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Date May 2017 Marks available 2 Reference code 17M.1.sl.TZ2.10
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 10 Adapted from N/A

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.

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.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 6 - Calculus » 6.2 » Derivative of \({x^n}\left( {n \in \mathbb{Q}} \right)\) , \(\sin x\) , \(\cos x\) , \(\tan x\) , \({{\text{e}}^x}\) and \(\ln x\) .
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