1. Find constants A and B such that \(\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\) for all x such that x ≠ -1 and x ≠ 2. Give your answer as the ordered pair (A,B).

2. Suppose that \(|a - b| + |b - c| + |c - d| + \dots + |m-n| + |n-o| + \cdots+ |x - y| + |y - z| + |z - a| = 20.\) What is the maximum possible value of \(|a - n|\)?

Guest Jun 19, 2020

#1**0 **

Q1)

\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac B{x + 1}\\\dfrac{x + 7}{(x- 2)(x+1)} = \dfrac{A}{x - 2} + \dfrac B{x + 1}\\\)

Multiply by \((x - 2)(x + 1)\) on both sides.

\(x+7=A(x+1)+B(x-2)\\ x+7=(A+B)x+(A-2B)\)

Comparing coefficients, \(\begin{cases}A + B = 1\\A - 2B = 7\end{cases}\)

Solving gives \((A, B) = \left(3,-2\right)\)

MaxWong Jun 19, 2020

#2**0 **

Q2)

By triangle inequality, \(|x| + |y| \geqslant |x + y|\)

By induction, we can generalise this to \(\displaystyle\sum_{i = 1}^n |x_i| \geqslant \left|\sum_{i = 1}^nx_i \right|\).

Therefore

\(|a- b| + |b-c|+|c-d|+\cdots+|m-n| \geqslant |a-n|\\ |n-o|+\cdots+|y-z|+|z-a| \geqslant |n - a| = |a - n|\)

Adding these inequalities gives

\(|a-b|+|b-c|+|c-d|+\cdots+|m-n|+|n-o|+\cdots+|x-y|+|y-z|+|z-a| \geqslant 2|a-n|\\ 20\geqslant 2|a-n|\\ |a-n|\leqslant 10\)

The maximum value of |a - n| is 10.

MaxWong Jun 19, 2020