Tag: mathematics and statistics

The director circle is the locus of the point of intersection of a pair of perpendicular tangents to a hyperbola.

Equation of the director circle of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2  b^2$ i.e. a circle whose center is origin and radius is $(a^2  b^2)$.

Hence, for the director circle to be a point circle, $a^2 = b^2$ .

$p^2 - 4 = p^2 + 4p + 3$ ---> $4p = -7$ ---> $p = -\dfrac{7}{4}$ . Hence, option (B).

If $A\subseteq B$ 
$\therefore A\cap B=A$ 
R is reflexive since for any integer $a$ we have $a-a=0$ and $0$ is divisible by $n$.
Hence $aRa\quad \forall a\in I$

R is symmetric, $aRb$. Then by definition of $R$, $a-b=nk$ where $k\in I$.
Hence $b-a=\left( -k \right) n$ where $-k\in I$ and so $bRa$.
Thus we shown that $aRb\Rightarrow bRa$

R is transitive, let $aRb$ and $bRc$. then by definition of $R$, we have
$a-b={ k } _{ 1 }n$ and $b-a={ nk } _{ 2 }$
where ${ k } _{ 1 },{ k } _{ 2 }\in I$
It follow that $a-c=\left( a-b \right) +\left( b-c \right) ={ k } _{ 1 }n+{ k } _{ 2 }n=\left( { k } _{ 1 }+{ k } _{ 2 } \right) n$ 

The only 2 common elements are 6 and 8.

$n(A \cup B)=n(A)+n(B)-n(A \cap B)$

Given: A = ${2, 4, 6, ...}$
     B = ${2, 3, 5, ...}$
$\displaystyle \therefore $ 2 is the only even prime number $\displaystyle A\cap B=\left { 2 \right }$

Let $P$ and $S$ represents the sets of hockey player wearing the prescribed hockey pants and shirts respectively.
Then $n(P) = 11$, $n(S) = 14$, $n$$\displaystyle \left ( P\cup S \right )$ $= 19$
$\displaystyle \Rightarrow $ $\displaystyle \left ( P\cap S \right )$ $=$ no of people wearing both pantand shirt
$= n(P) + n(S) - n$$\displaystyle \left ( P\cup S \right )$
$= 11 + 14 - 19 = 6$

Given:-

True as $\displaystyle \phi'\cap A=U\cap A=A$

Total no. of people$=500$