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
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 $ 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:
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 $ A = \{-2, -1, 0, 1, 2, 3\} $. Let $ R $ be a relation on $ A $ defined by $ (x, y) \in R $ if and only if $ |x| \le |y| $. Let $ m $ be the number of reflexive elements in $ R $ and $ n $ be the minimum number of elements required to be added in $ R $ to make it reflexive and symmetric relations, respectively. Then $ l + m + n $ is equal to
If the domain of the function \( f(x) = \frac{1}{\sqrt{3x + 10 - x^2}} + \frac{1}{\sqrt{x + |x|}} \) is \( (a, b) \), then \( (1 + a)^2 + b^2 \) is equal to:
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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}