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Date May 2012 Marks available 2 Reference code 12M.1.hl.TZ2.11
Level HL only Paper 1 Time zone TZ2
Command term Calculate Question number 11 Adapted from N/A

Question

Consider the following functions:

\[f(x) = \frac{{2{x^2} + 3}}{{75}},{\text{ }}x \geqslant 0\]

\[g(x) = \frac{{\left| {3x - 4} \right|}}{{10}},{\text{ }}x \in \mathbb{R}{\text{ }}.\]

State the range of f and of g .

[2]
a.

Find an expression for the composite function \(f \circ g(x)\) in the form \(\frac{{a{x^2} + bx + c}}{{3750}}\), where \(a,{\text{ }}b{\text{ and }}c \in \mathbb{Z}\) .

[4]
b.

(i)     Find an expression for the inverse function \({f^{ - 1}}(x)\) .

(ii)     State the domain and range of \({f^{ - 1}}\) .

[4]
c.

The domains of f and g are now restricted to {0, 1, 2, 3, 4} .

By considering the values of f and g on this new domain, determine which of f and g could be used to find a probability distribution for a discrete random variable X , stating your reasons clearly.

[6]
d.

Using this probability distribution, calculate the mean of X .

[2]
e.

Markscheme

\(f(x) \geqslant \frac{1}{{25}}\)     A1 

\(g(x) \in \mathbb{R},{\text{ }}g(x) \geqslant 0\)     A1

[2 marks]

a.

\(f \circ g(x) = \frac{{2{{\left( {\frac{{3x - 4}}{{10}}} \right)}^2} + 3}}{{75}}\)     M1A1

\( = \frac{{\frac{{2(9{x^2} - 24x + 16)}}{{100}} + 3}}{{75}}\)     (A1)

\( = \frac{{9{x^2} - 24x + 166}}{{3750}}\)     A1

[4 marks]

b.

(i)     METHOD 1

\(y = \frac{{2{x^2} + 3}}{{75}}\)

\({x^2} = \frac{{75y - 3}}{2}\)     M1

\(x = \sqrt {\frac{{75y - 3}}{2}} \)     (A1)

\( \Rightarrow {f^{ - 1}}(x) = \sqrt {\frac{{75x - 3}}{2}} \)     A1  

Note: Accept ± in line 3 for the (A1) but not in line 4 for the A1. 

Award the A1 only if written in the form \({f^{ - 1}}(x) = \) .

 

METHOD 2

\(y = \frac{{2{x^2} + 3}}{{75}}\)

\(x = \frac{{2{y^2} + 3}}{{75}}\)     M1

\(y = \sqrt {\frac{{75x - 3}}{2}} \)     (A1)

\( \Rightarrow {f^{ - 1}}(x) = \sqrt {\frac{{75x - 3}}{2}} \)     A1  

Note: Accept ± in line 3 for the (A1) but not in line 4 for the A1. 

Award the A1 only if written in the form \({f^{ - 1}}(x) = \) .

 

(ii)     domain: \(x \geqslant \frac{1}{{25}}\) ; range: \({f^{ - 1}}(x) \geqslant 0\)     A1

[4 marks]

 

c.

probabilities from \(f(x)\) :

     A2

 

Note: Award A1 for one error, A0 otherwise.

 

probabilities from \(g(x)\) :

     A2

Note: Award A1 for one error, A0 otherwise. 

 

only in the case of \(f(x)\) does \(\sum {P(X = x) = 1} \) , hence only \(f(x)\) can be used as a probability mass function     A2

[6 marks]

d.

\(E(x) = \sum {x \cdot {\text{P}}(X = x)} \)     M1

\( = \frac{5}{{75}} + \frac{{22}}{{75}} + \frac{{63}}{{75}} + \frac{{140}}{{75}} = \frac{{230}}{{75}}\left( { = \frac{{46}}{{15}}} \right)\)     A1

[2 marks]

e.

Examiners report

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

a.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

b.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

c.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

d.

In (a), the ranges were often given incorrectly, particularly the range of g where the modulus signs appeared to cause difficulty. In (b), it was disappointing to see so many candidates making algebraic errors in attempting to determine the expression for \(f \circ g(x)\). Many candidates were unable to solve (d) correctly with arithmetic errors and incorrect reasoning often seen. Since the solution to (e) depended upon a correct choice of function in (d), few correct solutions were seen with some candidates even attempting to use integration, inappropriately, to find the mean of X.

e.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.5 » Expected value (mean), mode, median, variance and standard deviation.
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