Math Problem Statement

Solve the following mathematical problems in the image provided: 1. Solve the inequality x^2 - 4 < 0. 2. Determine the number of positive signs for sine and cosine in each quadrant based on the given table. 3. Calculate h(1/2) for the function h(x) = x + 1/x. 4. Identify the truth or falsehood of statements about the function y = x^2 + 2x + 4.

Solution

Let's go through each question step-by-step and solve them.

Question 1

Solve the inequality: x24<0x^2 - 4 < 0 To solve, we rewrite it as: x2<4x^2 < 4 Taking the square root on both sides, we get: 2<x<2-2 < x < 2 So the solution is x(2,2)x \in (-2, 2), which corresponds to:

  • Answer: B) (2,2)(-2, 2)

Question 2

This question involves identifying the correct number of signs in the table for sine and cosine in each quadrant. Here's the general rule for trigonometric signs:

  • 1st Quadrant: Sine (+), Cosine (+)
  • 2nd Quadrant: Sine (+), Cosine (-)
  • 3rd Quadrant: Sine (-), Cosine (-)
  • 4th Quadrant: Sine (-), Cosine (+)

Counting the signs that are positive and negative across quadrants:

  • There are 3 instances where signs are positive (one for each "+" symbol in each quadrant).

So the answer is:

  • B) 3 ta

Question 3

For h(x)=x+1xh(x) = x + \frac{1}{x}, we need to find h(12)h\left(\frac{1}{2}\right).

Substitute x=12x = \frac{1}{2} into the function: h(12)=12+112h\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{\frac{1}{2}} =12+2=2.5= \frac{1}{2} + 2 = 2.5

So the answer is:

  • D) 2.5

Question 4

Determine the truth or falsehood of statements about the function y=x2+2x+4y = x^2 + 2x + 4.

Analysis:

  1. Domain: The function y=x2+2x+4y = x^2 + 2x + 4 is defined for all real values of xx, so the statement "Aniqlanish sohasi (,)(-∞, ∞) bo'ladi" is true.

  2. Minimum Value: This function is a parabola opening upwards, so it has a minimum value. Completing the square: y=(x+1)2+3y = (x+1)^2 + 3 The minimum value is y=3y = 3 when x=1x = -1, so the statement "Eng kichik qiymatga x=1x = -1 da erishadi" is true.

  3. No Zeros: The equation x2+2x+4=0x^2 + 2x + 4 = 0 has no real roots (its discriminant is negative), so the function has no real zeros. Therefore, the statement "Funksiya nollarga ega emas" is true.

The answer is:

  • A) to'g'ri, to'g'ri, to'g'ri

Let me know if you would like further explanations on any of the steps.


Additional Questions to Consider

  1. How would the solution change in Question 1 if the inequality was x240x^2 - 4 \leq 0?
  2. What are the general sign conventions for other trigonometric functions (like tangent or cotangent) in each quadrant?
  3. How does the function h(x)=x+1xh(x) = x + \frac{1}{x} behave as x0x \to 0?
  4. For Question 4, how would you verify the minimum of y=x2+2x+4y = x^2 + 2x + 4 by taking derivatives?
  5. How would the answer in Question 4 change if the function had a negative leading coefficient?

Tip

When solving inequalities involving squares, look for boundaries where the expression changes sign by identifying zero points.

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Math Problem Analysis

Mathematical Concepts

Algebra
Trigonometry
Quadratic Inequality
Function Evaluation

Formulas

Quadratic inequality formula x^2 - c < 0
Trigonometric sign rules for sine and cosine in each quadrant
Function evaluation for h(x) = x + 1/x
Completing the square for quadratic functions

Theorems

Quadratic inequality solution method
Trigonometric sign conventions by quadrant
Properties of parabolic functions

Suitable Grade Level

Grade 10