datavorous/entrance-exam-dataset huggingface dataset:
Let $N$ denote the setof natural numbers and $A=\left\{\mathrm{n}^{2}: \mathrm{n} \in N\right\}$ and $\mathrm{B}=\left\{\mathrm{n}^{3}: \mathrm{n} \in N\right\} .$ Which one of the following is incorrect?

Options:
A $A \cup B=N$
B The complement of $(\mathrm{A} \cup \mathrm{B})$ is an infinite set
C $\mathrm{A} \cap \mathrm{B}$ must be a finite set
D $A \cap B$ must be a proper subset of $\left\{m^{6}: m \in N\right\}$

The correct answer is:
$A \cup B=N$

Solution:
Let $\mathrm{A}=\left\{n^{2}: n \in \mathrm{N}\right\}$ and $\mathrm{B}=\left\{n^{3}: n \in \mathrm{N}\right\}$
$\mathrm{A}=\{1,4,9,16, \ldots . .\}$
and $\mathrm{B}=\{1,8,27,64, \ldots \ldots\}$
Now, $\mathrm{A} \cap \mathrm{B}=\{1\}$ which is a finite set.
Also, $\mathrm{A} \cup \mathrm{B}=\{1,4,8,9,27, \ldots . .\}$
So, complement of $\mathrm{A} \cup \mathrm{B}$ is infinite set. Hence, $\mathrm{A} \cup \mathrm{B} \neq \mathrm{N}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
Let $O(0,0,0), P(3,4,5), Q(m, n, r)$ and $R(1,1,1)$ be the vertices of a parallelogram taken in order. What is the value of $m+n+r ?$

Options:
A 6
B 12
C 15
D More than 15

The correct answer is:
15

Solution:
OPQR is a parallelogram and $\mathrm{OQ}, \mathrm{PR}$ are the diagonals of parallelogram. We know that in a parallelogram, diagonals bisect each other.
$\therefore\left(\frac{0+m}{2}, \frac{0+n}{2}, \frac{0+r}{2}\right) \equiv\left(\frac{1+3}{2}, \frac{1+4}{2}, \frac{1+5}{2}\right)$
$\therefore \quad \frac{m}{2}=\frac{4}{2}, \frac{n}{2}=\frac{5}{2}, \frac{r}{2}=\frac{6}{2}$
$\Rightarrow \quad m=4, n=5, r=6$
Hence, $m+n+r=4+5+6=15$

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 1)
Let $\mathrm{O}$ be the origin and $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ be the points such
that $\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}$. Then which one of the following
is correct?

Options:
A $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are the vertices of an equilateral triangle
B $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are the vertices of an isosceles triangle
C $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are collinear
D None of the above

The correct answer is:
$\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are collinear

Solution:
Consider $\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}$
$\Rightarrow \overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{QO}}=-\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OR}}$
$\Rightarrow \overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{OR}}+\overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OQ}}=\frac{1}{2}(\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OR}})$
Hence Q is the mid-point of P and $\mathrm{R}$. $\therefore \mathrm{P}, \mathrm{Q}, \mathrm{R}$ are collinear.

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 1)
Let $\mathrm{O}$ be the origin and point $\mathrm{A}$ be represented byz. If OA is rotated through an angle $\pi / 2$ in the anticlockwise direction keeping the length of OA same, then what represents the new point?

Options:
A $-\mathrm{iz}$
B $|z|$ i
C i
D $\mathrm{z}$

The correct answer is:
i

Solution:
Let $z=\cos \theta+i \sin \theta$
Now, on rotating through an angle $\frac{\pi}{2}, z$ becomes
$\mathrm{Z}=\cos \left(\frac{\pi}{2}+\theta\right)+\mathrm{i} \sin \left(\frac{\pi}{2}+\theta\right)$
$=-\sin \theta+\mathrm{i} \cos \theta=\mathrm{i}^{2} \sin \theta+\mathrm{i} \cos \theta$
$=\mathrm{i}(\cos \theta+\mathrm{i} \sin \theta)=\mathrm{i} \mathrm{z}$

Topic: Mathematics, Complex Number, NDA, NDA 2007 (Phase 1)
Let $\mathrm{P}=\{1,2,3\}$ and a relation on set $\mathrm{P}$ is given by the set $\mathrm{R}=\{(1,2),(1,3),(2,1),(1,1),(2,2),(3,3),(2,3)\}$. Then $\mathrm{R}$ is

Options:
A Reflexive, transitive but not symmetric
B Symmetric, transitive but notreflexive
C Symmetric, reflexive but not transitive
D None of the above

The correct answer is:
Reflexive, transitive but not symmetric

Solution:
Given relation $\mathrm{R}=\{(1,2),(1,3),(2,1),(1,1),(2,2),(3,3),(2,3)\}$ is
reflexive and transitive only but not symmetric $\because(3,1)$ and $(3,2) \notin \mathrm{R}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 2)
Let $\vec{p}$ and $\vec{q}$ be the position vectors of the points $P$ and $Q$ respectively with respect to origin $\mathrm{O}$. The points $\mathrm{R}$ and $\mathrm{S}$ divide PQ internally and externally respectively in the ratio $2: 3 .$ If $\overrightarrow{\mathrm{OR}}$ and $\overrightarrow{\mathrm{OS}}$ are perpendicular, then which one of the following is correct?

Options:
A $9 p^{2}=4 q^{2}$
B $4 \mathrm{p}^{2}=9 \mathrm{p}^{2}$
C $9 p=4 q$
D $4 p=9 q$

The correct answer is:
$9 p^{2}=4 q^{2}$

Solution:
R divides PQ internally in ratio $2: 3$
$\therefore \overrightarrow{\mathrm{OR}}=\frac{2 \overrightarrow{\mathrm{q}}+3 \overrightarrow{\mathrm{p}}}{5}$ ...(1)
S divides $\mathrm{PQ}$ externally in ratio $2: 3$
$\overrightarrow{\mathrm{OS}}=\frac{2 \overrightarrow{\mathrm{q}}-3 \overrightarrow{\mathrm{p}}}{2-3}=3 \overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}}$ ...(2)
Given, $\overrightarrow{\mathrm{OR}}$ and $\overrightarrow{\mathrm{OS}}$ are perpendicular.
$\therefore\left(\frac{3 \overrightarrow{\mathrm{p}}+2 \overrightarrow{\mathrm{q}}}{5}\right)(3 \overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}})=0$
$\Rightarrow 9 p^{2}-4 q^{2}=0 \Rightarrow 9 p^{2}=4 q^{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 1)
Let $\mathrm{P}=\left\{\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{p}_{4}\right\}$
$Q=\left\{q_{1}, q_{2}, q_{3}, q_{4}\right\}$ and
If $\begin{aligned} \mathrm{R}=\left\{\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}, \mathrm{r}_{4}\right\} \\ 10 &=\left\{\left(\mathrm{p}_{i}, \mathrm{q}_{\mathrm{f}}, \mathrm{r}_{\mathrm{k}}\right): \mathrm{i}+\mathrm{j}+\mathrm{k}=10\right\} \end{aligned}$
how many elements does $\mathrm{S}_{10}$ have?

Options:
A 2
B 6
C 4
D 8

The correct answer is:
4

Solution:
Given that $\mathrm{P}=\left\{\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{p}_{4}\right\}$
$Q=\left\{q_{1}, q_{2}, q_{3}, q_{4}\right\}$
and $\mathrm{R}=\left\{\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}, \mathrm{r}_{4}\right\}$
$\mathrm{S}_{10}=\left\{\mathrm{p}_{2}, \mathrm{q}_{4}, \mathrm{r}_{4}\right),\left(\mathrm{p}_{3}, \mathrm{q}_{3}, \mathrm{r}_{4}\right),\left(\mathrm{p}_{3}, \mathrm{q}_{4}, \mathrm{r}_{3}\right)$
$\left(\mathrm{p}_{4}, \mathrm{q}_{2}, \mathrm{r}_{4}\right)$
$\left.\left(\mathrm{p}_{4}, \mathrm{q}_{3}, \mathrm{r}_{3}\right),\left(\mathrm{p}_{4}, \mathrm{q}_{4}, \mathrm{r}_{2}\right)\right\}$
Total number of elements in $\mathrm{S}_{10}$ are 6 .

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Let $p, q, r$ and $s$ be respectively the magnitudes of the vectors
$3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}, 2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, 4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$. Which
one of the following is correct?

Options:
A $\quad r>s>q>p$
B $s>r>p>q$
C $\quad r>s>p>q$
D $\quad s>r>q>p$

The correct answer is:
$\quad r>s>p>q$

Solution:
Let $p=$ Magnitude of $3 \hat{i}-2 \hat{j}=\sqrt{9+4}=\sqrt{13}$
$q=$ Magnitude of $2 \hat{i}+2 \hat{j}+\hat{k}=\sqrt{4+4+1}=3$
$r=$ Magnitude of $4 \hat{i}-\hat{j}+\hat{k}=\sqrt{16+1+1}=\sqrt{18}=3 \sqrt{2}$
$s=$ Magnitude of $2 \hat{i}+2 \hat{j}+3 \hat{k}=\sqrt{4+4+9}=\sqrt{17}$
$\therefore \quad r>s>p>q$

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 1)
Let $\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{s}$ be the distances from origin of the points $(2,6)$, $(3,4),(4,5)$ and $(-2,5)$ respectively. Which one of the following is a whole number?

Options:
A $\mathrm{p}$x
B $\mathrm{q}$
C I
D $\mathrm{s}$

The correct answer is:
$\mathrm{q}$

Solution:
Let $\mathrm{A}(2,6) ; \mathrm{B}(3,4) ; \mathrm{C}(4,5)$ and $\mathrm{D}(-2,5)$ are the given points. Let $\mathrm{O}$ be the origin, i.e. $\mathrm{O}(0,0)$
$\mathrm{OA}=\sqrt{(2-0)^{2}+(6-0)^{2}}=\sqrt{40}=2 \sqrt{10}$ units
$O B=\sqrt{(3-0)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$ units
$\mathrm{OC}=\sqrt{(4-0)^{2}+(5-0)^{2}}=\sqrt{16+25}=\sqrt{41}$ units
$O D=\sqrt{(-2-0)^{2}+(5-0)^{2}}=\sqrt{4+25}=\sqrt{29}$ units
So, $\mathrm{q}=\mathrm{OB}=5$ units is the correct answer.

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
Let $Q$ be the image of the point $P(-2,1,-5)$ in the plane $3 x-2 y+2 z+1=0 .$ Consider the following:
1 . The coordinates of $\mathrm{Q}$ are $(4,-3,-1)$.
2 . PQ is of length more than 8 units.
3 . The point $(1,-1,-3)$ is the mid-point of the line segment $\mathrm{PQ}$ and lies on the given plane. Which of the above statements are correct?

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:
1,2 and 3

Solution:
Let $Q\left(x_{1}, y_{1}, z_{1}\right)$ be the image of the point $P$. The direction ratios of $P Q$ are $3,-2,2$
The Equation of line $P Q$ is $\frac{x+2}{3}=\frac{y-1}{-2}=\frac{z+5}{2}=r$ Coordinates of any point on the line $P Q$ is $3 r-2,-2 r+$ I and $2 r-5$.
Let $Q(3 r-2,-2 r+1,2 r-5)$ be such a point.
et $L$ be the mid point of $P Q, L=\left(\frac{3 r}{2}-2,1-r, r-5\right)$
Since $L$ lies on the plane $3 x-2 y+2 z+1=0$
So, $3\left(\frac{3 r}{2}-2\right)-2(1-r)+2(r-5)+1=0$
$\Rightarrow \frac{17}{2} r-17=0 \Rightarrow r=2$
So, coordinates of $Q$ are $(3 \times 2-2,-2 \times 2+1,2 \times 2-5)$
$=(4,-3,-1)$
Also the mid point of $P Q$ is $L=\left(\frac{3 \times 2}{2}-2,1-2,2-5\right)$ $\therefore P Q=\sqrt{(-2-4)^{2}+(1+3)^{2}+(-5+1)^{2}}=\sqrt{68}$
$\Rightarrow P Q=2 \sqrt{17}>8$
Option (d) is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2016 (Phase 2)
Let $Q$ be the image of the point $P(-2,1,-5)$ in the plane $3 x-2 y+2 z+1=0 .$ Consider the following:
1 . The direction ratios of the line segment $\mathrm{PQ}$ are $ < 3,-2,2>$.
2 . The sum of the squares of direction cosines of the line segment $\mathrm{PQ}$ is unity. Which of the above statements is/are correct?

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

The correct answer is:
Both 1 and 2

Solution:
From (i) above, 1 is correct. We know that, Sum of direction cosines of the line segment $\mathrm{PQ}=1$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2016 (Phase 2)
Let $\mathrm{R}$ be a relation on the set $\mathrm{N}$ of natural numbers defined
by 'nRM $\Leftrightarrow \mathrm{n}$ is a factor of $\mathrm{m}$ '. Then which one of the following is correct?

Options:
A $\mathrm{R}$ is reflexive, symmetric but not transitive
B $\mathrm{R}$ is transitive, symmetric but not reflexive
C $\mathrm{R}$ is reflexive, transitive but not symmetric
D $\mathrm{R}$ is an equivalence relation

The correct answer is:
$\mathrm{R}$ is reflexive, transitive but not symmetric

Solution:
$n R m \Leftrightarrow n$ is a factor of $m$ $\Rightarrow m$ is divisible by $n$. Reflexivity We know that
$n$ is divisible by $n \forall n \in N$
$(n, n) \in R \forall n \in N$
$R$ is reflexive. Symmetric
$n, m \in N$
Let $n=2, m=6$
$m$ is divisible by $n$ but $n$ is not divisible by $m$. Hence $R$ is not symmetric. Transitivity
Let $(n, m) \in R$ and $(m, p) \in R$ then $(n, m) \in R$ and
$(m, p) \in R \Rightarrow(\mathrm{n}, \mathrm{p}) \in \mathrm{R}$
or If $m$ is divisible by $n$ and $p$ is divisible by $m$. Hence $p$ is divisible by $n$.
$(n, p) \in R \forall n, p \in N$
$R$ is transitive relation on $N$. Hence $R$ is reflexive, transitive but not symmetric.
$\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x$ if $f(2 a-x)=f x$

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 1)
Let $\mathrm{R}$ be arelation from $\mathrm{A}=\{1,2,3,4\}$ to $\mathrm{B}=\{1,3,5\}$ suchthat $\mathrm{R}=[(\mathrm{a}, \mathrm{b}): \mathrm{a} < \mathrm{b}$, where $\mathrm{a} \in \mathrm{A}$ an $\mathrm{d} \mathrm{b} \in \mathrm{B}]$. What is $\mathrm{RoR}^{-1}$
equal to?

Options:
A $\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$
B $\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\}$
C $\{(3,3),(3,5),(5,3),(5,5)$
D $\{(3,3),(3,4),(4,5)\}$

The correct answer is:
$\{(3,3),(3,5),(5,3),(5,5)$

Solution:
$\mathrm{R}=\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$
$\Rightarrow \mathrm{R}^{-1}=\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\} .$
$\Rightarrow \operatorname{RoR}^{-1}=\{(3,3),(3,5),(5,3),(5,5)\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 2)
Let $R$ be the relation defined on the set of natural number $\mathrm{N}$ as $a \mathrm{Rb} ; \mathrm{a}, \mathrm{b} \in \mathrm{N}$, if a divides $\mathrm{b}$. Then, which one of the following is correct?

Options:
A $\mathrm{R}$ is reflexive only
B $\mathrm{R}$ is symmetric only
C $\mathrm{R}$ is transitive only
D $\mathrm{R}$ is reflexive and transitive

The correct answer is:
$\mathrm{R}$ is reflexive and transitive

Solution:
For reflexive: (d) For reflexive:
$\mathrm{aRa} \Rightarrow$ a divides a R is reflexive. For symmetric :
$\mathrm{aRb} \Rightarrow$ a divides $\mathrm{b}$
$\mathrm{bRa} \Rightarrow \mathrm{b}$ divides a
which may not be true $\Rightarrow \quad \mathrm{R}$ is not symmetric For transitive $\mathrm{aRB} \Rightarrow$ a divides $\mathrm{b} \Rightarrow \mathrm{b}=\mathrm{ka}$
$\mathrm{bRc} \Rightarrow \mathrm{b}$ divides $\mathrm{c} \Rightarrow \mathrm{c}=\mathrm{lb}$
Now, $\mathrm{c}=1 \mathrm{ka}$
$\Rightarrow$ a divides $\mathrm{c}$
$\Rightarrow$ a $R c$
$\Rightarrow \quad \mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{c} \mathrm{Ra}$
$\Rightarrow \mathrm{R}$ is transitive.

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 1)
Let $\mathrm{R}$ be the set of real numbers and let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a
function such that $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$. What is the range of $\mathrm{f}$ ?

Options:
A $\mathrm{R}$
B $\mathrm{R}-\{1\}$
C $[0,1]$
D $[0,1)$

The correct answer is:
$[0,1)$

Solution:
$\because \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$
Since, numerator $ < $ denominator $\mathrm{f}(\mathrm{x}) < 1$ for all values of $\mathrm{x}$ (negative or positive) and $\mathrm{f}(\mathrm{x})=0$ for $\mathrm{x}=0$
So, range of fis $[0,1)$.

Topic: Mathematics, Functions, NDA, NDA 2006 (Phase 1)
Let $R=\{x \mid x \in N, x$ is a multiple of 3 and $x \leq 100\}$ $S=\{x \mid x \in N, x$ is a multiple of 5 and $x \leq 100\}$
What is the number of elements in $(R \times S) \cap(S \times R)^{2}$

Options:
A 36
B 20
C 33
D 6

The correct answer is:
36

Solution:
Let $R=\{x: x \in N, x$ is a multiple of 3 and $x \leq 100\}$
and $S=\{x: x \in N, x$ is a multiple of 5 and $x \leq 100\}$ $R=\{3,6,9,12,15, \ldots, 99\}$
and $S=\{5,10,15, \ldots, 95,100\}$
Now, $(R \times S) \cap(S \times R)=(R \cap S) \times(S \cap R)$
$=(15,30,45,60,75,90) \times(15,30,45,60,75,90)$
Number of elements in $(R \times S) \cap(S \times R)$
$=6 \times 6=36$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
Let $S$ be a set of all distinct numbers of the form $\frac{\mathrm{P}}{\mathrm{Q}}$, where
$\mathrm{p}, \mathrm{q} \in\{1,2,3,4,5,6\}$. What is the cardinality of the set $\mathrm{S}$ ?

Options:
A 21
B 23
C 32
D 36

The correct answer is:
23

Solution:
Given that p, q: $(1,2,3,4,5,6$, )
For $\frac{\mathrm{p}}{\mathrm{q}}$ form, when $\mathrm{p}=1, \mathrm{q}=1,2,3,4,5,6$
thus, $\frac{\mathrm{p}}{\mathrm{q}}=1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ and $\frac{1}{6}$
$\mathrm{n}=\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=6$
When $p=2, q=1,3,5$
thus $\frac{\mathrm{p}}{\mathrm{q}}=2, \frac{2}{3}, \frac{2}{5}$ and $\mathrm{n}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=3$
When $\mathrm{p}=3, \mathrm{q}=1,2,4,5$
thus $\frac{\mathrm{p}}{\mathrm{q}}=3, \frac{3}{2}, \frac{3}{4}, \frac{3}{5}$ and $\mathrm{n}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=4$
When $\mathrm{p}=4, \mathrm{q}=1,3,5$
thus $\frac{\mathrm{p}}{\mathrm{q}}=4, \frac{4}{3}, \frac{4}{5}$ and $\mathrm{n}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=3$
When $p=5, q=1,2,3,4,6$
thus $\left(\frac{p}{q}\right)=5, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}$ and $\frac{5}{6}$ and $\mathrm{n}\left(\frac{\mathrm{p}}{q}\right)=5$
When $\mathrm{P}=6, \mathrm{q}=1,5$
thus $\left(\frac{p}{q}\right)=6, \frac{6}{5}$ and $n\left(\frac{p}{q}\right)=2$
Hence, cardinality of the set (s) $=6+3+4+3+5+2=23 .$

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 2)
Let $S$ be the set of all persons living in Delhi. We say that $x$, y in $\mathrm{S}$ are related if they were born in Delhi on the same day. Which one of the following is correct?

Options:
A The relation is an equivalent relation
B The relation is not reflexive but it is symmetric and transitive
C The relation is not symmetric but it is reflexive and transitive
D The relation is not transitive but it is reflexive and symmetric

The correct answer is:
The relation is an equivalent relation

Solution:
$\mathrm{S}=\{$ All persons living in Delhi\}
A relationship is said to be equivalence relation if it is reflexive, symmetric and transitive
$\Rightarrow$ Here, $(\mathrm{x}, \mathrm{y}) \in \mathrm{R} \Rightarrow$ reflexive relation
$\Rightarrow$ Since, $x \varepsilon y$ are born on the same day, $x R y \Rightarrow y R x$
So, it is symmeitric relation.
$\Rightarrow \mathrm{x} \mathrm{R} \mathrm{y}, \mathrm{y} \mathrm{R} \mathrm{Z} \Rightarrow \mathrm{xR} \mathrm{Z}$ (Date of births are same)
So, it is transitive relation. So, the given relation is an equivalent relation.

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 1)
Let $S$ denote set of all integers. Define a relation $R^{\circ}$
$' a R b$ if $a b \geq 0$ where $a, b \in S^{\prime}$. Then $R$ is:

Options:
A Reflexive but neither symmetric nor transitiverelation
B Reflexive, symmetric but not transitive relation
C An equivalence relation
D Symmetric but neither reflexivenor transitive relation

The correct answer is:
An equivalence relation

Solution:
$\mathrm{S}=$ Set of all integers and $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}), \mathrm{a}, \mathrm{b} \in \mathrm{S}$ and $\mathrm{ab} \geq 0\}$
For reflexive $: \mathrm{aRa} \Rightarrow \mathrm{a}, \mathrm{a}=\mathrm{a}^{2} \geq 0$
for all integers a. $\mathrm{a} \geq 0$ For symmetric $: \mathrm{aRb} \Rightarrow \mathrm{ab} \geq 0 \forall \mathrm{a}, \mathrm{b} \in \mathrm{S}$
If $a b \geq 0$, then $b a \geq 0 \Rightarrow b R a$
For transitive :
If $a b \geq 0, b c \geq 0$, then also ac $\geq 0$ Relation $\mathrm{R}$ is reflexive, symmetric and transitive

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 1)
Let $S_{n}$ denote the sum of the $n$ terms of an $\mathrm{AP}$ and $3 \mathrm{~S}_{\mathrm{n}}=\mathrm{S}_{2 \mathrm{n}}$.
What is $\mathrm{S}_{3 \mathrm{n}}=\mathrm{S}_{2 \mathrm{n}}$ equal to?

Options:
A $2: 1$
B $3: 1$
C $4: 1$
D $5: 1$

The correct answer is:
$2: 1$

Solution:
From explanation $\frac{\mathrm{S}_{3 \mathrm{n}}}{\mathrm{S}_{2 \mathrm{n}}}=\frac{6 \mathrm{n}^{2} \mathrm{~d}}{3 \mathrm{n}^{2} \mathrm{~d}}=\frac{2}{1}=2: 1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2014 (Phase 2)
Let $S_{n}$ denote the sum of the $n$ terms of an $\mathrm{AP}$ and $3 \mathrm{~S}_{\mathrm{n}}=\mathrm{S}_{2 \mathrm{n}}$.
What is $\mathrm{S}_{3 \mathrm{n}}=\mathrm{S}_{\mathrm{n}}$ equal to ?

Options:
A $4: 1$
B $6: 1$
C $8: 1$
D $10: 1$

The correct answer is:
$6: 1$

Solution:
Given, $\mathrm{S}_{\mathrm{n}}=$ Sum of first n terms of an AP.
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{h}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$ or $\mathrm{S}_{2 \mathrm{n}}=\frac{2 \mathrm{n}}{2}[2 \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}]$
Similarly, $\quad S_{3 n}=\frac{3 n}{2}[3 a+(3 n-1) d]$
According to direction, $3 \mathrm{~S}_{\mathrm{n}}=2 \mathrm{~S}_{2 \mathrm{n}}$
Putting the value of $\mathrm{S}_{\mathrm{n}}$ and $\mathrm{S}_{2 \mathrm{n}}$ in above equation. $3\left(\frac{\mathrm{n}}{2}\right)[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]=2\left(\frac{\mathrm{n}}{2}\right)[2 \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}]$
$6 a+3(n-1) d=4 a+2(n-1) d$
$2 \mathrm{a}=\mathrm{d}(\mathrm{n}+1)$
$\therefore \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[\mathrm{~d}(\mathrm{n}+1)+\mathrm{d}(\mathrm{n}-1)]$
$=\frac{\mathrm{n}}{2}[\mathrm{dn}+\mathrm{d}+\mathrm{dn}-\mathrm{d}]$
$=\frac{\mathrm{n}}{2}(2 \mathrm{dn})=\mathrm{n}^{2} \mathrm{~d}$
Now, $\mathrm{S}_{2 \mathrm{n}}=\mathrm{n}[\mathrm{d}(\mathrm{n}+1)+(2 \mathrm{n}-1) \mathrm{d}]=3 \mathrm{n}^{2} \mathrm{~d}$
$\mathrm{S}_{3 \mathrm{n}}=\frac{3 \mathrm{n}}{2}[\mathrm{~d}(\mathrm{n}+1+3 \mathrm{n}-1)]=6 \mathrm{n}^{2} \mathrm{~d}$
Hence, $\frac{\mathrm{S}_{3 \mathrm{n}}}{\mathrm{S}_{\mathrm{n}}}=\frac{6 \mathrm{n}^{2} \mathrm{~d}}{\mathrm{n}^{2} \mathrm{~d}}=\frac{6}{1}=6: 1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2014 (Phase 2)
Let $\sin (\mathrm{A}+\mathrm{B})=1$ and $\sin (\mathrm{A}-\mathrm{B})=\frac{1}{2}$ where $\mathrm{A}, \mathrm{B} \in\left[0, \frac{\pi}{2}\right]$.
What is the value of A?

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{3}$
C $\frac{\pi}{4}$
D $\frac{\pi}{8}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
$\begin{array}{l}\sin (\mathrm{A}+\mathrm{B})=1 \\ \Rightarrow \sin (\mathrm{A}+\mathrm{B})=\sin 90^{\circ} \\ \Rightarrow \mathrm{A}+\mathrm{B}=90^{\circ} \\ & \text { Given } \sin (\mathrm{A}-\mathrm{B})=\frac{1}{2}=\sin 30^{\circ} \\ \Rightarrow \mathrm{A}-\mathrm{B}=30^{\circ}\end{array}$
On solving $(1)$ and $(2)$, we get $\mathrm{A}=60$
$\mathrm{B}=30$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
Let $\mathrm{T}_{\mathrm{r}}$ be the $\mathrm{r}^{\text {th }}$ term of an AP for $\mathrm{r}=1,2,3, \ldots$. If for some distinct positive integers $\mathrm{m}$ and $\mathrm{n}$ we have $\mathrm{T}_{\mathrm{m}}=1 / \mathrm{n}$ and $\mathrm{T}_{\mathrm{n}}=1 / \mathrm{m}$, then what is $\mathrm{T}_{\mathrm{mn}}$ equal to?

Options:
A $(\mathrm{mn})^{-1}$
B $\mathrm{m}^{-1}+\mathrm{nn}^{-1}$
C 1
D 0

The correct answer is:
1

Solution:
$\quad \mathrm{T}_{\mathrm{n}}=\frac{1}{m}, \mathrm{~T}_{\mathrm{m}}=\frac{1}{n}$
$\Rightarrow \quad 1^{\text {st }}$ term $=\mathrm{c.d}=\frac{1}{m n}$
$\Rightarrow \quad \mathrm{T}_{m n}=\frac{1}{m n}+\frac{m n-1}{m n}=1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 2)
Let the coefficient of the middle term of the binomial expansion of $(1+x)^{2 n}$ be $\alpha$ and those of two middle terms of the binomial expansion of $(1+\mathrm{x})^{2 n-1}$ be $\beta$ and $\gamma$. Which one of the $\begin{array}{ll}\text { following relations is correct? } & \end{array}$

Options:
A $\alpha>\beta+\gamma$
B $\alpha < \beta+\gamma$
C $\alpha=\beta+\gamma$
D $\alpha=\beta \gamma$

The correct answer is:
$\alpha=\beta+\gamma$

Solution:
$\alpha={ }^{2 n} \mathrm{C}_{n}$
$\beta^{-2 n-1} \mathrm{C}_{n}$
$\gamma=^{2 n-1} \mathrm{C}_{n-1}$
$\beta+\gamma={ }^{2 n-1} \mathrm{C}_{n}+{ }^{2 n-1} \mathrm{C}_{n-1}={ }^{2 n} \mathrm{C}_{n}=\alpha$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2018 (Phase 2)
Let the coordinates of the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be $(1,8,4)$, $(0,-11,4)$ and $(2,-3,1)$ respectively. What are the coordinates of the point $D$ which is the foot of the perpendicular from $\mathrm{A}$ on $\mathrm{BC}$ ?

Options:
A $(3,4,-2)$
B $(4,-2,5)$
C $(4,5,-2)$
D $(2,4,5)$

The correct answer is:
$(4,5,-2)$

Solution:
$\mathrm{A}(1,8,4), \mathrm{B}(0,-11,4), \mathrm{C}(2,-3,1)$
Let $\mathrm{D}=(\alpha, \beta, \gamma)$
Direction ratios of $\mathrm{BC}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}, \mathrm{y}_{2}-\mathrm{y}_{1}, \mathrm{z}_{2}-\mathrm{z}_{1}\right)$
Let, $(\mathrm{a}, \mathrm{b}, \mathrm{c})=(2,8,-3)$
Direction 1 Let (a'. Since, $\mathrm{AD}$ is perpendicular to $\mathrm{BC}$,
e find option (c) is correct.
know, the equation of plane passing through 3

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 1)