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# Help please! Thank you <3

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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 $$|a - n|$$?

Jun 19, 2020

#1
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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 $$(A, B) = \left(3,-2\right)$$

Jun 19, 2020
#2
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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

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

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