Home / IB Math Analysis & Approaches Questionbank-Topic: SL 2.4 Determine key features of graphs SL Paper 1

IB Math Analysis & Approaches Questionbank-Topic: SL 2.4 Determine key features of graphs SL Paper 1

Question

Consider \(f(x) = 2k{x^2} – 4kx + 1\) , for \(k \ne 0\) . The equation \(f(x) = 0\) has two equal roots.

Find the value of k .[5]

a.

The line \(y = p\) intersects the graph of f . Find all possible values of p .[2]

b.
Answer/Explanation

Markscheme

valid approach     (M1)

e.g. \({b^2} – 4ac\) , \(\Delta = 0\) , \({( – 4k)^2} – 4(2k)(1)\)

correct equation     A1

e.g. \({( – 4k)^2} – 4(2k)(1) = 0\) , \(16{k^2} = 8k\) , \(2{k^2} – k = 0\)

correct manipulation     A1

e.g. \(8k(2k – 1)\) , \(\frac{{8 \pm \sqrt {64} }}{{32}}\)

\(k = \frac{1}{2}\)     A2     N3

[5 marks]

a.

recognizing vertex is on the x-axis     M1

e.g. (1, 0) , sketch of parabola opening upward from the x-axis

\(p \ge 0\)     A1     N1

[2 marks]

b.

Question

Let  \(f(x) = \frac{x}{{ – 2{x^2} + 5x – 2}}\) for \( – 2 \le x \le 4\) , \(x \ne \frac{1}{2}\) , \(x \ne 2\) . The graph of \(f\) is given below.


The graph of \(f\) has a local minimum at A(\(1\), \(1\)) and a local maximum at B.

Use the quotient rule to show that \(f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\) .[6]

a.

Hence find the coordinates of B.[7]

b.

Given that the line \(y = k\) does not meet the graph of f , find the possible values of k .[3]

c.
Answer/Explanation

Markscheme

correct derivatives applied in quotient rule     (A1)A1A1

\(1\), \( – 4x + 5\)

Note: Award (A1) for 1, A1 for \( – 4x\) and A1 for \(5\), only if it is clear candidates are using the quotient rule.

correct substitution into quotient rule     A1

e.g. \(\frac{{1 \times ( – 2{x^2} + 5x – 2) – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\) , \(\frac{{ – 2{x^2} + 5x – 2 – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)

correct working     (A1)

e.g. \(\frac{{ – 2{x^2} + 5x – 2 – ( – 4{x^2} + 5x)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)

expression clearly leading to the answer     A1

e.g. \(\frac{{ – 2{x^2} + 5x – 2 + 4{x^2} – 5x}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)

\(f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)    AG     N0

[6 marks]

a.

evidence of attempting to solve \(f'(x) = 0\)     (M1)

e.g. \(2{x^2} – 2 = 0\)

evidence of correct working     A1

e.g. \({x^2} = 1,\frac{{ \pm \sqrt {16} }}{4}{\text{, }}2(x – 1)(x + 1)\)

correct solution to quadratic     (A1)

e.g. \(x = \pm 1\)

correct x-coordinate \(x = – 1\) (may be seen in coordinate form \(\left( { – 1,\frac{1}{9}} \right)\) )    A1     N2

attempt to substitute \( – 1\) into f (do not accept any other value)     (M1)

e.g. \(f( – 1) = \frac{{ – 1}}{{ – 2 \times {{( – 1)}^2} + 5 \times ( – 1) – 2}}\)

correct working

e.g. \(\frac{{ – 1}}{{ – 2 – 5 – 2}}\)     A1

correct y-coordinate \(y = \frac{1}{9}\) (may be seen in coordinate form \(\left( { – 1,\frac{1}{9}} \right)\) )    A1     N2

[7 marks]

b.

recognizing values between max and min     (R1)

\(\frac{1}{9} < k < 1\)     A2     N3

[3 marks]

c.

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

The following diagram shows part of the graph of a quadratic function \(f\).

M16/5/MATME/SP1/ENG/TZ2/01

The vertex is at \((3,{\text{ }} – 1)\) and the \(x\)-intercepts at 2 and 4.

The function \(f\) can be written in the form \(f(x) = {(x – h)^2} + k\).

The function can also be written in the form \(f(x) = (x – a)(x – b)\).

Write down the value of \(h\) and of \(k\).[2]

a.

Write down the value of \(a\) and of \(b\).[2]

b.

Find the \(y\)-intercept.[2]

c.
Answer/Explanation

Markscheme

\(h = 3,{\text{ }}k =  – 1\)    A1A1     N2

[2 marks]

a.

\(a = 2,{\text{ }}b = 4{\text{ }}({\text{or }}a = 4,{\text{ }}b = 2)\)    A1A1     N2

[2 marks]

b.

attempt to substitute \(x = 0\) into their \(f\)     (M1)

eg\(\,\,\,\,\,\)\({(0 – 3)^2} – 1,{\text{ }}(0 – 2)(0 – 4)\)

\(y = 8\)    A1     N2

[2 marks]

c.
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