Absolute value

This week, we learn about the Absolute Value and Properties of the Absolute Value.

How would you use absolute value to solve the following? >

  1. If xx and yy are non-zero integers from 10-10 to 1010 (inclusive), what is the smallest positive value of x+yx+y? \frac{ | x+ y | } { |x| + |y| }?

  2. Share a problem which requires understanding of Absolute Value.

Note by Calvin Lin
5 years, 11 months ago

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Is it 119\frac{1}{19}?, taking x=10,y=9x=-10,y=9 because we need to minimize the numerator and at the same time we have to maximize the denominator.Please verify.

Kishan k - 5 years, 11 months ago

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Yep!

Ahaan Rungta - 5 years, 11 months ago

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For positive integers 1n100 1 \le n \le 100 , let f(n)=i=1100iin. f(n) = \sum_{i=1}^{100} i\left\lvert i-n \right\rvert. Compute f(54)f(55) f(54)-f(55) .

Proposed by Aaron Lin for the NIMO

Ahaan Rungta - 5 years, 11 months ago

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1234...5354+55+56+57...+99+100=2080 -1-2-3-4...-53-54+55+56+57...+99+100 = \boxed{2080}

Piyushkumar Palan - 5 years, 11 months ago

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Problem 1.39. Solve the equation x3+x+1=4|x-3|+|x+1|=4.

Problem 1.40. Show that the equation 2x3+x+1+5x=0.99|2x-3|+|x+1|+|5-x|=0.99 has no solutions.

Problem 1.41. Let a,b>0a,b>0. Find the values of mm for which the equation

xa+xb+x+a+x+b=m(a+b)|x-a|+|x-b|+|x+a|+|x+b|=m(a+b)

has at least one real solution.

Problem 1.42. Find all possible values of the expression

E(x,y,z)=x+yx+y+y+zy+z+z+xz+xE(x,y,z)=\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}

where x,y,zx,y,z are nonzero real numbers.

Problem 1.43. Find all positive real numbers x,x1,x2,,xnx,x_1,x_2,\dots,x_n such that

log(xx1)+log(x1x2)++log(xxn)+log(xx1)+log(xx2)++log(xxn)=logx1+logx2++logxn.|\log(xx_1)|+|\log(x_1x_2)|+\dots+|\log(xx_n)|\\+|\log\left(\frac{x}{x_1}\right)|+|\log\left(\frac{x}{x_2}\right)|+\dots+|\log\left(\frac{x}{x_n}\right)|\\=|\log x_1+\log x_2+\dots+\log x_n|.

Problem 1.44. Prove that for all real numbers a,ba,b, we have

a+b1+a+ba1+a+b1+b\frac{|a+b|}{1+|a+b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}

Problem 1.45. Let nn be an odd positive integer and let x1,x2,,xnx_1,x_2,\dots,x_n be distinct real numbers. Find all one-to-one functions

f:{x1,x2,,xn}{x1,x2,,xn}f:\{x_1,x_2,\dots,x_n\}\to\{x_1,x_2,\dots,x_n\}

such that

f(x1)x1=f(x2)x2==f(xn)xn|f(x_1)-x_1|=|f(x_2)-x_2|=\dots=|f(x_n)-x_n|

Problem 1.46. Suppose that the sequence a1,a2,,ana_1,a_2,\dots,a_n satisfies the following conditions:

a1=0,a2=a1+1,,an=an1+1.\begin{aligned}a_1=0,&|a_2|=|a_1+1|,\dots,&|a_n|=|a_{n-1}+1|.\end{aligned}

Prove that

a1+a2++ann12.\frac{a_1+a_2+\dots+a_n}{n}\ge-\frac{1}{2}.

Problem 1.47. Find real numbers a,b,ca,b,c such that

ax+by+cz+bx+cy+az+cx+ay+bz=x+y+z,|ax+by+cz|+|bx+cy+az|+|cx+ay+bz|=|x|+|y|+|z|,

for all real numbers x,y,zx,y,z.

Source: Mathematical Olympiad Treasures, Titu Andreescu

Cody Johnson - 5 years, 11 months ago

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(FUVEST 2012) Show the intervals of xx in which x210x+213x15|x^2-10x+21| \leq |3x-15| holds.

Guilherme Dela Corte - 5 years, 11 months ago

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The left expression is factored into (x - 3)(x - 7) and the right expression is factored into 3(x-5). Thus, we can get rid of the absolute values by looking at the intervals (,3),(3,5),(5,7),(7,)(-\infty, 3), (3, 5), (5, 7), (7, \infty) and applying the appropriate sign. Then we add the points x=3x = 3 and x=7x = 7 to the solution set because the LHS is 0. The rest of the solution is trivial.

Example (,3)(-\infty, 3) --> (x3)(x7)(x - 3)(x - 7) is positive, 3x153x - 15 is negative. x210x+21153xx27x6=(x1)(x6)0x^2 - 10x + 21 \leq 15 - 3x \rightarrow x^2 - 7x - 6 = (x - 1)(x - 6) \leq 0, so the interval is [1,6]x<3[1,3)[1, 6] \bigwedge x < 3 \rightarrow [1, 3).

Michael Tong - 5 years, 11 months ago

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Michael, haven't you forgot the other interval?

Guilherme Dela Corte - 5 years, 10 months ago

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sir i want to know that while integration and taking the limit from (-x to +x) we get area zero and how could it be so because we know that integration is area under the curve so how area could be subtracted from each other ?? plz solve my problem

Shujaat Khan - 5 years, 10 months ago

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