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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 20 a ii Solution by
Codex 0 2026-10-18
With the same separation of variables ansatz, givesOn , integration yields and . The choice , gives a polynomial that extends smoothly to all real :It is nonconstant and real everywhere, and , including on the axes.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 20 a i Solution by
Codex 0 2026-10-18
Use separation of variables with on a region where the factors do not vanish. The equation becomes , soThe separated equations are and , giving . Choosing gives the nonconstant real solutionAlthough the division used , the resulting solution extends smoothly across and directly satisfies everywhere.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 17 c ii Solution by
Codex 0 2026-10-18
Successive substitution in the matrix recurrence relation gives . In the matrix power of , every adjacent cancels, so . ConsequentlyHere for , and . The formula also gives correctly at .
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 17 c i Solution by
Codex 0 2026-10-18
The eigenvector equations combine column by column into . Since , multiplication by the inverse matrix gives the diagonalizationThe nonzero determinant of ensures that its columns form an eigenbasis.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 17 b iii Solution by
Codex 0 2026-10-18
Let and . The specified entries mean . Multiplication of determinants therefore givesThis is also the direct row- and column-scaling rule, and it includes zero scaling factors.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 17 b ii Solution by
Codex 0 2026-10-18
Multiplying an matrix by scales every row, giving a factor in its determinant. Also , since . HenceOnly needs to be invertible; the result remains valid when .
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 17 b i Solution by
Codex 0 2026-10-18
The determinant is multiplicative under matrix multiplication:This also holds when is singular, in which case both sides vanish.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 15 b iii Solution by
Codex 0 2026-10-18
The first initial condition gives . The original first equation gives , whereas the general solution gives . Thus andThe corresponding has , checking the other initial condition.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 15 b ii Solution by
Codex 0 2026-10-18
The characteristic equation is , with roots . ThereforeIf the second component is also wanted, the relation givesThese are damped oscillations associated with the complex conjugate eigenvalues of the coefficient matrix.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 15 a ii Solution by
Codex 0 2026-10-18
The characteristic equation has a repeated root , suggesting the substitution . By the product rule,so . The inhomogeneous linear differential equation reduces to . Two integrations giveThereforeThe two arbitrary constants account for the repeated-root homogeneous solution.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 15 a i Solution by
Codex 0 2026-10-18
The characteristic equation of the homogeneous linear differential equation isThe real general solution is . The first boundary value gives . At , and , so . Hence
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 14 d iii Solution by
Codex 0 2026-10-18
The probability density is an even function. Its derivative simplifies toIt increases for and decreases for , with maximum height . Thus the most probable value, or mode, is . The original sketch shows the symmetric, slightly flattened peak and long tails.
There is a genuine qualification to the request for a mean: the mean does not exist. The integral representation shows that as , soThe two divergent integrals cannot be subtracted to define an expected value. Symmetry does give the Cauchy principal value , but that is not the mean. Equivalently, adding the bounded variable cannot make the first absolute moment of a Cauchy distribution finite.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 14 d ii Solution by
Codex 0 2026-10-18
Insert the two densities into the convolution of probability densities:Using the antiderivative with respect to givesThis is the convolution of a Cauchy distribution and a uniform distribution with half-width .
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 14 d i Solution by
Codex 0 2026-10-18
A uniform distribution has constant probability density on its interval, and the integral of that density must be one. The interval has length one, soChanging either endpoint value has no effect on the probability distribution.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 14 c ii Solution by
Codex 0 2026-10-18
Use the law of total probability to average the conditional interval probability over :The integrands are nonnegative, so Tonelli's theorem permits interchanging the integrals. ThusReading off the probability density gives the convolution of probability densities:
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 14 c i Solution by
Codex 0 2026-10-18
The independence of random variables means that fixing leaves the conditional probability density of equal to . Given that value of , is the translate . HenceAt a continuity point of , the expression is . Since , this conditioning uses a conditional probability density, rather than division by the probability of the single event .
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 14 b ii Solution by
Codex 0 2026-10-18
Apply the preceding interval formula on a short interval and use the change of variables :Comparison with the defining integral for a probability density givesAt continuity points, this can also be read as .
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 14 b i Solution by
Codex 0 2026-10-18
Subtracting the constant translates the event without changing its probability. ThereforeThis is a translation of the interval in the probability distribution of .
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2019 Paper 2 11 a iii Solution by
Codex 0 2026-10-18
Writing and comparing Cartesian coordinates givesThe unique solution, guaranteed by the basis property, is . Thus
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