NIMCET Function Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Function PYQ

Qus : 1 NIMCET PYQ

The graph of function $f(x)=\log _e({x}^3+\sqrt[]{{x}^6+1})$ is symmetric about:

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Qus : 2 NIMCET PYQ

If f(x) is a polynomial of degree 4, f(n) = n + 1 & f(0) = 25, then find f(5) = ?

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Correct Shortcut Method — Find $ f(5) $

Step 1: Define a helper polynomial:

\[ g(x) = f(x) - (x + 1) \]

Given: $ f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 \Rightarrow g(1) = g(2) = g(3) = g(4) = 0 $

So, \[ g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1) \]

Step 2: Use $ f(0) = 25 $ to find A:

\[ f(0) = A(-1)(-2)(-3)(-4) + (0 + 1) = 24A + 1 = 25 \Rightarrow A = 1 \]

Step 3: Compute $ f(5) $:

\[ f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30} \]

✅ Final Answer: $ \boxed{f(5) = 30} $

Qus : 3 NIMCET PYQ

The maximum value of $f(x) = (x – 1)^2 (x + 1)^3$ is equal to $\frac{2^p3^q}{3125}$ then the ordered pair of (p, q) will be

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Maximum Value of $ f(x) = (x - 1)^2(x + 1)^3 $

Step 1: Let’s define the function:

\[ f(x) = (x - 1)^2 (x + 1)^3 \]

Step 2: Take derivative to find critical points

Use product rule:
Let $ u = (x - 1)^2 $, $ v = (x + 1)^3 $
\[ f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \] \[ f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)] \] \[ f'(x) = (x - 1)(x + 1)^2 (5x - 1) \]

Step 3: Find critical points

Set $ f'(x) = 0 $: \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5} \]

Step 4: Evaluate $ f(x) $ at these points

$ f(1) = 0 $
$ f(-1) = 0 $
$ f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3 $

\[ f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125} \]

Step 5: Compare with given form:

It is given that maximum value is $ \frac{3456}{3125} = 2^p \cdot 3^q / 3125 $

Factor 3456: \[ 3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3 \]

✅ Final Answer: $ \boxed{(p, q) = (7,\ 3)} $

Qus : 4 NIMCET PYQ

If $| x - 6|= | x - 4x | -| x^2- 5x +6 |$ , where x is a real variable

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Qus : 5 NIMCET PYQ

A real valued function f is defined as $f(x)=\begin{cases}{-1} & {-2\leq x\leq0} \\ {x-1} & {0\leq x\leq2}\end{cases}$. Which of the following statement is FALSE?

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Qus : 6 NIMCET PYQ

Number of onto (surjective) functions from A to B if n(A)=6 and n(B)=3, is

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Given: $ n(A)=6 $ and $ n(B)=3 $.

Formula: The number of onto (surjective) functions from a set of size $m$ to a set of size $n$ is \[ n! \, S(m,n) \] where $S(m,n)$ is the Stirling number of the second kind (number of ways to partition $m$ elements into $n$ non-empty subsets).

We can also use the Inclusion–Exclusion Principle: \[ n! \, S(m,n) = \sum_{k=0}^{n} (-1)^k \binom{n}{k}(n-k)^m \] For $m=6,\ n=3$: \[ N = 3^6 - 3\times 2^6 + 3\times 1^6 \]

Calculation: \[ 3^6 = 729,\quad 2^6 = 64 \] \[ N = 729 - 3(64) + 3(1) = 729 - 192 + 3 = 540. \]

Answer: The number of onto functions is \[ \boxed{540}. \]

Qus : 7 NIMCET PYQ

Let $X_i, i = 1,2,.. , n$ be n observations and $w_i = px_i +k, i = 1,2, ,n$ where p and k are constants. If the mean of $x_i 's$ is 48 and the standard deviation is 12, whereas the mean of $w_i 's$ is 55 and the standard deviation is 15, then the value of p and k should be

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Qus : 8 NIMCET PYQ

If the function $f:[1,\infty)\to[1,\infty)$ is defined by $f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

Given $2^{x(x-1)} = y$ Take $\log_2$: $x(x-1) = \log_2 y$ Quadratic: $x^{2}-x-\log_2 y = 0$ So $x = \dfrac{1 \pm \sqrt{1+4\log_2 y}}{2}$ Since $x \ge 1$, choose positive sign: $f^{-1}(x)=\dfrac12\left(1+\sqrt{1+4\log_2 x}\right)$

Qus : 9 NIMCET PYQ

Let S be the set $\{a\in Z^+:a\leq100\}$.If the equation $[tan^2 x]-tan x - a = 0$ has real roots (where [ . ] is the greatest integer function), then the number of elements is S is

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Qus : 10 NIMCET PYQ

The number of one - one functions f: {1,2,3} → {a,b,c,d,e} is

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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

Solution

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3) = 5 \times 4 \times 3 = 60$

✅ Final Answer: $\boxed{60}$

Qus : 11 NIMCET PYQ

The domain of the function $f(x)=\frac{{\cos }^{-1}x}{[x]}$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 12 NIMCET PYQ

The function $f(x)=\log (x+\sqrt[]{{x}^2+1})$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 13 NIMCET PYQ

The value of $f(1)$ for $f\Bigg{(}\frac{1-x}{1+x}\Bigg{)}=x+2$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

Solution

Given:
$$f\left(\frac{1 - x}{1 + x}\right) = x + 2$$

To Find: $ f(1) $

Let $ \frac{1 - x}{1 + x} = 1 \Rightarrow x = 0 $

Then, $ f(1) = f\left(\frac{1 - 0}{1 + 0}\right) = 0 + 2 = 2 $

Answer: $$\boxed{2}$$

Qus : 14 NIMCET PYQ

If $f(x)$ is a polynomial satisfying $f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$ and $f(3)=28$, then $f(4)$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

We know that $f(x) = 1 + x^n$

Finding $n$ using $f(3) = 28$:

$f(3) = 1 + 3^n = 28$

$3^n = 27$

$\therefore n = 3$

Thus: $f(x) = 1 + x^3$

Finding $f(4)$:

$f(4) = 1 + 4^3 = 1 + 64$

$$\boxed{f(4) = 65}$$

Qus : 15 NIMCET PYQ

If f(x)=cos[$\pi$^2]x+cos[-$\pi$^2]x, where [.] stands for greatest integer function, then $f(\pi/2)$=

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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

Solution

? Function with Greatest Integer and Cosine

Given:

\[ f(x) = \cos\left([\pi^2]x\right) + \cos\left([-\pi^2]x\right) \]

Find: \[ f\left(\frac{\pi}{2}\right) \]

Step 1: Estimate Floor Values

\[ \pi^2 \approx 9.8696 \Rightarrow [\pi^2] = 9,\quad [-\pi^2] = -10 \]

Step 2: Plug into the Function

\[ f\left(\frac{\pi}{2}\right) = \cos\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(-10 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{9\pi}{2}\right) + \cos(-5\pi) \]

Step 3: Simplify

\[ \cos\left(\frac{9\pi}{2}\right) = 0,\quad \cos(-5\pi) = -1 \]

✅ Final Answer:

\[ \boxed{-1} \]

Qus : 16 NIMCET PYQ

The number of functions $f$ from $A={0,1,2}$ into $B={0,1,2,3,4,5,6,7}$ such that $f(i) \le f(j)$ for $i
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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

Number of non-decreasing functions = combinations with repetition $= \binom{8+3-1}{3}=\binom{10}{3}$

Qus : 17 NIMCET PYQ

If $f(x)+f(1-x)=2$, then the value of $f\left(\dfrac{1}{2001}\right)+f\left(\dfrac{2}{2001}\right)+\cdots+f\left(\dfrac{2000}{2001}\right)$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

Terms pair as $f\left(\frac{k}{2001}\right)+f\left(1-\frac{k}{2001}\right)=2$ There are $1000$ such pairs. Sum $=1000\times2=2000$ Answer: $\boxed{2000}$

Qus : 18 NIMCET PYQ

The function $f(x) = \log\left(x + \sqrt{x^2 + 1}\right)$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 19 NIMCET PYQ

Identify the wrong statement.

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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

Solution

Check (2):

Left side = $(A-B)-C = A - (B \cup C)$

Right side = $(A-C)-(B-C)$ = $A - (B \cap C)$

These are not equal in general.

Qus : 20 NIMCET PYQ

Set $A$ has $3$ elements. Set $B$ has $4$ elements. Number of injections?

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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

Solution

Injection: choose 3 distinct elements from 4 and arrange. ${}^4P_3 = 4 \times 3 \times 2 = 24$

Qus : 21 NIMCET PYQ

Let $n$ be the number of injective functions $f:{1,2,3,4}\to {1,2,3,4,5,6,7,8}$ sending an even number to an even number. If $n=2^a3^b5^c$, then $a+b+c$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

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

Qus : 22 NIMCET PYQ

Let $R\subset \mathbb{N}\times \mathbb{N}$. Which of the following statement is necessarily true?

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

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.

Qus : 23 NIMCET PYQ

Which of the following function is the inverse of itself?

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 24 NIMCET PYQ

$f(x)=$

{	$1$	if	$\|x\|\le 1$
	$0$	if	$\|x\|>1$

, $g(x)=$

{	$2-x^2$	if	$\|x\|\le 2$
	$2$	if	$\|x\|>2$

If $h(x)=f[g(x)]$, then an interval in which $h(x)=1$ for all values of $x$ in that interval is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

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

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

Qus : 25 NIMCET PYQ

If the graph of y = (x – 2)² – 3 is shifted by 5 units up along y-axis and 2 units to the right along the x-axis, then the equation of the resultant graph is

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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

When y= f (x) is shifted by k units to the right along x

– axis, it become y= f (x - k )

Hence, new equation of

graph is y = (x - 4)² + 2

Qus : 26 NIMCET PYQ

Let $f:[0,\infty)\to \mathbb{R}$ be a function defined by

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

Then the value of $(f^{-1})'(2)$ is equal to:

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

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,

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

$=5$

Qus : 27 NIMCET PYQ

A function $f : (0,\pi) \to R$ defined by $f(x) = 2 sin x + cos 2x$ has

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 28 NIMCET PYQ

The number of points in $(-\infty,\infty)$, for which $x^2 - x\sin x - \cos x = 0$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ

Solution

Let $f(x) = x^2 - x\sin x - \cos x$
Differentiating with respect to $x$:
$f'(x) = 2x - \sin x - x\cos x + \sin x$
$f'(x) = 2x - x\cos x$
$f'(x) = x(2 - \cos x)$
Since $-1 \leq \cos x \leq 1$ $\Rightarrow$ $2 - \cos x \geq 1 > 0$ always
$\therefore$ Sign of $f'(x)$ depends on sign of $x$
$f'(x) < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$
$\therefore f(x)$ has minimum at $x = 0$
Minimum value $= f(0) = 0 - 0 - \cos 0 = -1 < 0$
Also, $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)$
$= x^2 - x\sin x - \cos x = f(x)$
$\therefore f(x)$ is an even function
As $x \rightarrow \pm\infty$, $f(x) \rightarrow +\infty$
Since $f(0) = -1 < 0$ is minimum and $f(x) \rightarrow +\infty$ on both sides
$\therefore$ Graph of $f(x)$ crosses $x$-axis at exactly two points
One in $(-\infty, 0)$ and one in $(0, \infty)$
$\therefore$ Total number of points $= \boxed{2}$

Qus : 29 NIMCET PYQ

The number of one-to-one functions from {1, 2, 3} to {1, 2, 3, 4, 5} is

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

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NIMCET Previous Year Questions (PYQs)

NIMCET Function PYQ

Solution

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Correct Shortcut Method — Find \( f(5) \)

Solution

Maximum Value of \( f(x) = (x - 1)^2(x + 1)^3 \)

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Solution

Solution

Solution

Solution

Solution

? Function with Greatest Integer and Cosine

Step 1: Estimate Floor Values

Step 2: Plug into the Function

Step 3: Simplify

✅ Final Answer:

Solution

Solution

Solution

Solution

Solution

Solution

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