NIMCET 2027 Function Previous Year Questions (PYQs) – Page 3 of 3

Even elements in the domain are $2$ and $4$.
Even elements in the codomain are $2,4,6,8$.

The two even elements of the domain must be mapped to even elements of the codomain injectively.

Number of ways:
${}^4P_2=4\times 3=12$

Now, $2$ elements of the codomain are already used, so $6$ elements are left.

The odd elements of the domain are $1$ and $3$.
They can be mapped injectively to the remaining $6$ elements.

Number of ways:
${}^6P_2=6\times 5=30$

So,
$n=12\times 30=360$

Now,
$360=2^3\times 3^2\times 5^1$

So,
$a=3,\ b=2,\ c=1$

Therefore,
$a+b+c=3+2+1=6$

Since
$R\subset \mathbb{N}\times \mathbb{N}$,

the relation $R$ can have at most countably infinite elements.

Now, suppose for some $a\in \mathbb{N}$, the set $R_a$ is infinite.

Since $R_a\subseteq \mathbb{N}$, it is countably infinite.

Also, for every element $b\in R_a$, the ordered pair $(a,b)$ belongs to $R$.

So, $R$ is also infinite. Since $R\subset \mathbb{N}\times \mathbb{N}$, $R$ is countably infinite.

Therefore, both $R_a$ and $R$ have the same cardinality.

Given,

$h(x)=f[g(x)]$

Now, $f(t)=1$ when

$|t|\le 1$

So, $h(x)=1$ when

$|g(x)|\le 1$

For $|x|\le 2$,

$g(x)=2-x^2$

So,

$|2-x^2|\le 1$

This gives

$-1\le 2-x^2\le 1$

Subtract $2$ from all sides:

$-3\le -x^2\le -1$

Multiplying by $-1$ reverses the inequalities:

$1\le x^2\le 3$

So,

$1\le |x|\le \sqrt{3}$

For $|x|>2$,

$g(x)=2$

So,

$|g(x)|=2>1$

Hence, this case is not valid.

Therefore, the required interval is

$1\le |x|\le \sqrt{3}$

We know that

$(f^{-1})'(y)=\frac{1}{f'(x)}$, where $f(x)=y$

Here, we need $(f^{-1})'(2)$.

So first find $x$ such that

$f(x)=2$

$\frac{3x^2+4x+1}{x^2+3x+2}=2$

$3x^2+4x+1=2x^2+6x+4$

$x^2-2x-3=0$

$(x-3)(x+1)=0$

Since domain is $[0,\infty)$,

$x=3$

Now,

$f(x)=\frac{3x^2+4x+1}{x^2+3x+2}$

Let

$N=3x^2+4x+1$

and

$D=x^2+3x+2$

Then,

$f'(x)=\frac{N'D-ND'}{D^2}$

Now at $x=3$,

$N=3(3)^2+4(3)+1=40$

$D=(3)^2+3(3)+2=20$

$N'=6x+4$

So,

$N'=22$

$D'=2x+3$

So,

$D'=9$

Therefore,

$f'(3)=\frac{22\cdot 20-40\cdot 9}{20^2}$

$f'(3)=\frac{440-360}{400}$

$f'(3)=\frac{80}{400}$

$f'(3)=\frac{1}{5}$

Hence,

$(f^{-1})'(2)=\frac{1}{f'(3)}$

$=5$