datavorous/entrance-exam-dataset huggingface dataset:
Let $\vec{a}$ and $\vec{b}$ be the position vectors of $A$ and $B$ respectively.
If $\mathrm{C}$ is the point $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$, then which one of the following is correct?

Options:
A $\mathrm{C}$ is in between $\mathrm{A}$ and $\mathrm{B}$
B A is in between $\mathrm{C}$ and $\mathrm{B}$
C $\mathrm{B}$ is in between $\mathrm{A}$ and $\mathrm{C}$
D A, B, C are not collinear

The correct answer is:
A is in between $\mathrm{C}$ and $\mathrm{B}$

Solution:
As given : $\vec{a}$ and $\vec{b}$ are position vectors of $A$ and $B$
respectively and position vector of $\mathrm{C}$ is $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$
$\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}$, where $\mathrm{O}$ is the origin and
$\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{3 \mathrm{a}}-2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=2 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$
$\Rightarrow \overrightarrow{\mathrm{AC}}=2(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=-2(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})=-2 \overrightarrow{\mathrm{AB}}$
So, $\overrightarrow{\mathrm{AC}}$ is opposite to $\overrightarrow{\mathrm{AB}}$ so
A is between $\mathrm{C}$ and $\mathrm{B}$ and position vector of $\mathrm{C}$ shows an external division by $\mathrm{C}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
Let $a$ and $b$ be two distinct roots of a polynomial equation $f(x)=0$. Then there exists at least one root lying between $a$ and $b$ of the polynomial equation.

Options:
A $f(x)=0$
B $f^{\prime}(x)=0$
C $f^{\prime \prime}(x)=0$
D None of these

The correct answer is:
$f^{\prime}(x)=0$

Solution:
Rolle's theorem says between any two roots of a polynomial $f(x)$, there is always a root of its derivative $f^{\prime}(x)$
Therefore between $a$ and $b$. There exist at least one root of the polynomial equation $f^{\prime}(x)=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
Let $\mathrm{A}$ and $\mathrm{B}$ be two events with $\mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{6}$ and
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{12} .$ What is $\mathrm{P}(\mathrm{B} \mid \overline{\mathrm{A}})$ equal to?

Options:
A $\frac{1}{5}$
B $\frac{1}{7}$
C $\frac{1}{8}$
D $\frac{1}{10}$

The correct answer is:
$\frac{1}{8}$

Solution:
$\mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{6}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{12}$
$\mathrm{P}(\mathrm{B} \mid \overline{\mathrm{A}})=\frac{\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{A}})}{\mathrm{P}(\overline{\mathrm{A}})}$
Now, $\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{A}})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}-\frac{1}{12}=\frac{1}{12}$
$\mathrm{P}(\overline{\mathrm{A}})=1-\frac{1}{3}=\frac{2}{3}$
$\therefore \mathrm{P}(\mathrm{B} \mid \overline{\mathrm{A}})=\frac{\frac{1}{12}}{\frac{2}{3}}=\frac{1}{12} \times \frac{3}{2}=\frac{1}{8}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)
Let A and B be two invertible square matrices each of order $\mathrm{n}$. What is adj $(\mathrm{AB})$ equal to?

Options:
A $(\mathrm{adj} \mathrm{A})(\mathrm{adj} \mathrm{B})$
B $(\operatorname{adj} \mathrm{A})+(\operatorname{adj} \mathrm{B})$
C $(\operatorname{adj} \mathrm{A})-(\operatorname{adj} \mathrm{B})$
D $(\operatorname{adj} B)(\operatorname{adjA})$

The correct answer is:
$(\operatorname{adj} B)(\operatorname{adjA})$

Solution:
Since, $\mathrm{A}$ and $\mathrm{B}$ be two invertible square matrices each of order $\mathrm{n}$, then $(\mathrm{AB})^{-1}=\left(\mathrm{B}^{-1}\right)\left(\mathrm{A}^{-1}\right)$
$\Rightarrow \frac{\operatorname{adj}(\mathrm{AB})}{|\mathrm{AB}|}=\frac{\operatorname{adj} \mathrm{B}}{|\mathrm{B}|} \cdot \frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}$
Since $|\mathrm{AB}|=|\mathrm{B}||\mathrm{A}|$
$\operatorname{adj}(\mathrm{AB})=(\operatorname{adj} \mathrm{B})(\operatorname{adj} \mathrm{A})$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
Let $A$ and $B$ be two matrices such that $A B=A$ and $B A=B$. Which of the following statements are correct? [2014-II]
$1.$ $A^{2}=A$
$2.$ $\quad B^{2}=B$
$3.$ $\quad(A B)^{2}=A B$
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:
1,2 and 3

Solution:
We have, $\mathrm{AB}=\mathrm{A}$
$\therefore \quad \mathrm{A}^{2}=(\mathrm{AB}) \cdot(\mathrm{AB})=\mathrm{A} .(\mathrm{BA}) \mathrm{B}$
=ABB $\quad(\because \mathrm{BA}=\mathrm{B})$
$=\mathrm{AB}=\mathrm{A}$ $\quad(\because \mathrm{AB}=\mathrm{A})$
Also, $B^{2}=(B A) \cdot(B A)=B(A B) \cdot A$
$=$ B.A.A $\quad(\because \mathrm{AB}=\mathrm{A})$
$=$ B.A $=B$ $\quad(\because \mathrm{BA}=\mathrm{B})$
Again, $(\mathrm{AB})^{2}=(\mathrm{AB}) \cdot(\mathrm{AB})=\mathrm{A}$. (BA) B
$=$ A.B.B $\quad(\because \mathrm{BA}=\mathrm{B})$
$=$ A. $\mathrm{B}=\mathrm{A}$ $\quad(\because \mathrm{AB}=\mathrm{A})$

Topic: Mathematics, Matrices, NDA, NDA 2014 (Phase 2)
Let A and $\mathrm{B}$ be two matrices such that $\mathrm{AB}$ is defined. If $\mathrm{AB}=0$, then which one of the following can be definitely concluded?

Options:
A $\mathrm{A}=0$ or $\mathrm{B}=0$
B $\mathrm{A}=0$ and $\mathrm{B}=0$
C A and B are non-zero square matrices
D A and B cannot both be non-singular

The correct answer is:
A and B are non-zero square matrices

Solution:
Since, $\mathrm{AB}$ is defined, neither A nor $\mathrm{B}$ is singular i.e., they are non-zero matrix and if $\mathrm{AB}=0$ both $\mathrm{A}$ and $\mathrm{B}$ are square matrix. So, Aand B are non-zerosquare matrices.

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
Let $A$ and $B$ be two non-empty subsets of a set $X$. If $(A-B) \cup(B-A)=A \cup B$, then which one of the following is correct?

Options:
A $\mathrm{A} \subset \mathrm{B}$
B $\mathrm{A} \subset(\mathrm{X}-\mathrm{B})$
C $\mathrm{A}=\mathrm{B}$
D $\mathrm{B} \subset \mathrm{A}$

The correct answer is:
$\mathrm{A} \subset(\mathrm{X}-\mathrm{B})$

Solution:
$\mathrm{A}$ and $\mathrm{B}$ are subsets of $\mathrm{X}$ and its, Venn diagram is shown.
$\mathrm{A}=\mathrm{B}$ indicates region a, $\mathrm{B}-\mathrm{A}$ indicates $\mathrm{c}, \mathrm{A} \cup \mathrm{B}$
indicates $\mathrm{a}, \mathrm{b}, \mathrm{c}$. If $a, c=a, b, c$ indicates, the $b$ is zero. $A$ and $B$ are mutually exclusive. So, when $(\mathrm{A}-\mathrm{B}) \cup(\mathrm{B}-\mathrm{A})=\mathrm{A} \cup \mathrm{B}$
$\mathrm{A} \subset(\mathrm{X}-\mathrm{B})$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Let $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be two unit vectors and $\alpha$ be the angle between
them. If $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$ is also the unit vectors, then what is the
value of $\alpha ?$

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{3}$
C $\frac{2 \pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{2 \pi}{3}$

Solution:
Let $\vec{a}$ and $\vec{b}$ be two unit vectors.
$\therefore \quad|a|=1$ and $|b|=1$
Since, $\alpha$ is the angle between $\vec{a}$ and $\vec{b}$
$\therefore \quad \cos \alpha=\frac{\vec{a} \cdot \vec{b}}{|a \| b|}$
$\cos \alpha=\frac{\vec{a} \cdot \vec{b}}{1}$
$\cos \alpha=\vec{a} \cdot \vec{b}$
Now, $|\vec{a}+\vec{b}|=1 \quad(\because \vec{a}+\vec{b}$ is unit vector $)$
Squaring both sides
$\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=1$
$\Rightarrow 1+1+2 \cos \alpha=1$
$\Rightarrow 2 \cos \alpha=-1$
$\Rightarrow \quad \cos \alpha=-\frac{1}{2}=\cos \frac{2 \pi}{3}$
$\Rightarrow \alpha=\frac{2 \pi}{3}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 1)
Let $\mathrm{a}, \mathrm{b} \in\{1,2,3\}$. What is the number of equations

Options:
A 1
B 2
C 5
D 3

The correct answer is:
3

Solution:
The given equation is $a x^{2}+b x+1=0$ This equation has real roots When discriminant $\geq 0$ $b^{2}-4 a \geq 0$
$\Rightarrow b^{2} \geq 4 a$
$\mathrm{a}, \mathrm{b}$ has to be selected from, three numbers, so total selections are possible when $(\mathrm{a}, \mathrm{b})$ are $(1,2),(1,3)$ and
$(2,3) .$
Thus, the number of equations of the form $a x^{2}$ $\mathrm{bx}+1=0$ having real root is 3

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 2)
Let $a, b$ and $c$ be the distinct non-negative numbers. If the vectors $a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+c \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}, c \hat{\mathbf{i}}+c \hat{\mathbf{j}}+b \hat{\mathbf{k}}$ lie on a plane, then
which one of the following is correct ?

Options:
A $c$ is the arithmetic mean of a and $b$
B $c$ is the geometric mean of a and $b$
C $c$ is the harmonic mean of a and $b$
D $c$ is equal to zero

The correct answer is:
$c$ is the geometric mean of a and $b$

Solution:
Given $a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$. and $c \hat{i}+c \hat{j}+b \hat{k}$ lie on a
plane.
$\Rightarrow$ vectors $a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$ are
coplanar.
$\therefore \quad\left|\begin{array}{lll}a & a & c \\ 1 & 0 & 1 \\ c & c & b\end{array}\right|=0$
$\Rightarrow a(-c)-a(b-c)+c(c)=0$
$\Rightarrow \quad-a c-a b+a c+c^{2}=0$
$\Rightarrow \quad c^{2}=a b$
$\Rightarrow c$ is the geometric mean of $a$ and $b$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
Let $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ be three mutually exclusive and exhaustive events associated with a random experiment. If $\mathrm{P}(\mathrm{B})=1.5$ $\mathrm{P}(\mathrm{A})$ and $\mathrm{P}(\mathrm{C})=0.5 \mathrm{P}(\mathrm{B})$, then $\mathrm{P}(\mathrm{A} 0$ is equal to

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

The correct answer is:
$\frac{4}{13}$

Solution:
As A, B and C are mutually exclusive and exhaustive event. $\therefore \quad \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})=1$
$\Rightarrow \mathrm{P}(\mathrm{A})+\frac{3}{2} \mathrm{P}(\mathrm{A})+\frac{1}{2} \times \frac{3}{2} \mathrm{P}(\mathrm{A})=1$
$\Rightarrow \quad \frac{13}{4} \mathrm{P}(\mathrm{A})=1$
$\Rightarrow \quad \mathrm{P}(\mathrm{A})=\frac{4}{13}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 2)
Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ be three mutually perpendicular vectors each
of unit magnitud. If $\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}, \quad \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$ and
$\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$, then which one of the following is correct?

Options:
A $|\overrightarrow{\mathrm{A}}|>|\overrightarrow{\mathrm{B}}|>|\overrightarrow{\mathrm{C}}|$
B $|\overrightarrow{\mathrm{A}}|=|\overrightarrow{\mathrm{B}}| \neq|\overrightarrow{\mathrm{C}}|$
C $|\overrightarrow{\mathrm{A}}|=|\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{C}}|$
D $|\overrightarrow{\mathrm{A}}| \neq|\overrightarrow{\mathrm{B}}| \neq|\overrightarrow{\mathrm{C}}|$

The correct answer is:
$|\overrightarrow{\mathrm{A}}|=|\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{C}}|$

Solution:
For simplicity let us take $\vec{a}, \vec{b}, \vec{c}$ as $\hat{i}, \hat{j}, \hat{k}$
Now magnitude of $\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{C}}$ will be $\sqrt{3}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\overrightarrow{\mathrm{a}}|=10,|\overrightarrow{\mathrm{b}}|=6$ and $|\overrightarrow{\mathrm{c}}|=14$
What is a. $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}$. $\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}$. a. equal to?

Options:
A $-332$
B $-166$
C 0
D 166

The correct answer is:
$-166$

Solution:
We have $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$
So $|\vec{a}+\vec{b}+\vec{c}|=0$
$\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow 0=(10)^{2}+(6)^{2}+(14)^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow 0=100+36+196+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow-\frac{332}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$
$\Rightarrow-166=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 2)
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\overrightarrow{\mathrm{a}}|=10,|\overrightarrow{\mathrm{b}}|=6$ and $|\overrightarrow{\mathrm{c}}|=14$
What is the angle between $\vec{a}$ and $\vec{b}$ ?

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
Since $\vec{a}+\vec{b}+\vec{c}=0$
$\Rightarrow \vec{a}+\vec{b}=-\vec{c}$
$\Rightarrow|\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|$
$\Rightarrow|\vec{a}+\vec{b}|^{2}=|\vec{c}|^{2}$
$\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2(\vec{a} \cdot \vec{b})=|\vec{c}|^{2}$
$\Rightarrow(10)^{2}+(6)^{2}+2(\vec{a} \cdot \vec{b})=(14)^{2}$
$\Rightarrow 2(\vec{a} \cdot \vec{b})=60$
$\Rightarrow \vec{a} \cdot \vec{b}=30$
$\Rightarrow \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{30}{10 \times 6}=\frac{1}{2}$
$\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow \theta=60^{\circ}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 2)
Let $\hat{a}, \hat{\mathrm{b}}$ be two unit vectors and $\theta$ be the angle between them.
What is $\cos \left(\frac{\theta}{2}\right)$ equal to?

Options:
A $\frac{|\hat{a}-\hat{b}|}{2}$
B $\frac{|\hat{a}+\hat{b}|}{2}$
C $\frac{|\hat{a}-\hat{b}|}{4}$
D $\frac{|\hat{a}+\hat{b}|}{4}$

The correct answer is:
$\frac{|\hat{a}+\hat{b}|}{2}$

Solution:
$\vec{a}$ and $\vec{b}$ are two unit vectors. . Hence.
$\begin{aligned}|\vec{a}+\vec{b}|^{2} &=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b}) \\ &=\hat{a} \cdot \hat{a}+\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{a}+\hat{b} \cdot \hat{b} \\ &=\hat{a} \cdot \hat{a}+\hat{b} \cdot \hat{b}+2 \hat{a} \cdot \hat{b} \\ &=|\hat{a}|^{2}+2|\hat{a}||\hat{b}| \cos \theta+|\hat{b}||\hat{b}| \end{aligned}$
$=1+2 \cos \theta+1 \quad[\because \hat{b}, \hat{a}$ are unit vector $]$
$=2(1+\cos \theta)$
$|\hat{a}+\hat{b}|^{2}=2 \cdot 2 \cos ^{2} \frac{\theta}{2}$
$\cos \frac{\theta}{2}=\frac{|\hat{a}+\hat{b}|}{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 1)
Let $\hat{a}, \hat{\mathrm{b}}$ be two unit vectors and $\theta$ be the angle between them.
What is $\sin \left(\frac{\theta}{2}\right)$ equal to?

Options:
A $\frac{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}{2}$
B $\frac{|\hat{a}+\hat{b}|}{2}$
C $\frac{|\hat{a}-\hat{b}|}{4}$
D $\frac{|\hat{a}+\hat{b}|}{4}$

The correct answer is:
$\frac{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}{2}$

Solution:
$\begin{aligned}|\hat{a}-\hat{b}|^{2} &=(\hat{a}-\hat{b}) \cdot(\hat{a}-\hat{b}) \\ &=\hat{a} \cdot \hat{a}-\hat{b} \cdot \hat{a}-\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{b} \\ &=|\hat{a}|^{2}-2|\hat{a}||\hat{b}| \cos \theta+|\hat{b}|^{2} \\ &=2-2 \cos \theta \\ &=2(1-\cos \theta) \\|\hat{a}-\hat{b}|^{2} &=2 \cdot 2 \sin ^{2} \frac{\theta}{2} \\ \sin \frac{\theta}{2}=\frac{|\hat{a}-\hat{b}|}{2} \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 1)
Let $a, b, c$ be in AP.
Consider the following statements:
1) $\frac{1}{a b}, \frac{1}{c a}$ and $\frac{1}{b c}$ are in AP.
2) $\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}$ and $\frac{1}{\sqrt{a}+\sqrt{b}}$ are in AP.
Which of the statements given above 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:
Let $\frac{1}{a b}, \frac{1}{c a}, \frac{1}{b c}$ are in AP.
$\Rightarrow \quad \frac{1}{c a}-\frac{1}{a b}=\frac{1}{b c}-\frac{1}{c a}$
$\Rightarrow \frac{1}{a}\left(\frac{1}{c}-\frac{1}{b}\right)=\frac{1}{c}\left(\frac{1}{b}-\frac{1}{a}\right)$
$\Rightarrow \frac{b-c}{a b c}=\frac{a-b}{a b c}$
$\Rightarrow b-c=a-b \Rightarrow 2 b=a+c$
$\Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{AP}$. Which is true
Now, $\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}$ are in A.P.
$\therefore \frac{2}{\sqrt{c}+\sqrt{a}}=\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{a}+\sqrt{b}}$
$\Rightarrow 2(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b})$
$=(\sqrt{c}+\sqrt{a})(\sqrt{a}+2 \sqrt{b}+\sqrt{c})$
$\Rightarrow \quad 2(\sqrt{a b}+b+\sqrt{a c}+\sqrt{b c})=\sqrt{a c}+2 \sqrt{b c}+c+a$
$+2 \sqrt{a b}+\sqrt{a c}$
$\Rightarrow \quad 2 \sqrt{a b}+2 b+2 \sqrt{a c}+2 \sqrt{b c}$
$=2 \sqrt{a c}+2 \sqrt{b c}+2 \sqrt{a b}+c+a$
$\Rightarrow \quad 2 b=a+c$
$\Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{A} . \mathrm{P}$. Which is true.
Hence. both the statements are correct.

Topic: Mathematics, Sequences and Series, NDA, NDA 2010 (Phase 1)
Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ be non-coplanar vectors and $\overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}}{[\overrightarrow{\mathrm{abc}}]}$,
$\overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}}{[\overrightarrow{\mathrm{abc}}]}, \overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{[\overrightarrow{\mathrm{abc}}]}$
What is the value of
$(\vec{a}-\vec{b}-\vec{c}) \cdot \vec{p}+(\vec{b}-\vec{c}-\vec{a}) \cdot \vec{q}+(\vec{c}-\vec{a}-\vec{b}) \cdot \vec{r} ?$

Options:
A 0
B $-3$
C 3
D $-9$

The correct answer is:
3

Solution:
As given $\overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}}{[\overrightarrow{\mathrm{abc}}]}, \overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}}{[\overrightarrow{\mathrm{abc}}]}$, and $\overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{[\overrightarrow{\mathrm{abc}}]}$
$\begin{aligned} \therefore &(\vec{a}-\vec{b}-\vec{c}) \cdot \vec{p}+(\vec{b}-\vec{c}-\vec{a}) \cdot \vec{q}+(\vec{c}-\vec{a}-\vec{b}) \cdot \vec{r} \\ &=\frac{a \cdot(\vec{b} \times \vec{c})}{[\overrightarrow{a b c}]}+\frac{\vec{b} \cdot(\vec{c} \times \vec{a})}{[a \vec{b} c]}+\frac{\vec{c} \cdot(\vec{a} \times \vec{b})}{[\overrightarrow{a b c}]} \end{aligned}$
$[$ Since $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=0, \overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=0, \overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})=0$
$\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})=0, \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=0$ and $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=0]$
$=\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}+\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}+\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}=3$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of points $A, B, C$ respectively. Under which one of the following conditions are the points $A, B, C$ collinear?

Options:
A $\quad \vec{a} \times \vec{b}=\overrightarrow{0}$
B $\vec{b} \times \vec{c}$ is parallel to $\vec{a} \times \vec{b}$
C $\vec{a} \times \vec{b}$ is perpendicular to $\vec{b} \times \vec{c}$
D $(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})=\overrightarrow{0}$

The correct answer is:
$(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})=\overrightarrow{0}$

Solution:
Points A, B and $\mathrm{C}$ are collinear, if
$(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})+(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})=\overrightarrow{0}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 1)
Let \(a, b \in \mathbf{R}-\{0\}\), and \(I_2\) be the identity matrix of order 2. Then the rank of the (block) matrix \(\left[\begin{array}{ll}a I_2 & b_2 \\ a I_2 & b I_2\end{array}\right]\) is

Options:
A 2
B 1
C 4
D 3

The correct answer is:
2

Solution:
Since \(a, b \in \mathbf{R}-\{0\}\) and \(I_2\) is a identify matrix of order 2 .
Then the block matrix \(\left[\begin{array}{ll}a I_2 & b I_2 \\ a I_2 & b I_2\end{array}\right]\) is a non-singular matrix, so the rank of the given block matrix is 2.

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 1)
Let A be a square matrix of order $\mathrm{n} \times \mathrm{n}$ where $\mathrm{n} \geq 2 .$ Let $\mathrm{B}$ be a matrix obtained from A with first and second rows interchanged. Then which one of the following is correct?

Options:
A $\operatorname{det} \mathrm{A}=\operatorname{det} \mathrm{B}$
B $\operatorname{det} \mathrm{A}=-\operatorname{det} \mathrm{B}$
C $\mathrm{A}=\mathrm{B}$
D $A=-B$

The correct answer is:
$\operatorname{det} \mathrm{A}=-\operatorname{det} \mathrm{B}$

Solution:
A be a square matrix of order $\mathrm{n} \times \mathrm{n}$ where $\mathrm{n} \geq 2 . \mathrm{B}$ be a matrix obtained from A with first and second rows interchanged. Then, det $A=-$ det $B$. Since interchanging any two rows makes the sign change with same value.

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 1)
Let $\mathrm{A}$ be an $\mathrm{m} \times \mathrm{n}$ matrix. Under which one of the following conditions does $\mathrm{A}^{-1}$ exist?

Options:
A $\mathrm{m}=\mathrm{n}$ only
B $\mathrm{m}=\mathrm{n}$ and $\operatorname{det} \mathrm{A} \neq 0$
C $\mathrm{m}=\mathrm{n}$ and $\mathrm{det} \mathrm{A}=0$
D $m \neq n$

The correct answer is:
$\mathrm{m}=\mathrm{n}$ and $\operatorname{det} \mathrm{A} \neq 0$

Solution:
Let a be an $\mathrm{m} \times \mathrm{n}$ matrix, then $\mathrm{A}^{-1}$ will exist if $\mathrm{m}=\mathrm{n}$ since only square matrix has determinant and det A $\neq 0$
$\left[\right.$ Since $\left.A^{-1}=\frac{\text { adjA }}{|A|}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 1)
Let $A$ be an $n \times n$ matrix. If det $(\lambda A)=\lambda^{s}$ det $(A)$, what is the value of $s ?$

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

The correct answer is:
$n$

Solution:
We know, if $A$ is an $n \times n$ matrix, then
$\operatorname{det}(\lambda A)=\lambda^{s} \operatorname{det}(A)$
But given det $(\lambda A)=\lambda^{s} \operatorname{det} A$
$\Rightarrow s=n$

Topic: Mathematics, Determinants, NDA, NDA 2010 (Phase 1)
Let a die be loaded in such a way that even faces are twice likely to occur as the odd faces. What is the probability that a prime number will show up when the die is tossed?

Options:
A $\frac{1}{3}$
B $\frac{2}{3}$
C $\frac{4}{9}$
D $\frac{5}{9}$

The correct answer is:
$\frac{4}{9}$

Solution:
Possible prime numbers on the dice are 2,3 and 5 .
Probability of getting prime number $=\frac{2}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{3}$.
$=\frac{2}{9}+\frac{2}{9}$
$=\frac{4}{9} .$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)