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

 

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Question

Let \(f(x) = 3 + \frac{{20}}{{{x^2} – 4}}\) , for \(x \ne \pm 2\) . The graph of f is given below.


The y-intercept is at the point A.

(i)     Find the coordinates of A.

(ii)    Show that \(f'(x) = 0\) at A.

[7]
a.

The second derivative \(f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}\) . Use this to

(i)     justify that the graph of f has a local maximum at A;

(ii)    explain why the graph of f does not have a point of inflexion.

[6]
b.

Describe the behaviour of the graph of \(f\) for large \(|x|\) .

[1]
c.

Write down the range of \(f\) .

[2]
d.
Answer/Explanation

Markscheme

(i) coordinates of A are \((0{\text{, }} – 2)\)     A1A1     N2

(ii) derivative of \({x^2} – 4 = 2x\) (seen anywhere)     (A1)

evidence of correct approach     (M1)

e.g. quotient rule, chain rule

finding \(f'(x)\)     A2

e.g. \(f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)\) , \(\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}\)

substituting \(x = 0\) into \(f'(x)\) (do not accept solving \(f'(x) = 0\) )     M1

at A \(f'(x) = 0\)     AG     N0

[7 marks]

a.

(i) reference to \(f'(x) = 0\) (seen anywhere)     (R1)

reference to \(f”(0)\) is negative (seen anywhere)     R1

evidence of substituting \(x = 0\) into \(f”(x)\)     M1

finding \(f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}\) \(\left( { = – \frac{5}{2}} \right)\)     A1

then the graph must have a local maximum     AG

(ii) reference to \(f”(x) = 0\) at point of inflexion     (R1)

recognizing that the second derivative is never 0     A1     N2

e.g. \(40(3{x^2} + 4) \ne 0\) , \(3{x^2} + 4 \ne 0\) , \({x^2} \ne  – \frac{4}{3}\) , the numerator is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

b.

correct (informal) statement, including reference to approaching \(y = 3\)     A1     N1

e.g. getting closer to the line \(y = 3\) , horizontal asymptote at \(y = 3\)

[1 mark]

c.

correct inequalities, \(y \le – 2\) , \(y > 3\) , FT from (a)(i) and (c)     A1A1     N2

[2 marks]

d.

Question

Let \(f(x) = \sqrt x \) . Line L is the normal to the graph of f at the point (4, 2) .

In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .


Show that the equation of L is \(y = – 4x + 18\) .

[4]
a.

Point A is the x-intercept of L . Find the x-coordinate of A.

[2]
b.

Find an expression for the area of R .

[3]
c.

The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .

[8]
d.
Answer/Explanation

Markscheme

finding derivative     (A1)

e.g. \(f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}\)

correct value of derivative or its negative reciprocal (seen anywhere)     A1

e.g. \(\frac{1}{{2\sqrt 4 }}\) , \(\frac{1}{4}\)

gradient of normal =  \(\frac{1}{{{\text{gradient of tangent}}}}\) (seen anywhere)     A1

e.g. \( – \frac{1}{{f'(4)}} = – 4\) , \( – 2\sqrt x \)

substituting into equation of line (for normal)     M1

e.g. \(y – 2 = – 4(x – 4)\)

\(y = – 4x + 18\)     AG     N0

[4 marks]

a.

recognition that \(y = 0\) at A     (M1)

e.g. \( – 4x + 18 = 0\)

\(x = \frac{{18}}{4}\) \(\left( { = \frac{9}{2}} \right)\)     A1     N2

[2 marks]

b.

splitting into two appropriate parts (areas and/or integrals)     (M1)

correct expression for area of R     A2     N3

e.g. area of R = \(\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( – 4x + 18){\rm{d}}x} \) , \(\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2\) (triangle)

Note: Award A1 if dx is missing.

[3 marks]

c.

correct expression for the volume from \(x = 0\) to \(x = 4\)     (A1)

e.g. \(V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x\) , \({\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x\) , \(\int_0^4 {\pi x{\rm{d}}x} \)

\(V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4\)     A1

\(V = \pi \left( {\frac{1}{2} \times 16 – \frac{1}{2} \times 0} \right)\)     (A1)

\(V = 8\pi \)     A1

finding the volume from \(x = 4\) to \(x = 4.5\)

EITHER

recognizing a cone     (M1)

e.g. \(V = \frac{1}{3}\pi {r^2}h\)

\(V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}\)     (A1)

\( = \frac{{2\pi }}{3}\)     A1

total volume is \(8\pi  + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\)     A1     N4

OR

\(V = \pi \int_4^{4.5} {{{( – 4x + 18)}^2}{\rm{d}}x} \)     (M1)

\( = \int_4^{4.5} {\pi (16{x^2} – 144x + 324){\rm{d}}x} \)

\( = \pi \left[ {\frac{{16}}{3}{x^3} – 72{x^2} + 324x} \right]_4^{4.5}\)     A1

\( = \frac{{2\pi }}{3}\)     A1

total volume is \(8\pi  + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\)     A1     N4

[8 marks]

d.

Question

Let \(f(x) = 6 + 6\sin x\) . Part of the graph of f is shown below.


The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.

Solve for \(0 \le x < 2\pi \)

(i)     \(6 + 6\sin x = 6\) ;

(ii)    \(6 + 6\sin x = 0\) .

[5]
a(i) and (ii).

Write down the exact value of the x-intercept of f , for \(0 \le x < 2\pi \) .

[1]
b.

The area of the shaded region is k . Find the value of k , giving your answer in terms of \(\pi \) .

[6]
c.

Let \(g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

[2]
d.

Let \(g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Given that \(\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x}  = k\) and \(0 \le p < 2\pi \) , write down the two values of p.

[3]
e.
Answer/Explanation

Markscheme

(i) \(\sin x = 0\)     A1

\(x = 0\) , \(x = \pi \)     A1A1     N2

(ii) \(\sin x = – 1\)     A1

\(x = \frac{{3\pi }}{2}\)    A1     N1

[5 marks]

a(i) and (ii).

\(\frac{{3\pi }}{2}\)     A1     N1

[1 mark]

b.

evidence of using anti-differentiation     (M1)

e.g. \(\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x} \)

correct integral \(6x – 6\cos x\) (seen anywhere)     A1A1

correct substitution     (A1)

e.g. \(6\left( {\frac{{3\pi }}{2}} \right) – 6\cos \left( {\frac{{3\pi }}{2}} \right) – ( – 6\cos 0)\) , \(9\pi  – 0 + 6\)

\(k = 9\pi + 6\)     A1A1     N3

[6 marks]

c.

translation of \(\left( {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right)\)     A1A1     N2

[2 marks]

d.

recognizing that the area under g is the same as the shaded region in f     (M1)

\(p = \frac{\pi }{2}\) , \(p = 0\)     A1A1     N3

[3 marks]

e.

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

Consider \(f(x) = {x^2} + qx + r\). The graph of \(f\) has a minimum value when \(x =  – 1.5\).

The distance between the two zeros of \(f\) is 9.

Show that the two zeros are 3 and \( – 6\).

[2]
a.

Find the value of \(q\) and of \(r\).

[4]
b.
Answer/Explanation

Markscheme

recognition that the \(x\)-coordinate of the vertex is \( – 1.5\) (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)axis of symmetry is \( – 1.5\), sketch, \(f'( – 1.5) = 0\)

correct working to find the zeroes     A1

eg\(\,\,\,\,\,\)\( – 1.5 \pm 4.5\)

\(x =  – 6\) and \(x = 3\)     AG     N0

[2 marks]

a.

METHOD 1 (using factors)

attempt to write factors     (M1)

eg\(\,\,\,\,\,\)\((x – 6)(x + 3)\)

correct factors     A1

eg\(\,\,\,\,\,\)\((x – 3)(x + 6)\)

\(q = 3,{\text{ }}r =  – 18\)    A1A1     N3

METHOD 2 (using derivative or vertex)

valid approach to find \(q\)     (M1)

eg\(\,\,\,\,\,\)\(f'( – 1.5) = 0,{\text{ }} – \frac{q}{{2a}} =  – 1.5\)

\(q = 3\)    A1

correct substitution     A1

eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0\)

\(r =  – 18\)    A1

\(q = 3,{\text{ }}r =  – 18\)    N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations     (M1)

eg\(\,\,\,\,\,\)\(9 + 3q + r = 0,{\text{ }}36 – 6q + r = 0\)

one correct value

eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r =  – 18\)     A1

correct substitution     A1

eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0,{\text{ }}{3^2} + 3q – 18 = 0,{\text{ }}36 – 6q – 18 = 0\)

second correct value     A1

eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r =  – 18\)

\(q = 3,{\text{ }}r =  – 18\)    N3

[4 marks]

b.

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.

Question

A quadratic function \(f\) can be written in the form \(f(x) = a(x – p)(x – 3)\). The graph of \(f\) has axis of symmetry \(x = 2.5\) and \(y\)-intercept at \((0,{\text{ }} – 6)\)

Find the value of \(p\).

[3]
a.

Find the value of \(a\).

[3]
b.

The line \(y = kx – 5\) is a tangent to the curve of \(f\). Find the values of \(k\).

[8]
c.
Answer/Explanation

Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an \(x\)-intercept     (M1)

eg\(\,\,\,\,\,\)\(x – 3 = 0\), M17/5/MATME/SP1/ENG/TZ1/09.a/M

valid approach     (M1)

eg\(\,\,\,\,\,\)\(3 – 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5\)

\(p = 2\)     A1     N2

METHOD 2 (expanding f (x)) 

correct expansion (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a{x^2} – a(3 + p)x + 3ap,{\text{ }}{x^2} – (3 + p)x + 3p\)

valid approach involving equation of axis of symmetry     (M1)

eg\(\,\,\,\,\,\)\(\frac{{ – b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}\)

\(p = 2\)     A1     N2

METHOD 3 (using derivative)

correct derivative (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a(2x – 3 – p),{\text{ }}2x – 3 – p\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f’(2.5) = 0\)

\(p = 2\)     A1     N2

[3 marks]

a.

attempt to substitute \((0,{\text{ }} – 6)\)     (M1)

eg\(\,\,\,\,\,\)\( – 6 = a(0 – 2)(0 – 3),{\text{ }}0 = a( – 8)( – 9),{\text{ }}a{(0)^2} – 5a(0) + 6a =  – 6\)

correct working     (A1)

eg\(\,\,\,\,\,\)\( – 6 = 6a\)

\(a =  – 1\)     A1     N2

[3 marks]

b.

METHOD 1 (using discriminant)

recognizing tangent intersects curve once     (M1)

recognizing one solution when discriminant = 0     M1

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 =  – {x^2} + 5x – 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} – 5x + kx + 1 = 0\)

correct discriminant (if seen explicitly, not just in quadratic formula)     A1

eg\(\,\,\,\,\,\)\({(k – 5)^2} – 4,{\text{ }}25 – 10k + {k^2} – 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(k – 5 =  \pm 2,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

METHOD 2 (using derivatives)

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 =  – {x^2} + 5x – 6\)

recognizing derivative/slope are equal     (M1)

eg\(\,\,\,\,\,\)\(f’ = {m_T},{\text{ }}f’ = k\)

correct derivative of \(f\)     (A1)

eg\(\,\,\,\,\,\)\( – 2x + 5\)

attempt to set up equation in terms of either \(x\) or \(k\)     M1

eg\(\,\,\,\,\,\)\(( – 2x + 5)x – 5 =  – {x^2} + 5x – 6,{\text{ }}k\left( {\frac{{5 – k}}{2}} \right) – 5 =  – {\left( {\frac{{5 – k}}{2}} \right)^2} + 5\left( {\frac{{5 – k}}{2}} \right) – 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} – 1 = 0,{\text{ }}{k^2} – 10k + 21 = 0\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(x =  \pm 1,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

[8 marks]

c.

Question

Let \(f(x) = {x^2}\). The following diagram shows part of the graph of \(f\).

M17/5/MATME/SP1/ENG/TZ2/10

The line \(L\) is the tangent to the graph of \(f\) at the point \({\text{A}}( – k,{\text{ }}{k^2})\), and intersects the \(x\)-axis at point B. The point C is \(( – k,{\text{ }}0)\).

The region \(R\) is enclosed by \(L\), the graph of \(f\), and the \(x\)-axis. This is shown in the following diagram.

M17/5/MATME/SP1/ENG/TZ2/10.d

Write down \(f'(x)\).

[1]
a.i.

Find the gradient of \(L\).

[2]
a.ii.

Show that the \(x\)-coordinate of B is \( – \frac{k}{2}\).

[5]
b.

Find the area of triangle ABC, giving your answer in terms of \(k\).

[2]
c.

Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).

[7]
d.
Answer/Explanation

Markscheme

\(f'(x) = 2x\)     A1     N1

[1 mark]

a.i.

attempt to substitute \(x =  – k\) into their derivative     (M1)

gradient of \(L\) is \( – 2k\)     A1     N2

[2 marks]

a.ii.

METHOD 1 

attempt to substitute coordinates of A and their gradient into equation of a line     (M1)

eg\(\,\,\,\,\,\)\({k^2} =  – 2k( – k) + b\)

correct equation of \(L\) in any form     (A1)

eg\(\,\,\,\,\,\)\(y – {k^2} =  – 2k(x + k),{\text{ }}y =  – 2kx – {k^2}\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(y = 0\)

correct substitution into \(L\) equation     A1

eg\(\,\,\,\,\,\)\( – {k^2} =  – 2kx – 2{k^2},{\text{ }}0 =  – 2kx – {k^2}\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  – {k^2}\)

\(x =  – \frac{k}{2}\)     AG     N0

METHOD 2

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{gradient}} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}},{\text{ }} – 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}\)

recognizing \(y = 0\) at B     (A1)

attempt to substitute coordinates of A and B into slope formula     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}},{\text{ }}\frac{{ – {k^2}}}{{x + k}}\)

correct equation     A1

eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}} =  – 2k,{\text{ }}\frac{{ – {k^2}}}{{x + k}} =  – 2k,{\text{ }} – {k^2} =  – 2k(x + k)\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  – {k^2}\)

\(x =  – \frac{k}{2}\)     AG     N0

[5 marks]

b.

valid approach to find area of triangle     (M1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)\)

area of \({\text{ABC}} = \frac{{{k^3}}}{4}\)     A1     N2

[2 marks]

c.

METHOD 1 (\(\int {f – {\text{triangle}}} \))

valid approach to find area from \( – k\) to 0     (M1)

eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ – k} f } \)

correct integration (seen anywhere, even if M0 awarded)     A1

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ – k}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(0 – \frac{{{{( – k)}^3}}}{3}\), area from \( – k\) to 0 is \(\frac{{{k^3}}}{3}\)

Note:     Award M0 for substituting into original or differentiated function.

attempt to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {f(x){\text{d}}x – {\text{ triangle}}} \)

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

METHOD 2 (\(\int {(f – L)} \))

valid approach to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ – k}^{ – \frac{k}{2}} {{x^2} – ( – 2kx – {k^2}){\text{d}}x + \int_{ – \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ – k}^{ – \frac{k}{2}} {(f – L) + \int_{ – \frac{k}{2}}^0 f } } } \)

correct integration (seen anywhere, even if M0 awarded)     A2

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ – k}^{ – \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ – \frac{k}{2}}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { – \frac{k}{2}} \right)}^2} + {k^2}\left( { – \frac{k}{2}} \right)} \right) – \left( {\frac{{{{( – k)}^3}}}{3} + k{{( – k)}^2} + {k^2}( – k)} \right) + (0) – \left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3}} \right)\)

Note:     Award M0 for substituting into original or differentiated function.

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} – \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} – \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} – {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

[7 marks]

d.

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