Maths Test-1 Number System

The power set of a set with n elements contains 2^n elements. In this case, the set contains 6 elements, so its power set will contain 2^6 = 64 elements.

The set of integers satisfies closure property under both addition and multiplication. This means that when two integers are added or multiplied, the result is always an integer. For addition, if you add two integers, the sum will always be an integer. Similarly, for multiplication, if you multiply two integers, the product will always be an integer. Therefore, both addition and multiplication satisfy closure property for the set of integers.

The diagram shows that A and B overlap, meaning they have some elements in common. It also shows that C is not a subset of B, meaning there are elements in C that are not in B. Additionally, the diagram shows that A and C are disjoint sets, meaning they have no elements in common. Therefore, all of the statements are true based on the given diagram.

The set A is defined as {0}, which means it only contains the number 0. The set W is defined as the set of whole numbers, which includes all positive integers and zero. When we subtract set A from set W (W-A), we are removing the elements of set A from set W. Since set A only contains the number 0, subtracting it from set W will not change the set W. Therefore, the answer is N, which represents the set of all whole numbers.

The symbol π represents the mathematical constant pi, which is the ratio of a circle's circumference to its diameter. Pi is an irrational number because it cannot be expressed as a fraction or a finite decimal. Its decimal representation goes on forever without repeating or terminating. Therefore, the correct answer is "Irrational number."

The correct answer is the Union of A and B. This is because the set {x/x E A or x E B or x E both A and B} includes all elements that are in either A or B or both. Therefore, it represents the union of A and B, which is the set that contains all elements that are in either A or B or both.

The given statement "{a=b and b = c a = c}" is an example of the transitive property. The transitive property states that if two quantities are equal to a third quantity, then they are also equal to each other. In this case, it is stated that a is equal to b and b is equal to c, which implies that a is also equal to c. Therefore, the transitive property applies to this statement.

The set {R} refers to the set of real numbers. The statement "closed under addition" means that if you add any two real numbers, the result will also be a real number. Similarly, the statement "closed under multiplication" means that if you multiply any two real numbers, the result will also be a real number. Since the set of real numbers satisfies both of these conditions, it is closed under both addition and multiplication. Therefore, the correct answer is "BOTH A AND B".

The correct answer is N⊂W⊂ Z⊂R. This is because the given statement shows a nested relationship between the sets. N is a subset of W, which is a subset of Z, which is a subset of R.

Since N is a subset of W, it means that every element in N is also in W. Therefore, the intersection of N' (complement of N) and N will result in an empty set because there are no elements that are in both N' and N. Hence, the correct answer is {}.

When we raise z to the power of 2, we double the angle θ, resulting in 2θ. Then, when we add z to the result, we simply add θ to the angle. Therefore, the argument of (z^2 + z) is θ + 2θ, which simplifies to 3θ. However, since we are asked for the argument of (z^2 + z), we divide 3θ by 2 to get the final answer of θ/2.

The set of rational numbers is a proper subset of the set of real numbers because the set of real numbers includes both rational and irrational numbers, while the set of rational numbers only includes numbers that can be expressed as a fraction. Therefore, the set of real numbers is a larger set that encompasses the set of rational numbers.

The set of natural numbers is a subset of the set of whole numbers. The intersection of the set of natural numbers (N) and the set of whole numbers (W) would therefore be the set of natural numbers itself (N).

The expression |z+z bar| represents the absolute value of the sum of z and its complex conjugate. In this case, z is given as a+ib. The complex conjugate of z is a-ib. When we add z and its complex conjugate, we get (a+ib) + (a-ib) = 2a. Therefore, the absolute value of 2a is equal to 2a.

The Venn diagram shows both sets and the operation on sets. It visually represents the relationship between different sets and how they intersect or overlap with each other. By using different shapes or colors, the Venn diagram helps to illustrate the concept of set theory and the various operations that can be performed on sets, such as union, intersection, and complement.

The given complex number z = √3 + i can be converted to its polar form by finding the magnitude and argument. The magnitude can be found using the formula √(a^2 + b^2), where a and b are the real and imaginary parts of the complex number respectively. In this case, the magnitude is √(√3^2 + 1^2) = √(3 + 1) = √4 = 2. The argument can be found using the formula tan^(-1)(b/a), where a and b are the real and imaginary parts of the complex number respectively. In this case, the argument is tan^(-1)(1/√3) = 30°. Therefore, the polar form of the complex number z is (2, 30°).

The value of z is 4 because in a complex plane, the vertices of a triangle can be represented by complex numbers. The area of a triangle formed by three complex numbers z1, z2, and z3 is given by the formula: A = 0.5 * Im((z2 - z1) * (z3 - z1)). In this case, we are given that the area is 2 units, so 2 = 0.5 * Im((iz - z) * (z + iz)). Simplifying this equation, we get Im(-2iz) = 4. Since the imaginary part of -2iz is 4, we can conclude that the value of z is 4.

If a decimal is non-recurring but terminating, it means that it has a finite number of digits after the decimal point and the digits do not repeat. This can only happen if the decimal can be expressed as a fraction of two integers. Therefore, the correct answer is "Rational number".

The set O doesn't satisfy the closure property under addition because if we add any two elements from the set O, the result will not be in the set O.

The given equation is l3zl = lz-4l. This equation represents the absolute value of the complex number z, which is equal to the absolute value of the complex number -4. The absolute value of a complex number is given by the square root of the sum of the squares of its real and imaginary parts. In this case, the real part is x and the imaginary part is y. Therefore, x^2 + y^2 = 16. Since the equation also states that x^2 + y^2 + x = 2, we can substitute the value of x^2 + y^2 into this equation to find that x = 2. Hence, the answer is 2.

The correct answer is (2^n)-1. This is because for a set with n elements, there are 2^n total subsets including the empty set and the set itself. However, the question asks for the number of proper subsets, which excludes the empty set and the set itself. Therefore, the number of proper subsets is 2^n - 1.

The set of irrational numbers is a subset of the real numbers because irrational numbers cannot be expressed as a fraction or a decimal that terminates or repeats. Real numbers include both rational and irrational numbers, so all irrational numbers are included in the set of real numbers.

If a number is recurring and non-terminating, it means that its decimal representation repeats indefinitely without ending. This type of number can only be a rational number because irrational numbers cannot be expressed as a fraction of two integers. Therefore, the correct answer is "Rational number."

The number of the form √(n) where n is a positive perfect square always belongs to the set of rational numbers (Q). This is because the square root of a perfect square is always a whole number, and whole numbers are a subset of rational numbers. Therefore, the answer is Q.

Since N is a subset of W, it means that all elements of N are also elements of W. Therefore, when we take the union of N' (complement of N) and N, we are essentially taking all the elements that are not in N and adding them to N. Since N already contains all the elements of N, the union of N' and N will still result in N.