💡Okay, so I have this problem here with two sets, S₁ and S₂, defined by these logarithmic inequalities. I need to find the ratio of the area of S₂ to the area of S₁. Hmm, let me try to unpack this step by step.First, let's look at S₁:[ S_1 = {(x,y) mid log_{10}(3 + x^2 + y^2) leq 2 + log_{10}(x+y) } ]And S₂:[ S_2 = {(x,y) mid log_{10}(5 + x^2 + y^2) leq 3 + log_{10}(x+y) } ]I remember that logarithmic inequalities can sometimes be simplified by exponentiating both sides to eliminate the logs. Let me try that for S₁ first.Starting with S₁:[ log_{10}(3 + x^2 + y^2) leq 2 + log_{10}(x+y) ]I can rewrite 2 as (log_{10}(10^2)) because (log_{10}(100) = 2). So the inequality becomes:[ log_{10}(3 + x^2 + y^2) leq log_{10}(100) + log_{10}(x+y) ]Using the logarithm property that (log_{10}(a) + log_{10}(b) = log_{10}(ab)), the right side becomes:[ log_{10}(100(x + y)) ]So now the inequality is:[ log_{10}(3 + x^2 + y^2) leq log_{10}(100(x + y)) ]Since the logarithm function is increasing, if (log_{10}(A) leq log_{10}(B)), then (A leq B). Therefore, we can drop the logs:[ 3 + x^2 + y^2 leq 100(x + y) ]Let me rearrange this inequality:[ x^2 + y^2 - 100x - 100y + 3 leq 0 ]Hmm, this looks like the equation of a circle. To write it in standard form, I need to complete the square for both x and y terms.Starting with the x terms: (x^2 - 100x). To complete the square, take half of -100, which is -50, square it to get 2500. Similarly for the y terms: (y^2 - 100y). Half of -100 is -50, squared is 2500.So, adding and subtracting 2500 for both x and y:[ (x^2 - 100x + 2500) + (y^2 - 100y + 2500) + 3 - 2500 - 2500 leq 0 ]Simplify:[ (x - 50)^2 + (y - 50)^2 + 3 - 5000 leq 0 ]Which is:[ (x - 50)^2 + (y - 50)^2 leq 4997 ]So, S₁ is a circle centered at (50, 50) with radius (sqrt{4997}).Now, let's do the same for S₂.Starting with S₂:[ log_{10}(5 + x^2 + y^2) leq 3 + log_{10}(x+y) ]Again, rewrite 3 as (log_{10}(10^3)) because (log_{10}(1000) = 3). So the inequality becomes:[ log_{10}(5 + x^2 + y^2) leq log_{10}(1000) + log_{10}(x+y) ]Using the same logarithm property:[ log_{10}(5 + x^2 + y^2) leq log_{10}(1000(x + y)) ]Dropping the logs:[ 5 + x^2 + y^2 leq 1000(x + y) ]Rearranging:[ x^2 + y^2 - 1000x - 1000y + 5 leq 0 ]Again, this looks like a circle equation. Let's complete the square for x and y.For x terms: (x^2 - 1000x). Half of -1000 is -500, squared is 250000. Similarly for y terms: (y^2 - 1000y). Half of -1000 is -500, squared is 250000.Adding and subtracting 250000 for both x and y:[ (x^2 - 1000x + 250000) + (y^2 - 1000y + 250000) + 5 - 250000 - 250000 leq 0 ]Simplify:[ (x - 500)^2 + (y - 500)^2 + 5 - 500000 leq 0 ]Which is:[ (x - 500)^2 + (y - 500)^2 leq 499995 ]So, S₂ is a circle centered at (500, 500) with radius (sqrt{499995}).Now, I need to find the ratio of the area of S₂ to the area of S₁.The area of a circle is (pi r^2). So, the area of S₁ is (pi times (sqrt{4997})^2 = pi times 4997). Similarly, the area of S₂ is (pi times (sqrt{499995})^2 = pi times 499995).So, the ratio is:[ frac{text{Area of } S₂}{text{Area of } S₁} = frac{499995}{4997} ]Let me compute this division.First, note that 4997 is approximately 5000, and 499995 is approximately 500000.But let's do it more accurately.Compute 499995 ÷ 4997.Let me see:4997 × 100 = 499700499995 - 499700 = 295So, 4997 × 100 = 499700Then, 4997 × 100.04 ≈ ?Wait, 4997 × 100 = 4997004997 × 0.04 = 199.88So, 4997 × 100.04 ≈ 499700 + 199.88 = 499899.88But we have 499995, which is 499995 - 499899.88 = 95.12 more.So, 4997 × x = 499995We have x ≈ 100.04 + (95.12 / 4997)Compute 95.12 / 4997 ≈ 0.01904So, x ≈ 100.04 + 0.01904 ≈ 100.05904Wait, that seems off because 4997 × 100.04 is 499899.88, and 499995 is 95.12 more, so 95.12 / 4997 ≈ 0.01904, so total x ≈ 100.04 + 0.01904 ≈ 100.05904But let me check:4997 × 100.05904 ≈ 4997 × 100 + 4997 × 0.05904 ≈ 499700 + 295 ≈ 499995Yes, that works.So, the ratio is approximately 100.05904.Looking at the options, the closest is C) 100.04.Wait, but my approximation gave me about 100.059, which is closer to 100.06, but the options don't have that. The closest is C) 100.04 and D) 100.10.Hmm, maybe I should compute it more precisely.Let me compute 499995 ÷ 4997.Let me write it as:499995 ÷ 4997 = ?Let me perform the division step by step.4997 × 100 = 499700Subtract: 499995 - 499700 = 295So, 4997 × 100 = 499700Now, 295 ÷ 4997 ≈ 0.05904So, total is 100 + 0.05904 ≈ 100.05904So, approximately 100.059.Looking at the options, the closest is C) 100.04 and D) 100.10.Since 100.059 is closer to 100.06, which is between 100.04 and 100.10. But the options don't have 100.06. Hmm.Wait, maybe I made a miscalculation earlier.Wait, 4997 × 100.04 = 4997 × 100 + 4997 × 0.04 = 499700 + 199.88 = 499899.88Then, 499995 - 499899.88 = 95.12So, 95.12 ÷ 4997 ≈ 0.01904So, total multiplier is 100.04 + 0.01904 ≈ 100.05904So, approximately 100.059, which is about 100.06.But the options are A) 99.90, B) 100.00, C) 100.04, D) 100.10, E) 100.20.So, 100.06 is closer to 100.10 than to 100.04.Wait, 100.06 is 0.06 above 100.00, and 100.10 is 0.04 above 100.06.Wait, no, 100.06 is 0.06 above 100.00, and 100.10 is 0.04 above 100.06.Wait, actually, 100.06 is 0.06 above 100.00, and 100.10 is 0.04 above 100.06.Wait, no, that's not correct. 100.06 is 0.06 above 100.00, and 100.10 is 0.10 above 100.00, so 100.06 is 0.04 below 100.10.So, 100.06 is closer to 100.10 by 0.04, and closer to 100.04 by 0.02.Wait, 100.06 - 100.04 = 0.02100.10 - 100.06 = 0.04So, 100.06 is closer to 100.04 by 0.02 than to 100.10 by 0.04.Therefore, the closest option is C) 100.04.But wait, my calculation gave me approximately 100.059, which is 100.06, which is closer to 100.04 by 0.02 than to 100.10 by 0.04.But let me check if I did the division correctly.Alternatively, maybe I can write 499995 ÷ 4997 as:499995 ÷ 4997 = (4997 × 100 + 295) ÷ 4997 = 100 + 295/4997 ≈ 100 + 0.05904 ≈ 100.05904So, 100.05904 is approximately 100.06, which is 100 and 0.06, which is 100.06.Given the options, the closest is C) 100.04 and D) 100.10.Since 100.06 is exactly halfway between 100.04 and 100.10? Wait, no.Wait, 100.04, 100.06, 100.10.The difference between 100.04 and 100.06 is 0.02.The difference between 100.06 and 100.10 is 0.04.So, 100.06 is closer to 100.04 than to 100.10.Therefore, the answer should be C) 100.04.But wait, let me check if I made a mistake in the calculation.Wait, 4997 × 100.04 = 4997 × 100 + 4997 × 0.04 = 499700 + 199.88 = 499899.88Then, 499995 - 499899.88 = 95.12So, 95.12 ÷ 4997 ≈ 0.01904So, total is 100.04 + 0.01904 ≈ 100.05904So, 100.05904 is approximately 100.06, which is 0.06 above 100.00.But the options are in increments of 0.04, 0.06 isn't an option.Wait, perhaps the exact value is 100.04999, which is approximately 100.05, which would round to 100.05, but since that's not an option, the closest is 100.04 or 100.10.But 100.05 is exactly halfway between 100.04 and 100.10? No, 100.04, 100.05, 100.06,..., 100.10.Wait, 100.04 to 100.10 is 0.06 difference.100.05 is 0.01 above 100.04 and 0.05 below 100.10.Wait, no, 100.05 is 0.01 above 100.04 and 0.05 below 100.10.Wait, no, 100.05 is 0.01 above 100.04 and 0.05 below 100.10.Wait, no, 100.05 is 0.01 above 100.04 and 0.05 below 100.10.Wait, no, 100.05 is 0.01 above 100.04 and 0.05 below 100.10.Wait, that can't be because 100.05 - 100.04 = 0.01100.10 - 100.05 = 0.05So, 100.05 is closer to 100.04 by 0.01 than to 100.10 by 0.05.Therefore, 100.05 would round to 100.04 if we are rounding to the nearest 0.04.But in our case, the ratio is approximately 100.059, which is 100.06, which is 0.02 above 100.04 and 0.04 below 100.10.Wait, no, 100.06 is 0.02 above 100.04 and 0.04 below 100.10.Wait, 100.04 + 0.02 = 100.06100.10 - 0.04 = 100.06So, 100.06 is exactly halfway between 100.04 and 100.10.But in reality, 100.06 is 0.02 above 100.04 and 0.04 below 100.10.So, it's closer to 100.04.Therefore, the answer should be C) 100.04.But wait, let me check the exact value.Compute 499995 ÷ 4997.Let me do this division more accurately.4997 × 100 = 499700Subtract: 499995 - 499700 = 295So, 295 ÷ 4997.Let me compute 295 ÷ 4997.4997 goes into 295 zero times. So, 0.But we can write it as 295.000...So, 4997 goes into 295000 how many times?Compute 4997 × 5 = 249854997 × 5 = 24985Subtract from 29500: 29500 - 24985 = 4515Bring down a zero: 451504997 goes into 45150 how many times?4997 × 9 = 44973Subtract: 45150 - 44973 = 177Bring down a zero: 17704997 goes into 1770 zero times. Bring down another zero: 177004997 × 3 = 14991Subtract: 17700 - 14991 = 2709Bring down a zero: 270904997 × 5 = 24985Subtract: 27090 - 24985 = 2105Bring down a zero: 210504997 × 4 = 19988Subtract: 21050 - 19988 = 1062Bring down a zero: 106204997 × 2 = 9994Subtract: 10620 - 9994 = 626Bring down a zero: 62604997 × 1 = 4997Subtract: 6260 - 4997 = 1263Bring down a zero: 126304997 × 2 = 9994Subtract: 12630 - 9994 = 2636Bring down a zero: 263604997 × 5 = 24985Subtract: 26360 - 24985 = 1375Bring down a zero: 137504997 × 2 = 9994Subtract: 13750 - 9994 = 3756Bring down a zero: 375604997 × 7 = 34979Subtract: 37560 - 34979 = 2581Bring down a zero: 258104997 × 5 = 24985Subtract: 25810 - 24985 = 825Bring down a zero: 82504997 × 1 = 4997Subtract: 8250 - 4997 = 3253Bring down a zero: 325304997 × 6 = 29982Subtract: 32530 - 29982 = 2548Bring down a zero: 254804997 × 5 = 24985Subtract: 25480 - 24985 = 495Bring down a zero: 49504997 × 0 = 0So, we can see that the division is non-terminating, but up to this point, we have:295 ÷ 4997 ≈ 0.05904...So, the total is 100 + 0.05904 ≈ 100.05904So, approximately 100.059, which is 100.06 when rounded to two decimal places.But the options are in increments of 0.04, so 100.04, 100.08, 100.12, etc., but the options given are A) 99.90, B) 100.00, C) 100.04, D) 100.10, E) 100.20.So, 100.06 is closer to 100.04 (difference of 0.02) than to 100.10 (difference of 0.04). Therefore, the closest option is C) 100.04.But wait, let me check if I made a mistake in the initial steps.Wait, when I completed the square for S₁:Starting with:[ x^2 + y^2 - 100x - 100y + 3 leq 0 ]Completing the square:[ (x - 50)^2 + (y - 50)^2 leq 4997 ]Yes, because:[ (x - 50)^2 = x^2 - 100x + 2500 ]Similarly for y.So, 2500 + 2500 = 5000, so 3 - 5000 = -4997, hence:[ (x - 50)^2 + (y - 50)^2 leq 4997 ]Similarly for S₂:[ x^2 + y^2 - 1000x - 1000y + 5 leq 0 ]Completing the square:[ (x - 500)^2 + (y - 500)^2 leq 499995 ]Yes, because:[ (x - 500)^2 = x^2 - 1000x + 250000 ]Similarly for y.So, 250000 + 250000 = 500000, so 5 - 500000 = -499995, hence:[ (x - 500)^2 + (y - 500)^2 leq 499995 ]So, the areas are:Area of S₁: π * 4997Area of S₂: π * 499995Ratio: 499995 / 4997 ≈ 100.059So, approximately 100.06, which is closer to 100.04 than to 100.10.Therefore, the answer should be C) 100.04.But wait, let me check if the problem is about the ratio of areas, which is (π * 499995) / (π * 4997) = 499995 / 4997.Yes, that's correct.So, the ratio is 499995 / 4997 ≈ 100.059, which is approximately 100.06, but since the options are in 0.04 increments, the closest is C) 100.04.Wait, but 100.06 is closer to 100.04 by 0.02 and to 100.10 by 0.04, so yes, C is the closest.Alternatively, maybe the exact value is 100.04999, which would be approximately 100.05, which is still closer to 100.04.Therefore, the answer is C) 100.04.

💡Okay, so I have this problem here where I'm given that c is greater than 1, and then two expressions for a and b. a is the square root of (c + 1) minus the square root of c, and b is the square root of c minus the square root of (c - 1). I need to figure out whether a is greater than b, less than b, equal to b, or if it depends on the value of c.Alright, let's start by writing down what we know:c > 1a = sqrt(c + 1) - sqrt(c)b = sqrt(c) - sqrt(c - 1)I need to compare a and b. Hmm, both a and b are differences of square roots, but a is the difference between sqrt(c + 1) and sqrt(c), while b is the difference between sqrt(c) and sqrt(c - 1). So, a is like moving forward by one in the square roots, and b is moving backward by one.I remember that when you have expressions like sqrt(n + 1) - sqrt(n), they get smaller as n increases. Is that right? Let me think. For example, sqrt(2) - sqrt(1) is about 1.414 - 1 = 0.414. Then sqrt(3) - sqrt(2) is about 1.732 - 1.414 = 0.318. So, yes, as n increases, the difference between consecutive square roots decreases. So, if a is moving forward and b is moving backward, maybe a is smaller than b?But wait, let's not jump to conclusions. Maybe I should rationalize these expressions to make them easier to compare. I remember that multiplying by the conjugate can help simplify expressions like this.Let's try that for a:a = sqrt(c + 1) - sqrt(c)Multiply numerator and denominator by the conjugate, which is sqrt(c + 1) + sqrt(c):a = [sqrt(c + 1) - sqrt(c)] * [sqrt(c + 1) + sqrt(c)] / [sqrt(c + 1) + sqrt(c)]The numerator becomes (sqrt(c + 1))^2 - (sqrt(c))^2 = (c + 1) - c = 1So, a = 1 / [sqrt(c + 1) + sqrt(c)]Similarly, let's do the same for b:b = sqrt(c) - sqrt(c - 1)Multiply numerator and denominator by the conjugate, which is sqrt(c) + sqrt(c - 1):b = [sqrt(c) - sqrt(c - 1)] * [sqrt(c) + sqrt(c - 1)] / [sqrt(c) + sqrt(c - 1)]The numerator becomes (sqrt(c))^2 - (sqrt(c - 1))^2 = c - (c - 1) = 1So, b = 1 / [sqrt(c) + sqrt(c - 1)]Now, both a and b are expressed as 1 divided by the sum of two square roots. So, a is 1 over [sqrt(c + 1) + sqrt(c)], and b is 1 over [sqrt(c) + sqrt(c - 1)].To compare a and b, since both are reciprocals of sums, the one with the larger denominator will be smaller. So, if [sqrt(c + 1) + sqrt(c)] is larger than [sqrt(c) + sqrt(c - 1)], then a will be smaller than b.Is [sqrt(c + 1) + sqrt(c)] larger than [sqrt(c) + sqrt(c - 1)]? Let's see.Compare sqrt(c + 1) and sqrt(c - 1). Since c > 1, c + 1 > c > c - 1, so sqrt(c + 1) > sqrt(c) > sqrt(c - 1). Therefore, sqrt(c + 1) + sqrt(c) is definitely larger than sqrt(c) + sqrt(c - 1).Therefore, the denominator of a is larger than the denominator of b, which means a is less than b.Wait, let me double-check. If I have two fractions, both with numerator 1, and denominators D1 and D2, then if D1 > D2, 1/D1 < 1/D2. Yes, that's correct.So, since sqrt(c + 1) + sqrt(c) > sqrt(c) + sqrt(c - 1), it follows that 1/[sqrt(c + 1) + sqrt(c)] < 1/[sqrt(c) + sqrt(c - 1)], which means a < b.Therefore, the correct answer should be B: a < b.But just to make sure, let's plug in a specific value for c and see if this holds. Let's take c = 2.Compute a = sqrt(3) - sqrt(2) ≈ 1.732 - 1.414 ≈ 0.318Compute b = sqrt(2) - sqrt(1) ≈ 1.414 - 1 ≈ 0.414Indeed, 0.318 < 0.414, so a < b.Let's try another value, say c = 5.a = sqrt(6) - sqrt(5) ≈ 2.449 - 2.236 ≈ 0.213b = sqrt(5) - sqrt(4) ≈ 2.236 - 2 ≈ 0.236Again, 0.213 < 0.236, so a < b.One more, c = 10.a = sqrt(11) - sqrt(10) ≈ 3.316 - 3.162 ≈ 0.154b = sqrt(10) - sqrt(9) ≈ 3.162 - 3 ≈ 0.162Still, a < b.Okay, so this seems consistent. Therefore, I can confidently say that a is less than b for c > 1.

💡Alright, so I've got this complex number problem here: Z equals the square root of 2 minus i cubed, all over 1 minus the square root of 2 times i. And I need to figure out which quadrant or axis the corresponding point of Z lies in. The options are first or third quadrant, second or fourth quadrant, positive half of the x-axis, or positive half of the y-axis.Okay, let's start by recalling what complex numbers are. A complex number has a real part and an imaginary part, usually written as a + bi, where a is the real part and b is the imaginary part. The position of this point in the complex plane is determined by these two components. If both a and b are positive, it's in the first quadrant; if a is negative and b is positive, it's in the second; both negative is the third; and a positive and b negative is the fourth.So, my goal is to simplify this expression for Z and find its real and imaginary parts. Once I have those, I can determine which quadrant or axis it lies on.First, let's look at the numerator: sqrt(2) minus i cubed. I remember that i squared is -1, so i cubed should be i squared times i, which is -1 times i, so that's -i. So, i cubed is -i. Therefore, the numerator becomes sqrt(2) minus (-i), which simplifies to sqrt(2) plus i.So now, Z is equal to (sqrt(2) + i) divided by (1 minus sqrt(2)i). Hmm, complex numbers in the denominator can be tricky, but I remember that we can get rid of the imaginary part in the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.The conjugate of 1 minus sqrt(2)i is 1 plus sqrt(2)i. So, I'll multiply both the numerator and the denominator by that.Let's write that out:Z = (sqrt(2) + i) * (1 + sqrt(2)i) / [(1 - sqrt(2)i)(1 + sqrt(2)i)]Now, let's multiply out the numerator and the denominator.Starting with the numerator: (sqrt(2) + i)(1 + sqrt(2)i)Using the distributive property (FOIL method):First: sqrt(2) * 1 = sqrt(2)Outer: sqrt(2) * sqrt(2)i = (sqrt(2))^2 * i = 2iInner: i * 1 = iLast: i * sqrt(2)i = sqrt(2)i^2 = sqrt(2)*(-1) = -sqrt(2)So, adding all these up:sqrt(2) + 2i + i - sqrt(2)Simplify:sqrt(2) - sqrt(2) cancels out, and 2i + i is 3i. So, the numerator simplifies to 3i.Now, the denominator: (1 - sqrt(2)i)(1 + sqrt(2)i)Again, using the distributive property:First: 1 * 1 = 1Outer: 1 * sqrt(2)i = sqrt(2)iInner: -sqrt(2)i * 1 = -sqrt(2)iLast: -sqrt(2)i * sqrt(2)i = - (sqrt(2))^2 * i^2 = -2*(-1) = 2Adding these up:1 + sqrt(2)i - sqrt(2)i + 2Simplify:sqrt(2)i - sqrt(2)i cancels out, and 1 + 2 is 3. So, the denominator is 3.Putting it all together, Z = 3i / 3 = i.Wait, so Z simplifies to i? That means the real part is 0 and the imaginary part is 1. So, in the complex plane, this would be the point (0,1), which lies on the positive half of the y-axis.Looking back at the options: A is first or third quadrant, B is second or fourth quadrant, C is positive half of the x-axis, and D is positive half of the y-axis. So, the correct answer should be D.But just to make sure I didn't make any mistakes, let me go through the steps again.Starting with Z = (sqrt(2) - i^3)/(1 - sqrt(2)i)i^3 is indeed -i, so numerator becomes sqrt(2) + i.Denominator is 1 - sqrt(2)i.Multiplying numerator and denominator by the conjugate of the denominator, which is 1 + sqrt(2)i.Numerator: (sqrt(2) + i)(1 + sqrt(2)i)As before, this gives sqrt(2)*1 + sqrt(2)*sqrt(2)i + i*1 + i*sqrt(2)iWhich simplifies to sqrt(2) + 2i + i - sqrt(2) = 3i.Denominator: (1 - sqrt(2)i)(1 + sqrt(2)i) = 1 + sqrt(2)i - sqrt(2)i - (sqrt(2)i)^2 = 1 - (-2) = 3.So, Z = 3i / 3 = i.Yes, that seems correct. So, Z is purely imaginary with a positive coefficient, meaning it lies on the positive y-axis.Therefore, the answer is D.

💡Okay, so I have two trigonometry problems here, and I need to solve them step by step. Let me start with the first one.**Problem (1):** Given that 90° < α < 180°, and a point P on the terminal side of angle α is P(x, √5), with cosα = (√2 / 4)x, find the values of sinα and tanα.Alright, let's break this down. First, I know that in the unit circle, the coordinates of a point on the terminal side of an angle are (x, y), and the distance from the origin to that point is the radius r, which is √(x² + y²). So, in this case, the point is (x, √5), so r = √(x² + (√5)²) = √(x² + 5).Given that cosα = x / r, and they also tell us that cosα = (√2 / 4)x. So, I can set up the equation:x / r = (√2 / 4)xHmm, let me write that out:x / √(x² + 5) = (√2 / 4)xWait, so both sides have an x. If I divide both sides by x (assuming x ≠ 0), I get:1 / √(x² + 5) = √2 / 4That's a simpler equation. Let me solve for x.First, take reciprocals on both sides:√(x² + 5) = 4 / √2Simplify 4 / √2. Multiply numerator and denominator by √2 to rationalize:4 / √2 = (4√2) / 2 = 2√2So, √(x² + 5) = 2√2Now, square both sides:x² + 5 = (2√2)² = 8So, x² = 8 - 5 = 3Therefore, x = ±√3But wait, the angle α is between 90° and 180°, which is in the second quadrant. In the second quadrant, the x-coordinate is negative, so x must be negative. Therefore, x = -√3.Now, let's find r. Since x = -√3, r = √(x² + 5) = √(3 + 5) = √8 = 2√2.Okay, so now we can find sinα and tanα.sinα is y / r. Here, y is √5, so sinα = √5 / (2√2). Let me rationalize the denominator:√5 / (2√2) = (√5 * √2) / (2 * 2) = √10 / 4So, sinα = √10 / 4.Next, tanα is y / x. So, tanα = √5 / (-√3) = -√5 / √3. Rationalizing the denominator:-√5 / √3 = (-√15) / 3So, tanα = -√15 / 3.Alright, that seems to make sense. Let me just double-check my steps.1. Found r correctly as √(x² + 5).2. Set up the equation for cosα correctly.3. Solved for x, got x = ±√3, but since α is in the second quadrant, x is negative, so x = -√3.4. Calculated r as 2√2.5. Found sinα as √5 / (2√2) and simplified to √10 / 4.6. Found tanα as √5 / (-√3) and simplified to -√15 / 3.Everything seems to check out. I think I did that correctly.**Problem (2):** It is known that there is a point P(x, -1) (x ≠ 0) on the terminal side of angle θ, and tanθ = -x, find sinθ and cosθ.Alright, so here, the point is (x, -1). So, the coordinates are (x, -1). The radius r is √(x² + (-1)²) = √(x² + 1).Given that tanθ = -x. But tanθ is also y / x. So, tanθ = (-1) / x.So, set up the equation:(-1) / x = -xSimplify:(-1) / x = -xMultiply both sides by x (assuming x ≠ 0, which is given):-1 = -x²Multiply both sides by -1:1 = x²So, x² = 1, which means x = ±1.So, x can be 1 or -1.Now, let's consider both cases.**Case 1: x = 1**So, the point is (1, -1). Then, r = √(1² + (-1)²) = √(1 + 1) = √2.Now, sinθ = y / r = (-1) / √2 = -√2 / 2.cosθ = x / r = 1 / √2 = √2 / 2.**Case 2: x = -1**So, the point is (-1, -1). Then, r = √((-1)² + (-1)²) = √(1 + 1) = √2.Now, sinθ = y / r = (-1) / √2 = -√2 / 2.cosθ = x / r = (-1) / √2 = -√2 / 2.Wait, so in both cases, sinθ is the same, but cosθ changes sign depending on x.But let me think about the quadrants. Since the point is (x, -1), the y-coordinate is negative, so θ is in either the third or fourth quadrant.If x is positive (1), then θ is in the fourth quadrant.If x is negative (-1), then θ is in the third quadrant.In the fourth quadrant, sinθ is negative and cosθ is positive.In the third quadrant, both sinθ and cosθ are negative.So, the results we got make sense.For x = 1: sinθ = -√2 / 2, cosθ = √2 / 2.For x = -1: sinθ = -√2 / 2, cosθ = -√2 / 2.Wait, hold on. In the third quadrant, both sin and cos are negative, so sinθ is negative, which matches, and cosθ is negative, which also matches.In the fourth quadrant, sinθ is negative, and cosθ is positive, which also matches.So, both cases are valid.Therefore, depending on the value of x, we get the respective sinθ and cosθ.Let me just recap the steps.1. Given point (x, -1), so r = √(x² + 1).2. tanθ = y / x = (-1) / x.3. Given tanθ = -x, so set (-1)/x = -x.4. Solve for x: x² = 1, so x = ±1.5. For each x, compute sinθ and cosθ.6. Check quadrants to ensure signs make sense.Yep, that all adds up. I think I did that correctly.**Final Answer**For problem (1): sinα = boxed{dfrac{sqrt{10}}{4}} and tanα = boxed{-dfrac{sqrt{15}}{3}}.For problem (2): When x = 1, sinθ = boxed{-dfrac{sqrt{2}}{2}} and cosθ = boxed{dfrac{sqrt{2}}{2}}; When x = -1, sinθ = boxed{-dfrac{sqrt{2}}{2}} and cosθ = boxed{-dfrac{sqrt{2}}{2}}.

💡Alright, so I've got this problem where Jana and Viviane are playing a game with dandelions arranged in an m x n grid. They take turns mowing down entire rows or columns, and the person who mows the last dandelion wins. I need to figure out for which pairs (m, n) Jana has a winning strategy.First, let me try to understand the game mechanics. Each turn, a player can choose any row or column and mow down all the dandelions in that line. So, each move reduces the number of remaining rows or columns by one. The game ends when all dandelions are mowed down, and the person who makes the last move wins.Hmm, this reminds me a bit of the game of Nim, where players take turns removing objects from heaps. In Nim, the winning strategy often involves binary representations and XOR operations, but I'm not sure if that applies here directly. Maybe I need to think in terms of parity or something simpler.Let me consider some small cases to get a feel for the problem.Case 1: m = 1 or n = 1. If the grid is 1 x n or m x 1, then the first player can just mow down the entire row or column in one move and win immediately. So, for any (1, n) or (m, 1), Jana can win on her first turn.Case 2: m = 2, n = 2. Let's see. If Jana starts, she can mow a row or a column. Suppose she mows a row, then there's one row left. Viviane can then mow the remaining row and win. Alternatively, if Jana mows a column, Viviane can mow the remaining column and win. So, in a 2x2 grid, Viviane can win if Jana starts.Wait, is that right? Let me double-check. If Jana mows a row, leaving a 1x2 grid. Then Viviane can mow the remaining row and win. Similarly, if Jana mows a column, Viviane can mow the remaining column. So yes, in 2x2, Viviane can win.Case 3: m = 3, n = 3. Let's see. If Jana mows a row, then we have a 2x3 grid. Then Viviane can mow a column, leaving a 2x2 grid. Then Jana is forced to mow a row or column, leaving a 1x2 or 2x1 grid, and Viviane can finish it. Hmm, so maybe Viviane can win here too? Wait, no, maybe Jana can choose a different strategy.Alternatively, if Jana mows a row, leaving 2x3. Viviane mows a column, leaving 2x2. Then Jana mows a row, leaving 1x2, and Viviane mows the last row. So Viviane still wins. Maybe Jana should mow a column instead? Let's see. If Jana mows a column, leaving 3x2. Viviane can mow a row, leaving 2x2. Then Jana is back to the same situation. Hmm, seems like Viviane can still win.Wait, maybe I'm missing something. Is there a way for Jana to force a win in 3x3? Maybe not, because whatever she does, Viviane can mirror her moves or something. Maybe the parity of m + n matters here.Wait, in the 2x2 case, m + n = 4, which is even, and Viviane wins. In the 3x3 case, m + n = 6, which is also even, and Viviane wins. In the 1x1 case, m + n = 2, which is even, but Jana wins because she can take the last one. Wait, that contradicts. So maybe the parity isn't the only factor.Wait, in 1x1, Jana can win. In 1x2, Jana can win. In 2x2, Viviane can win. In 3x3, Viviane can win. Hmm, maybe it's when m + n is odd? Let's check.If m + n is odd, like 1x2 (1+2=3), Jana can win. 2x3 (2+3=5), let's see. If Jana mows a row, leaving 1x3. Then Viviane can mow the remaining row and win. Wait, no, if Jana mows a row in 2x3, leaving 1x3, Viviane can mow the remaining row and win. Alternatively, if Jana mows a column, leaving 2x2, then Viviane can mow a row or column, leaving 1x2 or 2x1, and Jana can finish it. Wait, so in 2x3, if Jana mows a column, leaving 2x2, then Viviane mows a row, leaving 1x2, and Jana can mow the last row and win. So in 2x3, Jana can win.Wait, so in 2x3, Jana can win, but in 2x2, she can't. So maybe it's not just about m + n being odd or even. Maybe it's something else.Wait, let me think again. In 2x2, Jana mows a row, leaving 1x2. Viviane mows the last row and wins. Similarly, if Jana mows a column, Viviane mows the last column. So Viviane can win in 2x2.In 2x3, if Jana mows a column, leaving 2x2, then Viviane is in a losing position because whatever she does, Jana can finish. Wait, no, if Viviane mows a row, leaving 1x2, then Jana can mow the last row. If Viviane mows a column, leaving 2x1, Jana can mow the last column. So in 2x3, Jana can win.Wait, so maybe when m and n are both even, Viviane can win, and when at least one is odd, Jana can win? Let's test that.In 1x1, both are odd, Jana wins. In 1x2, one odd, one even, Jana wins. In 2x2, both even, Viviane wins. In 2x3, one even, one odd, Jana wins. In 3x3, both odd, Jana can win? Wait, earlier I thought Viviane could win in 3x3, but maybe not.Wait, in 3x3, if Jana mows a row, leaving 2x3. Then Viviane mows a column, leaving 2x2. Then Jana is forced to mow a row or column, leaving 1x2 or 2x1, and Viviane can finish. So Viviane wins. But 3x3 has both m and n odd, but Viviane wins. So my previous hypothesis is incorrect.Hmm, maybe it's about the parity of m + n. Let's see:- 1x1: m + n = 2, even, Jana wins.- 1x2: m + n = 3, odd, Jana wins.- 2x2: m + n = 4, even, Viviane wins.- 2x3: m + n = 5, odd, Jana wins.- 3x3: m + n = 6, even, Viviane wins.So it seems like when m + n is odd, Jana can win, and when m + n is even, Viviane can win. But wait, in 1x1, m + n = 2, even, but Jana wins. So maybe there's an exception when m or n is 1.Wait, in 1x1, Jana can win because she can take the last one. In 1x2, m + n = 3, odd, Jana wins. In 2x2, m + n = 4, even, Viviane wins. In 2x3, m + n = 5, odd, Jana wins. In 3x3, m + n = 6, even, Viviane wins.So maybe the rule is: If m + n is odd, Jana can win, except when both m and n are 1, which is a special case where Jana still wins. Wait, no, 1x1 is m + n = 2, even, but Jana wins. So maybe the rule is: If m + n is odd, or if either m or n is 1, Jana can win. Otherwise, Viviane can win.Wait, let's test that:- 1x1: m + n = 2 (even), but m = 1, so Jana wins.- 1x2: m + n = 3 (odd), Jana wins.- 2x2: m + n = 4 (even), neither m nor n is 1, so Viviane wins.- 2x3: m + n = 5 (odd), Jana wins.- 3x3: m + n = 6 (even), neither m nor n is 1, Viviane wins.- 1x3: m + n = 4 (even), but m = 1, so Jana wins.- 3x1: same as above.Yes, that seems to fit. So the rule is: Jana can win if either m + n is odd, or if either m or n is 1.Wait, but in 1x1, m + n is even, but Jana still wins because she can take the last one. So maybe the rule is: Jana wins if m + n is odd, or if either m or n is 1, regardless of the parity.But wait, in 1x2, m + n is odd, and Jana wins. In 1x3, m + n is even, but since m = 1, Jana wins. Similarly, in 2x1, same thing.So, to generalize, Jana can win if either:1. m + n is odd, or2. Either m or n is 1.But wait, in 1x1, m + n is even, but since m = 1, Jana wins. So the rule holds.Let me test another case: 4x5. m + n = 9, which is odd, so Jana should win. Let's see:Jana can mow a row, leaving 3x5. Then Viviane mows a column, leaving 3x4. Jana mows a row, leaving 2x4. Viviane mows a column, leaving 2x3. Jana mows a row, leaving 1x3. Viviane mows the last row and wins. Wait, that contradicts. So maybe my rule is wrong.Wait, no, maybe Jana can choose a better move. Instead of mowing a row, maybe she should mow a column. Let's see:Jana mows a column, leaving 4x4. Then Viviane mows a row, leaving 3x4. Jana mows a column, leaving 3x3. Viviane mows a row, leaving 2x3. Jana mows a column, leaving 2x2. Viviane mows a row, leaving 1x2. Jana mows the last row and wins.Wait, so in 4x5, Jana can win by choosing to mow a column first. So my initial thought was correct.Wait, but in my first attempt, I had Jana mow a row, leading to a loss, but if she chooses to mow a column, she can win. So the key is that Jana can choose the optimal move to force a win when m + n is odd.Similarly, in 3x3, m + n = 6, even, so Viviane can win. Let's see:Jana mows a row, leaving 2x3. Viviane mows a column, leaving 2x2. Jana mows a row, leaving 1x2. Viviane mows the last row and wins.Alternatively, if Jana mows a column first, leaving 3x2. Viviane mows a row, leaving 2x2. Jana mows a row, leaving 1x2. Viviane mows the last row and wins.So in 3x3, Viviane can win.So, putting it all together, it seems that Jana can win if either m + n is odd, or if either m or n is 1. Otherwise, Viviane can win.Wait, but in 1x1, m + n is even, but Jana wins because she can take the last one. So the rule is: Jana wins if m + n is odd, or if either m or n is 1, regardless of the parity.But wait, in 1x2, m + n is odd, and Jana wins. In 1x3, m + n is even, but since m = 1, Jana wins. Similarly, in 2x1, same thing.So, the final rule is: Jana has a winning strategy if and only if m + n is odd, or if either m or n is 1.Wait, but in 1x1, m + n is even, but Jana still wins because she can take the last one. So maybe the rule is: Jana wins if m + n is odd, or if either m or n is 1, regardless of the parity.Yes, that seems to cover all cases.So, to summarize:- If m + n is odd, Jana can win.- If either m or n is 1, Jana can win.- Otherwise, Viviane can win.Therefore, the pairs (m, n) for which Jana has a winning strategy are those where m + n is odd or where either m or n is 1.

💡Okay, so I've got this system of equations here, and I need to find the range of values for 'a' such that the solution satisfies a certain inequality. Let me try to break this down step by step.First, the system of equations is:1. ( x + y = 3a + 4 )2. ( x - y = 7a - 4 )And the inequality that the solution must satisfy is:( 3x - 2y < 11 )Alright, so I think the first thing I need to do is solve this system of equations for x and y in terms of 'a'. Once I have expressions for x and y, I can substitute them into the inequality and solve for 'a'.Let me start by solving the system of equations. I have two equations with two variables, x and y. It looks like I can use either substitution or elimination. Since both equations are already solved for expressions involving x and y, elimination might be straightforward.If I add the two equations together, the y terms will cancel out because one is positive and the other is negative. Let's try that.Adding equation 1 and equation 2:( (x + y) + (x - y) = (3a + 4) + (7a - 4) )Simplifying the left side:( x + y + x - y = 2x )And the right side:( 3a + 4 + 7a - 4 = 10a )So, ( 2x = 10a )Divide both sides by 2:( x = 5a )Okay, so x is 5a. Now, I can substitute this value back into one of the original equations to find y. Let's use equation 1:( x + y = 3a + 4 )Substitute x with 5a:( 5a + y = 3a + 4 )Now, solve for y:Subtract 5a from both sides:( y = 3a + 4 - 5a )Simplify:( y = -2a + 4 )So, now I have expressions for both x and y in terms of 'a':( x = 5a )( y = -2a + 4 )Great, now that I have x and y in terms of 'a', I can substitute these into the inequality ( 3x - 2y < 11 ) to find the range of 'a'.Let's do that substitution:( 3x - 2y < 11 )Substitute x and y:( 3(5a) - 2(-2a + 4) < 11 )Now, let's simplify this expression step by step.First, multiply out the terms:( 15a - 2(-2a) - 2(4) < 11 )Which simplifies to:( 15a + 4a - 8 < 11 )Combine like terms:( 19a - 8 < 11 )Now, solve for 'a':Add 8 to both sides:( 19a < 19 )Divide both sides by 19:( a < 1 )So, according to this, 'a' must be less than 1 for the inequality to hold.Wait, let me double-check my steps to make sure I didn't make any mistakes.Starting from the substitution:( 3(5a) - 2(-2a + 4) < 11 )Which is:( 15a - 2*(-2a) - 2*4 < 11 )So, that's:( 15a + 4a - 8 < 11 )Combining like terms:( 19a - 8 < 11 )Adding 8 to both sides:( 19a < 19 )Dividing by 19:( a < 1 )Yes, that seems correct. So, the range of values for 'a' is all real numbers less than 1.But just to be thorough, let me test a value of 'a' that's less than 1 and see if the inequality holds.Let's choose a = 0.Then, x = 5*0 = 0y = -2*0 + 4 = 4Now, plug into the inequality:( 3x - 2y = 3*0 - 2*4 = 0 - 8 = -8 )Is -8 < 11? Yes, that's true.Now, let's test a value greater than 1, say a = 2, to see if it doesn't satisfy the inequality.x = 5*2 = 10y = -2*2 + 4 = -4 + 4 = 0Now, plug into the inequality:( 3x - 2y = 3*10 - 2*0 = 30 - 0 = 30 )Is 30 < 11? No, it's not. So, that's consistent with our solution.What about a = 1? Let's check the boundary.x = 5*1 = 5y = -2*1 + 4 = -2 + 4 = 2Plug into the inequality:( 3x - 2y = 3*5 - 2*2 = 15 - 4 = 11 )Is 11 < 11? No, it's equal. So, the inequality is strict, meaning a must be less than 1, not equal to.Therefore, the solution is correct.So, the range of values for 'a' is all real numbers less than 1.