NIMCET 2016 — Mathematics PYQ

Concept:
Set theory:
- A ∪ B means set of all the values in the set A and B.
- A ∩ B is the set of common elements of A and B.
Binomial theorem:
- $(a+b)n=nC0a^n{b}^0+nC1a^{(}n-1)b^1+nC2a^{(}n-2)b^2+...+nCn-2a^2{b}^{(}n-2)+nCn-1a^1{b}^{(}n-1)+nCn{a}^0{b}^n$

Calculation:

${\text{Given X= { 4}}^{\mathrm{n}}-3$n- 1, n$\in$N$\}$ and Y= $\{$9$n-9,n\in N\}$ O C. co ${\text{X= 4}}^{\mathrm{n}}-3$n- 1= ( 3+ 1${)}^{\mathrm{n}}$- 3$\mathrm{n}-1$

$\Rightarrow X={(}^{\mathrm{n}}{C}_0$ $3^{\mathrm{n}}{1}^0{+}^{\mathrm{n}}{C}_1{3}^{\mathrm{n}-\mathrm{l}}{1}^{\mathrm{l}}{+}^{\mathrm{n}}{C}_2{3}^{\mathrm{n}-2}{1}^2+\dots {+}^{\mathrm{n}}{C}_{\mathrm{n}-2}{3}^2{1}^{\mathrm{n}-2}{+}^{\mathrm{n}}{C}_{\mathrm{n}-1}{3}^{\mathrm{l}}{1}^{\mathrm{n}-1}+C_{\mathrm{n}}{3}^0{1}^{\mathrm{n}})-3n-1$

$({\because }^{\mathrm{n}}{\mathbb{C}}_{\mathrm{n}-1}=$n and n${\mathbb{C}}_{\mathrm{n}}=1)$

$\Rightarrow X={(}^{\mathrm{n}}{C}_0$ $3^{\mathrm{n}}{1}^0{+}^{\mathrm{n}}{C}_1{3}^{\mathrm{n}-1}{1}^{\mathrm{l}}{+}^{\mathrm{n}}{C}_2{3}^{\mathrm{n}-2}{1}^2+\dots {+}^{\mathrm{n}}{C}_{\mathrm{n}-2}{3}^2{1}^{\mathrm{n}-2})+3\mathbf{n}+1-3n-1$

$\Rightarrow \mathbf{X}=3^2\left({\mathrm{nC}}_0{3}^{\mathrm{n}-2}+{\mathrm{nC}}_1{3}^{\mathrm{n}-3}+{\mathrm{nC}}_2{3}^{\mathrm{n}-4}+\dots +{\mathrm{nC}}_{\mathrm{n}-2}\right)$

$\Rightarrow$X= 9$\left({}^n{\mathrm{C}}_0{3}^{n-2}+{{}^{\mathrm{n}}\mathrm{C}}_1{3}^{n-3}+{{}^{\mathrm{n}}\mathrm{C}}_2{3}^{n-4}+\dots +{{}^{\mathrm{n}}\mathrm{C}}_{n-2}\right)$

For n $\ge 2,\text{Xis some multiple of 9}$

$\mathrm{And}{X}_{\mathrm{n}=1}=0$

$...(\mathbf{i})$

$\therefore \mathbf{X}=\{0$, some multiples of 9 but not all}

Y= 9n- 9= 9( n- 1) 

$\therefore \mathbf{Y}=\{\mathbf{All}\mathbf{multiple}\mathbf{of}9\mathbf{starting}\mathbf{from}0\}$

$...(\mathbf{ii})$

From (i) and (ii) we can say, XcY

$\therefore \mathbf{X}\cup \mathbf{Y}=\mathbf{Y}$