Division by Zero — Solutions James Cranch March 7, 2014 1. For a STEP question, candidates need to evaluate the integral π Z 0 n X cos(r − 1)xdx. r=1 One student (who is unfortunately not attending the Sheffield STEP Preparation Classes) writes the following: Z 0 π n X cos(r − 1)xdx = n Z X π cos(r − 1)xdx r=1 0 n X r=1 π sin(r − 1)x = r−1 0 r=1 n X sin(r − 1)π sin(r − 1)0 = − r−1 r−1 r=1 = = n X r=1 n X (0 − 0) (since sin kπ = 0 for every integer k) 0 = 0. r=1 Comment on their working. Solution The manipulation is all valid except for the term of the sum where r = 1. Here the student has divided by zero. A correct analysis of this term would observe that cos 0x = 1, and its integral is therefore π. Since all other terms are zero, the whole integral is therefore also π. 2. Which quadratic polynomials x2 + ax + b have roots at x = a and x = b? Solution We’re asking when we have x2 + ax + b = (x − a)(x − b). Equating coefficients, we get that −(a + b) = a, ab = b. 1 and The latter equation gives us that b = 0 or a = 1. In the former case, we get a = 0 from the former equation, and hence the solution a = b = 0. In the latter case, we get that −b − 1 = 1, and hence the solution a = 1, b = −2. (We’d have missed the former solution if we cancelled b from the second equation). 3. Classify all solutions to the system of simultaneous equations x + yz = y + xz = z + xy. Solution Note that x − y + yz − xz = 0. But that factorises, giving us (x − y)(z − 1) = 0. As a result, either x = y or z = 1. We’ll proceed in each case separately. If x = y, then the equations reduce to x + xz = z + x2 . This gives a quadratic equation in x, namely x2 − (1 + z)x + z = 0. This has the solutions x = 1 and x = z. Again, we’ll split into cases. In the first case, we have the solution x = y = 1, z arbitrary, and it’s easy to see that that always works. In the second case we have x = y = z arbitrary, and again it’s easy to see that that too always works. Now consider what happens if z = 1. Then the equations become x + y = 1 + xy. This factorises as (x − 1)(y − 1) = 0, meaning that either x = 1 or y = 1. In the former case we get x = z = 1 and y arbitrary, which clearly works. In the latter we get y = z = 1 and x arbitrary, which again clearly works. Hence there are four families of solutions: where two of the variables are equal to 1, and when all variables are equal. 4. Show that there are no solutions in real numbers to the equation 1 1 1 = + . a+b a b Show that the equation 1 1 1 1 = + + a+b+c a b c is satisfied if and only if a, b and c obey at least one of the equations a + b = 0, b + c = 0 or c + a = 0. 2 Solution to obtain We take the first equation and multiply through by ab(a + b) ab = b(a + b) + a(a + b), which rearranges to a2 + ab + b2 = 0. However, this is nonzero, being equal to equation does have no solutions. 1 2 a2 + (a + b)2 + b2 . So the As for the second equation, we multiply through by abc(a + b + c) to get abc = (a + b + c)(ab + bc + ca). This rearranges to a2 b + a2 c + b2 c + b2 a + c2 a + c2 b + 2abc = 0, which factorises to (a + b)(b + c)(c + a) = 0, which is satisfied if any of those three equations hold. Conversely, if any of those equations hold, and a, b, c are all nonzero then a + b + c is nonzero too, so we can divide out to get the original equation. 5. Let a and b be positive constants. Show that Z 1 1 dx = . 2 (ax + b(1 − x)) ab 0 Solution The integral becomes Z 1 0 dx (b + (a − b)x)2 which, in the case that a 6= b, becomes Z a 1 1 1 1 1 du = − −− = 2 a−b b u a−b a b ab under the substitution u = b + (a − b)x. In the case a = b, the integral just becomes Z 1 dx 1 1 = 2 = . 2 b b ab 0 6. Show that x3 − 3bcx + b3 + c3 can be written in the form (x + b + c)Q(x), where Q(x) is a quadratic expression. Show that 2Q(x) can be written as the sum of three expressions, each of which is a perfect square. It is given that that the equations ay 2 + by + c = 0 and by 2 + cy + a = 0 have a common root k. The coefficients a, b and c are real, and a is not zero. Show that (ac − b2 )k = bc − a2 and find a similar equation involving k 2 . Hence show that (ac − b2 )(ab − c2 ) = (bc − a2 )2 and that a3 − 3bca + b3 + c3 = 0. Deduce that either k = 1 or the two equations are identical. 3 Solution We can check that x = −(b + c) is a root of the cubic so that we know such a factorisation exists. To find Q(x), first note the sum of cubes formula b3 + c3 = (b + c)(b2 − bc + c2 ) so we look for a factorisation of the form x3 − 3bcx + b3 + c3 = (x + b + c)(x2 + Ax + b2 − bc + c2 ). By looking at coefficients in x2 we see that see that we must have 0 = A + b + c, so Q(x) = x2 − (b + c)x + b2 − bc + c2 . We see that 2Q(x) = 2x2 − 2bx − 2cx + 2b2 − 2bc + 2c2 = (x2 − 2bx + b2 ) + (x2 − 2cx + c2 ) + (b2 − 2bc + c2 ) = (x − b)2 + (x − c)2 + (b − c)2 . Since k is a simultaneous solution, ak 2 + bk + c = 0 bk 2 + ck + a = 0. By multiplying the first by b and the second by a and subtracting (to get rid of terms in k 2 ), we get b2 k + cb − ack − a2 = 0, or, in other words, (ac − b2 )k = cb − a2 as required. Multiplying the first by c and the second by b and subtracting, we get cak 2 + c2 − b2 k 2 − ba = 0 which gives us (ac − b2 )k 2 = ab − c2 . Therefore (ac − b2 )(ab − c2 ) = (ac − b2 )(ac − b2 )k 2 = (ac − b2 )2 k 2 = (cb − a2 )2 . Multiplying all this out and simplifying (including a perfectly permissible division by a), we get the required relation a3 − 3bca + b3 + c3 = 0. Therefore, we get (a + b + c)Q(a) = 0. Thus either a + b + c = 0 or Q(a) = 0. In the second case, Q(a) = 0, note that (a − b)2 + (a − c)2 + (b − c)2 = 0, giving a = b = c, which means ‘the equations are identical’. In the first case, a + b + c = 0, so y = 1 is indeed a simultaneous root. We must show that it is the only root so that we can deduce that k = 1. Assume β is another common root. Then ay 2 + by + c = a(y − 1)(y − β) = ay 2 − a(1 + β)y + aβ. Thus b = −a(1 + β) and c = aβ. Therefore since bβ 2 +cβ +a = 0 we get −a(1+β)β 2 +aβ 2 +a = 0 so −β 2 −β 3 +β 2 +1 = 0, i.e. β 3 = 1, to which the only solution is β = 1. Therefore, y = 1 is the only simultaneous solution if it is one, so k = 1. 4 7. (a) Given real numbers a, b, c with a + b + c = 0, show that a3 + b3 + c3 > 0 if and only if a5 + b5 + c5 > 0. (b) Given real numbers a, b, c, d with a + b + c + d = 0, show that a3 + b3 + c3 + d3 > 0 if and only if a5 + b5 + c5 + d5 > 0. 8. Show that the set of real numbers x which satisfy the inequality 70 X k=1 k 5 ≥ x−k 4 is a disjoint union of intervals, the sum of whose lengths is 1988. 5