Let A={x∈(0,π)−{2π}:log(2/π)∣sinx∣+log(2/π)∣cosx∣=2} and B={x⩾0:x(x−4)−3∣x−2∣+6=0}. Then n(A∪B) is equal to:
Get the complete, step-by-step math solution for: "Let {A} = \{x (0, π) - \{(π)/(2)\} : _{(2 / π)} |sin x| + _{(2 / π)} |cos x| = 2\} and {B} = \{x 0 : √(x) (√(x) - 4) - 3 |√(x) - 2| + 6 = 0\}. Then n(...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Solve for set A
We start by solving the equation for set A. Using the logarithm property logbM+logbN=logb(MN), we can combine the two logarithmic terms. The base of the logarithm is 2/π, which is between 0 and 1.
Step 2: Simplify and solve for sin(2x)
Applying the logarithm property, we get log(2/π)∣sinxcosx∣=2. Converting this to exponential form gives ∣sinxcosx∣=(2/π)2. We know that sin(2x)=2sinxcosx, so ∣sinxcosx∣=21∣sin(2x)∣. Substituting this, we find ∣sin(2x)∣=π28. Since π≈3.14, π2≈9.86, so 8/π2≈0.81, which is less than 1.
Step 3: Find solutions for x in set A
Let sinα=8/π2 where α∈(0,π/2). Since x∈(0,π)−{π/2}, we have 2x∈(0,2π)−{π}. The equation ∣sin(2x)∣=8/π2 means sin(2x)=8/π2 or sin(2x)=−8/π2. This gives four distinct solutions for 2x in (0,2π), which are α, π−α, π+α, and 2π−α. None of these values are π. Therefore, there are 4 distinct values for x in set A.
Step 4: Solve for set B (Case 1: x−2≥0)
Now we solve the equation for set B. Let y=x. The equation becomes y(y−4)−3∣y−2∣+6=0. We consider two cases for ∣y−2∣.
Case 1: y−2≥0⟹y≥2. The equation becomes y2−4y−3(y−2)+6=0, which simplifies to y2−7y+12=0.
Step 5: Find solutions for y in Case 1
Factoring the quadratic equation y2−7y+12=0, we get (y−3)(y−4)=0. This gives solutions y=3 or y=4. Both solutions satisfy the condition y≥2.
Step 6: Solve for x in Case 1
Since y=x, we have x=3⟹x=9 and x=4⟹x=16. Both x=9 and x=16 are valid solutions for set B.
Step 7: Solve for set B (Case 2: x−2<0)
Case 2: y−2<0⟹0≤y<2. The equation becomes y2−4y−3(−(y−2))+6=0, which simplifies to y2−4y+3y−6+6=0, or y2−y=0.
Step 8: Find solutions for y in Case 2 and final count for B
Factoring y2−y=0, we get y(y−1)=0. This gives solutions y=0 or y=1. Both solutions satisfy the condition 0≤y<2. Therefore, x=0⟹x=0 and x=1⟹x=1. So, set B has elements {0,1,9,16}. Thus, n(B)=4.
Step 9: Calculate n(A∪B)
We found n(A)=4 and n(B)=4. Now we need to check for common elements. The elements of A are of the form x=21α, 21(π−α), 21(π+α), 21(2π−α), where α=arcsin(8/π2). Since 8/π2≈0.81, α≈0.94 radians. The values for x in A are approximately 0.47,1.10,2.08,2.65. None of these are integers. The elements of B are {0,1,9,16}, which are all integers. Therefore, A∩B=∅, so n(A∩B)=0. Thus, n(A∪B)=4+4−0=8.