax}{In b}\)
(c)   (i) Write down the value of a and of b
(ii) Hence on graph paper, sketch the graph of f, for -5≤x≤5,-5≤y≤5.
(iii) Write down the equation of the asymptote.
(d) Write down the value of \(f^{-1}(0)\)
The point A lies on the graph of f. At A, x = 4.5.
(e) On your diagram, sketch the graph of \(f^{-1}\), noting the image of point A.

Answer/Explanation

Ans
(a) \(log_3\frac{1}{2}x+log_34=log_3\frac{4x}{2}=log_32x\)
(b) f(0.5) = 0, f(4.5)=2
(c) (i) a = 2, b = 3
(ii)

(iii) x = 0  (must be an equation)
(d) \(f^{-}(0)=\)
(e)

Question

There were 1420 doctors working in a city on 1 January 1994. After n years the number of doctors, D, working in the city is given by D = 1420 + 100n.
(a)    (i)  How many doctors were there working in the city at the start of 2004?
(ii) In what year were there first more than 2000 doctors working in the city?
At the beginning of 1994 the city had a population of 1.2 million. After n years, the population, P, of the city is given by P = 1 200 000(1.025)n.
(b)   (i) Find the population P at the beginning of 2004.
(ii) Calculate the percentage growth in population between 1 January 1994 and 1 January 2004.
(c)    (i) What was the average number of people per doctor at the beginning of 1994?
(ii) After how many complete years will the number of people per doctor first fall below 600?

Answer/Explanation

Ans
(a) (i) 2420
(ii) 1420 + 100n>2000
n>5.8
1999 (accept 6th year or n = 6)
(b) (i) 1 200 000(1.025)10=1 536 101
(accept 1 540 000 or 1.54(million))
(ii) \(\frac{1 536 101 – 1 200 000}{1200000}\times 100\)
28.0% (accept 28.3% from 1 540 000)
(iii) 1 200 000(1.025)n>2 000 000 (accept an equation)
\(\Rightarrow n>20.69\)
2014 (accept 21st year or n = 21)
(c) (i) \(\frac{1 200 000}{1420}=845\)
(ii) \(\frac{1 200 000(1.025)^n}{1420+100n}<600 \Rightarrow n>14.197\)
15 years.

Question

A city is concerned about pollution, and decides to look at the number of people using taxis. At the end of the year 2000, there were 280 taxis in the city. After n years the number of taxis, T, in the city is given by
\(T=280 \times 1.12^n\)
(a) (i) Find the number of taxis in the city at the end of 2005.
(ii) Find the year in which the number or taxis is double the number of taxis there were at the end of 2000.
(b) At the end of 2000 there were 25 600 people in the city who used taxis. After n years the number of people, P, in the city who used taxis is given by
\(P=\frac{2560000}{10+90e^{-0.1n}}\)
(i) Find the value of P at the end of 2005, giving your answer to the nearest whole number.
(ii) After seven complete years, will the value of P be double its value at the end of 2000? Justify your answer.
(c) Let R be the ratio of the number of taxis in the city to the number of taxis. The city will reduce the number of taxis if R<70.
(i) Find the value of R at the end of 2000.
(ii) After how many complete years will the city first reduce the number of taxis?

Answer/Explanation

Ans
(a) (i) n = 5
(ii) \(T = 280 \times 1.12^5 = 493\)
n = 6.116….
in the year 2007
(b) (i) \(P=\frac{2560000}{10+90e^{-1.1(5)}}=39 635.993…=39636\)
(ii) \(P=\frac{2560000}{10+90e^{-0.1(7)}}=46 806.997…\) not doubled  (P<51200)
(c) (i) \(\frac{25600}{280}\) OR 91.4 OR 640:7
(ii) value 9.31…
after 10 years

Question

Initially a tank contains 10 000 litres of liquid. At the time t = 0 minutes a tap is opened, and liquid then flows out of the tank. The volume of liquid, V litres, which remains in the tank after t minutes is given by
V = 10000(0.933t)
(a) Find the value of V after 5 minutes.
(b) Find how long, to the nearest second, it takes for half of the initial amount of liquid to flow out of the tank.
(c) The tank is regarded as effectively empty when 95% of the liquid has flowed out. Show that it takes almost three-quarters of an hour for this to happen.
(d) (i) Find the value of 10 000 – V when t = 0.001 minutes.
(ii) Hence or otherwise, estimate the initial flow rate of the liquid.
Given your answer in litres per minute, correct to two significant figures.

Answer/Explanation

Ans
(a) \(V(5)=10000 \times (0.933^5)=7069.8….=7070 (3 sf)\)
(b) \(5000 = 10000 \times (0.933)^t \Leftrightarrow 0.5 = 0.933^t \Leftrightarrow t = \frac{In(0.5)}{In(0.933)}=9.9949\)
After 10 minutes 0 seconds, to nearest second (or 600 seconds).
(c) \(0.05 = 0.933^t \Leftrightarrow t = 46.197\) minutes ≈ 3/4 hour
(d) (i) \(10000-10000(0.966)^{0.001}=0.693\)
(ii) Initial flow rate =\(\frac{0.693}{0.001}=693=690(2 sf)\)
        OR
Later on, we may use derivatives to find the rate: \(\frac{dV}{dr}=690\) (when t = 0)

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