💡First, I need to identify the elements of set ( M ), which are (-4, -3, -2, -1, 0, 1).Next, I'll determine the elements of set ( N ) by solving the inequality ( x^2 + 3x < 0 ). Factoring the quadratic expression gives ( x(x + 3) < 0 ). This inequality holds true when ( x ) is between (-3) and (0), so ( N = {-3 < x < 0} ).Now, I'll find the intersection of ( M ) and ( N ) by identifying the common elements in both sets. From ( M ), the elements within the range (-3 < x < 0) are (-2) and (-1).Therefore, the intersection ( M cap N ) is ({-2, -1}).

💡First, I need to understand the custom operation ( a spadesuit b ), which is defined as:[a spadesuit b = frac{3a}{b} cdot frac{b}{a}]Simplifying this expression:[a spadesuit b = frac{3a cdot b}{b cdot a} = frac{3ab}{ab} = 3]This shows that regardless of the values of ( a ) and ( b ) (as long as ( b neq 0 )), the result of ( a spadesuit b ) is always 3.Next, I'll evaluate the expression ( (7 spadesuit (4 spadesuit 9)) spadesuit 2 ) step by step.1. **Calculate the inner operation ( 4 spadesuit 9 ):** [ 4 spadesuit 9 = 3 ] 2. **Substitute this result back into the expression:** [ 7 spadesuit 3 = 3 ] 3. **Finally, calculate the outer operation with 2:** [ 3 spadesuit 2 = 3 ] Therefore, the value of the entire expression is 3.

💡Okay, so I've got this equation to solve: [cos (arctan( sin (arccos z) ) ) = z.]Hmm, that's a bit of a mouthful. Let me try to break it down step by step. I remember that when dealing with inverse trigonometric functions, it's helpful to introduce substitution variables to simplify the expressions. Maybe I can let some angle be equal to the innermost function and work my way outwards.Let me start by letting (alpha = arccos z). That means (z = cos alpha), right? And since (arccos z) is defined for (z) in the interval ([-1, 1]), (alpha) must be between (0) and (pi). So, (0 leq alpha leq pi).Now, the next part inside the equation is (sin(arccos z)), which is (sin alpha). I remember that (sin^2 alpha + cos^2 alpha = 1), so (sin alpha = sqrt{1 - cos^2 alpha}). Since (cos alpha = z), this becomes (sin alpha = sqrt{1 - z^2}). So, (sin(arccos z) = sqrt{1 - z^2}). That simplifies the equation a bit. Now, the equation becomes:[cos (arctan( sqrt{1 - z^2} )) = z.]Alright, let's tackle the next part: (arctan(sqrt{1 - z^2})). Let me denote this as another angle, say (beta), so:[beta = arctan(sqrt{1 - z^2}).]This means that (tan beta = sqrt{1 - z^2}). Since (arctan) returns values between (-frac{pi}{2}) and (frac{pi}{2}), (beta) must be in that interval. But since (sqrt{1 - z^2}) is non-negative (because square roots are always non-negative), (beta) is actually between (0) and (frac{pi}{2}).Now, I need to find (cos beta). I know that in a right triangle, (tan beta = frac{text{opposite}}{text{adjacent}} = sqrt{1 - z^2}). So, if I imagine a right triangle where the opposite side is (sqrt{1 - z^2}) and the adjacent side is 1, then the hypotenuse would be (sqrt{1 + (1 - z^2)} = sqrt{2 - z^2}).Wait, hold on, that doesn't seem right. Let me think again. If (tan beta = sqrt{1 - z^2}), then the opposite side is (sqrt{1 - z^2}) and the adjacent side is 1. Therefore, the hypotenuse is (sqrt{1 + (1 - z^2)} = sqrt{2 - z^2}). So, (cos beta = frac{text{adjacent}}{text{hypotenuse}} = frac{1}{sqrt{2 - z^2}}).But wait, the original equation is (cos(arctan(sqrt{1 - z^2})) = z), which is (cos beta = z). So, putting it all together, we have:[frac{1}{sqrt{2 - z^2}} = z.]Hmm, okay, so now we have an equation in terms of (z):[frac{1}{sqrt{2 - z^2}} = z.]Let me solve this equation for (z). First, I can square both sides to eliminate the square root:[left(frac{1}{sqrt{2 - z^2}}right)^2 = z^2][frac{1}{2 - z^2} = z^2]Now, multiply both sides by (2 - z^2) to get rid of the denominator:[1 = z^2 (2 - z^2)][1 = 2z^2 - z^4]Let me rearrange this equation:[z^4 - 2z^2 + 1 = 0]Hmm, this looks like a quadratic in terms of (z^2). Let me set (y = z^2). Then the equation becomes:[y^2 - 2y + 1 = 0]This factors as:[(y - 1)^2 = 0]So, (y = 1). Since (y = z^2), this means (z^2 = 1), so (z = pm 1). But wait, in the original equation, (z) is inside an (arccos), which requires (z) to be in ([-1, 1]). So, (z = 1) and (z = -1) are both possible solutions. But the question asks for the number of positive solutions, so we only consider (z = 1).But hold on a second, let me check if (z = 1) actually satisfies the original equation. Let's plug (z = 1) back into the equation:First, compute (arccos(1)). Since (arccos(1) = 0), because (cos(0) = 1).Then, compute (sin(arccos(1)) = sin(0) = 0).Next, compute (arctan(0) = 0), since (tan(0) = 0).Finally, compute (cos(0) = 1).So, the left-hand side is 1, and the right-hand side is also 1. So, (z = 1) is indeed a solution.But wait, earlier when I solved for (z), I got (z = pm 1), but (z = -1) is negative, so we don't consider it for positive solutions. So, does that mean there's only one positive solution, (z = 1)?But hold on, when I squared both sides of the equation, I might have introduced an extraneous solution. Let me check if (z = 1) is the only solution or if there are others.Wait, when I squared both sides, I went from:[frac{1}{sqrt{2 - z^2}} = z]to[frac{1}{2 - z^2} = z^2]But squaring can sometimes introduce solutions that don't satisfy the original equation. Let me check if there are any other solutions.Looking back, after substituting (alpha = arccos z), we had:[cos(sqrt{1 - z^2}) = z]Wait, no, that was a different approach. Let me see. Maybe I made a mistake in my substitution earlier.Wait, let me go back to the step where I set (beta = arctan(sqrt{1 - z^2})). Then, I found that (cos beta = frac{1}{sqrt{2 - z^2}}), which equals (z). So, (frac{1}{sqrt{2 - z^2}} = z).But when I squared both sides, I got (z^4 - 2z^2 + 1 = 0), which factors to ((z^2 - 1)^2 = 0), giving (z^2 = 1), so (z = pm 1). So, only (z = 1) is positive.But wait, let me think again. If (z = 1), then (arccos(1) = 0), so (sin(0) = 0), then (arctan(0) = 0), and (cos(0) = 1). So, it works.But what if (z = frac{sqrt{2}}{2})? Let me test that. If (z = frac{sqrt{2}}{2}), then (arccos(z)) is (frac{pi}{4}), since (cos(frac{pi}{4}) = frac{sqrt{2}}{2}). Then, (sin(frac{pi}{4}) = frac{sqrt{2}}{2}). Then, (arctan(frac{sqrt{2}}{2})). Hmm, what is (arctan(frac{sqrt{2}}{2}))? It's some angle whose tangent is (frac{sqrt{2}}{2}). Let's call that angle (gamma). Then, (tan gamma = frac{sqrt{2}}{2}), so (gamma) is approximately 35.264 degrees.Then, (cos(gamma)) would be (frac{1}{sqrt{1 + (frac{sqrt{2}}{2})^2}} = frac{1}{sqrt{1 + frac{1}{2}}} = frac{1}{sqrt{frac{3}{2}}} = frac{sqrt{2}}{sqrt{3}} = frac{sqrt{6}}{3}). But (frac{sqrt{6}}{3}) is approximately 0.816, which is not equal to (z = frac{sqrt{2}}{2} approx 0.707). So, that doesn't satisfy the equation.Wait, so maybe (z = frac{sqrt{2}}{2}) is not a solution. But earlier, I thought that was the solution. Hmm, perhaps I made a mistake in my substitution.Wait, let's go back. When I set (alpha = arccos z), then (sin alpha = sqrt{1 - z^2}). Then, (arctan(sin alpha)) is (arctan(sqrt{1 - z^2})), which I called (beta). Then, (cos beta = z).But I think I might have confused the substitution. Let me try a different approach.Let me consider the equation again:[cos (arctan( sin (arccos z) ) ) = z.]Let me denote (theta = arccos z), so (z = cos theta), and (theta in [0, pi]). Then, (sin theta = sqrt{1 - z^2}).So, the equation becomes:[cos (arctan( sqrt{1 - z^2} )) = z.]Let me denote (phi = arctan( sqrt{1 - z^2} )). Then, (tan phi = sqrt{1 - z^2}), and since (sqrt{1 - z^2} geq 0), (phi in [0, frac{pi}{2})).Now, (cos phi = frac{1}{sqrt{1 + tan^2 phi}} = frac{1}{sqrt{1 + (1 - z^2)}} = frac{1}{sqrt{2 - z^2}}).So, the equation becomes:[frac{1}{sqrt{2 - z^2}} = z.]Which is the same equation I had before. Squaring both sides:[frac{1}{2 - z^2} = z^2][1 = z^2 (2 - z^2)][z^4 - 2z^2 + 1 = 0][(z^2 - 1)^2 = 0][z^2 = 1][z = pm 1]So, only (z = 1) is positive. Therefore, there's only one positive solution.Wait, but earlier I thought (z = frac{sqrt{2}}{2}) was a solution, but when I tested it, it didn't satisfy the equation. So, maybe I was wrong about that.Let me double-check by plugging (z = 1) into the original equation:[cos (arctan( sin (arccos 1) ) ) = cos (arctan( sin 0 ) ) = cos (arctan 0 ) = cos 0 = 1 = z.]Yes, that works.What about (z = 0)? Let me test that. If (z = 0), then (arccos 0 = frac{pi}{2}), (sin(frac{pi}{2}) = 1), (arctan(1) = frac{pi}{4}), and (cos(frac{pi}{4}) = frac{sqrt{2}}{2} neq 0). So, (z = 0) is not a solution.What about (z = frac{sqrt{2}}{2})? As I did earlier, plugging in:[arccosleft(frac{sqrt{2}}{2}right) = frac{pi}{4}][sinleft(frac{pi}{4}right) = frac{sqrt{2}}{2}][arctanleft(frac{sqrt{2}}{2}right) approx 0.6155 text{ radians}][cos(0.6155) approx 0.816 neq frac{sqrt{2}}{2} approx 0.707]So, that doesn't work. Therefore, the only positive solution is (z = 1).But wait, let me think again. When I squared both sides, I might have lost some solutions or introduced extraneous ones. Let me check if there are any other solutions.Looking back at the equation:[frac{1}{sqrt{2 - z^2}} = z]Since both sides are positive (because (z) is positive and the left side is always positive), squaring both sides is valid and doesn't introduce extraneous solutions in terms of sign, but it might still introduce solutions that don't satisfy the original equation.But in this case, the only solution we got was (z = 1), and it does satisfy the original equation. So, I think that's the only positive solution.Wait, but let me consider the domain of (z). Since (z = cos theta), and (theta in [0, pi]), (z) is in ([-1, 1]). But we're only looking for positive solutions, so (z in (0, 1]).Also, in the equation (frac{1}{sqrt{2 - z^2}} = z), the left side is a decreasing function of (z) because as (z) increases, the denominator increases, making the whole expression decrease. The right side is an increasing function of (z). So, they can intersect at most once. Since we found (z = 1) as the intersection point, that must be the only solution.Therefore, the number of positive solutions is 1.

💡First, I need to determine how to divide the total sum of ₹2743 into two parts such that the interest from the first part over 8 years at 3% per annum equals the interest from the second part over 3 years at 5% per annum.Let’s denote the first part of the sum as ( x ) and the second part as ( 2743 - x ).Next, I'll calculate the interest for both parts using the simple interest formula:[text{Interest} = text{Principal} times text{Rate} times text{Time}]For the first part:[text{Interest} = x times frac{3}{100} times 8]For the second part:[text{Interest} = (2743 - x) times frac{5}{100} times 3]According to the problem, these two interests are equal:[x times frac{3}{100} times 8 = (2743 - x) times frac{5}{100} times 3]Simplifying the equation:[frac{24x}{100} = frac{15(2743 - x)}{100}]Multiplying both sides by 100 to eliminate the denominators:[24x = 15(2743 - x)]Expanding the right side:[24x = 41145 - 15x]Combining like terms:[24x + 15x = 41145][39x = 41145]Solving for ( x ):[x = frac{41145}{39} = 1055]Finally, calculating the second part:[text{Second part} = 2743 - 1055 = 1688]

💡Alright, so I'm trying to tackle this geometry problem about a parallelogram ABCD with point M as the intersection of the diagonals AC and BD. There's a point N inside triangle AMB such that angle AND is equal to angle BNC. I need to prove that angle MNC equals angle NDA and angle MND equals angle NCB.First, let me draw a diagram to visualize the problem. ABCD is a parallelogram, so opposite sides are equal and parallel. Diagonals AC and BD intersect at M, which should be the midpoint of both diagonals because in a parallelogram, diagonals bisect each other. So, AM = MC and BM = MD.Point N is inside triangle AMB. That means N is somewhere within the area bounded by points A, M, and B. The condition given is that angle AND is equal to angle BNC. Hmm, that seems important. Maybe I can use some properties of triangles or cyclic quadrilaterals here?Let me recall that in a parallelogram, opposite angles are equal, and consecutive angles are supplementary. Also, since M is the midpoint, triangles AMB and CMD are congruent, as are triangles BMC and AMD.Given that angle AND equals angle BNC, perhaps there's some similarity or congruence between triangles AND and BNC? But wait, AND and BNC are angles, not triangles. Maybe if I can find some similar triangles involving these angles?Alternatively, maybe I can use the concept of isogonal conjugates or reflection. Since M is the midpoint, reflecting points across M might preserve some angles or lines.Let me consider reflecting point N across M. Let's call the reflection N'. Since M is the midpoint, reflecting N over M would place N' such that M is the midpoint of NN'. So, coordinates might help here, but maybe I can do it without coordinates.If I reflect N over M to get N', then line AN would reflect to line CN', and line DN would reflect to line BN'. So, angle AND would reflect to angle CN'B. Given that angle AND equals angle BNC, perhaps angle CN'B equals angle BNC?Wait, that might not directly help. Let me think again. Since angle AND = angle BNC, maybe triangles AND and BNC have some relationship. But they share angle at N, but I don't know if the sides are proportional.Alternatively, maybe quadrilateral ANNC is cyclic? If angle AND equals angle BNC, perhaps points A, N, N', C lie on a circle? Not sure.Wait, maybe I can use Ceva's theorem or something similar. Ceva's theorem relates the concurrency of lines in a triangle, but I'm not sure if that applies here.Alternatively, since ABCD is a parallelogram, maybe I can use vector methods or coordinate geometry. Let me assign coordinates to the points to make it more concrete.Let me place point A at (0,0). Since it's a parallelogram, let me assign coordinates as follows: Let’s say point B is at (a,0), point D is at (0,b), so point C would be at (a,b). Then, the diagonals AC and BD intersect at M, which would be the midpoint of both diagonals. So, coordinates of M would be ((a/2), (b/2)).Point N is inside triangle AMB. Let me denote coordinates of N as (x,y). Since N is inside triangle AMB, x and y must satisfy certain conditions, like being between 0 and a/2 for x, and between 0 and b/2 for y, but I need to confirm that.Given that angle AND equals angle BNC. Let me express these angles in terms of coordinates.First, angle AND is the angle at point N between points A and D. Similarly, angle BNC is the angle at point N between points B and C.Expressing these angles might be complicated, but perhaps I can use vector dot products or slopes to find the condition.Alternatively, maybe I can use the concept of similar triangles. If angle AND equals angle BNC, perhaps triangles AND and BNC are similar? Let me check.For triangles AND and BNC to be similar, their corresponding angles must be equal. We know angle AND equals angle BNC. If another pair of angles is equal, then the triangles would be similar.Looking at triangle AND: angles at A, N, D.Triangle BNC: angles at B, N, C.Is there a relationship between angles at A and B or D and C? Since ABCD is a parallelogram, angle at A equals angle at C, and angle at B equals angle at D. But I'm not sure if that directly helps.Alternatively, maybe I can use the Law of Sines in triangles AND and BNC.In triangle AND:sin(angle at A)/ND = sin(angle AND)/ADIn triangle BNC:sin(angle at B)/NC = sin(angle BNC)/BCSince angle AND = angle BNC, and AD = BC (because ABCD is a parallelogram), then:sin(angle at A)/ND = sin(angle at B)/NCBut in a parallelogram, angle at A and angle at B are supplementary, so sin(angle at A) = sin(angle at B). Therefore, ND = NC.So, from this, we get that ND = NC. That's interesting. So, point N is equidistant from points C and D.Wait, that means N lies on the perpendicular bisector of segment CD. But in a parallelogram, the perpendicular bisector of CD would pass through M, since M is the midpoint of CD as well.Wait, no, in a parallelogram, diagonals bisect each other, so M is the midpoint of both AC and BD, but CD is a side, so the perpendicular bisector of CD would pass through the midpoint of CD, which is also M because in a parallelogram, diagonals bisect each other, so M is the midpoint of all sides? Wait, no, M is the midpoint of the diagonals, not necessarily the sides.Wait, in a parallelogram, the diagonals bisect each other, so M is the midpoint of AC and BD, but not necessarily the midpoint of the sides. The midpoints of the sides would be different points.So, the perpendicular bisector of CD would pass through the midpoint of CD, which is different from M unless the parallelogram is a rhombus. Since we don't know that, maybe it's not necessarily passing through M.But we have that N lies on the perpendicular bisector of CD, meaning that N is equidistant from C and D. So, NC = ND.That's a useful piece of information. So, point N is such that NC = ND.Now, I need to prove that angle MNC = angle NDA and angle MND = angle NCB.Let me see. Since NC = ND, triangle NCD is isoceles with NC = ND. Therefore, angles at C and D are equal. Wait, but in triangle NCD, angles at C and D are equal, so angle NCD = angle NDC.But how does that help with the angles involving M?Alternatively, maybe I can consider triangles MNC and NDA.Wait, angle MNC is the angle at N between points M and C, and angle NDA is the angle at D between points N and A.Hmm, not sure.Alternatively, maybe I can use some cyclic quadrilateral properties. If I can show that certain points lie on a circle, then the angles subtended by the same chord would be equal.Let me consider quadrilateral ANNC. Wait, that's not a standard notation. Maybe quadrilateral ANDC? If I can show that ANDC is cyclic, then angle AND would equal angle ACD, but not sure.Alternatively, since NC = ND, maybe reflecting N over M would help, as I thought earlier.Let me try reflecting N over M to get N'. Since M is the midpoint of NN', coordinates might help here.If I assign coordinates as before: A(0,0), B(a,0), D(0,b), C(a,b), M(a/2, b/2). Let N be (x,y). Then N' would be (a - x, b - y).Given that NC = ND, so distance from N to C equals distance from N to D.Distance NC: sqrt[(a - x)^2 + (b - y)^2]Distance ND: sqrt[(0 - x)^2 + (b - y)^2]Setting them equal:sqrt[(a - x)^2 + (b - y)^2] = sqrt[x^2 + (b - y)^2]Squaring both sides:(a - x)^2 + (b - y)^2 = x^2 + (b - y)^2Simplify:(a - x)^2 = x^2Expanding:a^2 - 2ax + x^2 = x^2Simplify:a^2 - 2ax = 0So, a(a - 2x) = 0Since a ≠ 0 (as it's the length of the side AB), we have a - 2x = 0 => x = a/2.So, the x-coordinate of N is a/2. That means N lies on the vertical line passing through M, which is at (a/2, b/2). So, N is somewhere along the line x = a/2, between y=0 and y=b/2, since it's inside triangle AMB.So, coordinates of N are (a/2, y), where 0 < y < b/2.Now, let me find the angles MNC and NDA.First, angle MNC is the angle at N between points M and C.Points:M(a/2, b/2)N(a/2, y)C(a, b)So, vector NM is from N to M: (0, b/2 - y)Vector NC is from N to C: (a/2, b - y)So, angle MNC is the angle between vectors NM and NC.Similarly, angle NDA is the angle at D between points N and A.Points:D(0, b)N(a/2, y)A(0,0)So, vector DN is from D to N: (a/2, y - b)Vector DA is from D to A: (0, -b)So, angle NDA is the angle between vectors DN and DA.I need to show that these two angles are equal.Let me compute the slopes or use dot products to find the angles.First, for angle MNC:Vectors NM = (0, b/2 - y)Vector NC = (a/2, b - y)The angle between them can be found using the dot product:cos(theta) = (NM . NC) / (|NM| |NC|)Compute NM . NC = 0*(a/2) + (b/2 - y)(b - y) = (b/2 - y)(b - y)|NM| = sqrt(0^2 + (b/2 - y)^2) = |b/2 - y||NC| = sqrt((a/2)^2 + (b - y)^2)So,cos(theta) = [(b/2 - y)(b - y)] / [ |b/2 - y| * sqrt((a/2)^2 + (b - y)^2) ]Simplify numerator and denominator:Note that (b/2 - y) is positive since y < b/2 (as N is inside triangle AMB). So, |b/2 - y| = b/2 - y.Thus,cos(theta) = (b/2 - y)(b - y) / [ (b/2 - y) * sqrt((a/2)^2 + (b - y)^2) ) ]Cancel out (b/2 - y):cos(theta) = (b - y) / sqrt((a/2)^2 + (b - y)^2 )Similarly, for angle NDA:Vectors DN = (a/2, y - b)Vector DA = (0, -b)Compute the angle between DN and DA.Dot product:DN . DA = (a/2)*0 + (y - b)(-b) = -b(y - b) = b(b - y)|DN| = sqrt( (a/2)^2 + (y - b)^2 ) = sqrt( (a/2)^2 + (b - y)^2 )|DA| = sqrt(0^2 + (-b)^2 ) = bSo,cos(phi) = (DN . DA) / (|DN| |DA| ) = [b(b - y)] / [ sqrt( (a/2)^2 + (b - y)^2 ) * b ]Simplify:cos(phi) = (b - y) / sqrt( (a/2)^2 + (b - y)^2 )So, cos(theta) = cos(phi), which implies that theta = phi or theta = -phi. Since angles are between 0 and 180, they must be equal. Therefore, angle MNC = angle NDA.Similarly, now let's compute angle MND and angle NCB.Angle MND is the angle at N between points M and D.Points:M(a/2, b/2)N(a/2, y)D(0, b)So, vector NM = (0, b/2 - y)Vector ND = (-a/2, b - y)Angle MND is the angle between vectors NM and ND.Compute the dot product:NM . ND = 0*(-a/2) + (b/2 - y)(b - y) = (b/2 - y)(b - y)|NM| = |b/2 - y| = b/2 - y|ND| = sqrt( ( -a/2 )^2 + (b - y)^2 ) = sqrt( (a/2)^2 + (b - y)^2 )So,cos(theta') = [(b/2 - y)(b - y)] / [ (b/2 - y) * sqrt( (a/2)^2 + (b - y)^2 ) ) ] = (b - y)/sqrt( (a/2)^2 + (b - y)^2 )Now, angle NCB is the angle at C between points N and B.Points:C(a, b)N(a/2, y)B(a, 0)So, vector CN = (-a/2, y - b)Vector CB = (0, -b)Angle NCB is the angle between vectors CN and CB.Compute the dot product:CN . CB = (-a/2)*0 + (y - b)(-b) = -b(y - b) = b(b - y)|CN| = sqrt( (-a/2)^2 + (y - b)^2 ) = sqrt( (a/2)^2 + (b - y)^2 )|CB| = sqrt(0^2 + (-b)^2 ) = bSo,cos(phi') = (CN . CB) / (|CN| |CB| ) = [b(b - y)] / [ sqrt( (a/2)^2 + (b - y)^2 ) * b ] = (b - y)/sqrt( (a/2)^2 + (b - y)^2 )Thus, cos(theta') = cos(phi'), so theta' = phi', meaning angle MND = angle NCB.Therefore, both required angle equalities are proven.

💡Okay, so I have this problem here: I need to determine whether "a < -4" is a sufficient condition, necessary condition, or both for the function f(x) = ax + 3 to have a zero within the interval [-1, 1]. The options are A, B, C, D, with A being sufficient but not necessary, B necessary but not sufficient, C both, and D neither.Alright, first, I need to understand what a sufficient condition and a necessary condition mean. From what I remember, a sufficient condition means that if it's met, then the outcome must happen. But it doesn't mean that it's the only way the outcome can happen. On the other hand, a necessary condition means that for the outcome to happen, the condition must be met, but meeting the condition doesn't guarantee the outcome.So, in this case, the outcome is that the function f(x) = ax + 3 has a zero in the interval [-1, 1]. I need to see if "a < -4" is sufficient, necessary, both, or neither for this outcome.First, let's recall that a zero of a function is a point where the function crosses the x-axis, meaning f(x) = 0. So, for f(x) = ax + 3, setting it equal to zero gives ax + 3 = 0, which simplifies to x = -3/a. So, the zero of this function is at x = -3/a.Now, we want this zero to lie within the interval [-1, 1]. That means -1 ≤ -3/a ≤ 1.Hmm, okay, so let's write that inequality down:-1 ≤ -3/a ≤ 1I need to solve this inequality for a.But before I proceed, I should consider the sign of a because when dealing with inequalities involving reciprocals, the direction of the inequality can change depending on whether a is positive or negative.So, let's consider two cases: a > 0 and a < 0.Case 1: a > 0If a is positive, then -3/a is negative. So, the zero is at a negative x-value. Now, we want this x-value to be greater than or equal to -1.So, -1 ≤ -3/aMultiplying both sides by a (which is positive, so inequality direction remains the same):-1 * a ≤ -3Which simplifies to:-a ≤ -3Multiplying both sides by -1 (which reverses the inequality):a ≥ 3So, in this case, if a is greater than or equal to 3, the zero is within [-1, 1].Case 2: a < 0If a is negative, then -3/a is positive. So, the zero is at a positive x-value. We want this x-value to be less than or equal to 1.So, -3/a ≤ 1Multiplying both sides by a (which is negative, so inequality direction reverses):-3 ≥ aWhich simplifies to:a ≤ -3So, in this case, if a is less than or equal to -3, the zero is within [-1, 1].Putting both cases together, the function f(x) = ax + 3 has a zero in the interval [-1, 1] if and only if a ≥ 3 or a ≤ -3.Now, the condition given is "a < -4". Let's see how this relates to our solution.From our earlier analysis, the function has a zero in the interval if a ≤ -3. So, "a < -4" is a subset of this condition because -4 is less than -3. Therefore, if a is less than -4, it certainly satisfies a ≤ -3, which means the function will have a zero in the interval.But is "a < -4" the only way for the function to have a zero in the interval? No, because a could also be between -4 and -3, or even greater than or equal to 3. So, "a < -4" is not necessary because there are other values of a that also satisfy the condition.Therefore, "a < -4" is a sufficient condition because it guarantees the function has a zero in the interval, but it's not necessary because there are other values of a that also work.Wait, but let me double-check. If a is less than -4, then a is certainly less than -3, so the zero is at x = -3/a. Since a is negative, x is positive. And since a < -4, then -3/a is less than 3/4, which is less than 1. So, x is within [0, 1), which is within [-1, 1]. So, yes, it works.But if a is between -4 and -3, say a = -3.5, then x = -3/(-3.5) = 3/3.5 ≈ 0.857, which is still within [-1, 1]. So, even when a is between -4 and -3, the zero is still in the interval. Therefore, "a < -4" is not necessary because a could be greater than -4 but still less than or equal to -3, and the zero would still be in the interval.Similarly, if a is greater than or equal to 3, the zero is at x = -3/a, which is negative. For a = 3, x = -1, which is the endpoint. For a > 3, x is greater than -1, so still within [-1, 1].Therefore, the condition "a < -4" is sufficient because it ensures the zero is in the interval, but it's not necessary because there are other values of a that also ensure the zero is in the interval.So, the answer should be A: Sufficient but not necessary condition.Wait, but let me think again. If a is exactly -3, then x = -3/(-3) = 1, which is the endpoint. So, a = -3 is also acceptable. Similarly, a = 3 gives x = -1, which is also acceptable.So, the function has a zero in the interval if a ≥ 3 or a ≤ -3. Therefore, "a < -4" is a subset of a ≤ -3, so it's sufficient but not necessary.Yes, that seems correct.