n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n(A ∩ B) = 5% of 10,000 = 500

n(B ∩ C) = 3% of 10,000 = 300

n(C ∩ A) = 4% of 10,000 = 400

n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find the number of families which buy only A = n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)]

=4000 - [500 + 400 - 200] = 4000 - 700 = 3300

Since
$4^n-3n-1=(3+1{)}^n-3n-1=3^n+{}^n{C}_1{3}^{n-1}+{}^n{C}_2{3}^{n-2}+.....+{}^n{C}_{n-1}3+{}^n{C}_n-3n-1={}^n{C}_2{3}^2+{}^n{C}_3{.3}^3+...+{}^n{C}_n{3}^n,({}^n{C}_0={}^n{C}_n,{}^n{C}_{n-1}={}^n{C}_1.....soon.)=9[{}^n{C}_2+{}^n{C}_3\left(3\right)+......+{}^n{C}_4{3}^{n-1}]$
$\therefore 4^n-3n-1$ is a multiple of 9 for $n\ge 2$.
$Forn=1,4^n-3n-1=4-3-1=0Forn=2,4^n-3n-1=16-6-1=9\therefore 4^n-3n-1$ is a multiple of 9 for all $nϵN$
$\therefore X$ contains elements, which are multiples of 9, and clearly $Y$ contains all multiples of 9.
$\therefore X\subset Yi.e.,X\cup Y=Y$

n($A^c$ ∩ $B^c$) = n(U) - n(A ∪ B)

= n(U) - [n(A) + n(B) - n(A ∩ B)]

= 700 - [200 + 300 - 100] = 300.

Since y =   $\frac{1}{x}$, y = -x meet when -x =   $\frac{1}{x}$  ⇒ $x^2$ = -1,

which does not give any real value of x.

Hence, A ∩ B = ∅.

A ∩ $(A\cap B{)}^c$) = A ∩ ($A^c$ ∪ $B^c$)

                             = (A ∩ ($A^c$ ) ∪ (A  ∩ ($B^c$)

                             = ∅ ∪ (A ∩ ($B^c$) = A ∩ ($B^c$).

A = {x: x ∈ R, -1 < x < 1}

B = {x: x ∈ R:x-1 ≤ -1 or x-1 ≥ 1}

   = {x : x ∈ R:x ≤ 0 or  x ≥ 2}

∴ A ∪ B = R - D, where D = {x : x ∈ R, 1 ≤ x < 2}.

$x^2=16\Rightarrow x=\pm 4$
$2x=6\Rightarrow x=3$

There is no value of $x$ which satisfies both the given equations. The set A is an empty set or a null set.

Thus, A = {}.

Number of subsets of A = ${}^n{C}_0$ + ${}^n{C}_1$  + .............+ ${}^n{C}_n$ = $2^n$

Since $x^2$ + 1 = 0, gives $x^2=-1$

⇒ $x=\pm i$

∴ $x$ is not real but $x$ has to be is real (given)

∴ No real value of $x$ is possible in this case.

Since, y =  $e^x$ and y = x do not meet for any x ∈ R

$\therefore$ A ∩ B = ∅ .