MCA NIMCET Differentibility Previous Year Question Paper and Solution with PDF

Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

MCA NIMCET
Previous Year Questions

MCA NIMCET
Previous Year Questions
MCA NIMCET

MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Differentibility PYQ

Qus : 1 MCA NIMCET PYQ

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x)=\begin{cases} x \sin\left(\frac{1}{x}\right), & x>0,\\ 0, & x \le 0. \end{cases}$

Then

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 2 MCA NIMCET PYQ

Let $f(x)=\begin{cases}{{x}^2\sin \frac{1}{x}} & {,\, x\ne0} \\ {0} & {,x=0}\end{cases}$

Then which of the follwoing is true

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Qus : 3 MCA NIMCET PYQ

Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f(0)=\frac{1}{\pi}$ and $f(x)=\frac{x}{e^{\pi x}-1}$ for $x\ne0$, then

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Analysis of Continuity and Differentiability

Function:

\[ f(x) = \begin{cases} \dfrac{x}{e^{\pi x} - 1}, & x \neq 0 \\ \dfrac{1}{\pi}, & x = 0 \end{cases} \]

✅ Continuity at $ x = 0 $:

\[ \lim_{x \to 0} f(x) = \frac{1}{\pi} = f(0) \quad \Rightarrow \quad \text{Function is continuous at } x = 0 \]

✏️ Differentiability at $ x = 0 $:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = -\frac{1}{2} \]

✅ Final Result:

Function is continuous at $ x = 0 $
Function is differentiable at $ x = 0 $
$ f'(0) = \boxed{-\frac{1}{2}} $

Qus : 4 MCA NIMCET PYQ

If $y=\sec^{-1}\left(\frac{x+1}{x-1}\right)+\sin^{-1}\left(\frac{x-1}{x+1}\right)$, $x\in[0,\infty)$ and $x\ne1$, then $\dfrac{dy}{dx}$ is equal to

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Recall the identity:
$\sec^{-1}(\theta) = \cos^{-1}\left(\frac{1}{\theta}\right)$
Applying this:
$\sec^{-1}\left(\frac{x+1}{x-1}\right) = \cos^{-1}\left(\frac{x-1}{x+1}\right)$

Substituting back:
$y = \cos^{-1}\left(\frac{x-1}{x+1}\right) + \sin^{-1}\left(\frac{x-1}{x+1}\right)$

Using the standard identity:
$\sin^{-1}(\theta) + \cos^{-1}(\theta) = \dfrac{\pi}{2}$

Therefore:
$y = \dfrac{\pi}{2}$

Differentiating:

$\dfrac{dy}{dx} = \dfrac{d}{dx}\left(\dfrac{\pi}{2}\right)$

$\boxed{\dfrac{dy}{dx} = 0}$

Qus : 5 MCA NIMCET PYQ

The set of points, where $f(x)=\frac{x}{1+|x|}$ is differentiable in

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 6 MCA NIMCET PYQ

Let $f:\mathbb{R}\to \mathbb{R}$ be a function defined by $f(x)=|x+1|e^{-x^2}$. Then which of the following statement is true?

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

Given,

$f(x)=|x+1|e^{-x^2}$

For $x<-1$,

$f(x)=-(x+1)e^{-x^2}$

Differentiate:

$f'(x)=e^{-x^2}(2x^2+2x-1)$

For critical points,

$2x^2+2x-1=0$

Using quadratic formula,

$x=\frac{-2\pm\sqrt{4+8}}{4}$

$x=\frac{-2\pm 2\sqrt{3}}{4}$

$x=\frac{-1\pm\sqrt{3}}{2}$

For $x<-1$, the valid critical point is

$x=\frac{-1-\sqrt{3}}{2}$

This lies in the interval $(-2,-1)$.

At this point, $f$ has a point of maxima.

Qus : 7 MCA NIMCET PYQ

For $a\in \mathbb{R}$, consider the real valued function defined on $(-1,1)$ as follows:

For $x\neq 0$,

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

and for $x=0$,

$f(x)=a$

If $f$ is differentiable at $x=0$, then the value of $a+f'(0)$ is equal to:

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

Using expansion near $x=0$,

$(1+x)^{\frac{1}{3}}=1+\frac{x}{3}-\frac{x^2}{9}+O(x^3)$

Also,

$(1+2x)^{\frac{1}{4}}=1+\frac{x}{2}-\frac{3x^2}{8}+O(x^3)$

Now,

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

$=\left(1+\frac{x}{3}-\frac{x^2}{9}\right)-\left(1+\frac{x}{2}-\frac{3x^2}{8}\right)+O(x^3)$

$=-\frac{x}{6}+\left(-\frac{1}{9}+\frac{3}{8}\right)x^2+O(x^3)$

$=-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)$

Therefore,

$f(x)=\frac{-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)}{x}$

$f(x)=-\frac{1}{6}+\frac{19x}{72}+O(x^2)$

For differentiability at $x=0$, function must be continuous at $x=0$.

So,

$a=\lim_{x\to 0}f(x)=-\frac{1}{6}$

Also,

$f'(0)=\frac{19}{72}$

Hence,

$a+f'(0)=-\frac{1}{6}+\frac{19}{72}$

$=-\frac{12}{72}+\frac{19}{72}$

$=\frac{7}{72}$

Qus : 8 MCA NIMCET PYQ

Let $\mathbb{R}\rightarrow\mathbb{R}$ be any function defined as $f(x)=\begin{cases}{{x}^{\alpha}\sin \frac{1}{{x}^{\beta}}} & {,x\ne0} \\ {0} & {,x=0}\end{cases}$, $\alpha , \beta \in \mathbb{R}$. Which of the following is true? ($\mathbb{R}$ denotes the set of all real numbers)

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

The function is $f(x) = x^{\alpha}\sin\left(\dfrac{1}{x^{\beta}}\right)$ for $x \ne 0$, and $f(0)=0$.

To check continuity at $x=0$, consider: $\displaystyle \lim_{x\to 0} x^{\alpha}\sin\left(\frac{1}{x^{\beta}}\right)$.

Since $|\sin \theta| \le 1$ for all $\theta$

$|x^{\alpha}\sin(1/x^{\beta})| \le |x^{\alpha}|$.

Now, $\displaystyle \lim_{x\to 0} x^{\alpha} = 0 \quad \text{iff } \alpha > 0$.

Therefore, $\displaystyle \lim_{x\to 0} x^{\alpha}\sin(1/x^{\beta}) = 0 = f(0)$ iff $\alpha > 0$.

The value of $\beta$ does not matter because $\sin(1/x^{\beta})$ is always bounded.

Hence: ${f(x)\text{ is continuous at }x=0\text{ for all }\alpha>0\text{ and }\beta\in\mathbb{R}.}$

Qus : 9 MCA NIMCET PYQ

The slope of the function \[ f(x) = \begin{cases} x^2 \sin\!\left(\dfrac{1}{x}\right), & \text{if } x \ne 0, \\[8pt] 0, & \text{if } x = 0 \end{cases} \]

1
2
3
4
Go to Discussion

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

$f'(0)=\lim_{x\to0}\dfrac{f(x)-f(0)}{x}$

$=\lim_{x\to0}\dfrac{x^2\sin(1/x)}{x}$

$=\lim_{x\to0}x\sin(\frac{1}{x})=0$

For $x\ne0$,

$f'(x)=2x\sin\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)$

Answer: $\boxed{f'(0)=0}$ ✅

MCA NIMCET

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

MCA NIMCET

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

Please provide more information - at least 10 characters

Aspire's Library

MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Differentibility PYQ

Solution

Solution

Solution

Analysis of Continuity and Differentiability

✅ Continuity at \( x = 0 \):

✏️ Differentiability at \( x = 0 \):

✅ Final Result:

Solution

Solution

Solution

Solution

Solution

Solution

MCA NIMCET

MCA NIMCET

Feedback

Info.

Information

Aspire's Library

MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Differentibility PYQ

Solution

Solution

Solution

Analysis of Continuity and Differentiability

✅ Continuity at \( x = 0 \):

✏️ Differentiability at \( x = 0 \):

✅ Final Result:

Solution

Solution

Solution

Solution

Solution

Solution

MCA NIMCET

MCA NIMCET

Game Changer NIMCET Test Series 2026

Info.

Confirm Delete

Edit