The correct answer is: 26

S represents the shaded region shown in the diagram.
Clearly, z1 will be the point of intersection of PA and given circle.
PA : 2x + y = 4 and given circle has equation
(x – 2)2 + y2 = 1.
On solving, we get
\(z_1=()+\frac{2}{5}i=|z_1|^2=5-\frac{4}{\sqrt5}\)
z2 will be either B or C.
So
\(5(|z_1|^2+|z_2|^2=30-4√5\)
Clearly α = 30 and β = –4 ⇒α + β = 26
If the domain of the function $ f(x) = \log_7(1 - \log_4(x^2 - 9x + 18)) $ is $ (\alpha, \beta) \cup (\gamma, \delta) $, then $ \alpha + \beta + \gamma + \delta $ is equal to
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Let $ A = \{0, 1, 2, 3, 4, 5, 6\} $ and $ R_1 = \{(x, y): \max(x, y) \in \{3, 4 \}$. Consider the two statements:
Statement 1: Total number of elements in $ R_1 $ is 18.
Statement 2: $ R $ is symmetric but not reflexive and transitive.
Let $ A $ be the set of all functions $ f: \mathbb{Z} \to \mathbb{Z} $ and $ R $ be a relation on $ A $ such that $$ R = \{ (f, g) : f(0) = g(1) \text{ and } f(1) = g(0) \} $$ Then $ R $ is:
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:

A relation in mathematics defines the relationship between two different sets of information. If two sets are considered, the relation between them will be established if there is a connection between the elements of two or more non-empty sets. Therefore, we can say, ‘A set of ordered pairs is defined as a relation.’
Read Also: Relation and Function
There are 8 main types of relations which are:
There are two ways by which a relation can be represented-
The roster form and set-builder for for a set integers lying between -2 and 3 will be-
I= {-1,0,1,2}
I= {x:x∈I,-2<x<3}