Date | November 2015 | Marks available | 5 | Reference code | 15N.2.hl.TZ0.13 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Show that | Question number | 13 | Adapted from | N/A |
Question
The following diagram shows a vertical cross section of a building. The cross section of the roof of the building can be modelled by the curve f(x)=30e−x2400, where −20≤x≤20.
Ground level is represented by the x-axis.
Find f″.
Show that the gradient of the roof function is greatest when x = - \sqrt {200} .
The cross section of the living space under the roof can be modelled by a rectangle CDEF with points {\text{C}}( - a,{\text{ }}0) and {\text{D}}(a,{\text{ }}0), where 0 < a \le 20.
Show that the maximum area A of the rectangle CDEF is 600\sqrt 2 {{\text{e}}^{ - \frac{1}{2}}}.
A function I is known as the Insulation Factor of CDEF. The function is defined as I(a) = \frac{{P(a)}}{{A(a)}} where {\text{P}} = {\text{Perimeter}} and {\text{A}} = {\text{Area of the rectangle}}.
(i) Find an expression for P in terms of a.
(ii) Find the value of a which minimizes I.
(iii) Using the value of a found in part (ii) calculate the percentage of the cross sectional area under the whole roof that is not included in the cross section of the living space.
Markscheme
f'(x) = 30{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}} \bullet - \frac{{2x}}{{400}}\;\;\;\left( { = - \frac{{3x}}{{20}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}} \right) M1A1
Note: Award M1 for attempting to use the chain rule.
f''(x) = - \frac{3}{{20}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}} + \frac{{3{x^2}}}{{4000}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}\;\;\;\left( { = \frac{3}{{20}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}\left( {\frac{{{x^2}}}{{200}} - 1} \right)} \right) M1A1
Note: Award M1 for attempting to use the product rule.
[4 marks]
the roof function has maximum gradient when f''(x) = 0 (M1)
Note: Award (M1) for attempting to find f''\left( { - \sqrt {200} } \right).
EITHER
= 0 A1
OR
f''(x) = 0 \Rightarrow x = \pm \sqrt {200} A1
THEN
valid argument for maximum such as reference to an appropriate graph or change in the sign of f''(x) eg f''( - 15) = 0.010 \ldots ( > 0) and f''( - 14) = - 0.001 \ldots ( < 0) R1
\Rightarrow x = - \sqrt {200} AG
[3 marks]
A = 2a \bullet 30{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}}\;\;\;\left( { = 60a{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}} = - 400g'(a)} \right) (M1)(A1)
EITHER
\frac{{{\text{d}}A}}{{{\text{d}}a}} = 60a{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}} \bullet - \frac{a}{{200}} + 60{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}} = 0 \Rightarrow a = \sqrt {200} {\text{ }}\left( { - 400f''(a) = 0 \Rightarrow a = \sqrt {200} } \right) M1A1
OR
by symmetry eg a = - \sqrt {200} found in (b) or {A_{{\text{max}}}} coincides with f''(a) = 0 R1
\Rightarrow a = \sqrt {200} A1
Note: Award A0(M1)(A1)M0M1 for candidates who start with a = \sqrt {200} and do not provide any justification for the maximum area. Condone use of x.
THEN
{A_{{\text{max}}}} = 60 \bullet \sqrt {200} {{\text{e}}^{ - \frac{{200}}{{400}}}} M1
= 600\sqrt 2 {{\text{e}}^{ - \frac{1}{2}}} AG
[5 marks]
(i) perimeter = 4a + 60{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}} A1A1
Note: Condone use of x.
(ii) I(a) = \frac{{4a + 60{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}}}}{{60a{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}}}} (A1)
graphing I(a) or other valid method to find the minimum (M1)
a = 12.6 A1
(iii) area under roof = \int_{ - 20}^{20} {30{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}} {\text{d}}x M1
= 896.18 \ldots (A1)
area of living space = 60 \cdot (12.6...) \cdot e - {\frac{{(12.6...)}}{{400}}^2} = 508.56...
percentage of empty space = 43.3\% A1
[9 marks]
Total [21 marks]