Comparing the magnitudes of real numbers forms the foundation of mathematical logic. On the number line, real numbers correspond one-to-one with points. By observing their positions, we can intuitively perceive 'inequality'.
Fundamental Facts:
Fundamental Facts:
- If $a - b$ is positive, then $a > b$;
- If $a - b$ equals 0, then $a = b$;
- If $a - b$ is negative, then $a < b$.
Core Properties of Inequalities:
1. Transitivity: $a > b, b > c \Rightarrow a > c$
2. Addition: $a > b \iff a + c > b + c$
3. Multiplication: $c > 0 \Rightarrow ac > bc$; $c < 0 \Rightarrow ac < bc$
1. Transitivity: $a > b, b > c \Rightarrow a > c$
2. Addition: $a > b \iff a + c > b + c$
3. Multiplication: $c > 0 \Rightarrow ac > bc$; $c < 0 \Rightarrow ac < bc$
$$a > b \iff a - b > 0$$
1. Gather polynomial terms: one $x^2$ square, three $x$ rectangular strips, and two $1\times1$ unit squares.
2. Begin geometrically assembling them.
3. They perfectly form a larger continuous rectangle! Width is $(x+2)$, height is $(x+1)$.
QUESTION 1
Which of the following representations of inequality relationships is incorrect?
Speed limit of $40\text{ km/h}$ on a road represented as $v \le 40$
Yogurt fat content $f$ not less than $2.5\%$ represented as $f > 2.5\%$
Sum of two sides of a triangle greater than the third side represented as $a + b > c$
Perpendicular segment $d_{\text{perp}}$ not greater than oblique segment $d_{\text{obq}}$ represented as $d_{\text{perp}} \le d_{\text{obq}}$
Correct! 'Not less than' means 'greater than or equal to,' so it should be represented as $f \ge 2.5\%$.
Pay attention to the keyword: 'not less than' includes equality. Please recheck the meaning of the symbols in each option.
QUESTION 2
The result of comparing $(x+3)(x+7)$ and $(x+4)(x+6)$ is:
$(x+3)(x+7) > (x+4)(x+6)$
$(x+3)(x+7) = (x+4)(x+6)$
$(x+3)(x+7) < (x+4)(x+6)$
Cannot be determined—it depends on the value of $x$
Correct. Subtracting gives: $(x^2 + 10x + 21) - (x^2 + 10x + 24) = -3 < 0$, so the first term is smaller than the second.
Hint: Use the method of difference. Expand both polynomials, subtract them, and observe the constant term.
QUESTION 3
The most fundamental theoretical basis for proving inequality properties 1, 3, 4, and 6 is:
The fundamental fact of real number comparison ($a > b \iff a - b > 0$)
Symmetry and transitivity of equations
Monotonicity of functions
Area relationships of geometric figures
Correct. All basic properties of inequalities are derived by taking differences and analyzing the sign of real number operations.
Recall the beginning of the course: all property derivations start from the sign of $a - b$.
QUESTION 4
If $x$ is a real number, the condition for $\sqrt{x^2 + x - 12}$ to be meaningful is:
$x > 3$ or $x < -4$
$x \ge 3$ or $x \le -4$
$-4 \le x \le 3$
$x \in \mathbf{R}$
Correct. For a square root to be meaningful, the radicand must be non-negative: $x^2 + x - 12 \ge 0$. Solving gives $(x+4)(x-3) \ge 0$, so $x \ge 3$ or $x \le -4$.
The expression inside the square root must satisfy $\ge 0$. This is a problem involving a quadratic inequality.
QUESTION 5
If $a > b$ and $\frac{1}{a} > \frac{1}{b}$, then it must be true that:
$ab > 0$
$ab < 0$
$a > 0, b < 0$
$a < 0, b > 0$
Correct. From $\frac{1}{a} - \frac{1}{b} > 0$, we get $\frac{b - a}{ab} > 0$. Since $a > b$, $b - a < 0$. For the fraction to be positive, the denominator $ab$ must be negative.
Hint: Combine fractions and subtract for $\frac{1}{a} > \frac{1}{b}$, then use the sign of $a - b$ to determine the sign of the denominator $ab$.
QUESTION 6
If $a, b > 0$ and $ab = a + b + 3$, find the range of values for $ab$.
$ab \ge 4$
$ab \ge 9$
$ab > 3$
$ab \ge 6$
Correct. From $a + b \ge 2\sqrt{ab}$, we get $ab - 3 \ge 2\sqrt{ab}$. Let $t = \sqrt{ab}$, then $t^2 - 2t - 3 \ge 0 \Rightarrow t \ge 3$, so $ab \ge 9$.
Use the basic inequality $a + b \ge 2\sqrt{ab}$ for substitution and transformation.
QUESTION 7
Which of the following statements about inequality properties is correct?
If $a > b$ and $c > d$, then $ac > bd$
If $a > b$, then $ac^2 > bc^2$
If $a > b > 0$, then $\frac{1}{a^2} < \frac{1}{b^2}$
If $a > b$ and $c < d$, then $a - c < b - d$
Correct. Since $a^2 > b^2 > 0$, taking reciprocals reverses the inequality direction.
Option A lacks the premise of positivity; Option B holds equality when $c = 0$; Option D should be $a - c > b - d$.
QUESTION 8
Given $a > b$, the correct logical steps to prove $\frac{a + b}{2} > b$ are:
Since $a > b$, so $a + b > 2b$, thus $\frac{a + b}{2} > b$
Since $b < a$, so $\frac{a}{2} < b$, thus it does not hold
Directly derived from the basic inequality
Equality holds if and only if $a = b$
Correct. Using Property 3 (Addition): add $b$ to both sides of $a > b$ to get $a + b > 2b$, then apply Property 4 (Multiplication) by dividing by 2.
This is a simple derivation based on the addition property of inequalities.
QUESTION 9
On a certain highway, the total height $h$ of vehicles and cargo must not exceed $4\text{ m}$, mathematically expressed as:
$h < 4$
$h \le 4$
$h > 4$
$0 < h \le 4$
Correct. 'Cannot exceed' includes the case where $h = 4$. Although physically $h > 0$, the pure mathematical description is $h \le 4$.
Key phrase: 'cannot exceed'.
QUESTION 10
Compare the areas $S_1$ and $S_2$ of a circle (circumference $L$) and a square (circumference $L$):
$S_1 = S_2$
$S_1 > S_2$
$S_1 < S_2$
Cannot be compared—it depends on the value of $L$
Correct. $S_1 = L^2 / 4\pi$, $S_2 = L^2 / 16$. Since $4\pi \approx 12.56 < 16$, the smaller the denominator, the larger the value, so the circle has a larger area.
Calculate and compare the sizes of $\frac{L^2}{4\pi}$ and $\frac{L^2}{16}$.
Challenge: Optimal Design for Reservoir Construction Cost
Modeling and Integrated Application of Inequalities
Construct a rectangular open-top reservoir with a volume of $1200\text{ m}^3$ and depth of $6\text{ m}$. The cost of walls is 95 yuan/$\text{m}^2$, and the cost of the bottom is 135 yuan/$\text{m}^2$. How should the length and width be designed to keep the total cost within 70,000 yuan?
Task 1
Establish an inequality model relating total cost $y$ to the base side length $x$.
Let one base side be $x$ meters, then the other side is $\frac{1200 / 6}{x} = \frac{200}{x}$ meters.
Base area is $200\text{ m}^2$, costing $200 \times 135 = 27,000$ yuan.
Total wall area is $2 \times (x \times 6 + \frac{200}{x} \times 6) = 12(x + \frac{200}{x})$.
Total cost $y = 27,000 + 95 \times 12(x + \frac{200}{x}) = 27,000 + 1140(x + \frac{200}{x})$.
Require $y \le 70,000$.
Base area is $200\text{ m}^2$, costing $200 \times 135 = 27,000$ yuan.
Total wall area is $2 \times (x \times 6 + \frac{200}{x} \times 6) = 12(x + \frac{200}{x})$.
Total cost $y = 27,000 + 95 \times 12(x + \frac{200}{x}) = 27,000 + 1140(x + \frac{200}{x})$.
Require $y \le 70,000$.
Task 2
Solve the inequality to determine the range of length and width (accurate to $0.1\text{ m}$).
$27,000 + 1140(x + \frac{200}{x}) \le 70,000$
$1140(x + \frac{200}{x}) \le 43,000 \Rightarrow x + \frac{200}{x} \le \frac{4300}{114} \approx 37.72$
Rearranging gives $x^2 - 37.72x + 200 \le 0$.
Using the quadratic formula, $x \approx 6.4$ or $x \approx 31.3$.
Thus, the length and width should be between $6.4\text{ m}$ and $31.3\text{ m}$.
$1140(x + \frac{200}{x}) \le 43,000 \Rightarrow x + \frac{200}{x} \le \frac{4300}{114} \approx 37.72$
Rearranging gives $x^2 - 37.72x + 200 \le 0$.
Using the quadratic formula, $x \approx 6.4$ or $x \approx 31.3$.
Thus, the length and width should be between $6.4\text{ m}$ and $31.3\text{ m}$.
✨ Key Points
Method of Difference,Determine Sign,Magnitude RelationshipBecomes Clear.When Multiplying by Negative,Sign Changes,Logic Is StrictMust Not Be Missed!
💡 Three Steps of the Method of Difference
Step 1: 'Take the difference'; Step 2: 'Transform' (often via factorization or completing the square); Step 3: 'Determine the sign'.
💡 Beware of Negative Numbers!
When multiplying or dividing both sides of an inequality by a negative number, always remember to reverse the inequality sign. This is the most common mistake.
💡 Conditions for Basic Inequalities
To use $\sqrt{ab} \le \frac{a + b}{2}$, the conditions are: one positive ($a, b > 0$), two fixed (product or sum is constant), and three equality (equality holds when $a = b$).
💡 Equivalent Thinking
$a > b \iff a - b > 0$ is bidirectionally equivalent and commonly used as the first step in proof transformations.
💡 Translation from Everyday Language
'At most' corresponds to $\le$, 'at least' to $\ge$, 'exceeds' to $>$, 'insufficient' to $<$.