Sets

$f\left(A,B\right)=\frac{1-sinB+sinA-sinAsinB+cosAcosB}{cosA\left(1-sinB\right)}$

$\begin{array}{l}\Rightarrow f\left(A,B\right)\left(cosA-cosAsinB-cosB+sinAcosB\right)\\ =2sinA-2sinB\end{array}$

$\Rightarrow f\left(A,B\right)=\frac{2\left(sinA-sinB\right)}{cosA-cosB+sin\left(A-B\right)}=2g\left(A,B\right)$

$n(A\cup B\cup C)+n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=50$

$n(\cup )=50$

$n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=n(\cup )-n(A\cup B\cup C)\Rightarrow 50-29\Rightarrow 21$

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

$\Rightarrow \mathrm{}10+15+20-5-6-10-5\Rightarrow 29$

$n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=21$

A = {1,2,3,4,5} B = {0,1,2,3,4}
No. of elements common in A&B = 4.
∴ No. of elements common in A × B and B × A = 4 × 4 = 16

Out of numbers $\frac{1}{2},\frac{2}{4},\frac{3}{6},\frac{1}{3},\frac{2}{6},\frac{2}{3},\frac{4}{6}$ will result only 3 distinct rational numbers.

Total numbers =   

2a² + 3b² = 35
a=0, b≠I
a=±1, b²=11 ⇒ b≠I
a=±2, b=9 ⇒ b=3 or -3
a=±3, b²=17/3 ⇒ b≠I
a=±4, b²=1 ⇒ b=±1

Sum of elements $A\cap \left(B\cup C\right)=274\times 400$  

In set B numbers of the form 9k + 2 are {101, 109, .992}

  $\therefore sum=\frac{100}{2}\left(101+992\right)=\frac{100\times 1093}{2}......\left(i\right)$         

Another possible number is 9k + 5 forms are {104, .995}

  $\therefore sum=\frac{100}{2}\left(104+995\right)=\frac{100}{2}\times 1099.........\left(ii\right)$

$\therefore Total=\frac{100}{2}\times \left[1093+1099\right]=100\times 1096=274\times 4\times 100=274\times 400$           

$\therefore$ possible value of $\mathcal{l}$ = 5

A= {1,2,3,4,5.100}
B= {4,7,10,13,16,19.}
C= {2,4,6,8,10.}
B−C= {7,13,19, .97}
A∩ (B−C)= {7,13,19, .97}
sum of elements=832