Answer :
To determine which points lie in the specified quadrants, let's evaluate each point one by one. The four quadrants on the Cartesian plane are defined as follows:
- In the 1st quadrant, both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are positive.
- In the 2nd quadrant, [tex]\(x\)[/tex] is negative and [tex]\(y\)[/tex] is positive.
- In the 3rd quadrant, both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are negative.
- In the 4th quadrant, [tex]\(x\)[/tex] is positive and [tex]\(y\)[/tex] is negative.
### 1. Point [tex]\(\left(\cos 10^\circ - \cos 13^\circ, \tan\frac{\pi}{9} - \tan\frac{\pi}{7}\right)\)[/tex]
- Calculate [tex]\(x_1 = \cos 10^\circ - \cos 13^\circ\)[/tex]:
[tex]\[ \cos 10^\circ \approx 0.9848 \quad \text{and} \quad \cos 13^\circ \approx 0.9744 \quad \Rightarrow \quad x_1 \approx 0.9848 - 0.9744 = 0.0104 \][/tex]
Thus, [tex]\(x_1 > 0\)[/tex].
- Calculate [tex]\(y_1 = \tan\frac{\pi}{9} - \tan\frac{\pi}{7}\)[/tex]:
[tex]\[ \tan\frac{\pi}{9} \approx 0.364 \quad \text{and} \quad \tan\frac{\pi}{7} \approx 0.481 \quad \Rightarrow \quad y_1 \approx 0.364 - 0.481 = -0.117 \][/tex]
Thus, [tex]\(y_1 < 0\)[/tex].
Since [tex]\(x_1 > 0\)[/tex] and [tex]\(y_1 < 0\)[/tex], the point [tex]\(\left(\cos 10^\circ - \cos 13^\circ, \tan\frac{\pi}{9} - \tan\frac{\pi}{7}\right)\)[/tex] lies in the 4th quadrant.
### 2. Point [tex]\(\left(\cos 18^\circ - \sin 15^\circ, \cos 105^\circ - \tan 15^\circ\right)\)[/tex]
- Calculate [tex]\(x_2 = \cos 18^\circ - \sin 15^\circ\)[/tex]:
[tex]\[ \cos 18^\circ \approx 0.9511 \quad \text{and} \quad \sin 15^\circ \approx 0.2588 \quad \Rightarrow \quad x_2 \approx 0.9511 - 0.2588 = 0.6923 \][/tex]
Thus, [tex]\(x_2 > 0\)[/tex].
- Calculate [tex]\(y_2 = \cos 105^\circ - \tan 15^\circ\)[/tex]:
[tex]\[ \cos 105^\circ \approx -0.2588 \quad \text{and} \quad \tan 15^\circ \approx 0.2679 \quad \Rightarrow \quad y_2 \approx -0.2588 - 0.2679 = -0.5267 \][/tex]
Thus, [tex]\(y_2 < 0\)[/tex].
Since [tex]\(x_2 > 0\)[/tex] and [tex]\(y_2 < 0\)[/tex], the point [tex]\(\left(\cos 18^\circ - \sin 15^\circ, \cos 105^\circ - \tan 15^\circ\right)\)[/tex] lies in the 4th quadrant.
### 3. Point [tex]\(\left(\sec^2 9^\circ - \tan^2 9^\circ, \cos 7^\circ - 1\right)\)[/tex]
- Use the Pythagorean identity: [tex]\(\sec^2 \theta - \tan^2 \theta = 1\)[/tex].
[tex]\[ x_3 = \sec^2 9^\circ - \tan^2 9^\circ = 1 \][/tex]
Since [tex]\(x_3 = 1\)[/tex], which is positive.
- Calculate [tex]\(y_3 = \cos 7^\circ - 1\)[/tex]:
[tex]\[ \cos 7^\circ \approx 0.9925 \quad \Rightarrow \quad y_3 \approx 0.9925 - 1 = -0.0075 \][/tex]
Thus, [tex]\(y_3 < 0\)[/tex].
Since [tex]\(x_3 > 0\)[/tex] and [tex]\(y_3 < 0\)[/tex], the point [tex]\(\left(\sec^2 9^\circ - \tan^2 9^\circ, \cos 7^\circ - 1\right)\)[/tex] actually lies in the 4th quadrant, not the 3rd.
### 4. Point [tex]\(\left(\log_2\left(\frac{x}{x^2 + 1}\right)\cdot \log_{\frac{1}{2}}(x^2 + x + 1), \forall x \in \mathbb{R}^+\right)\)[/tex]
- For the product [tex]\(\log_2\left(\frac{x}{x^2 + 1}\right) \cdot \log_{\frac{1}{2}}(x^2 + x + 1)\)[/tex]:
- Consider the sign of each logarithm:
[tex]\[ \log_2\left(\frac{x}{x^2 + 1}\right) \quad \text{and} \quad \log_{\frac{1}{2}}(x^2 + x + 1) \][/tex]
- For [tex]\(x > 0\)[/tex], [tex]\(x^2 + 1 > x\)[/tex], so [tex]\(\frac{x}{x^2 + 1} < 1\)[/tex] and [tex]\(\log_2\left(\frac{x}{x^2 + 1}\right) < 0\)[/tex].
- For [tex]\(x^2 + x + 1 > 1\)[/tex], and since [tex]\(\log_{\frac{1}{2}}\)[/tex] is decreasing, [tex]\(\log_{\frac{1}{2}}(x^2 + x + 1) < 0\)[/tex].
- The product of two negative numbers is positive.
Since both logarithms are less than zero and their product is positive, there is no intersection in the third quadrant here for these points as implied by [tex]\(x > 0\)[/tex].
### Conclusion
The points that are in the correct quadrants are:
- First point lies in the 4th quadrant.
- Second point lies in the 4th quadrant.
Therefore, options 1 and 2 are correct, option 3 does not lie in the 3rd quadrant but the 4th quadrant, and option 4 is invalid.
- In the 1st quadrant, both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are positive.
- In the 2nd quadrant, [tex]\(x\)[/tex] is negative and [tex]\(y\)[/tex] is positive.
- In the 3rd quadrant, both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are negative.
- In the 4th quadrant, [tex]\(x\)[/tex] is positive and [tex]\(y\)[/tex] is negative.
### 1. Point [tex]\(\left(\cos 10^\circ - \cos 13^\circ, \tan\frac{\pi}{9} - \tan\frac{\pi}{7}\right)\)[/tex]
- Calculate [tex]\(x_1 = \cos 10^\circ - \cos 13^\circ\)[/tex]:
[tex]\[ \cos 10^\circ \approx 0.9848 \quad \text{and} \quad \cos 13^\circ \approx 0.9744 \quad \Rightarrow \quad x_1 \approx 0.9848 - 0.9744 = 0.0104 \][/tex]
Thus, [tex]\(x_1 > 0\)[/tex].
- Calculate [tex]\(y_1 = \tan\frac{\pi}{9} - \tan\frac{\pi}{7}\)[/tex]:
[tex]\[ \tan\frac{\pi}{9} \approx 0.364 \quad \text{and} \quad \tan\frac{\pi}{7} \approx 0.481 \quad \Rightarrow \quad y_1 \approx 0.364 - 0.481 = -0.117 \][/tex]
Thus, [tex]\(y_1 < 0\)[/tex].
Since [tex]\(x_1 > 0\)[/tex] and [tex]\(y_1 < 0\)[/tex], the point [tex]\(\left(\cos 10^\circ - \cos 13^\circ, \tan\frac{\pi}{9} - \tan\frac{\pi}{7}\right)\)[/tex] lies in the 4th quadrant.
### 2. Point [tex]\(\left(\cos 18^\circ - \sin 15^\circ, \cos 105^\circ - \tan 15^\circ\right)\)[/tex]
- Calculate [tex]\(x_2 = \cos 18^\circ - \sin 15^\circ\)[/tex]:
[tex]\[ \cos 18^\circ \approx 0.9511 \quad \text{and} \quad \sin 15^\circ \approx 0.2588 \quad \Rightarrow \quad x_2 \approx 0.9511 - 0.2588 = 0.6923 \][/tex]
Thus, [tex]\(x_2 > 0\)[/tex].
- Calculate [tex]\(y_2 = \cos 105^\circ - \tan 15^\circ\)[/tex]:
[tex]\[ \cos 105^\circ \approx -0.2588 \quad \text{and} \quad \tan 15^\circ \approx 0.2679 \quad \Rightarrow \quad y_2 \approx -0.2588 - 0.2679 = -0.5267 \][/tex]
Thus, [tex]\(y_2 < 0\)[/tex].
Since [tex]\(x_2 > 0\)[/tex] and [tex]\(y_2 < 0\)[/tex], the point [tex]\(\left(\cos 18^\circ - \sin 15^\circ, \cos 105^\circ - \tan 15^\circ\right)\)[/tex] lies in the 4th quadrant.
### 3. Point [tex]\(\left(\sec^2 9^\circ - \tan^2 9^\circ, \cos 7^\circ - 1\right)\)[/tex]
- Use the Pythagorean identity: [tex]\(\sec^2 \theta - \tan^2 \theta = 1\)[/tex].
[tex]\[ x_3 = \sec^2 9^\circ - \tan^2 9^\circ = 1 \][/tex]
Since [tex]\(x_3 = 1\)[/tex], which is positive.
- Calculate [tex]\(y_3 = \cos 7^\circ - 1\)[/tex]:
[tex]\[ \cos 7^\circ \approx 0.9925 \quad \Rightarrow \quad y_3 \approx 0.9925 - 1 = -0.0075 \][/tex]
Thus, [tex]\(y_3 < 0\)[/tex].
Since [tex]\(x_3 > 0\)[/tex] and [tex]\(y_3 < 0\)[/tex], the point [tex]\(\left(\sec^2 9^\circ - \tan^2 9^\circ, \cos 7^\circ - 1\right)\)[/tex] actually lies in the 4th quadrant, not the 3rd.
### 4. Point [tex]\(\left(\log_2\left(\frac{x}{x^2 + 1}\right)\cdot \log_{\frac{1}{2}}(x^2 + x + 1), \forall x \in \mathbb{R}^+\right)\)[/tex]
- For the product [tex]\(\log_2\left(\frac{x}{x^2 + 1}\right) \cdot \log_{\frac{1}{2}}(x^2 + x + 1)\)[/tex]:
- Consider the sign of each logarithm:
[tex]\[ \log_2\left(\frac{x}{x^2 + 1}\right) \quad \text{and} \quad \log_{\frac{1}{2}}(x^2 + x + 1) \][/tex]
- For [tex]\(x > 0\)[/tex], [tex]\(x^2 + 1 > x\)[/tex], so [tex]\(\frac{x}{x^2 + 1} < 1\)[/tex] and [tex]\(\log_2\left(\frac{x}{x^2 + 1}\right) < 0\)[/tex].
- For [tex]\(x^2 + x + 1 > 1\)[/tex], and since [tex]\(\log_{\frac{1}{2}}\)[/tex] is decreasing, [tex]\(\log_{\frac{1}{2}}(x^2 + x + 1) < 0\)[/tex].
- The product of two negative numbers is positive.
Since both logarithms are less than zero and their product is positive, there is no intersection in the third quadrant here for these points as implied by [tex]\(x > 0\)[/tex].
### Conclusion
The points that are in the correct quadrants are:
- First point lies in the 4th quadrant.
- Second point lies in the 4th quadrant.
Therefore, options 1 and 2 are correct, option 3 does not lie in the 3rd quadrant but the 4th quadrant, and option 4 is invalid.
