MCQ
If $A \subset B$, then
  • A
    $A^e \subset B^e$
  • B
    $B^e \not \subset A^c$
  • C
    $A^c=B^c$
  • $B^c \subset A^c$

Answer

Correct option: D.
$B^c \subset A^c$
Let $A \subset$ B
To prove $B^C \subset A^c$, it is enough to show that $x \in B^c \Rightarrow x \in A^c$
$\text { Let } x \in B^C$
$\Rightarrow x \notin B$
$\Rightarrow x \notin A \text { since } A \subset B$
$\Rightarrow x \in A^c$
Hence $B ^{ C } \subset A ^{ C }$
 

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

Similar questions

The coefficient of the middle term in the expansion of $(2 + 3x)4$ is:
Let $a_1, a_2, a_3, a_4$ be real numbers such that $a_1+a_2+a_3+a_4=0$ and $a_1^2+a_2^2+a_3^2+a_4^2=1$. Then, the smallest possible value of the expression $\left(a_1-a_2\right)^2+\left(a_2-a_3\right)^2+\left(a_3-a_4\right)+\left(a_4-a_1\right)^2$ lies in the interval
Dialing a telephone number an old man forgets the last two digits remembering only that these are different dialled at random. The probability that the number is dialled correctly, is
The sum of the series $\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{2^{2}}{x^{4}+1}+\ldots . .+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is :
Point from which two distinct tangents can be drawn on two different branches of the hyperbola $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = \,1$ but no two different tangent can be drawn to the circle $x^2 + y^2 = 36$ is
If $\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ and $\text{g(x)}=\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2},$ then $f(g(x))$ is equal to:
If $\text{f(x)}=\frac{2^{\text{x}}+2^{-\text{x}}}{2},$ then $f(x + y)f(x - y)$ is equal to:
$\mathop {\lim }\limits_{x \to 0} \,\frac{{{{(1 + x)}^{1/x}} - e}}{x}$ equals
The locus of the mid point of a chord of the circle $x^2 + y^2 = 4$ which subtends a right angle at the origin is
Let $a$ be the sum of all coefficients in the expansion of $\left(1-2 x+2 x^2\right)^{2023}\left(3-4 x^2+2 x^3\right)^{2024}$ and $b=\lim _{x \rightarrow 0}\left(\frac{\int_0^x \frac{\log (1+t)}{t^{2024}+1} d t}{x^2}\right)$. If the equations $\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0$ and $2 \mathrm{bx}^2+\mathrm{ax}+4=0$ have a common root, where $c, d, e \in R$, then $d: c: e$ equals