Waste less time on Facebook — follow Brilliant.
×

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 \(x\) and \(y\) are non-zero integers from \(-10\) to \(10\) (inclusive), what is the smallest positive value of \[ \frac{ | x+ y | } { |x| + |y| }? \]

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

Note by Calvin Lin
3 years, 9 months ago

No vote yet
11 votes

Comments

Sort by:

Top Newest

Is it \(\frac{1}{19}\)?, taking \(x=-10,y=9\) because we need to minimize the numerator and at the same time we have to maximize the denominator.Please verify. Kishan K · 3 years, 9 months ago

Log in to reply

@Kishan K Yep! Ahaan Rungta · 3 years, 9 months ago

Log in to reply

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

Proposed by Aaron Lin for the NIMO Ahaan Rungta · 3 years, 9 months ago

Log in to reply

@Ahaan Rungta \( -1-2-3-4...-53-54+55+56+57...+99+100 = \boxed{2080} \) Piyushkumar Palan · 3 years, 9 months ago

Log in to reply

Problem 1.39. Solve the equation \(|x-3|+|x+1|=4\).

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

Problem 1.41. Let \(a,b>0\). Find the values of \(m\) for which the equation

\[|x-a|+|x-b|+|x+a|+|x+b|=m(a+b)\]

has at least one real solution.

Problem 1.42. Find all possible values of the expression

\[E(x,y,z)=\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}\]

where \(x,y,z\) are nonzero real numbers.

Problem 1.43. Find all positive real numbers \(x,x_1,x_2,\dots,x_n\) such that

\[|\log(xx_1)|+|\log(x_1x_2)|+\dots+|\log(xx_n)|\\+|\log\left(\frac{x}{x_1}\right)|+|\log\left(\frac{x}{x_2}\right)|+\dots+|\log\left(\frac{x}{x_n}\right)|\\=|\log x_1+\log x_2+\dots+\log x_n|.\]

Problem 1.44. Prove that for all real numbers \(a,b\), we have

\[\frac{|a+b|}{1+|a+b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}\]

Problem 1.45. Let \(n\) be an odd positive integer and let \(x_1,x_2,\dots,x_n\) be distinct real numbers. Find all one-to-one functions

\[f:\{x_1,x_2,\dots,x_n\}\to\{x_1,x_2,\dots,x_n\}\]

such that

\[|f(x_1)-x_1|=|f(x_2)-x_2|=\dots=|f(x_n)-x_n|\]

Problem 1.46. Suppose that the sequence \(a_1,a_2,\dots,a_n\) satisfies the following conditions:

\[\begin{align*}a_1=0,&|a_2|=|a_1+1|,\dots,&|a_n|=|a_{n-1}+1|.\end{align*}\]

Prove that

\[\frac{a_1+a_2+\dots+a_n}{n}\ge-\frac{1}{2}.\]

Problem 1.47. Find real numbers \(a,b,c\) such that

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

for all real numbers \(x,y,z\).

Source: Mathematical Olympiad Treasures, Titu Andreescu Cody Johnson · 3 years, 9 months ago

Log in to reply

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 · 3 years, 8 months ago

Log in to reply

(FUVEST 2012) Show the intervals of \(x\) in which \(|x^2-10x+21| \leq |3x-15|\) holds. Guilherme Dela Corte · 3 years, 9 months ago

Log in to reply

@Guilherme Dela Corte 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 \((-\infty, 3), (3, 5), (5, 7), (7, \infty)\) and applying the appropriate sign. Then we add the points \(x = 3\) and \(x = 7\) to the solution set because the LHS is 0. The rest of the solution is trivial.

Example \((-\infty, 3)\) --> \((x - 3)(x - 7)\) is positive, \(3x - 15\) is negative. \(x^2 - 10x + 21 \leq 15 - 3x \rightarrow x^2 - 7x - 6 = (x - 1)(x - 6) \leq 0\), so the interval is \([1, 6] \bigwedge x < 3 \rightarrow [1, 3)\). Michael Tong · 3 years, 9 months ago

Log in to reply

@Michael Tong Michael, haven't you forgot the other interval? Guilherme Dela Corte · 3 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...