Derive version 6.10 DfW file saved on 28 Jun 2005
f(x):=e7c0^(-x) - 1
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\viewkind4\uc1\pard\b\i\f0\fs28 Mathematical Methods (CAS) 2002 Examination 1 part 1 sample solutions Q 14 to 27\b0\i0\fs24\par
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\viewkind4\uc1\pard\f0\fs22 Note: To use \b\i Derive \b0\i0 efficiently, students should be familiar with the 'tick plus equals' and 'tick plus approximately equals' evaluation buttons. These simultaneously 'author' and 'evaluate' expressions exactly and numerically respectively. Students should also be familiar with the use of defined functions in the form f(x):= \i rule of the function\i0 , such as in the sample solutions for Question 10.\par
\par
Some questions are conceptual in nature, that is, technology will not be of assistance, for example, Question 7. For other questions, such as Question 1, the facility of \b\i Derive \b0\i0 to quickly produce scaled graphs, and the like, means that such problems could be tackled by inspection of each alternative, although this is not a recommended approach.In many cases students will reason what is likely to be the answer, and then confirm this with \b\i Derive\b0\i0 . \par
\lang3081\f1\fs24\par
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\viewkind4\uc1\pard\b\i\f0\fs24 Question 14\par
\par
\b0\i0\fs22 B is the correct answer.This can be identified by application of the chain rule pattern for the composite function log(f(x)). The corresponding derivative is f'(x)/f(x), as f(x) = cos(2x), this means that A and B are the only possibilities. As the derivative of cos(2x) is -2sin(2x) B is correct (the negative is sufficient to discriminate here). Alternatively,differentiation by \b\i Derive \b0\i0 gives:\fs24\par
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\viewkind4\uc1\pard\f0\fs24\par
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CExpnObj8User
LOG(COS(2*x))8 Dif(#1,x)DIF(LOG(COS(2*x)),x)XSimp(Dif(#1,x)){Gz?-2*TAN(2*x)r{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 15\par
\par
\b0\i0\fs22 E can be eliminated since the normal is a linear function. As each of A to D has a different gradient, this information is sufficent to identify the correct alternative.The gradient of the normal is -1/f'(\'b9), where f(x) = x*sin(x). Hence the correct alternative is D, since the gradient is 1/\'b9.\par
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h
Simp(User)x*SIN(x)8User -1/f'(pi)Simp(#5)1/pi"D:{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\f0\fs22 alternatively, the rule of the normal, and hence the correct alternative, can be obtained in a single evaluation using the form y - f(\'b9) = -1/f'(\'b9)(x-\'b9) and solving for y:\par
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8ThUser!SOLVE(y-f(pi)=-1/f'(pi)*(x-pi),y)`Simp(#7)y=(x-pi)/pi {\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 16\par
\par
\b0\i0\fs22 This can be evaluated directly using a defined function, to identify C as the correct alternative:\fs24\par
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89User f(x):=#e^(-x)-18I`YUser
f'(0)iy Simp(#10)-1F{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 17\b0\i0\par
\par
\fs22 This is a formulation question, no evaluation is required. With x = 3, h = 0.02, so by substitution (without evaluation) the correct form is A.\par
\fs24\par
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\viewkind4\uc1\pard\b\i\f0\fs24 Question 18\par
\par
\b0\i0\fs22 A conceptual question. All graphs include required zeroes of the function, C can be eliminated as it does not have the correct location for stationary points.As the derivative is required to be positive for x < -1, only B and D are possibilities. B is the only alternative for which the derivative is negative (except at x = 0) for\par
x > -1.\fs24\par
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\viewkind4\uc1\pard\b\i\f0\fs24 Question 19\par
\par
\b0\i0\fs22 This question tests knowledge of the chain rule, the derivative of g(x) = e^(f(x)) is g'(x) = f'(x)*e^(f(x). since g'(x) = -2xe^(-x^2))this matches f(x) with -x^2. Thus the correct alternative is A. Another approach is to anti-differentiate g'(x) and hence identify f(x) = -x^2:\fs24\par
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Simp(User)-2*x*#e^(-x^2)8!Q
Int(#12,x)
INT(-2*x*#e^(-x^2),x)xaSimp(Int(#12,x)) #e^(-x^2)/{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 20\par
\par
\b0\i0\fs22 The correct alternative, B, can be identified by direct evaluation of an anti-derivative, and inclusion of an arbitrary constant c:\par
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8User
2*COS(5*x)8'
Int(#15,x)INT(2*COS(5*x),x)X7gSimp(Int(#15,x)){Gz?2*SIN(5*x)/5w<{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs22 Question 21\par
\par
\b0\i0 This is a formulation and evaluation question, where distance travelled is modelled by a definite integral. As there are no turning points (ie where v(t) = 0)during the time interval specified, and the function is non-negative over this interval (see graph), the definite integral can be used directly without having to partition the interval or use modulus, to identify D as the correct alternative:\fs24\par
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8UserSIN(pi*t/30)^2CDispOleObj-$ CDispItemD>
!"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntryF@${Ole
CONTENTS1CompObjj
F$Picture (Device Independent Bitmap)
StaticDib9q
!"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~BM16(^1CONTENTS8 T
Int(#18,t)INT(SIN(pi*t/30)^2,t,0,30)d t Simp(Int(#18,t))15 {\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 22\par
\par
\b0\i0\fs22 This is a conceptual question, involving recognition that where a function changes sign over an interval, the definite integral does not give the area bounded by the function and the horizontal axis over that interval. The required area is given by the definite integral from a to b, plus the \b\i negative \b0\i0 of the definite integral from b to c. The latter effect can be achieved by changing the order of the terminals in the second definite integral, thus, D is the correct alternative.\fs24\par
}
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\viewkind4\uc1\pard\b\i\f0\fs24 Question 23\par
\par
\b0\i0\fs22 The anti-differentiation for this question could be carried out by hand, the required definite integral being evaluated as -x^2 + 8x from 3.5 to 4 (students should be able to do this by hand):\fs24\par
}
8c
User-16+32-(-(3.5)^2)-8*(3.5)x
Approx(#21)0.25
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\viewkind4\uc1\pard\f0\fs22 to obtain the correct answer B. Alternatively, it may be evaluated directly using \b\i Derive\b0\i0 :\fs24\par
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8
User-2*x+88
$
Int(#23,x)INT(-2*x+8,x,3.5,4)4dSimp(Int(#23,x)){Gz?1/4t{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 24\par
\par
\b0\i0\fs22 This is a formulation question, no evaluation is required. Students should recognise that the hypergeometric distribution applies, and that Pr(at least one chip of value $10) in the sample = 1 - Pr(no chip of value $10) in the sample. Thus the answer is 1 -\fs16 15\fs22 C\fs16 4\fs22 /\fs16 20\fs22 C\fs16 4\fs22 , that is, alternative A.\par
}
2{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 25\par
\par
\b0\i0\fs22 This question requires the solution of two simultaneous equations in n and p, which can be readily done without any calculation. The mean \'b5 = 10 = np and the variance (standard deviation squared), \'de^2 = 9 = np(1-p). Thus 9 = 10(1-p), so p = 0.1, and A is the correct alternative. These can also be solved, for both n and p, using \b\i Derive\b0\i0 :\par
}
8BRSolve(User,[n,p])!SOLVE([n*p=10,n*p*(1-p)=9],[n,p])8bSimp(Solve(User,[n,p])){Gz?[n=100 AND p=1/10]
M{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 26\par
\par
\b0\i0\fs22 This question tests understanding of the transformation between the standard normal distribution X~N(0,1), some other normal distributions, X~N(u,\'de^2), and the symmetry of a normal distribution. So, if \'b5 = 4.7 and \'de = 1.2 then Pr( X < 3.5) = \par
Pr(Z < -1) since Z = (x - \'b5)/\'de = (3.5 - 4.7)/1.2 = -1. By symmetry of the standard normal distribution, Pr(Z < -1)is the same as Pr(Z > 1), alternative B.\par
}
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<{\rtf1\ansi\ansicpg1252\deff0\deflang3081{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}}
\viewkind4\uc1\pard\b\i\f0\fs24 Question 27\b0\i0\fs22\par
The middle 95% will be in the range \'b5 - 6 to \'b5 + 6, thus the upper 97.5 % will be in the range \'b5 - 6 to infinity. For each of the alternatives A - D the largest of these values of \'b5, u = 254 gives 2.5% of packets less than 254 - 6 = 248, so alternative E is the only one with a large enough value for \'b5 that could work anyway. The problem can also be solved directly using \b\i Derive\b0\i0 :\par
}
8o
P
Solve(User,)&NSOLVE(NORMAL(250,03bc,3)=0.01,03bc)P
Simp(Solve(User,))Q?03bc=256.9790436
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\viewkind4\uc1\pard\f0\fs22 thus, as before the correct alternative is E.\fs24\par
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\viewkind4\uc1\pard\f0\fs24\'a9 \f1 2004 Draft sample Derive working solutions by David Leigh-Lancaster & Dr. Pam Norton. These solutions may be used by schools on a not for profit basis. Care has been taken in produding these materials but they have not been checked. No responsibility is accepted by the authors for any errors they may contain.\par
}
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