datavorous/entrance-exam-dataset huggingface dataset:
If a line makes $30^{\circ}$ with the positive direction of $\mathrm{x}$ -axis, angle $\beta$ with the positive direction of $y$ -axis and angle $\gamma$ with the positive direction of $\mathrm{z}$ -axis, then what is $\cos ^{2} \beta+\cos ^{2} \gamma$ equal to ?

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

The correct answer is:
$1 / 4$

Solution:
Direction cosines are $\cos 30^{\circ}, \cos \beta$ and $\cos \gamma$
Since we know $\cos ^{2} 30+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\Rightarrow \cos ^{2} \beta+\cos ^{2} \gamma=\frac{1}{4} \quad\left(\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 1)
If a line makes the angles $\alpha, \beta, \gamma$ with the axes, then what is the value of $1+\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma$ equal to

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

The correct answer is:
0

Solution:
We have
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
Consider $1+\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma$
$=1+\left(2 \cos ^{2} \alpha-1\right)+\left(2 \cos ^{2} \beta-1\right)+\left(2 \cos ^{2} \gamma-1\right)$
$=2 \cos ^{2} \alpha+2 \cos ^{2} \beta+2 \cos ^{2} \gamma-2$
$=2\left[\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right]-2=2(1)-2=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 1)
If a line OP oflength $\mathrm{r}$ (where ' $\mathrm{O}$ ' is the origin) makes an angle $\alpha$ with $\mathrm{x}$ -axis and lies in the $\mathrm{xz}$ -plane, then what are the coordinates of $\mathrm{P} ?$

Options:
A $(r \cos \alpha, 0, r \sin \alpha)$
B $(0,0, r \sin \alpha)$
C $(\mathrm{r} \cos \alpha, 0,0)$
D $(0,0, r \cos \alpha)$

The correct answer is:
$(r \cos \alpha, 0, r \sin \alpha)$

Solution:
Since line OP of length 'r'which makes an angle ' $\alpha$ ' with $\mathrm{x}$ -axis lies in $\mathrm{xz}$ -plane. Therefore y-coordinate of $\mathrm{P}$ is zero. Now, from $\Delta \mathrm{OAP}$, we have $\mathrm{OA}=\mathrm{r} \cos \alpha, \mathrm{PA}=\mathrm{r} \sin \alpha$
$\therefore \quad \mathrm{P}=(\mathrm{r} \cos \alpha, 0, \mathrm{r} \sin \alpha)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
If a matrix A has inverses $\mathrm{B}$ and $\mathrm{C}$, then which one of the following is correct?

Options:
A $\mathrm{B}$ may not be equal to $\mathrm{C}$
B $\mathrm{B}$ should be equal to $\mathrm{C}$
C $\mathrm{B}$ and $\mathrm{C}$ should be unit matrices
D None of the above

The correct answer is:
$\mathrm{B}$ should be equal to $\mathrm{C}$

Solution:
Since, Inverse is unique.
$\therefore \mathrm{B}$ should be equal to $\mathrm{C}$.

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 1)
If a matrix $A$ is such that
$3 A^{3}+2 A^{2}+5 A+I=0$
Then what is $A^{-1}$ equal to?

Options:
A $-\left(3 A^{2}+2 A+5\right)$
B $3 A^{2}+2 A+5 I$
C $3 A^{2}-2 A-5 I$
D $\left(3 A^{2}+2 A-5 I\right)$

The correct answer is:
$-\left(3 A^{2}+2 A+5\right)$

Solution:
Let A bea matrix such that $3 \mathrm{~A}^{3}+2 \mathrm{~A}^{2}+5 \mathrm{~A}+\mathrm{I}=0$
Post multiply by $\mathrm{A}^{-1}$ on both the sides, we get $3 \mathrm{~A}^{3} \mathrm{~A}^{-1}+2 \mathrm{~A}^{2} \mathrm{~A}^{-1}+5 \mathrm{AA}^{-1}+\mathrm{IA}^{-1}=0$
$\Rightarrow 3 \mathrm{~A}^{2}+2 \mathrm{~A}+5 \mathrm{I}+\mathrm{A}^{-1}=0$
$\Rightarrow \mathrm{A}^{-1}=-\left(3 \mathrm{~A}^{2}+2 \mathrm{~A}+5 \mathrm{I}\right)$

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 2)
If a matrix $A$ is symmetric as well as anti-symmetric, then which one of the following is correct?

Options:
A $A$ is a diagonal matrix
B $A$ is a null matrix
C $A$ is a unit matrix
D $A$ is a triangular matrix

The correct answer is:
$A$ is a null matrix

Solution:
Since, matrix $\mathrm{A}$ is symmetric and anti-symmetric therefore
$A^{\prime}=A$ and $A^{\prime}=-A$
$\Rightarrow A=-A \Rightarrow 2 A=0$
$\Rightarrow A$ is a null matrix

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 2)
If a matrix $\mathrm{B}$ is obtained from a square matrix $\mathrm{A}$ by interchanging any two of its rows, then what is $|\mathrm{A}+\mathrm{B}|$ equal to

Options:
A $2|\mathrm{~A}|$
B $2|\mathrm{~B}|$
C 0
D $|\mathrm{A}|-|\mathrm{B}|$

The correct answer is:
0

Solution:
Let $A=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]$
Let rows 1 and 2 be interchanged.
and $B=\left[\begin{array}{lll}\mathrm{d} & \mathrm{e} & \mathrm{f} \\ \mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{g} & \mathrm{h} & \mathrm{i}\end{array}\right]$
$A+B=\left[\begin{array}{ccc}a+d & b+e & c+f \\ a+d & b+e & c+f \\ 2 g & 2 h & 2 i\end{array}\right]$
$|A+B|=\left|\begin{array}{ccc}a+d & b+e & c+f \\ a+d & b+e & c+f \\ 2 g & 2 h & 2 i\end{array}\right|$
$=0 \quad$ (since two rows are identical)

Topic: Mathematics, Determinants, NDA, NDA 2006 (Phase 1)
If $\left|\mathrm{A}_{\mathrm{n} \times \mathrm{n}}\right|=3$ and $|\operatorname{adj} \mathrm{A}|=243$, what is the value of $\mathrm{n}$?

Options:
A 4
B 5
C 6
D 7

The correct answer is:
6

Solution:
As given: $\left|\mathrm{A}_{\mathrm{nxn}}\right|=3$ and $|\operatorname{adjA}|=243$
Dterminant of adjoint $\mathrm{A}$ is given by:
$|\operatorname{adj} \mathrm{A}|=\left|\mathrm{A}_{\mathrm{n} \times \mathrm{n}}\right|^{\mathrm{n}-1}$
$\Rightarrow 243=3^{n-1} \Rightarrow 3^{5}=3^{n-1} \Rightarrow n-1=5 \Rightarrow n=6$

Topic: Mathematics, Matrices, NDA, NDA 2008 (Phase 1)
If a non-empty set A contains $n$ elements, then its power set contains how many elements?

Options:
A $\mathrm{n}^{2}$
B $2^{n}$
C $2 n$
D $n+1$

The correct answer is:
$2^{n}$

Solution:
No. of elements in power set of $\mathrm{A}=2^{\mathrm{n}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 2)
If $A=P(\{1,2\})$ where $P$ denotes the power set, then which one of the following is correct?

Options:
A $\{1,2\} \subset A$
B $1 \in A$
C $\phi \notin A$
D $\{1,2\} \in A$

The correct answer is:
$\{1,2\} \in A$

Solution:
$A=P(\{1,2\})=\{\phi,\{1\},\{2\},\{1,2\}\}$
(\because: Power set is the collection of all subsets of the set A) From above it is clear that $\{1,2\} \in \mathrm{A}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
If a polygon has 20 diagonals, then what is the number of sides?

Options:
A 6
B 10
C 12
D 8

The correct answer is:
8

Solution:
Number of diagonals $=\frac{n(n-3)}{2}$ where $n$ is the number of sides of polygon
$\therefore \quad 20=\frac{n(n-3)}{2}$
$\Rightarrow \quad 40=n^{2}-3 n$
$\Rightarrow \quad n^{2}-3 n-40=0$
$\Rightarrow \quad n^{2}-8 n+5 n-40=0$
$\Rightarrow \quad(n-8)(n+5)=0$
$\Rightarrow \quad \quad n-8=0, n+5=0$
$\therefore$ since, thenumber of diagonals, $n$ cannot benegative.
$\Rightarrow$
$n=8$

Topic: Mathematics, Permutation Combination, NDA, NDA 2008 (Phase 2)
If a primary root continues to grow, the type of root system will be known as:

Options:
A Secondary
B Fibrous
C Tap
D Stilt

The correct answer is:
Tap

Solution:
Root develops from the radical, the root which develops from radical continuous to grow as primary root and produces branches called secondary and tertiary root than such root system is called as tap root system and is characteristics of dicots like mustard.

Topic: Biology, Morphology of Flowering Plants, NEET
If a progressive wave is represented as $y=2 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ where $\mathrm{x}$ is in metre and $\mathrm{t}$ is in second, then the distance travelled by the wave in $5 \mathrm{~s}$ is

Options:
A $5 \mathrm{~m}$
B $10 \mathrm{~m}$
C $25 \mathrm{~m}$
D $32 \mathrm{~m}$

The correct answer is:
$10 \mathrm{~m}$

Solution:
$\begin{array}{l}
\text { Given, } y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right) \\
=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)
\end{array}$
Comparing it with standard equation
$\begin{array}{l}
y=r \sin \frac{2 \pi}{\lambda}(v t-x) \\
=r \sin \left(\frac{2 \pi v t}{\lambda}-\frac{2 \pi x}{\lambda}\right)
\end{array}$
We have, $\frac{2 \pi \mathrm{v}}{\lambda}=\frac{\pi}{2}$ or $\mathrm{v}=\frac{\lambda}{4}$
and $\frac{2 \pi}{\lambda}=\frac{\pi}{4}$ or $\lambda=8 \mathrm{~m} \therefore \mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{s}$
So, the distance travelled by wave in $t$ second $=\mathrm{vt}=2 \times 5=10 \mathrm{~m}$

Topic: Physics, Waves and Sound, VITEEE, VITEEE 2013
If a reaction is carried out under non-standard conditions, how does the enthalpy of reaction compare to the standard enthalpy of reaction (∆H)?

Options:
A It is always greater than ∆H
B It is always smaller than ∆H
C It can be greater or smaller than ∆H depending on the conditions
D It is equal to ∆H regardless of the conditions

The correct answer is:
It can be greater or smaller than ∆H depending on the conditions

Solution:
The enthalpy of a reaction under non-standard conditions can be greater or smaller than ∆H depending on the specific conditions of the reaction.

Topic: Chemistry, Thermodynamics (C), NEET
If a secretary and a joint secretary are to be selected from a committee of 11 members, then in how many ways can they be selected?

Options:
A 110
B 55
C 22
D 11

The correct answer is:
55

Solution:
Selection of 2 members out of 11 has ${ }^{11} \mathrm{C}_{2}$ number of ways ${ }^{11} \mathrm{C}_{2}=55$

Topic: Mathematics, Permutation Combination, NDA, NDA 2006 (Phase 1)
If a set $A$ contains 4 elements, then what is the number of elements in $A \times P(A)$ ?

Options:
A 16
B 32
C 64
D 128

The correct answer is:
64

Solution:
Since, set $A$ contains 4 elements, then number of elements in $P(A)=2^{4}=16$
So, the number of elements in $A \times P(A)=4 \times 16=64$

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
If a set $\mathrm{X}$ contains $\mathrm{n}(\mathrm{n}>5)$ elements, then what is the number of subsets of $X$ containing less than 5 elements?

Options:
A $\mathrm{C}(\mathrm{n}, 4)$
B $\bar{C}(\mathrm{n}, 5)$
C $\sum_{r=0}^{5} \mathrm{C}(\mathrm{n}, \mathrm{r})$
D $\sum_{r=0}^{4} C(n, r)$

The correct answer is:
$\sum_{r=0}^{4} C(n, r)$

Solution:
Number of subsets of $\mathrm{X}$ containing less than 5 elements is given by
${ }^{\mathrm{n}} \mathrm{C}_{0}+{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}+{ }^{\mathrm{n}} \mathrm{C}_{3}+{ }^{\mathrm{n}} \mathrm{C}_{4}$
$\sum_{\mathrm{r}=0}^{4}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\sum_{\mathrm{r}=0}^{4} \mathrm{C}(\mathrm{n}, \mathrm{r})$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
If a simple harmonic motion is represented by d 2 x d t 2 + α x = 0 , its time period is

Options:
A 2 π α
B 2 π α
C 2 π / α
D 2 π / α

The correct answer is:
2 π / α

Solution:
Standard differential equation of SHM is d 2 x d t 2 + ω 2 x = 0
Given equation is d 2 x d t 2 + α x = 0
∴ ω 2 = α
or ω = α
∴ T = 2 π ω = 2 π α .

Topic: Physics, Oscillations, NEET
If $\mathrm{A}=\sin ^{2} \theta+\cos ^{4} \theta$, the for all real $\theta$, which one of the $\begin{array}{ll}\text { following is correct? }

Options:
A $1 \leq \mathrm{A} \leq 2$
B $\frac{3}{4} \leq \mathrm{A} \leq 1$
C $\frac{13}{16} \leq \mathrm{A} \leq 1$
D $\frac{3}{4} \leq \mathrm{A} \leq \frac{13}{16}$

The correct answer is:
$\frac{3}{4} \leq \mathrm{A} \leq 1$

Solution:
$\mathrm{A}=\sin ^{2} \theta+\cos ^{4} \theta$
$=\sin ^{2} \theta+\left(1-\sin ^{2} \theta\right)^{2}$
$=1+\sin ^{4} \theta-\sin ^{2} \theta$
$=1-\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)$
$=1-\sin ^{2} \theta \cdot \cos ^{2} \theta$
$\quad=\frac{4-4 \sin ^{2} \theta \cdot \cos ^{2} \theta}{4}=\frac{4-\sin ^{2}(2 \theta)}{4}$
As, we know, $0 \leq \sin ^{2} 2 \theta \leq 1$
$\therefore \quad A=\frac{4-0}{4}$ or $\frac{4-1}{4} \Rightarrow \frac{3}{4} \leq A \leq 1$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2018 (Phase 2)
If $\mathrm{A}=\{\lambda,(\lambda, \mu)\}$, then the power set of A is

Options:
A $\{\phi,\{\phi\},\{\lambda\},\{\lambda, \mu\}\}$
B $\{\phi,\{\lambda\},\{\{\lambda, \mu\}\},\{\lambda,\{\lambda, \mu\}\}\}$
C $\{\phi,\{\lambda\},\{\lambda, \mu\},\{\lambda,\{\lambda, \mu\}\}\}$
D $\{\{\lambda\},\{\lambda, \mu\},\{\lambda,\{\lambda, \mu\}\}\}$

The correct answer is:
$\{\phi,\{\lambda\},\{\{\lambda, \mu\}\},\{\lambda,\{\lambda, \mu\}\}\}$

Solution:
$\mathrm{A}=\{\lambda,\{\lambda, \mu\}$
Power set $=\{\phi,\{\lambda\},\{\{\lambda, \mu\}\},\{\lambda,\{\lambda, \mu\}\}\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2019 (Phase 1)
If a variable takes values $0,1,2,3, \ldots, \mathrm{n}$ with frequencies $1, \mathrm{C}(\mathrm{n}, 1), \mathrm{C}(\mathrm{n}, 2), \mathrm{C}(\mathrm{n}, 3), \ldots, \mathrm{C}(\mathrm{n}, \mathrm{n})$ respectively, then the
$\begin{array}{ll}\text { arithmetic mean is } & {}\end{array}$

Options:
A $2 \mathrm{n}$
B $\mathrm{n}+1$
C $\mathrm{n}$
D $\frac{\mathrm{n}}{2}$

The correct answer is:
$\mathrm{n}+1$

Solution:
The arithmetic mean is always between minimum and maximum value.
So, $\frac{\mathrm{n}}{2}$ is arithmetic mean.

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 2)
If a vertex of a triangle is $(1,1)$ and the midpoints of two sides of the triangle through this vertex are $(-1,2)$ and $(3,2)$, then the centroid of the triangle is

Options:
A $\left(-\frac{1}{3}, \frac{7}{3}\right)$
B $\left(-1, \frac{7}{3}\right)$
C $\left(\frac{1}{3}, \frac{7}{3}\right)$
D $\left(1, \frac{7}{3}\right)$

The correct answer is:
$\left(1, \frac{7}{3}\right)$

Solution:
Midpoint of $\mathrm{AB}=(-1,2)$
$\Rightarrow\left(\frac{\mathrm{x}_{2}+1}{2}, \frac{\mathrm{y}_{2}+1}{2}\right)=(-1,2)$
$\Rightarrow \frac{\mathrm{x}_{2}+1}{2}=-1 ; \frac{\mathrm{y}_{2}+1}{2}=2$
$\Rightarrow \mathrm{x}_{2}+1=-2 ; \mathrm{y}_{2}+1=4$
$\Rightarrow \mathrm{x}_{2}=-3, \mathrm{y}_{2}=3$
Midpoint of $\mathrm{AC}=(3,2)$
$\Rightarrow\left(\frac{\mathrm{x}_{1}+1}{2}, \frac{\mathrm{y}_{1}+1}{2}\right)=(3,2)$
$\Rightarrow \mathrm{x}_{1}+1=6 ; \mathrm{y}_{1}+1=4$
$\Rightarrow \mathrm{x}_{1}=5, \mathrm{y}_{1}=3$
So, vertices of triangle $\mathrm{ABC}$ are $(1,1),(-3,3),(5,3)$
So, centroid $=\left(\frac{1-3+5}{3}, \frac{1+3+3}{3}\right)=\left(\frac{3}{3}, \frac{7}{3}\right)=\left(1, \frac{7}{3}\right)$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 1)
If $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$ represents a joint equation of directrices
of the hyperbola $16 x^{2}-9 y^{2}=144$, then $g+f-c=$

Options:
A $-81$
B $-25$
C $81$
D $25$

The correct answer is:
$81$

Solution:
Equation of the hyperbola: $16 x^{2}-9 y^{2}=-144$
This can be rewritten in the following way:
$$
\frac{x^{2}}{9}-\frac{y^{2}}{16}=-1
$$
This is the standard equation of a hyperbola, where
$$
\begin{array}{l}
a^{2}=9 \text { and } b^{2}=16 \\
\Rightarrow a^{2}=b^{2}\left(e^{2}-1\right) \\
\Rightarrow 9=16\left(e^{2}-1\right) \\
\Rightarrow e^{2}-1=\frac{9}{16} \\
\Rightarrow e^{2}=\frac{25}{16} \\
\Rightarrow e=\frac{5}{4}
\end{array}
$$
Coordinates of foci are given by $(0, \pm a e)$, i.e.
$(0, \pm 5)$.
Equation of the directrices: $y=\pm \frac{a}{e}$
$\Rightarrow y=\pm \frac{4}{\frac{5}{4}}$ $\Rightarrow 5 y \pm 16=0$
Length of the latus rectum of the hyperbola $=\frac{2 a^{2}}{b}$
Length of the latus rectum $=\frac{2 \times 9}{4}=\frac{9}{2}$.

Topic: Mathematics, Hyperbola, MHT CET, MHT CET 2020 (15 Oct Shift 1)
If $a^{x}=b, b^{y}=c, c^{2}=a$, then what is the value of
$\frac{1}{(x y+y z+z x)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) ?$

Options:
A 0
B $a b c$
C 1
D $-1$

The correct answer is:
1

Solution:
Given that $a^{x}=b, b^{y}=c, c^{z}=a$
$\Rightarrow c^{2}=b^{y z}=a \Rightarrow b^{y z}=a^{x y z}=a$
$\Rightarrow x y z=1$
Now, $\frac{1}{(x y+y z+z x)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$
$=\frac{1}{(x y+y z+z x)}\left(\frac{x y+y z+z x}{x y z}\right)=\frac{1}{x y z}=1$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
If $\mathrm{A}=\mathrm{x} \in$
$\mathrm{B}=\left\{\mathrm{x} \in \mathbb{R}: \mathrm{x}^{2}+9 \mathrm{x}+14>0\right\}$, then which of the following
is are correct?
1.$(A \cap B)=(-2,1)$
2.$(\mathrm{A} \mathrm{B})=(-7,-2)$
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:
1 only

Solution:
$x^{2}+6 x-7 < 0$
$\Rightarrow(x+7)(x-1) < 0$
$\Rightarrow \mathrm{x}=(-7,1)$
$\Rightarrow \mathrm{A}=\{-6,-5,-4,-3,-2,-1,0\}$
$\Rightarrow x^{2}+9 x+14>0$
$\Rightarrow(x+7)(x+2)>0$
$\Rightarrow \mathrm{x}=(-\infty,-7) \cup(-2, \infty)$
So $A \cap B=(-2,1)$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 2)
If $\mathrm{A}=\left\{\mathrm{x} \in \mathrm{R}: \mathrm{x}^{2}+6 \mathrm{x}-7 < 0\right\}$ and
$\mathrm{B}=\left\{\mathrm{x} \in \mathrm{R}: \mathrm{x}^{2}+9 \mathrm{x}+14>0\right\}$, then which of the following
is/are correct?
1.$\mathrm{A} \cap \mathrm{B}=\{\mathrm{X} \in \mathrm{R}:-2 < \mathrm{x} < 1\}$
2.$A \backslash B=\{x \in R:-7 < x < -2\}$
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:
Both 1 and 2

Solution:
We have :
$x^{2}+6 x-7 < 0 \quad$ \& $\quad x^{2}+9 x+14>0$
$\Rightarrow(x-1)(x+7) < 0 \& \quad \Rightarrow(x+2)(x+7)>0$
$\Rightarrow x \in(-7,1) \quad \& \quad \Rightarrow x \in(-\infty,-7) \cup(-2, \infty)$
$A \cap B=\{x \in R:-2 < x < 1\} \rightarrow$ It is true.
$A \backslash B=A-B=\{x \in R:-7 < x < -2\} \rightarrow$ It is also true.

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 2)
If $a=x+\sqrt{x^{2}+1}$, then what is $x$ equal to?

Options:
A $(1 / 2)\left(a+a^{-1}\right)$
B $(1 / 2)\left(a-a^{-1}\right)$
C $a+a^{-1}$
D $a=a^{-}$

The correct answer is:
$(1 / 2)\left(a-a^{-1}\right)$

Solution:
$\begin{aligned} & a=x+\sqrt{x^{2}+1} \Rightarrow a-x=\sqrt{x^{2}+1} \\ & \Rightarrow x^{2}+1=(a-x)^{2} \Rightarrow x^{2}+1=a^{2}+x^{2}-2 a x \\ & \Rightarrow 2 a x=a^{2}-1 \Rightarrow 2 x=a-\frac{1}{a} \\ & \Rightarrow x=\frac{1}{2}\left(a-a^{-1}\right) \end{aligned}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 2)
If $A=\{x: x$ is a multiple of 3$\}$ and $B=\{x: x$ is a multiple of 4$\}$ and $C=\{x: x$ is a multiple of 12$\}$, then which one of the following
is a null set?

Options:
A $(A / B) \cup C$
B $(A / B) / C$
C $(A \cap B) \cap C$
D $(A \cap B) / C$

The correct answer is:
$(A \cap B) / C$

Solution:
$A=\{x$ $\mathrm{A}=\{3,6,9,12,15,18,24, \ldots .\}$
$\frac{\mathrm{A} \cap \mathrm{B}}{\mathrm{C}}$
$\mathrm{B}=$
$\mathrm{C}=\{\mathrm{x}: \mathrm{x}$ is a multiple of 12$\}$
$\mathrm{C}=\{12,24,36,48,60,72,84,96, \ldots . .\}$
$\mathrm{A} \cap \mathrm{B}=\{12,24, \ldots . .\}=\frac{\mathrm{A} \cap \mathrm{B}}{\mathrm{C}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
If $\mathrm{A}=\{\mathrm{x}: \mathrm{x}$ is amultiple of 2$\}, \mathrm{B}=\{\mathrm{x}: \mathrm{x}$ is a multiple of 5$\}$ and $\mathrm{C}=\{\mathrm{x}: \mathrm{x}$ is a multiple of 10$\}$, then $\mathrm{A} \cap(\mathrm{B} \cap \mathrm{C})$ is equal to

Options:
A $\mathrm{A}$
B $\bar{B}$
C C
D $\{\mathrm{x}: \mathrm{x}$ is a multiple of 100$\}$

The correct answer is:
C

Solution:
$\mathrm{A}=\{\mathrm{x}: \mathrm{x}$ ismultiple of 2$\}$ $\mathrm{B}=\{\mathrm{x}: \mathrm{x}$ ismultiple of 5$\}$
$C=\{x: x$ is multiple of 10
We know, multiples of 2 include multiples of 10 . $\therefore \mathrm{C} \subset \mathrm{A}$
Also, multiples of 5 include multiples of 10 . $\therefore \mathrm{C} \subset \mathrm{B}$
Also, $\mathrm{C}=\mathrm{A} \cap \mathrm{B}$
But $B=A \cap C, B \cap C=B$
$A \cap(B \cap C)=A \cap B=C$

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 2)
If $\mathrm{A}=\{\mathrm{x}, \mathrm{y}, \mathrm{B}=\{2,3\}, \mathrm{C}=\{3,4\}$, then what is the number
of elements in $\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})$?

Options:
A 2
B 4
C 6
D 8

The correct answer is:
6

Solution:
$\quad$ Let $B=\{2,3\}$ and $C=\{3,4\}$
Now, $\mathrm{B} \cup \mathrm{C}=\{2,3,4\}$ consider $\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})$
$=\{\mathrm{x}, \mathrm{y}\} \times\{2,3,4\}$
$=\{(\mathrm{x}, 2),(\mathrm{x}, 3),(\mathrm{x}, 4),(\mathrm{y}, 2),(\mathrm{y}, 3),(\mathrm{y}, 4)\}$
Hence, number of element in $\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=6$.

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{lll}\mathrm{x} & \mathrm{x}^{2} & 1+\mathrm{x}^{2} \\ \mathrm{y} & \mathrm{y}^{2} & 1+\mathrm{y}^{2} \\ \mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{2}\end{array}\right]$ where $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are distinct what is $\mid \mathrm{A}|$?

Options:
A 0
B $x^{2} y-y^{2} x+x y z$
C $(x-y)(y-z)(z-x)$
D $x y z$

The correct answer is:
$(x-y)(y-z)(z-x)$

Solution:
$A=\left[\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right]$
$|A|=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$
$|\mathrm{~A}|=\left|\begin{array}{ccc}\mathrm{x}-\mathrm{y} & (\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}) & (\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}) \\ \mathrm{y}-\mathrm{z} & (\mathrm{y}-\mathrm{z})(\mathrm{y}+\mathrm{z}) & (\mathrm{y}-\mathrm{z})(\mathrm{y}+\mathrm{z}) \\ \mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{2}\end{array}\right|$
$=(x-y)(y-z)\left|\begin{array}{ccc}1 & x+y & x+y \\ 1 & y+z & y+z \\ z & z^{2} & 1+z^{2}\end{array}\right|$
Applying $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}$
$|\mathrm{~A}|=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})\left|\begin{array}{ccc}1 & \mathrm{x}+\mathrm{y} & 0 \\ 1 & \mathrm{y}+\mathrm{z} & 0 \\ \mathrm{z} & \mathrm{z}^{2} & 1\end{array}\right|$
$=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})[1\{\mathrm{y}+\mathrm{z}-(\mathrm{x}+\mathrm{y})]$
$=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 1)
If $\mathrm{A}=\left\{\mathrm{x} \in \mathrm{Z}: \mathrm{x}^{2}-1=0\right\}$ and $\mathrm{B}=\left\{\mathrm{x} \in \mathrm{Z}: \mathrm{x}^{2}+\mathrm{x}+1=0\right\}$, where
Z is set of complex numbers, then what is $A \cap B$ equal to ?

Options:
A Null set
B $\left\{\frac{-1+\sqrt{3} \mathrm{i}}{2}, \frac{-1-\sqrt{3} \mathrm{i}}{2}\right\}$
C $\left\{\frac{-1+\sqrt{3} \mathrm{i}}{4}, \frac{-1-\sqrt{3} \mathrm{i}}{4}\right\}$
D $\left\{\frac{1+\sqrt{3} i}{2}, \frac{1-\sqrt{3} i}{2}\right\}$

The correct answer is:
$\left\{\frac{-1+\sqrt{3} \mathrm{i}}{2}, \frac{-1-\sqrt{3} \mathrm{i}}{2}\right\}$

Solution:
Given, $\mathrm{A}=\left\{\mathrm{x} \in \mathrm{z}: \mathrm{x}^{3}-1=0\right\}$
$\mathrm{B}=\left\{\mathrm{x} \in \mathrm{z}: \mathrm{x}^{2}+\mathrm{x}+1=0\right\}$
The roots of $\mathrm{x}^{3}-1=0$ are $1, \omega, \omega^{2}$ The roots of $x^{2}+x+1=0$ are $\omega, \omega^{2}$ $\therefore A \cap B=\left\{1, \omega, \omega^{2}\right\} \cap\left\{\omega, \omega^{2}\right\}=\left\{\omega, \omega^{2}\right\}$
$=\left\{\frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}\right\}$

Topic: Mathematics, Complex Number, NDA, NDA 2019 (Phase 1)
If $\mathrm{AB}=\left[\begin{array}{cc}4 & 11 \\ 4 & 5\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ll}3 & 2 \\ 1 & 2\end{array}\right]$, then what is the value of the determinant of the matrix $\mathrm{B}$?

Options:
A 4
B $-6$
C $-\frac{1}{4}$
D $-28$

The correct answer is:
$-6$

Solution:
Given that, $\mathrm{AB}=\left[\begin{array}{cc}4 & 11 \\ 4 & 5\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ll}3 & 2 \\ 1 & 2\end{array}\right]$
Let $B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}3 & 2 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]=\left[\begin{array}{cc}4 & 11 \\ 4 & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}3 a+2 c & 3 b+2 d \\ a+2 c & b+2 d\end{array}\right]=\left[\begin{array}{cc}4 & 11 \\ 4 & 5\end{array}\right]$
$\Rightarrow 3 \mathrm{a}+2 \mathrm{c}=4$ and $\mathrm{a}+2 \mathrm{c}=4$ ...(1)
and $3 b+2 d=11$ and $b+2 d=5$ ...(2)
From equation set $(1) \mathrm{a}=0$ and $\mathrm{c}=2$ and from equation $\operatorname{set}(2), \mathrm{b}=3$ and $\mathrm{d}=1$
$\Rightarrow B=\left[\begin{array}{ll}0 & 3 \\ 2 & 1\end{array}\right]$
Hence $|\mathrm{B}|=\left|\begin{array}{ll}0 & 3 \\ 2 & 1\end{array}\right|=0-6=-6$

Topic: Mathematics, Determinants, NDA, NDA 2007 (Phase 2)