NIMCET 2020 — Mathematics PYQ

1. Finding Elements of Set $X$

The general term for set $X$ is given by $4^n-3n-1$. Let's substitute values of $n$:

• For $n=1$:

  $$4^1-3(1)-1=4-3-1=0$$

• For $n=2$:

  $$4^2-3(2)-1=16-6-1=9$$

• For $n=3$:

  $$4^3-3(3)-1=64-9-1=54$$

• For $n=4$:

  $$4^4-3(4)-1=256-12-1=243$$

Thus, the elements of set $X$ are:

$$X=\{0,9,54,243,\dots \}$$

2. Finding Elements of Set $Y$

The general term for set $Y$ is given by $9(n-1)$. Let's substitute values of $n$:

• For $n=1$:

  $$9(1-1)=9(0)=0$$

• For $n=2$:

  $$9(2-1)=9(1)=9$$

• For $n=3$:

  $$9(3-1)=9(2)=18$$

• For $n=4$:

  $$9(4-1)=9(3)=27$$

• For $n=7$:

  $$9(7-1)=9(6)=54$$

Thus, the elements of set $Y$ are all non-negative multiples of $9$:

$$Y=\{0,9,18,27,36,45,54,\dots \}$$

3. Comparison and Conclusion

By observing both sets:

• $X=\{0,9,54,243,\dots \}$

• $Y=\{0,9,18,27,36,45,54,\dots \}$

Every element present in set $X$ is also an element of set $Y$, but set $Y$ contains extra elements (like $18,27,36$, etc.) that are not in set $X$.

Therefore, $X$ is a proper subset of $Y$.

$$X\subset Y$$

Correct Option

The correct option is A) X ⊂ Y.