Sets A and B have 3 and 6 elements respectively. What can be the … - Vidyadip

Question

Sets $A$ and $B$ have $3$ and $6$ elements respectively. What can be the minimum number of elements in $A \cup B$

✓

Answer

b
(b) $n(A \cup B) = n(A) + n(B) -n(A \cap B) $ = $3 + 6 - n(A \cap B)$

Since, maximum number of elements in $A \cap B = 3$

$\therefore $ Minimum number of elements in $A \cup B = 9 - 3 = 6$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{4 a^{2}}=1$ and the co-ordinate axis is $kab,$ then $\mathrm{k}$ is equal to ..... .

The coefficient of $x ^7$ in $\left(1-x+2 x^3\right)^{10}$ is $........$.

If $|a \times b|\, = 4$ and $|a\,.\,b|\, = 2$, then $|a{|^2}\,\,|b{|^2} = $

$\int_{\pi /4}^{\pi /2} {{\rm{cose}}{{\rm{c}}^2}xdx = } $

If $(\alpha,\beta,\gamma)$ be intersection point of lines $x -3y + 2z + 4 = 0 = 2x + y + 4z + 1$ and $\frac{x-\frac{1}{3}}{8}=\frac{y}{3}=\frac{z}{-1}$, then $\alpha+\beta+\gamma$ is -

For the function $f(x)\, = \frac{{{{\log }_e}(1 + x) - {{\log }_e}(1 - x)}}{x}$ to be continuous at $x = 0,$ the value of $f(0),$ should be

If the points of intersection of two distinct conics $x^2+y^2=4 b$ and $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve $y^2=3 x^2$, then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is ______________ .

Let $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ be two vectors such that $|2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}|$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$. If $\frac{1}{8} \vec{a}$ is a unit vector, then $|\vec{b}|$ is equal to :

If $\vec w = \alpha \left( {\vec a \times \vec b} \right) + \beta \left( {\vec b \times \vec c} \right) + \gamma \left( {\vec c \times \vec a} \right),$ $\left[ {\vec a,\vec b,\vec c} \right] = 2$ and $\vec w.\left( {\vec a + \vec b + \vec c} \right) = 8$, then $\alpha + \beta + \gamma =$

Let $[ x ]$ denote greatest integer less than or equal to $x .$ If for $n \in N ,\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}$ is equal to