datavorous/entrance-exam-dataset huggingface dataset:
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a function whose inverse is $\frac{\mathrm{x}+5}{3}$. What is
$\begin{array}{ll}\mathrm{f}(\mathrm{x}) \text { equal to } ? & {[2012-\mathrm{II}]}\end{array}$

Options:
A $\quad f(x)=3 x+5$
B $\mathrm{f}(\mathrm{x})=3 \mathrm{x}-5$
C $f(x)=5 x-3$
D $\mathrm{f}(\mathrm{x})$ does not exist

The correct answer is:
$\mathrm{f}(\mathrm{x})=3 \mathrm{x}-5$

Solution:
Let $\mathrm{y}=\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}+5}{3}$
$\Rightarrow 3 y-5=x$
Now, $y=f^{-1}(x)$
$\Rightarrow x=f(y)$
$\Rightarrow x=3 y-5$
Hence, $f(x)=3 x-5$

Topic: Mathematics, Functions, NDA, NDA 2012 (Phase 2)
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}, \mathrm{a}, \mathrm{b}, \mathrm{c}$ being fixed
non-zero real numbers. Which one of the following statements is correct, in general?

Options:
A If $b^{2}-4 a c>0$, then $f^{-1}(0)$ does not contain 0
B If b $^{2}-4$ ac $ < 0$, then $\mathrm{f}^{-1}(0)$ must contain 0
C If $b^{2}-4 a c>0$, then $f^{-1}(0)$ may contain 0
D If b $^{2}-4 a c < 0$, then $\mathrm{f}^{-1}(0)$ may contain 0

The correct answer is:
If $b^{2}-4 a c>0$, then $f^{-1}(0)$ does not contain 0

Solution:
As given $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$, Let $\mathrm{y}=\mathrm{f}(\mathrm{x})$. To get $\mathrm{f}^{-1}(\mathrm{x})$
we express $x$ is terms of y in equation $y=a x^{2}+b x+c$.
$\frac{y}{a}=x^{2}+\frac{b}{a} x+\frac{c}{a}=\left(x+\frac{b}{2 a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}$
$\Rightarrow x+\frac{b}{2 a}=\sqrt{\frac{y}{a}+\frac{b^{2}}{4 a^{2}}-\frac{c}{a}}=\frac{\pm \sqrt{4 a y+b^{2}-4 a c}}{2 a}$
$x=\frac{-b \pm \sqrt{4 a y+b^{2}-4 a c}}{2 a}$
or, $f^{1}(x)=\frac{-b \pm \sqrt{4 a x+b^{2}-4 a c}}{2 a}$
Putting $\mathrm{x}=0$
$\mathrm{f}^{-1}(0)=\frac{-\mathrm{b} \pm \sqrt{0+\mathrm{b}^{2}-4 \mathrm{ac}}}{2 \mathrm{a}} \neq 0 \quad\left[\right.$ if $\left.\mathrm{b}^{2}-4 \mathrm{ac}>0\right]$
so, $\mathrm{f}^{-1}(0) \neq 0$. i.e. if $b^{2}-4 a c>0$
$\mathrm{f}^{-1}(0)$ does not contain 0, if $b^{2}-4 a c>0 .$

Topic: Mathematics, Functions, NDA, NDA 2007 (Phase 1)
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $\mathrm{f}(\mathrm{x})=\sin (|\mathrm{x}|)$
Which one of the following is correct?

Options:
A fis not differentiable only at 0
B fis differentiable at 0 only
C fis differentiable everywhere
D fis non-differentiable at many points

The correct answer is:
fis not differentiable only at 0

Solution:
Given function is $: \mathrm{f}(\mathrm{x})=\sin |\mathrm{x}|$
$=\left\{\begin{array}{ll}\sin (\mathrm{x}), & \mathrm{x} \geq 0 \\ \sin (-\mathrm{x}), & \mathrm{x} < 0\end{array}\right.$
$=\left\{\begin{array}{cl}\sin \mathrm{x}, & \mathrm{x} \geq 0 \\ -\sin \mathrm{x}, & \mathrm{x} < 0\end{array}\right.$
$\mathrm{LHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{0-\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-\sin (-h)-0}{-h}=-1$
$\mathrm{RHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{0+\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\sin (\mathrm{h}-0)}{\mathrm{h}}=1$
$\mathrm{LHD} \neq \mathrm{RHD}$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2008 (Phase 1)
Let $f: R \rightarrow R$ be defined by $f(x)=|x| / x, x \neq 0, f(0)=2$ What is the range of $f$ ?

Options:
A {1,2}
B {1,-1}
C {-1,1,2}
D {1}

The correct answer is:
{-1,1,2}

Solution:
Let $f: R \rightarrow R$ be defined as $f(x)=\frac{|x|}{x}, x \neq 0$
Also, $f(0)=2$ ie. value of function at $x=0$ is 2 .
Consider, $f(x)=\left\{\begin{array}{l}\frac{x}{x}=1 \text { if } x>0 \\ \frac{-x}{x}=-1 \text { if } x < 0\end{array}\right.$
because we have $\mid x \vDash\left\{\begin{array}{ll}x, & x \geq 0 \\ -x, & x < 0\end{array}\right.$
Thus, Range of $f(x)=\{1,-1,2\}$

Topic: Mathematics, Functions, NDA, NDA 2009 (Phase 2)
Let \(f: R \rightarrow R\) be the function defined by \(f(x)=\left\{\begin{array}{cl}5, & \text { if } x \leq 1 \\ a+b x, & \text { if } 1 < x < 3 \\ b+5 x, & \text { if } 3 \leq x < 5 \\ 30, & \text { if } x \geq 5\end{array}\right.\) then \(f\) is

Options:
A continuous if \(a=5\) and \(b=5\)
B continuous if \(a=0\) and \(b=5\)
C continuous if \(a=-5\) and \(b=10\)
D not continuous for any values of \(a\) and \(b\)

The correct answer is:
not continuous for any values of \(a\) and \(b\)

Solution:
Given function \(f: R \rightarrow R\), such that
\(f(x)=\left[\begin{array}{cl}
5, & \text { if } x \leq 1 \\
a+b x, & \text { if } 1 < x < 3 \\
b+5 x, & \text { if } 3 \leq x < 5 \\
30, & \text { if } x \geq 5
\end{array}\right.\)
If \(f\) is continuous at \(x=1\), then
\(a+b=5\) ...(i)
If \(f\) is continuous at \(x=3\), then
\(\Rightarrow \quad \begin{gathered}
a+3 b=b+15 \\
a+2 b=15 \quad \ldots (ii)
\end{gathered}\)
and if \(f\) is continuous at \(x=5\), then
\(b+25=30 \Rightarrow b=5\) ...(iii)
From Eqs. (ii) and (iii), we get
\(a=5\) ...(iv)
but \(a=5\) and \(b=5\) doesn't satisfy the Eq. (i). so, \(f: R \rightarrow R\) is not continuous for any values of \(a\) and \(b\).
Hence, option (4) is correct.

Topic: Mathematics, Continuity and Differentiability, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
Let $f(x)=\frac{1}{1-|1-x|} .$ Then, what is $\lim _{x \rightarrow 0} f(x)$ equal to

Options:
A 0
B of
C 1
D -1

The correct answer is:
1

Solution:
$\begin{aligned} & \text { LHL }=\lim _{h \rightarrow 0} \frac{1}{1-|1-(1-h)|} \\ &=\lim _{h \rightarrow 0} \frac{1}{1-|h|}=\lim _{h \rightarrow 0} \frac{1}{1-h}=1 \\ & \text { RHL }=\lim _{h \rightarrow 0} \frac{1}{1-|1-(1+h)|} \\ &=\lim _{h \rightarrow 0} \frac{1}{1-|-h|}=\lim _{h \rightarrow 0} \frac{1}{1-h}=1 \\ & \therefore \lim _{x \rightarrow 0} f(x)=1 \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2008 (Phase 2)
Let $f(x)=\left\{\begin{array}{ll}-2, & -3 \leq x \leq 0 \\ x-2, & 0 < x \leq 3\end{array}\right.$ and $g(x)=f(|x|)+|f(x)|$
What is the value of the differential coefficient of $g(x)$ at $x=-2$ ?

Options:
A -1
B 0
C 1
D 2

The correct answer is:
0

Solution:
For $x=-2$ $g(x)=-2+|-2|=-2+2$
$\Rightarrow g(x)=0$
$\Rightarrow$ differential coefficient at $x=-2$ is given as :
$g^{\prime}(x)=\lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=\lim _{h \rightarrow 0} \frac{0-0}{4}=0 .$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 2)
Let $f(x)=\left\{\begin{array}{ll}-2, & -3 \leq x \leq 0 \\ x-2, & 0 < x \leq 3\end{array}\right.$ and $g(x)=f(|x|)+|f(x)|$
Which of the following statements is/are correct?
1- $\mathrm{g}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$.
2- $\mathrm{g}(\mathrm{x})$ is differentiable at $\mathrm{x}=2$. Select the correct answer using the code given below:

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
$f(x)=\left\{\begin{array}{ll}-2, & -3 \leq x \leq 0 \\ x-2, & 0 < x \leq 3 \text { and }\end{array}\right.$
$g(x)=f(|x|)+|f(x)|$
At $x=0$
For LHD $: g(x)=-2+|-2|=-2+2=0 \Rightarrow g(x)=0$
$\begin{array}{l}\text { LHD }=\lim _{x \rightarrow 0^{-}} \frac{g(x)-g(0)}{x-0}=\lim _{h \rightarrow 0} \frac{g(-h)-g(0)}{-h} \\ \quad= & \lim _{h \rightarrow 0} \frac{0-0}{-h}=\lim _{h \rightarrow 0} 0 \\ \text { LHD }=0\end{array}$
For RHD : $g(x)=|x|-2+|x-2|$
$g(x)=x-2-(x-2) \quad x>0$ (and just greater
than zero)
$g(x)=x-2-x+2=0$
Now $g(x)$ is not continuous at $x=0$, hence $g(x)$ is not differentiable at $\mathrm{x}=0$ At $x=2$ For LHD :
$g(x)=|x|-2+|x-2|=x-2-(x-2)$
$\quad=x-2-x+2=0$
$\therefore \mathrm{LHD}=\lim _{x \rightarrow 2^{-}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{0}{x-2}=0$
For RHD:
$g(x)=|x|-2+|x-2|=x-2+x-2=2 x-4$
$\begin{aligned} \Rightarrow \mathrm{RHD} &=\lim _{x \rightarrow 2^{+}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{2 x-4-2(2)+4}{x-2} \\ &=\lim _{x \rightarrow 2} \frac{2(x-2)}{x-2}=2 \\ \Rightarrow \mathrm{LHD} & \neq \mathrm{RHD} \end{aligned}$
Thus $g(x)$ is not differentiable at $x=2$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 2)
Let $f(x)=\left\{\begin{array}{ll}3 x-4, & 0 \leq x \leq 2 \\ 2 x+\ell, & 2 < x \leq 9\end{array}\right.$
If $\mathrm{f}$ is continuous at $\mathrm{x}=2$, then what is the value of $\ell$ ?

Options:
A 0
B 2
C -2
D -1

The correct answer is:
-2

Solution:
Given function is :
$f(x)=\left\{\begin{array}{ll}3 x-4, & 0 \leq x \leq 2 \\ 2 x+\ell, & 2 < x \leq 9\end{array}\right.$
and also given that $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$ For a function to be continuous at a point $\mathrm{LHL}=\mathrm{RHL}$ $=\mathrm{V} . \mathrm{F}$. at that point. $\mathrm{f}(2)=2=\mathrm{V}$.F.
$\Rightarrow \mathrm{RHL}: \lim _{\mathrm{x} \rightarrow 2}(2 \mathrm{x}+\ell)=3(2)-4$
$\Rightarrow \lim _{\mathrm{h} \rightarrow 0}\{2(2+\mathrm{h})+\ell\}=6-4$
$\Rightarrow 4+\ell=2$
$\Rightarrow \ell=-2$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2008 (Phase 1)
$\quad \operatorname{Let} f(x)=\frac{1}{\sqrt{18-x^{2}}}$.
What is the value of $\lim _{x \rightarrow 3} \frac{f(x)-f(3)}{x-3}$ ?

Options:
A 0
B $-\frac{1}{9}$
C $\frac{1}{3}$
D $\frac{1}{9}$

The correct answer is:
$\frac{1}{9}$

Solution:
The given functions is $\mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{18-\mathrm{x}^{2}}}$
So, $\mathrm{f}(3)=\frac{1}{\sqrt{18-9}}=\frac{1}{3}$
$\Rightarrow \lim _{x \rightarrow 3} \frac{f(x)-f(3)}{x-3}=\lim _{x \rightarrow 3} \frac{\frac{1}{\sqrt{18-x^{2}}}-\frac{1}{3}}{x-3}$
Putting $\mathrm{x}=3$ makes the that of $\frac{0}{0}$ form
$\lim _{x \rightarrow 3}-\frac{1}{2}\left(18-x^{2}\right)^{-3 / 2}(-2 x)$
(Applying L' Hospital's Rule)
$=-\frac{1}{2}(9)^{-3 / 2}(-2 \times 3)=\frac{1}{27} \times 3=\frac{1}{9}$

Topic: Mathematics, Limits, NDA, NDA 2006 (Phase 1)
Let $f(x)=a x^{2}+b x+c$ such that $f(1)=f(-1)$ and $a, b, c$ are in Arithmetic Progression.
$\mathrm{f}^{\prime}(\mathrm{a}), \mathrm{f}^{\prime}(\mathrm{b}), \mathrm{f}^{\prime}(\mathrm{c})$ are

Options:
A A.P.
B GP.
C H.P.
D Arithmetico-geometric progression

The correct answer is:
A.P.

Solution:
We have $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{ax}$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{a})=2 \mathrm{a}^{2}, \mathrm{f}^{\prime}(\mathrm{b})=2 \mathrm{ab}=0$
and $\mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{ac} \quad(\because \mathrm{b}=0)$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{a})=2 \mathrm{a}^{2}$
$\mathrm{f}^{\prime}(\mathrm{b})=0$
and $\mathrm{f}^{\prime}(\mathrm{c})=-2 \mathrm{a}^{2} \quad(\because 2 \mathrm{~b}=\mathrm{a}+\mathrm{c} \Rightarrow \mathrm{c}=-\mathrm{a})$
Hence $\mathrm{f}^{\prime}(\mathrm{a}), \mathrm{f}^{\prime}(\mathrm{b})$ and $\mathrm{f}^{\prime}(\mathrm{c})$ are in $\mathrm{AP}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2014 (Phase 2)
Let $f(x)=a x^{2}+b x+c$ such that $f(1)=f(-1)$ and $a, b, c$ are in Arithmetic Progression.
$\mathrm{f}^{\prime \prime}(\mathrm{a}), \mathrm{f}^{\prime \prime}(\mathrm{b}), \mathrm{f}^{\prime \prime}(\mathrm{c})$ are

Options:
A in A.P. only
B in GP. only
C in both A.P. and GP.
D neither in A.P. nor in G.P.

The correct answer is:
in both A.P. and GP.

Solution:
$\mathrm{f}^{\prime \prime}(\mathrm{x})=2 \mathrm{a}$
$\therefore \mathrm{f}^{\prime \prime}(\mathrm{a})=\mathrm{f}^{\prime \prime}(\mathrm{b})=\mathrm{f}^{\prime \prime}(\mathrm{c})$
Hence, $\mathrm{f}^{\prime \prime}(\mathrm{a}), \mathrm{f}^{\prime \prime}(\mathrm{b})$ and $\mathrm{f}^{\prime \prime}(\mathrm{c})$ are in both $\mathrm{AP}$ and $\mathrm{GP}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2014 (Phase 2)
Let $f(x)=a x^{2}+b x+c$ such that $f(1)=f(-1)$ and $a, b, c$ are in Arithmetic Progression.
What is the value of b?

Options:
A $-1$
B 0
C 1
D Cannot be determined deu to insufficient data

The correct answer is:
0

Solution:
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c} \\ & \therefore \mathrm{f}(1)=\mathrm{a}+\mathrm{b}+\mathrm{c} \\ & \text { and } \mathrm{f}(-1)=\mathrm{a}-\mathrm{b}+\mathrm{c} \\ & \therefore \mathrm{f}(1)=\mathrm{f}(-1) \\ & \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{a}-\mathrm{b}+\mathrm{c} \Rightarrow \mathrm{b}=0 \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2014 (Phase 2)
Let $\mathrm{f}(\mathrm{x})$ be a function such that $\mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{x}^{3} \mathrm{f}^{\prime}(\mathrm{x})=0$, What is
$\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ equal to? $\quad$

Options:
A $2 \mathrm{f}(1)$
B 0
C $2 \mathrm{f}(-1)$
D $4 \mathrm{f}(1)$

The correct answer is:
$2 \mathrm{f}(-1)$

Solution:
No Solution

Topic: Mathematics, Differential Equations, NDA, NDA 2016 (Phase 2)
Let $f(x)$ be an indefinite integral of $\sin ^{2} x .$ Consider the following statements:
Statement $1:$ The function $f(x)$ satisfies $f(x+\pi)$ $=f(\mathrm{x})$ for all real $\mathrm{x}$. Statement $2: \sin ^{2}(x+\pi)=\sin ^{2} x$ for all real $x$. Which one of the following is correct in respect of the above statements?

Options:
A Both the statements are true and Statement 2 is the correct explanation of Statement 1
B Both the statements are true but $\mathrm{Statement} 2$ is not the correct explanation of Statement 1
C Statement 1 is true but Statement 2 is false
D Statement 1 is false but Statement 2 is true

The correct answer is:
Both the statements are true but $\mathrm{Statement} 2$ is not the correct explanation of Statement 1

Solution:
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\int \sin ^{2} \mathrm{x} \cdot \mathrm{dx}=\int \frac{1-\cos 2 \mathrm{x}}{2} \mathrm{~d} \mathrm{x} \\ &=\frac{1}{2} \mathrm{x}-\frac{\sin 2 \mathrm{x}}{2} \cdot \frac{1}{2}+\mathrm{c}=\frac{1}{2} \mathrm{x}-\frac{1}{4} \sin 2 \mathrm{x}+\mathrm{c} \\ \text { (i) } & \mathrm{f}(\pi+\mathrm{x})=\frac{1}{2}(\pi+\mathrm{x})-\frac{1}{4} \sin 2(\pi+\mathrm{x}) \\ &=\frac{1}{2} \pi+\frac{1}{2} \mathrm{x}-\frac{1}{2} \sin (2 \pi+2 \mathrm{x})+\mathrm{c} \\ &=\frac{\mathrm{x}}{2}-\frac{1}{2} \sin 2 \mathrm{x}+\mathrm{c}=\mathrm{f}(\mathrm{x}) \end{aligned}$
So, statement 1 is true.
(ii) $\sin ^{2}(\pi+x)=\sin ^{2} x$
$(-\sin x)^{2}=\sin ^{2} x$
$\Rightarrow \sin ^{2} x=\sin ^{2} x$
So, statement 2 is true

Topic: Mathematics, Indefinite Integration, NDA, NDA 2017 (Phase 1)
Let $f(\mathrm{x})$ be defined as follows :
$f(x)=\left\{\begin{array}{ll}2 x+1, & -3 < x < -2 \\ x-1, & -2 \leq x < 0 \\ x+2, & 0 \leq x < 1\end{array}\right.$
Which one of the following statements is correct in respect of the above function?

Options:
A It is discontinuous at $\mathrm{x}=-2$ but continuous at every other point.
B It is continuous only in the interval $(-3,-2)$.
C It is discontinuous at $\mathrm{x}=0$ but continuous at every other point.
D It is discontinuous at every point.

The correct answer is:
It is discontinuous at $\mathrm{x}=0$ but continuous at every other point.

Solution:
$f(x)=\left\{\begin{array}{cc}2 x+1, & -3 < x < -2 \\ x-1, & -2 \leq x < 0 \\ x+2, & 0 \leq x < 1\end{array}\right.$
Here, $\mathrm{f}\left(0^{-}\right)=-1$ and $\mathrm{f}\left(0^{+}\right)=2$
So, $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$ and continuous at other points.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 1)
Let $\mathrm{f}(\mathrm{x})$ be the greatest integer function and $\mathrm{g}(\mathrm{x})$ be the modulus function.
What is $(\mathrm{f} \circ \mathrm{f})\left(-\frac{9}{5}\right)+(\mathrm{g} \circ \mathrm{g})(-2)$ equal to?

Options:
A $-1$
B 0
C 1
D 2

The correct answer is:
0

Solution:
$f(x) \rightarrow$ greatest integer function
$f(x)=[x]$
$g(x) \rightarrow$ modulus function $g(x)=|x|$
$f o f(x)=f(f(x))=f([x])$
$\begin{array}{l}\text { fof }\left(-\frac{9}{5}\right)=f(-2) \quad \because\left[-\frac{9}{5}\right]=-2 \\ =[-2]=-2 & \\ \operatorname{gog}(x)=g(|x|) & \\ =|| x|| & |x| \\ \quad=|x| & \\ \operatorname{gog}(-2)=|-2|=2 & \\ (\text { fof })\left(-\frac{9}{5}\right)+\operatorname{gog}(-2)=-2+2=0\end{array}$

Topic: Mathematics, Functions, NDA, NDA 2016 (Phase 1)
Let $\mathrm{f}(\mathrm{x})$ be the greatest integer function and $\mathrm{g}(\mathrm{x})$ be the modulus function.
What is $(\mathrm{g} \circ \mathrm{f})\left(-\frac{5}{3}\right)-(\mathrm{f} \circ \mathrm{g})\left(-\frac{5}{3}\right)$ equal to?

Options:
A -1
B 0
C 1
D 2

The correct answer is:
1

Solution:
$f(x) \rightarrow$ greatest integer function
$f(x)=[x]$
$g(x) \rightarrow$ modulus function $g(x)=|x|$
$\begin{aligned} \operatorname{gof}(x)=g(f(x)) & \quad \operatorname{fog}(x)=f(g(x)) \\=g([x]) &=f(|x|) \\=|[x]| &=[|x|] \\ \operatorname{gof}\left(-\frac{5}{3}\right)=\left|\left[-\frac{5}{3}\right]\right| ; f o g\left(-\frac{5}{3}\right)=\left[\left|-\frac{5}{3}\right|\right] \\=|-2| ;=\left[\frac{5}{3}\right] \\=2 ;=1 \\ \operatorname{gof}\left(-\frac{5}{3}\right)-\operatorname{fog}\left(-\frac{5}{3}\right)=2-1=1 \end{aligned}$

Topic: Mathematics, Functions, NDA, NDA 2016 (Phase 1)
Let $f(x)=p x+q$ and $g(x)=m x+n$. Then $f(g(x))=g(f(x))$ is equivalent to

Options:
A $f(\mathrm{p})=\mathrm{g}(\mathrm{m})$
B $f(\mathrm{q})=\mathrm{g}(\mathrm{n})$
C $f(\mathrm{n})=\mathrm{g}(\mathrm{q})$
D $f(\mathrm{~m})=\mathrm{g}(\mathrm{p})$

The correct answer is:
$f(\mathrm{n})=\mathrm{g}(\mathrm{q})$

Solution:
$\begin{array}{l}\mathrm{f}(\mathrm{x})=\mathrm{px}+\mathrm{q}, \mathrm{g}(\mathrm{x})=\mathrm{mx}+\mathrm{n} \\ \mathrm{f}(\mathrm{g}(\mathrm{x}))=\mathrm{g}(\mathrm{f}(\mathrm{x})) \\ \Rightarrow \mathrm{f}(\mathrm{mx}+\mathrm{n})=\mathrm{g}(\mathrm{px}+\mathrm{q}) \\ & \Rightarrow \mathrm{p}(\mathrm{mx}+\mathrm{n})+\mathrm{q}=\mathrm{m}(\mathrm{px}+\mathrm{q})+\mathrm{n} \\ & \Rightarrow \mathrm{pmx}+\mathrm{pn}+\mathrm{q}=\mathrm{pmx}+\mathrm{mq}+\mathrm{n} \\ & \Rightarrow \mathrm{pn}+\mathrm{q}=\mathrm{mq}+\mathrm{n} \\ & \Rightarrow \mathrm{f}(\mathrm{n})=\mathrm{g}(\mathrm{q})\end{array}$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 1)
Let $f(\mathrm{x}):\left\{\begin{array}{ll}\mathrm{x}, & \mathrm{x} \text { is rational } \\ 0, & \mathrm{x} \text { is irrational }\end{array}\right.$
and $g(x):\left\{\begin{array}{ll}0, & x \text { is rational } \\ x, & x \text { is irrational }\end{array}\right.$
If $f: \mathrm{R} \rightarrow \mathrm{R}$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$, then $(f-\mathrm{g})$ is

Options:
A one-one and into
B neither one-one nor onto
C many-one and onto
D one-one and onto

The correct answer is:
one-one and onto

Solution:
$\quad(\mathrm{f}-\mathrm{g})(\mathrm{x})=\left\{\begin{array}{c}\mathrm{x}-0=\mathrm{x}, \mathrm{x} \text { is rational } \\ 0-\mathrm{x}=-\mathrm{x}, \mathrm{x} \text { is irrational }\end{array}\right.$
Clearly, $\mathrm{f}-\mathrm{g}$ is one $-$ one and onto.

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 1)
Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}, & \mathrm{x}>0 \\ 0, & \mathrm{x}=0\end{array}\right.$ be a real valued function.
Which of the following statements is/are correct?
1- $\mathrm{f}(\mathrm{x})$ is right continuous at $\mathrm{x}=0$.
2- $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=1$. Select the correct answer using the code given below:

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
For right hand continuity at $x=0$ $\begin{aligned} \mathrm{RHL} &=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=\lim _{h \rightarrow 0} \frac{\left(1+h+\frac{h^{2}}{2 !}+\ldots\right)-1}{h} \\=& \lim _{h \rightarrow 0} \frac{h+\frac{h^{2}}{2 !}+\frac{h^{3}}{3 !}+\ldots .}{h} \\=& \lim _{h \rightarrow 0} 1+\frac{h}{2 !}+\frac{h^{2}}{3 !}+\ldots . .=1 \end{aligned}$
$f(0)=0$
$\Rightarrow f(x)$ is not right continuous at $x=0$. For discontinuity at $x=1$
$\begin{aligned} \mathrm{RHL} &=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &=\lim _{h \rightarrow 0} \frac{e^{1+h}-1}{1+h} \\ &=\lim _{h \rightarrow 0} \frac{\left(1+(1+h)+\frac{(1+h)^{2}}{2}+\ldots\right)-1}{1+h} \\=& \lim _{h \rightarrow 0} \frac{(1+h)+\frac{(1+h)^{2}}{2 !}+\frac{(1+h)^{3}}{3 !}+\ldots .}{(1+h)} \\=& \lim _{h \rightarrow 0} 1+\frac{(1+h)}{2 !}+\frac{(1+h)^{2}}{3 !}+\ldots . \end{aligned}$
$\begin{aligned}=& 1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \\ \mathrm{LHL} &=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} \frac{e^{(1-h)}-1}{1-h} \\ &=\lim _{h \rightarrow 0} \frac{\left(1+(1-h)+\frac{(1-h)^{2}}{2 !}+\ldots .\right)-1}{(1-h)} \\ &=\lim _{h \rightarrow 0} 1+\frac{(1-h)}{2 !}+\frac{(1-h)^{2}}{3 !}+\ldots . \\ &=1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots . . \end{aligned}$
$f(1)=\frac{e^{1}-1}{1}=\left(1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots .\right)-1$
$\quad=\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots$
$\Rightarrow \mathrm{RHL} \neq f(1), \mathrm{LHL} \neq f(1)$
So $f$ is discontinuous.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 2)
Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}, & \mathrm{x}>0 \\ 0, & \mathrm{x}=0\end{array}\right.$ be a real valued function.
Which one of the following statements is correct?

Options:
A $\mathrm{f}(\mathrm{x})$ is a strictly decreasing function in $(0, \mathrm{x})$,
B $\mathrm{f}(\mathrm{x})$ is a strictly increasing function in $(0, \mathrm{x})$,
C $\mathrm{f}(\mathrm{x})$ is neither increasing nor decreasing in $(0, \mathrm{x})$
D $\mathrm{f}(\mathrm{x})$ is not decreasing in $(0, \mathrm{x})$.

The correct answer is:
$\mathrm{f}(\mathrm{x})$ is a strictly increasing function in $(0, \mathrm{x})$,

Solution:
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}>0$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{xe}^{\mathrm{x}}-\left(\mathrm{e}^{\mathrm{x}}-1\right)}{\mathrm{x}^{2}}=\frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+1}{\mathrm{x}^{2}}$
$=\left(\frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{2}}\right)$, which is a strictly increasing function.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 2)
Let $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, where $[.]$ is the greatest integer function and $\mathrm{g}(\mathrm{x})=\sin \mathrm{x}$ be two real valued functions over R.
Which of the following statements are correct?
1- (fof)(x)=f(x)
2- (gog)(x)=g(x) only when $x=0$.
3- $(\mathrm{go}(\mathrm{fog}))(\mathrm{x})$ can take only three values. Select the correct answer using the code given below:

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1 and 3 only

Solution:
$($ fof $)(x)=[[x]]$ and $f(x)=[x]$
Suppose $x=0.2$
$\Rightarrow(f o f)(x)=[[0.2]]=[0]=0$
$f(x)=[0.2]=0$
$\Rightarrow f o f(x)=f(x)$
Now, $(\operatorname{gog})(x)=\sin \sin x$ and $g(x)=\sin x$
At $x=0$
$(\operatorname{gog})(x)=\sin \sin 0=\sin 0=0$
$g(x)=\sin 0=0$
$\Rightarrow(\operatorname{gog})(x)=g(x)$ at $x=0$
and this is true for $\mathrm{x}=\mathrm{n} \pi$, where $\mathrm{n}=0,1,2,3,4 \ldots \ldots$
$\because(f o g)(x)=[\sin x]$
$(g o(f o g))(x)=\sin [\sin x]$
$\because \sin x$ has value from $-1$ to 1
If $-1 \leq \sin x < 0$
$(g o(f o g))(x)=\sin (-1)-\sin (-1)$
If $0 \leq \sin x < 1$
$\therefore(g o(f o g))(x)=\sin (0)=\sin 0$ and
If $\sin x=1$
$\therefore(g o(f o g))(x)=\sin (1)=\sin 1$

Topic: Mathematics, Functions, NDA, NDA 2016 (Phase 2)