💡To determine how many zeros the function ( g(x) = sin(log_2 x) ) has on the interval ( 1 < x < 2 ), we need to find the values of ( x ) for which ( sin(log_2 x) = 0 ).1. **Understanding the function**: The function ( g(x) = sin(log_2 x) ) involves the sine function composed with the logarithm base 2. To find the zeros, we need to solve ( sin(log_2 x) = 0 ).2. **Range of ( log_2 x ) for ( 1 < x < 2 )**: Since ( log_2 x ) is a strictly increasing function, as ( x ) ranges from 1 to 2, ( log_2 x ) ranges from 0 to 1.3. **Zeros of ( sin(t) )**: The sine function ( sin(t) ) is zero whenever ( t = kpi ) for any integer ( k ).4. **Finding relevant ( k ) values**: We need to find integers ( k ) such that ( 0 leq kpi leq 1 ). The only integer ( k ) that satisfies this condition is ( k = 0 ).5. **Conclusion**: Since the only value of ( t ) in the interval ([0, 1]) for which ( sin(t) = 0 ) is ( t = 0 ), corresponding to ( x = 2^0 = 1 ). However, ( x = 1 ) is not included in the open interval ( 1 < x < 2 ).Therefore, there are no zeros of ( g(x) ) in the interval ( 1 < x < 2 ).The final answer is (boxed{A}).

💡First, I need to find the coordinates of point ( Q ) by reflecting point ( P(5, 3) ) over the ( x )-axis. Reflecting over the ( x )-axis changes the sign of the ( y )-coordinate, so ( Q ) will be ( (5, -3) ).Next, I reflect point ( Q(5, -3) ) over the line ( y = x ). Reflecting over the line ( y = x ) swaps the ( x ) and ( y ) coordinates, so ( R ) will be ( (-3, 5) ).Now, I have the coordinates of all three points: ( P(5, 3) ), ( Q(5, -3) ), and ( R(-3, 5) ). To find the area of triangle ( PQR ), I can use the formula for the area of a triangle given three vertices:[text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|]Plugging in the coordinates:[text{Area} = frac{1}{2} |5(-3 - 5) + 5(5 - 3) + (-3)(3 - (-3))|][= frac{1}{2} |5(-8) + 5(2) + (-3)(6)|][= frac{1}{2} |-40 + 10 - 18|][= frac{1}{2} |-48|][= frac{1}{2} times 48 = 24]Therefore, the area of triangle ( PQR ) is 24 square units.

💡Okay, so I have this problem here: Given the function f(x) = x² + mx + 1, and the proposition "There exists an x₀ > 0 such that f(x₀) < 0" is true. I need to find the range of values for m. The options are A: (-∞, -2], B: [2, +∞), C: (-∞, -2), and D: (2, +∞). Hmm, let me try to figure this out step by step.First, I know that f(x) is a quadratic function. Its graph is a parabola. Since the coefficient of x² is positive (1), the parabola opens upwards. That means the vertex is the minimum point of the parabola. So, if the parabola opens upwards, for f(x) to be less than zero somewhere, it must cross the x-axis at two points. Otherwise, if it doesn't cross the x-axis, f(x) would always be positive because it's opening upwards.So, for the quadratic equation x² + mx + 1 = 0 to have real roots, the discriminant must be positive. The discriminant D is given by D = m² - 4*1*1 = m² - 4. So, for real roots, D > 0, which implies m² - 4 > 0. Solving this inequality, m² > 4, so m > 2 or m < -2. That gives me two intervals: (-∞, -2) and (2, +∞). So, m must be either less than -2 or greater than 2.But wait, the proposition is that there exists an x₀ > 0 such that f(x₀) < 0. So, it's not just about having real roots, but specifically having a root where x is positive. Because if the roots are both negative, then for all x > 0, f(x) would be positive since the parabola opens upwards. So, I need to ensure that at least one of the roots is positive.How can I determine the signs of the roots? I remember that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a. In this case, the sum of the roots is -m/1 = -m, and the product is 1/1 = 1. Since the product is positive, both roots are either positive or both are negative. If the sum of the roots is positive, then both roots are positive. If the sum is negative, then both roots are negative.So, the sum of the roots is -m. For the roots to be positive, the sum must be positive, which implies -m > 0, so m < 0. Alternatively, if the sum is negative, then m > 0, which would mean both roots are negative.Wait, so if m < 0, the sum of the roots is positive, meaning both roots are positive. If m > 0, the sum is negative, meaning both roots are negative. So, if m > 0, both roots are negative, which means for all x > 0, f(x) is positive because the parabola opens upwards and there are no roots in the positive x region. Therefore, to have f(x₀) < 0 for some x₀ > 0, we need at least one positive root, which happens when m < 0.But earlier, we found that m must be either less than -2 or greater than 2 for the quadratic to have real roots. So, combining these two results: m must be less than -2 because if m is between -2 and 0, the quadratic doesn't have real roots, so f(x) is always positive. Therefore, m must be less than -2.Wait, let me double-check that. If m is between -2 and 0, say m = -1, then the discriminant D = (-1)² - 4 = 1 - 4 = -3, which is negative. So, no real roots, meaning f(x) is always positive. So, indeed, for m between -2 and 0, f(x) is always positive, so there's no x₀ > 0 where f(x₀) < 0.Therefore, m must be less than -2. So, the range is (-∞, -2). Looking back at the options, that's option C.But hold on, let me think again. If m = -2, what happens? Then f(x) = x² - 2x + 1. That factors as (x - 1)², which is a perfect square. So, f(x) = 0 at x = 1, and it's positive everywhere else. So, at x = 1, f(x) = 0, but it's not less than zero. So, for m = -2, there is no x₀ > 0 where f(x₀) < 0 because the function touches the x-axis at x = 1 but doesn't go below it. Therefore, m must be strictly less than -2, not including -2. So, the range is (-∞, -2), which is option C.I think that makes sense. So, the answer should be C.**Final Answer**boxed{C}

💡Okay, so I have this problem here about an inverse proportion function, y equals negative three over x, and I need to find the range of y when x is greater than 3. The options are A: y is greater than negative one, B: y is less than one, C: negative one is less than y which is less than zero, and D: zero is less than y which is less than one. Hmm, let me try to figure this out step by step.First, I remember that an inverse proportion function generally looks like y equals k over x, where k is a constant. In this case, k is negative three, so the function is y equals negative three over x. That means as x increases, y should decrease, right? But since there's a negative sign, the behavior might be a bit different.Let me think about what happens when x is greater than 3. If x is, say, 4, then y would be negative three over four, which is negative three-fourths. That's -0.75. If x is 5, then y is negative three over five, which is -0.6. If x is 6, y is negative three over six, which is -0.5. So, as x increases beyond 3, y is becoming less negative, approaching zero from below.Wait, so when x is just a little bit bigger than 3, like 3.1, y would be negative three over 3.1, which is approximately -0.967. That's pretty close to -1. So, when x is just over 3, y is just under -1. And as x increases, y becomes less negative, getting closer to zero but never reaching it.So, putting that together, when x is greater than 3, y is between -1 and 0. That means y is greater than -1 and less than 0. Looking back at the options, that's option C: -1 < y < 0.Let me double-check to make sure I didn't make a mistake. If x is 3, y would be negative three over three, which is -1. But the problem says x is greater than 3, not equal to 3, so y is greater than -1. And since x is positive and increasing, y approaches zero from the negative side, so y is less than 0. Yeah, that makes sense.I think I got it right. So, the range of y when x is greater than 3 is between -1 and 0, which is option C.

💡Alright, so I have this equation here: ( x^{2} + 4x + 1 = 0 ). I need to factor it and see which of the given options it matches. The options are:A: ( (x + 4)^{2} = 3 )B: ( (x + 2)^{2} = -3 )C: ( (x + 2)^{2} = 3 )D: ( (x + 2)^{2} = -5 )Okay, first, I remember that factoring a quadratic equation can sometimes involve completing the square. Maybe that's what I need to do here. Let me try that.Starting with the original equation:( x^{2} + 4x + 1 = 0 )I think the first step is to move the constant term to the other side. So, subtract 1 from both sides:( x^{2} + 4x = -1 )Now, to complete the square, I need to add a number to both sides that will make the left side a perfect square trinomial. The formula for completing the square is to take half of the coefficient of x, square it, and add it to both sides.The coefficient of x here is 4. Half of 4 is 2, and squaring that gives 4. So, I'll add 4 to both sides:( x^{2} + 4x + 4 = -1 + 4 )Simplifying the right side:( x^{2} + 4x + 4 = 3 )Now, the left side is a perfect square. It can be written as:( (x + 2)^{2} = 3 )Hmm, that looks like option C. But wait, let me double-check my steps to make sure I didn't make a mistake.Starting again:1. Original equation: ( x^{2} + 4x + 1 = 0 )2. Move the constant term: ( x^{2} + 4x = -1 )3. Complete the square by adding ( (4/2)^2 = 4 ) to both sides: ( x^{2} + 4x + 4 = -1 + 4 )4. Simplify: ( (x + 2)^{2} = 3 )Yes, that seems correct. So, the transformed equation is ( (x + 2)^{2} = 3 ), which is option C.But wait, let me look at the other options to make sure none of them could also be correct or if I missed something.Option A: ( (x + 4)^{2} = 3 )If I expand this, it would be ( x^{2} + 8x + 16 = 3 ), which simplifies to ( x^{2} + 8x + 13 = 0 ). That's not the same as the original equation, so A is out.Option B: ( (x + 2)^{2} = -3 )Expanding this gives ( x^{2} + 4x + 4 = -3 ), which simplifies to ( x^{2} + 4x + 7 = 0 ). Again, not the same as the original equation.Option C: ( (x + 2)^{2} = 3 )Expanding: ( x^{2} + 4x + 4 = 3 ), which simplifies to ( x^{2} + 4x + 1 = 0 ). That's exactly the original equation. So, C is correct.Option D: ( (x + 2)^{2} = -5 )Expanding: ( x^{2} + 4x + 4 = -5 ), which simplifies to ( x^{2} + 4x + 9 = 0 ). Not matching the original equation.Okay, so after checking all options, C is the correct one. I think I did everything right, but just to be thorough, let me solve both the original equation and option C to see if they have the same solutions.Original equation: ( x^{2} + 4x + 1 = 0 )Using the quadratic formula: ( x = frac{-b pm sqrt{b^{2} - 4ac}}{2a} )Here, a = 1, b = 4, c = 1.So, discriminant: ( 16 - 4 = 12 )Solutions: ( x = frac{-4 pm sqrt{12}}{2} = frac{-4 pm 2sqrt{3}}{2} = -2 pm sqrt{3} )Now, option C: ( (x + 2)^{2} = 3 )Taking square roots: ( x + 2 = pm sqrt{3} )So, ( x = -2 pm sqrt{3} )Same solutions as the original equation. Therefore, option C is indeed the correct transformation.I think I'm confident now that C is the right answer.

💡First, I need to compute the matrix operation ( mathbf{M} (3 mathbf{u} - 2 mathbf{z}) ).I recall that matrix multiplication is linear, which means I can distribute the matrix ( mathbf{M} ) over the linear combination of vectors ( 3 mathbf{u} - 2 mathbf{z} ).So, I can rewrite the expression as:[mathbf{M} (3 mathbf{u} - 2 mathbf{z}) = 3 mathbf{M} mathbf{u} - 2 mathbf{M} mathbf{z}]Next, I substitute the given values:[mathbf{M} mathbf{u} = begin{pmatrix} 2 -3 end{pmatrix} quad text{and} quad mathbf{M} mathbf{z} = begin{pmatrix} 4 1 end{pmatrix}]Multiplying each matrix by its scalar:[3 mathbf{M} mathbf{u} = 3 begin{pmatrix} 2 -3 end{pmatrix} = begin{pmatrix} 6 -9 end{pmatrix}][-2 mathbf{M} mathbf{z} = -2 begin{pmatrix} 4 1 end{pmatrix} = begin{pmatrix} -8 -2 end{pmatrix}]Finally, I subtract the second resulting vector from the first:[begin{pmatrix} 6 -9 end{pmatrix} - begin{pmatrix} 8 2 end{pmatrix} = begin{pmatrix} 6 - 8 -9 - 2 end{pmatrix} = begin{pmatrix} -2 -11 end{pmatrix}]Therefore, the result of ( mathbf{M} (3 mathbf{u} - 2 mathbf{z}) ) is ( begin{pmatrix} -2 -11 end{pmatrix} ).