Mathematical Methods for Physics and Engineering - Matematica.NET

1.9 HINTS AND ANSWERS1.27 Establish the values of k for which the binomial coefficient p C k is divisible by pwhen p is a prime number. Use your result and the method of induction to provethat n p − n is divisible by p for all integers n and all prime numbers p. Deducethat n 5 − n isdivisibleby30foranyintegern.1.28 An arithmetic progression of integers a n is one in which a n = a 0 + nd, wherea 0and d are integers and n takes successive values 0, 1, 2,....(a) Show that if any one term of the progression is the cube of an integer thenso are infinitely many others.(b) Show that no cube of an integer can be expressed as 7n + 5 for some positiveinteger n.1.29 Prove, by the method of contradiction, that the equationx n + a n−1 x n−1 + ···+ a 1 x + a 0 =0,in which all the coefficients a i are integers, cannot have a rational root, unlessthat root is an integer. Deduce that any integral root must be a divisor of a 0 andhence find all rational roots of(a) x 4 +6x 3 +4x 2 +5x +4=0,(b) x 4 +5x 3 +2x 2 − 10x +6=0.Necessary and sufficient conditions1.30 Prove that the equation ax 2 + bx + c =0,inwhicha, b and c are real and a>0,has two real distinct solutions IFF b 2 > 4ac.1.31 For the real variable x, show that a sufficient, but not necessary, condition forf(x) =x(x + 1)(2x + 1) to be divisible by 6 is that x is an integer.1.32 Given that at least one of a and b, and at least one of c and d, are non-zero,show that ad = bc is both a necessary and sufficient condition for the equationsax + by =0,cx + dy =0,to have a solution in which at least one of x and y is non-zero.1.33 The coefficients a i in the polynomial Q(x) =a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x are allintegers. Show that Q(n) is divisible by 24 for all integers n ≥ 0 if and only if allof the following conditions are satisfied:(i) 2a 4 + a 3 is divisible by 4;(ii) a 4 + a 2 is divisible by 12;(iii) a 4 + a 3 + a 2 + a 1 is divisible by 24.1.9 Hints and answers1.1 (b) The roots are 1, 1 (−7+√ 33) = −0.1569, 1 (−7 − √ 33) = −1.593. (c) −5 and8 87are the values of k that make f(−1) and f( 1 ) equal to zero.4 21.3 (a) a =4,b= 3 23and c = are all positive. Therefore 8 16 f′ (x) > 0forallx>0.(b) f(1) = 5, f(0) = −2 andf(−1) = 5, and so there is at least one root in eachof the ranges 0