/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 111 (a) The sum of two positive numb... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) The sum of two positive numbers equals 5. Can their product be equal to \(7 ?\) (b) (Arithmetic mean, Geometric mean, Harmonic mean, Quadratic mean) Prove that, if \(a, b>0,\) then $$ \begin{array}{c} \frac{2}{\left[\frac{1}{a}+\frac{1}{b}\right]}=\frac{2 a b}{a+b} \leqslant \sqrt{a b} \leqslant \frac{a+b}{2} \leqslant \sqrt{\frac{a^{2}+b^{2}}{2}} \\ (\mathrm{HM} \leqslant \mathrm{GM} \leqslant \mathrm{AM} \leqslant \mathrm{QM}) \end{array} $$

Short Answer

Expert verified
(a) No, their product cannot be 7. (b) All mean inequalities hold true.

Step by step solution

01

Identify the Conditions for the Product of Numbers

We need two positive numbers, let's call them \(x\) and \(y\), such that their sum is 5, i.e., \(x + y = 5\). The question asks if their product can be 7, i.e., \(xy = 7\). From the equations, we can form the quadratic equation by substituting \(y = 5 - x\) into the product equation to get \(x(5-x) = 7\). Thus, \(-x^2 + 5x - 7 = 0\). This quadratic equation will help determine the possibility.
02

Solve the Quadratic Equation

The quadratic equation derived is \(-x^2 + 5x - 7 = 0\). Rearranging gives \(x^2 - 5x + 7 = 0\). To solve this, we use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -5\), and \(c = 7\). Calculating the discriminant: \(b^2 - 4ac = (-5)^2 - 4(1)(7) = 25 - 28 = -3\). Since the discriminant is negative, this implies no real solutions for \(x\) and \(y\), so two positive numbers adding to 5 cannot have a product of 7.
03

Prove the Inequalities for Mean Relationships

We need to show that the Harmonic Mean (HM) is less than or equal to the Geometric Mean (GM), which is less than or equal to the Arithmetic Mean (AM), which in turn is less than or equal to the Quadratic Mean (QM). Start with: \[ \text{HM} = \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b} \] and \[ \text{GM} = \sqrt{ab} \] and \[ \text{AM} = \frac{a+b}{2} \] and \[ \text{QM} = \sqrt{\frac{a^2 + b^2}{2}} \].
04

Prove HM ≤ GM

The inequality \( \frac{2ab}{a+b} \leq \sqrt{ab} \) can be rewritten as \(4a^2b^2 \leq ab(a+b)^2\). This simplifies to \(4ab \leq (a+b)^2\) because we divide by \(ab\) assuming \(a, b > 0\). Expanding \((a+b)^2\) gives \(a^2 + 2ab + b^2\), leading to \(0 \leq (a-b)^2\), which is true.
05

Prove GM ≤ AM

The inequality \(\sqrt{ab} \leq \frac{a+b}{2}\) can be squared on both sides to get \(ab \leq \frac{(a+b)^2}{4}\). Expanding the right side, we have \(ab \leq \frac{a^2 + 2ab + b^2}{4}\) which simplifies to \(0 \leq (a-b)^2/4\), again true due to the perfect square.
06

Prove AM ≤ QM

The inequality \(\frac{a+b}{2} \leq \sqrt{\frac{a^2 + b^2}{2}}\) can also be squared to get \(\frac{(a+b)^2}{4} \leq \frac{a^2 + b^2}{2}\). Simplifying the left side gives \(\frac{a^2 + 2ab + b^2}{4} \leq \frac{a^2 + b^2}{2}\), leading to \(0 \leq (a-b)^2/4\) which is clearly true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations are mathematical expressions where the highest power of the unknown variable is squared, or raised to the power of two. They generally take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable we solve for. These equations are critical in algebra because many real-world problems can be modeled as quadratics. You can solve them using different methods like factoring, completing the square, or the quadratic formula. The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

This allows us to find the roots or solutions of the equation. The term under the square root, \(b^2 - 4ac\), is called the discriminant.
  • If the discriminant is positive, the equation has two real and distinct solutions.
  • If it's zero, there is exactly one real solution.
  • If it's negative, like in our exercise's quadratic, the equation has no real solutions, indicating no real numbers satisfy the condition set by the equation.
Arithmetic Mean
The arithmetic mean is often simply referred to as the 'average.' It is calculated by dividing the sum of a set of values by their number. Suppose you have numbers \(a\) and \(b\), the arithmetic mean \(\text{AM}\) is given by \(\frac{a+b}{2}\). This measure is widely used because it provides a central value for a data set.

Here’s why it's important:
  • It's a simple and quick way to summarize a set of data.
  • It's commonly used in statistics and other fields to represent a typical value.
This mean gives us a baseline to compare other means, like GM, HM, and QM, offering a foundational concept to understand inequalities in means.
Geometric Mean
The geometric mean, denoted as \(\text{GM}\), is the nth root of the product of n numbers. In our context with two numbers \(a\) and \(b\), it is \(\sqrt{ab}\). It is especially useful in contexts where numbers are multiplied together or are exponentially related.

Here are some insights into why nature tends to prefer the geometric mean:
  • Maintains the multiplicative relationship of the numbers.
  • Useful in growth rates and financial scenarios, like compound interest.
  • Less sensitive to extreme values compared to the arithmetic mean.
This mean also fits within the framework of mean inequalities, illustrating its position between other types of means like HM and AM.
Mean Inequalities
Mean inequalities describe the relationships between different types of means, highlighting their order. Mathematically, it can be illustrated as: \[ \text{HM} \leq \text{GM} \leq \text{AM} \leq \text{QM} \]. Each inequality piece proves to be true through the argumentation involving basic algebra and properties of squares.

Here's a breakdown of the inequalities:
  • The Harmonic Mean (HM) is least. It balances the harmonic properties of numbers and is useful in rates.
  • The Geometric Mean (GM) considers the multiplicative traits, sitting between HM and AM.
  • The Arithmetic Mean (AM) is an equally averaged value.
  • The Quadratic Mean (QM) or root mean square considers squared values, reducing the impact of larger numbers. Thus, it is always the largest in this series.
Understanding these inequalities helps in data analysis and decision-making when dealing with varying datasets.

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Most popular questions from this chapter

(a) Mark on the coordinate line all those points \(x\) for which two of the following inequalities are true, and five are false: $$ |x|>1,|x|>2,|x|>3,|x|>4,|x|>5,|x|>6,|x|>7 $$ (b) Mark on the coordinate line all those points \(x\) for which two of the following inequalities are true, and five are false: $$ |x-1|>1,|x-2|>2,|x-3|>3,|x-4|>4,|x-5|>5,|x-6|>6,|x-7|>7 $$

(a) Write down the coordinates of the midpoint \(M\) of the line segment joining \(Y=(a, b)\) and \(Z=(c, d)\). Justify your answer. (b) Position a general triangle \(X Y Z\) so that the vertex \(X\) lies at the origin \((0,0) .\) Suppose that \(Y\) then has coordinates \((a, b)\) and \(Z\) has coordinates \((c, d) .\) Let \(M\) be the midpoint of \(X Y,\) and \(N\) be the midpoint of \(X Z\). Prove the Midpoint Theorem, namely that \(" M N\) is parallel to \(Y Z\) and half its length" (c) Given any quadrilateral \(A B C D\), let \(P\) be the midpoint of \(A B,\) let \(Q\) be the midpoint of \(B C\), let \(R\) be the midpoint of \(C D,\) and let \(S\) be the midpoint of \(D A .\) Prove that \(P Q R S\) is always a parallelogram.

What is \(|-x|\) equal to: \(x\) or \(-x ?\) (What if \(x\) is negative?)

(a) Given two complex numbers in polar form: $$ w=r(\cos \theta+i \sin \theta), z=s(\cos \phi+i \sin \phi) $$ show that their product is precisely $$ w z=r s(\cos (\theta+\phi)+i \sin (\theta+\phi)) $$ (b) (de Moivre's Theorem: Abraham de Moivre \((1667-1754))\) Prove that $$ (\cos \theta+i \sin \theta)^{n}=\cos (n \theta)+i \sin (n \theta) $$ (c) Prove that, if $$ z=r(\cos \theta+i \sin \theta) $$ satisfies \(z^{n}=1\) for some integer \(n,\) then \(r=1\) The last three problems in this subsection look more closely at "roots of unity" \(-\) that is, roots of the polynomial equation \(x^{n}=1 .\) In the real domain, we know that: (i) when \(n\) is odd, the equation \(x^{n}=1\) has exactly one root, namely \(x=1 ;\) and (ii) when \(n\) is even, the equation \(x^{n}=1\) has just two solutions, namely \(x=\pm 1\). In contrast, in the complex domain, there are \(n " n^{\text {th }}\) roots of unity" . Problem \(\mathbf{1 3 0}(\mathrm{c})\) shows that these "roots of unity" all lie on the unit circle, centered at the origin. And if we put \(n \theta=2 k \pi\) in Problem \(130(\mathrm{~b})\) we see that the \(n n^{\text {th }}\) roots of unity include the point \(" 1=\cos 0+i \sin 0\) ", and are then equally spaced around that circle with \(\theta=\frac{2 k \pi}{n}(1 \leqslant k \leqslant n-1),\) and form the vertices of a regular \(n\) -gon.

(a)(i) Factorise \(a^{3}-b^{3}\). (ii) Factorise \(a^{4}-b^{4}\) as a product of one linear factor and one factor of degree \(3,\) and as a product of two linear factors and one quadratic factor. develops the ideas that were implicit in Problem 113. The clue lies in Problem \(113(\mathrm{a}),\) and in the comment made in the main text in Chapter 1 (after Problem 4 in Chapter 1 ), which we repeat here: "The last part [of Problem \(113(\mathrm{a})]\) is included to emphasise a frequently neglected message: Words and images are part of the way we communicate. But most of us cannot calculate with words and images. To make use of mathematics, we must routinely translate words into symbols. So "numbers" need to be represented by symbols, and points in a geometric diagram need to be properly labelled before we can begin to calculate, and to reason, effectively." As soon as one reads the words "one less than a square", one should instinctively translate this into the form " \(x^{2}-1\) ". Bells will then begin to ring; for it is impossible to forget the factorisation $$ x^{2}-1=(x-1)(x+1) $$ And it follows that: for a number that factorises in this way to be prime, the smaller factor \(x-1\) must be equal to 1 ; \(\therefore x=2,\) so there is only one such prime. The integer factorisations in Problem \(113(\mathrm{c})-\) namely $$ 3^{3}-1=2 \times 13,4^{3}-1=3 \times 21,5^{3}-1=4 \times 31,6^{3}-1=5 \times 43, \ldots $$ may help one to remember (or to discover) the related factorisation $$ x^{3}-1=(x-1)\left(x^{2}+x+1\right) $$ \(\therefore\) For a number that factorises in this way to be prime, the smaller factor " \(x-1\) " must be equal to 1 ; \(\therefore x=2,\) so there is only one such prime. Problem 113 parts (a) and (c) highlight the completely general factorisation (Problem \(115(\) a \()(\) iii \()\) : $$ x^{n}-1=(x-1)\left(x^{n-1}+x^{n-2}+\cdots+x^{2}+x+1\right) $$ This family of factorisations also shows that we should think about the factorisation of \(x^{2}-1\) as \((x-1)(x+1),\) with the uniform factor \((x-1)\) first (rather than as \((x+1)(x-1))\). Similarly, the results of Problem \(\mathbf{1 1 5}\) show that we should think of the familiar factorisation of \(a^{2}-b^{2}\) as \((\boldsymbol{a}-\boldsymbol{b})(a+b),(\) not as \((a+b)(a-b),\) but always with the factor \((\boldsymbol{a}-\boldsymbol{b})\) first). The integer factorisations in Problem \(113(\) d \()-\) namely $$ 3^{3}+1=4 \times 7,4^{3}+1=5 \times 13,5^{3}+1=6 \times 21,6^{3}+1=7 \times 31,7^{3}+1=8 \times 43, \ldots $$ may help one to remember (or to discover) the related factorisation $$ x^{3}+1=(x+1)\left(x^{2}-x+1\right) $$ \(\therefore\) For such a number to be prime, one of the factors must be equal to 1 This time one has to be more careful, because the first bracket may not be the "smaller factor" - so there are two cases to consider: (i) if \(x+1=1,\) then \(x=0,\) and \(x^{3}+1=1\) is not prime; (ii) if \(x^{2}-x+1=1,\) then \(x=0\) or \(x=1,\) so \(x=1\) and we obtain the prime 2 as the only solution. The factorisation for \(x^{3}+1\) works because " 3 is odd", which allows the alternating \(+/-\) signs to end in a "+" as required. Hence Problem \(113(\mathrm{~d})(\mathrm{iii})\) highlights the completely general factorisation for odd powers: $$ x^{2 n+1}+1=(x+1)\left(x^{2 n}-x^{2 n-1}+x^{2 n-2}-\cdots+x^{2}-x+1\right) $$ You probably know that there is no standard factorisation of \(x^{2}+1,\) or of \(x^{4}+1\) (but see Problem 114 above).

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