datavorous/entrance-exam-dataset huggingface dataset:
Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}$, where $\vec{c}$ is parallel to
$\vec{a}$ and $\vec{d}$ is perpendicular to $\vec{a}$
If $\overrightarrow{\mathrm{d}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$, then which of the following equations
is/are correct?
$1.$ $\quad \mathrm{y}-\mathrm{x}=4$
$2.$ $\quad 2 z-3=0$
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:
Since $z=4$ and $x=\frac{3}{2}, y=-\frac{3}{2}$.
So, neither 1 nor 2 is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 2)
Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}$, where $\vec{c}$ is parallel to
$\vec{a}$ and $\vec{d}$ is perpendicular to $\vec{a}$
What is $\vec{c}$ equal to?

Options:
A $\frac{3(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{2}$
B $\frac{2(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{3}$
C $\frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{2}$
D $\frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{3}$

The correct answer is:
$\frac{3(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{2}$

Solution:
Since $\vec{c}$ is parallel to $\vec{a}$
$\vec{c}=\lambda \vec{a}$
Now $\vec{b}=\vec{c}+\vec{d}=\lambda \vec{a}+\vec{d}$
$=\lambda(\hat{i}+\hat{j})+x \hat{i}+y \hat{j}+z \hat{k}$
$3 \hat{i}+4 \hat{k}=(\lambda+x) \hat{i}+(\lambda+y) \hat{j}+z \hat{k}$
Comparing we get $z=4, \lambda+y=0, \lambda+x=3 \Rightarrow-y+x=3 \quad($ From $(1))$
$\Rightarrow \lambda=-y \quad \ldots(1) \Rightarrow x-y=3$
Now $\vec{d}$ is $\perp \mathrm{r}$ to $\vec{a}$
So, $\cos \theta=0$
$\Rightarrow x+y=0$
Solving $(2)$ and $(3)$ we get
$2 x=3$
$\Rightarrow x=\frac{3}{2}, y=-\frac{3}{2}$
$\Rightarrow \vec{c}=\lambda(\vec{a})=\frac{3}{2}(\hat{i}+\hat{j})$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 2)
Let $\mathrm{A}=\{(\mathrm{n}, 2 \mathrm{n}): \mathrm{n} \in \mathrm{N}\}$ and $\mathrm{B}=\{(2 \mathrm{n}, 3 \mathrm{n}): \mathrm{n} \in \mathrm{N}\} .$ What is A $\cap$ B equal to ?

Options:
A $\{(\mathrm{n}, 6 \mathrm{n}): \mathrm{n} \in \mathrm{N}\}$
B $\{(2 n, 6 n): n \in N\}$
C $\{(\mathrm{n}, 3 \mathrm{n}): \mathrm{n} \in \mathrm{N}\}$
D $\phi$

The correct answer is:
$\phi$

Solution:
$\mathrm{A}=\{(\mathrm{n}, 2 \mathrm{n}): \mathrm{n} \in \mathrm{N}\}$ and $\mathrm{B}=\{(2 \mathrm{n}, 3 \mathrm{n})\}: \mathrm{n} \in \mathrm{N}$
Listing few members of each set $\mathrm{A}=\{(1,2),(2,4),(3,6), \ldots\}$
$\mathrm{B}=\{(2,3),(4,6),(6,9) \ldots \ldots\}$
There is no member common to both these sets, hence $A \cap B=\phi$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Let a vector $\bar{r}$ make angle $60^{\circ}, 30^{\circ}$ with $\mathrm{x}$ and y-axes respectively.
What angle does $\bar{r}$ make with $z$ -axis?

Options:
A $\quad 30^{\circ}$
B $60^{\circ}$
C $90^{\circ}$
D $120^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\cos ^{2} 60^{\circ}+\cos ^{2} 30^{\circ}+\cos ^{2} \gamma=1$
$\frac{1}{4}+\frac{3}{4}+\cos ^{2} \gamma=1$
$\cos ^{2} \gamma=0 \Rightarrow \gamma=90^{\circ}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
Let a vector $\bar{r}$ make angle $60^{\circ}, 30^{\circ}$ with $\mathrm{x}$ and y-axes respectively.
What are the direction cosines of $\bar{r}$ ?

Options:
A $\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right\rangle$
B $\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right\rangle$
C $\left\langle\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right\rangle$
D $\left\langle-\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right\rangle$

The correct answer is:
$\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right\rangle$

Solution:
$\left.\mathrm{r}= < 1, \mathrm{~m}, \mathrm{n}>; \mathrm{r}= < \cos 60^{\circ}, \cos 30^{\circ}, \cos 90^{\circ}\right\rangle$
Direction cosines of $\overrightarrow{\mathrm{r}}= < \frac{1}{2}, \frac{\sqrt{3}}{2}, 0>$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
Let $a x^{3}+b x^{2}+c x+d=\left|\begin{array}{lcc}x+1 & 2 x & 3 x \\ 2 x+3 & x+1 & x \\ 2-x & 3 x+4 & 5 x-1\end{array}\right|$ then
What is the value of $a+b+c+d$ ?

Options:
A 62
B 63
C 65
D 68

The correct answer is:
63

Solution:
$a+b+c+d=63$

Topic: Mathematics, Determinants, NDA, NDA 2016 (Phase 2)
Let $a x^{3}+b x^{2}+c x+d=\left|\begin{array}{lcc}x+1 & 2 x & 3 x \\ 2 x+3 & x+1 & x \\ 2-x & 3 x+4 & 5 x-1\end{array}\right|$ then
What is the value of $\mathrm{c}$ ?

Options:
A $-1$
B 34
C 35
D 50

The correct answer is:
35

Solution:
$a x^{3}+b x^{2}+c x+d$
$=(x+1)[(x+1)(5 x-1)-x(3 x+4)]-2 x[(2 x+3)(5 x-1)$
$-x(2-x)]+3 x[(2 x+3)(3 x+4)-(2-x)(x+1)]$
$\Rightarrow a x^{3}+b x^{2}+c x+d=x^{3}+28 x^{2}+35 x-1$
$\Rightarrow c=35$

Topic: Mathematics, Determinants, NDA, NDA 2016 (Phase 2)
Let $\mathrm{A}=(\mathrm{x} \in \mathrm{R}:-1 \leq \mathrm{x} \leq 1)$ and $\mathrm{S}$ be the subset of $\mathrm{A} \times \mathrm{B}$
defined by $\mathrm{S}=\left[(\mathrm{x}, \mathrm{y}) \in \mathrm{A} \times \mathrm{B}: \mathrm{x}^{2}+\mathrm{y}^{2}=1\right]$
Which one of the following is correct?

Options:
A $\mathrm{S}$ is a one-one function from A into $\mathrm{B}$
B $\mathrm{S}$ is a many-one function from $\mathrm{A}$ into $\mathrm{B}$
C $\mathrm{S}$ is a bijective mapping from A into $\mathrm{B}$
D $\mathrm{S}$ is not a function

The correct answer is:
$\mathrm{S}$ is not a function

Solution:
$\mathrm{S}$ is not a function (By vertical line test)

Topic: Mathematics, Functions, NDA, NDA 2018 (Phase 2)
Let $\mathrm{A}=\{\mathrm{x} \in \mathrm{R} \mid-9 \leq \mathrm{x} < 4\} ; \mathrm{B}=\{\mathrm{x} \in \mathrm{R} \mid-13 < \mathrm{x} \leq 5\}$ and
$\mathrm{C}=\{\mathrm{x} \in \mathrm{R} \mid-7 \leq \mathrm{x} \leq 8\}$
Then, which one of the following is correct?

Options:
A $-9 \in(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$
B $-7 \in(A \cap B \cap C)$
C $4 \in(A \cap B \cap C)$
D $5 \in(A \cap B \cap C)$

The correct answer is:
$-7 \in(A \cap B \cap C)$

Solution:
Given sets in set builder form are:
$A=\{x \in R:-9 \leq x < 4\}$
$B=\{x \in R:-13 < x 5\}$
and $C=\{x \in R:-7 \leq x \leq 8\}$
$\Rightarrow A \cap B \cap C=\{x \in R:-7 \leq x < 4\}$
Hence, $-7 \in(A \cap B \cap C)$

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 1)
Let $A=\{x \in R \mid x \geq 0\}$.
A function $f: A \rightarrow A$ is defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$.
Which one of the following is correct?

Options:
A The function does not have inverse $\quad[2012-I I]$
B $\mathrm{f}$ is its own inverse
C The function has an inverse but $\mathrm{f}$ is not its own inverse
D None of the above

The correct answer is:
The function has an inverse but $\mathrm{f}$ is not its own inverse

Solution:
Let $f: A \rightarrow$ A defined as $f(x)=x^{2}$. Let $f(x)=y \quad \therefore x=f^{-1}(y)$
Now $\Rightarrow x^{2}=y$
$x=\sqrt{y}$
Thus, $f^{-1}(y)=\sqrt{y}$
Hence, the function has an inverse but $\mathrm{f}$ is not its own inverse.

Topic: Mathematics, Functions, NDA, NDA 2012 (Phase 2)
Let $A=\{x \in W$, the set of whole numbers and $x < 3\}$, $\mathrm{B}=\{\mathrm{x} \in \mathrm{N}$, the set of natural numbers and $2 \leq \mathrm{x} < 4\}$ and
$\mathrm{C}=\{3,4\}$, then how many elements will $(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}$
contain?

Options:
A 6
B 8
C 10
D 12

The correct answer is:
8

Solution:
We have $\mathrm{A}=\{0,1,2\}$
$\mathrm{B}=\{2,3\}$
$\mathrm{C}=\{3,4\}$
$\mathrm{A} \cup \mathrm{B}=\{0,1,2,3\}$
$(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}=\{(0,3),(0,4),(1,3),(1,4) ;(2,3)$
$(2,4),(3,3) ;(3,4)\}$
$\therefore \mathrm{n}[(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}]=8$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 2)
Let $A=\{x \mid x \leq 9, x \in N\} .$ Let $B=\{a, b, c\}$ be the subset of $A$ where $(a+b+c)$ is a multiple of $3 .$ What is the largest possible number of subsets like B ?

Options:
A 12
B 21
C 27
D 30

The correct answer is:
30

Solution:
Given
$\mathrm{A}=\{\mathrm{x}: \mathrm{x} \leq 9, \mathrm{x} \in \mathrm{N}\}=\{1,2,3,4,5,6,7,8,9\}$
Total possible multiple of 3 are
$3,6,9,12,15,18,21,24,27$
But 3 and 27 are not possible because 3 and 27 can not be express as such that $a+b+c$ is multiple of 3 $6 \rightarrow 1+2+3$
$9 \rightarrow 2+3+4,5+3+1,6+2+1$
$12 \rightarrow 9+2+1,8+3+1,7+1+4,7+2+3$
$6+4+2,6+5+1,5+4+3$
$15 \rightarrow 9+4+2,9+5+1,8+6+1,8+5+2$,
$8+4+3,7+6+2,7+5+3,6+5+4$
$18 \rightarrow 9+8+1,9+7+2,9+6+3$
$9+5+4,8+7+3,8+6+4,7+6+5$
$21 \rightarrow 9+8+4,9+7+5,8+7+6$
$24 \rightarrow 9+8+7$
Hence, total largest possible subsets are 30 .

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
Let $\mathrm{A}=\{\mathrm{x}: \mathrm{x}$ is a square of anatural number and $\mathrm{x}$ is less than $100\}$ and $B$ is a set of even natural numbers. What is the cardinality of $\mathrm{A} \cap \mathrm{B}$?

Options:
A 4
B 5
C 9
D None of the above

The correct answer is:
4

Solution:
Here $\mathrm{A}=\{1,4,9,16,25,36,49,64,81\}$
$\mathrm{B}=$ \{even natural numbers $\mathrm{A} \cap \mathrm{B}=\{4,16,36,64\}$
So, cardinal ity of $\mathrm{A} \cap \mathrm{B}=4$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
Let
$A=\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right], B=\left[\begin{array}{c}2 \\ -1\end{array}\right]$ and $C=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
If $A B=C$, then what is $A^{2}$ equal to?

Options:
A $\left[\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]$
B $\left[\begin{array}{cc}-10 & 5 \\ 4 & 24\end{array}\right]$
C $\left[\begin{array}{cc}-5 & -6 \\ -4 & -20\end{array}\right]$
D $\left[\begin{array}{ll}-5 & -7 \\ -5 & 20\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]$

Solution:
$A=\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right]$
$B=\left[\begin{array}{c}2 \\ -1\end{array}\right]$ and $C=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
Here $\mathrm{AB}=\mathrm{C}$
$\therefore \quad\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & \mathrm{y} \\ 2 \mathrm{x} & \mathrm{x}-\mathrm{y}\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}2(\mathrm{x}+\mathrm{y}) & -\mathrm{y} \\ 4 \mathrm{x} & -\mathrm{x}+\mathrm{y}\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
$2 x+y=3$ ...(i)
$3 x+y=2$ ...(ii)
From equations (i) and (ii), we get $x=-1$ and $y=5$
$\therefore A=\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]$
Now, $A^{2}=\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]\left[\begin{array}{cc}4 & 5 \\ -2 & -6\end{array}\right]$
$=\left[\begin{array}{cc}16-10 & 20-30 \\ -8+12 & -10+36\end{array}\right]=\left[\begin{array}{cc}6 & -10 \\ 4 & 26\end{array}\right]$
$\therefore$ Option (a) is correct.

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 1)
Let $A=\{x, y, z\}$ and $B=\{p, q, r, s\} .$ What is the number of distinct relations from B to A?

Options:
A 4096
B 4094
C 128
D 126

The correct answer is:
4096

Solution:
$\mathrm{A}=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}$
$\mathrm{B}=(\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{s})$
$\mathrm{n}(\mathrm{A})=3$
$\mathrm{n}(\mathrm{B})=4$
Number of distinct relations $=2^{n(A) \times n(B)}=2^{3 \times 4}=2^{12}=4096$
$\therefore$ option (a) is correct.

Topic: Mathematics, Permutation Combination, NDA, NDA 2015 (Phase 1)
Let $\mathrm{a}, \mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{b}$ be in $\mathrm{AP}$, where $\mathrm{x}+\mathrm{y}+\mathrm{z}=15$. Let $\mathrm{a}, \mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{b}$ be
in HP, where $\mathrm{p}^{-1}+\mathrm{q}^{-1}+\mathrm{r}^{-1}=\frac{5}{3}$
What is the value of ab?

Options:
A 10
B 9
C 8
D 6

The correct answer is:
9

Solution:
$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l)$
$\Rightarrow \mathrm{a}+\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{b}=\frac{5}{2}(a+b)$
$a+b+15=\frac{5}{2}(a+b)$
$\Rightarrow \mathrm{a}+\mathrm{b}=10$ ...(1)
$\& \frac{1}{a}+\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{b}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\Rightarrow \quad \frac{1}{a}+\frac{1}{b}+\frac{5}{3}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\Rightarrow \frac{3(a+b)}{a b}=\frac{10}{3}$ ...(2)
$\Rightarrow \quad \frac{3 \times 10}{a b}=\frac{10}{3}$
$\Rightarrow \quad a b=9$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
Let $\mathrm{a}, \mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{b}$ be in $\mathrm{AP}$, where $\mathrm{x}+\mathrm{y}+\mathrm{z}=15$. Let $\mathrm{a}, \mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{b}$ be
in HP, where $\mathrm{p}^{-1}+\mathrm{q}^{-1}+\mathrm{r}^{-1}=\frac{5}{3}$
What is the value of pqr?

Options:
A $35 / 243$
B $81 / 35$
C $243 / 35$
D Cannot be determined

The correct answer is:
$243 / 35$

Solution:
since a, $\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{b}$ or $1, \mathrm{p}, \mathrm{q}, \mathrm{r}, 9$ are in H.P.
$\begin{aligned} \Rightarrow & \frac{1}{1+4 \mathrm{~d}}=9 \Rightarrow d=-\frac{2}{9} . \\ & \frac{1}{\mathrm{p}}=1-\frac{2}{9}=\frac{7}{9} \Rightarrow p=\frac{9}{7} \\ & \frac{1}{\mathrm{q}}=\frac{7}{9}-\frac{2}{9}=\frac{5}{9} \Rightarrow q=\frac{9}{5} \\ & \& \frac{1}{\mathrm{r}}=\frac{5}{9}-\frac{2}{9}=\frac{3}{9} \Rightarrow r=\frac{9}{3} \\ \Rightarrow & \mathrm{p} \times \mathrm{q} \times \mathrm{r}=\frac{243}{35} \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
Let $\mathrm{a}, \mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{b}$ be in $\mathrm{AP}$, where $\mathrm{x}+\mathrm{y}+\mathrm{z}=15$. Let $\mathrm{a}, \mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{b}$ be
in HP, where $\mathrm{p}^{-1}+\mathrm{q}^{-1}+\mathrm{r}^{-1}=\frac{5}{3}$
What is the value of $x y z$ ?

Options:
A 120
B 105
C 90
D Cannot be determined

The correct answer is:
105

Solution:
On solving eq (1) \& (2), we get
(i) $a=1 \& b=9 \Rightarrow a+4 d=9 \Rightarrow d=2$
(ii) $a=9 \& b=1 \Rightarrow a+4 d=1 \Rightarrow d=-2$.
For $a=1 \& d=2$, $x=3, y=5 \& z=7$
For $a=9 \& d=-2$
$x=7, y=5 \& z=3$
$\Rightarrow \quad x y z=7 \times 5 \times 3=105$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
Let Aand $\mathrm{B}$ be two matrices of order $\mathrm{n} \times \mathrm{n}$. Let $\mathrm{A}$ be nonsingular and $\mathrm{B}$ be singular. Consider the following:
$1. \quad \mathrm{AB}$ is singular
$2. \quad \mathrm{AB}$ is non-singular
$3. \quad \mathrm{A}^{-1} \mathrm{~B}$ is singular
$4. \quad \mathrm{A}^{-1} \mathrm{~B}$ is non singular
Which of the above is/ are correct?

Options:
A 1 and 3
B 2 and 4 only
C 1 only
D 3 only

The correct answer is:
2 and 4 only

Solution:
If A is non-singular and B is singular, then $\mathrm{AB}$ and $\mathrm{A}^{-1} \mathrm{~B}$ are non-singular. Statements (2) and (4) are correct.

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 1)
Let $\mathrm{ABCD}$ be a parallelogram whose diagonals intersect at $\mathrm{P}$ and let $\mathrm{O}$ be the origin, then what is
$\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}$ equal to $?$

Options:
A $\overrightarrow{\mathrm{OP}}$
B $2 \overrightarrow{\mathrm{OP}}$
C $3 \overrightarrow{\mathrm{OP}}$
D $4 \overline{\mathrm{OP}}$

The correct answer is:
$4 \overline{\mathrm{OP}}$

Solution:
Diagonals of a parallelogram bisect each other. Therefore, $\mathrm{P}$ is the mid point of $\mathrm{AC}$ and $\mathrm{BD}$ both.
So, in $\Delta \mathrm{OAC}, \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=2 \overrightarrow{\mathrm{OP}}$
and in $\Delta \mathrm{ODB}, \overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OD}}=2 \overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=2 \overrightarrow{\mathrm{OP}}$ and $\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OD}}=2 \overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}=4 \overrightarrow{\mathrm{OP}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 2)
Let $\mathrm{ABCD}$ be a parallelogram whose diagonals intersect at $\mathrm{P}$ and let $\mathrm{O}$ be the origin. What is $\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}$
equal to?

Options:
A $2 \overrightarrow{\mathrm{OP}}$
B $4 \overrightarrow{\mathrm{OP}}$
C $6 \overrightarrow{\mathrm{OP}}$
D $8 \overrightarrow{\mathrm{OP}}$

The correct answer is:
$4 \overrightarrow{\mathrm{OP}}$

Solution:
From the figure, observe that
$\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{OP}}$
Similarly, $\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{BP}}=\overrightarrow{\mathrm{OP}}$
$\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{CP}}=\overrightarrow{\mathrm{OP}}$
$\overrightarrow{\mathrm{OD}}+\overrightarrow{\mathrm{DP}}=\overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}+\overrightarrow{\mathrm{AP}}$
$+\overrightarrow{\mathrm{BP}}+\overrightarrow{\mathrm{CP}}+\overrightarrow{\mathrm{DP}}=4 \overrightarrow{\mathrm{OP}} \quad \rightarrow$ (i)
In a parallelogram, diagonals bisect each other.
$\therefore \overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{PC}} \quad(\therefore$ Since $\mathrm{AP}=\mathrm{CP})$
$\Rightarrow \overrightarrow{\mathrm{AP}}=-\overrightarrow{\mathrm{PC}}$
$\Rightarrow \overrightarrow{\mathrm{AP}}+\overrightarrow{\mathrm{PC}}=\mathrm{O}$
Similarly $\overrightarrow{\mathrm{BP}}+\overrightarrow{\mathrm{DP}}=\mathrm{O}$
$\therefore(\mathrm{i}) \Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}=4 \overrightarrow{\mathrm{OP}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 1)
Let $\mathrm{ABCD}$ be a square and let $\mathrm{P}$ be a point on $\mathrm{AB}$ such that $\mathrm{AP}: \mathrm{PB}=1: 2 .$ If $\angle \mathrm{APD}=\theta$, then what is the value of
$\cos \theta ?$

Options:
A $\frac{1}{\sqrt{10}}$
B $\frac{1}{\sqrt{5}}$
C $\frac{2}{\sqrt{10}}$
D $\frac{2}{\sqrt{5}} $

The correct answer is:
$\frac{1}{\sqrt{10}}$

Solution:
Let, ABCD be a square having length of side $=\mathrm{x}$ $\therefore \quad$ So, $A D=A B=x$
As given : $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}+\mathrm{AP}}=\frac{1}{2+1}$
$\Rightarrow \mathrm{AP}=\frac{1}{1+2} \mathrm{x}=\frac{\mathrm{x}}{3}$
$\cos \theta=\frac{\mathrm{AP}}{\mathrm{PD}}$
Now, $\mathrm{PD}^{2}=\mathrm{AP}^{2}+\mathrm{AD}^{2}=\left(\frac{x}{3}\right)^{2}+x^{2}$
$=\frac{x^{2}}{9}+\mathrm{x}^{2}=\frac{10 \mathrm{x}^{2}}{9}$
$\Rightarrow \mathrm{PD}=\frac{\sqrt{10} \mathrm{x}}{3}$
$\cos \theta=\frac{\mathrm{x} / 3}{\sqrt{10} \mathrm{x} / 3}=\frac{1}{\sqrt{10}}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 2)
Let \(A=\left(\begin{array}{ccc}2 & 0 & 3 \\ 4 & 7 & 11 \\ 5 & 4 & 8\end{array}\right)\). Then

Options:
A \(\operatorname{det} A\) is divisible by 11
B \(\operatorname{det} A\) is not divisible by 11
C \(\operatorname{det} A=0\)
D A is orthogonal matrix

The correct answer is:
\(\operatorname{det} A\) is divisible by 11

Solution:
\(\begin{aligned}
& \text {Hint: } \operatorname{det}(A)=2(56-44)+0+3(16-35) \\
& =24-57 \\
& =-33 \text { which is divisible by } 11
\end{aligned}\)

Topic: Mathematics, Determinants, WBJEE, WBJEE 2023
Let \(A=\left(\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right), B=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right)\) and \(P=\left(\begin{array}{lll}0 & 1 & 0 \\ x & 0 & 0 \\ 0 & 0 & y\end{array}\right)\) be an orthogonal matrix such that \(B=PAP^{-1}\) holds. Then

Options:
A \(x=1=y\)
B \(x=1, y=0\)
C \(x=0, y=1\)
D \(x=-1, y=0\)

The correct answer is:
\(x=1=y\)

Solution:
Hint : \(B=P A P^{-1}\)
\(P^{-1} B P=P^{-1} P A P^{-1} P\)
\(\begin{aligned}
& {\left[\begin{array}{ccc}
0 & 0 & x y \\
x & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=A=\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]} \\
& \therefore x=1=y
\end{aligned}\)

Topic: Mathematics, Matrices, WBJEE, WBJEE 2023
Let \(\alpha, \beta\) be the roots of the equation \(a x^2+b x+c=0, a, b, c\) real and \(s_n=\alpha^n+\beta^n\) and \(\left|\begin{array}{ccc}3 & 1+s_1 & 1+s_2 \\ 1+s_1 & 1+s_2 & 1+s_3 \\ 1+s_2 & 1+s_3 & 1+s_4\end{array}\right|=k \frac{(a+b+c)^2}{a^4}\) then \(k=\)

Options:
A \(b^2-4 a c\)
B \(b^2+4 a c\)
C \(b^2+2 a c\)
D \(4 a c-b^2\)

The correct answer is:
\(b^2-4 a c\)

Solution:
Hint: \(\left|\begin{array}{ccc}3 & 1+s_1 & 1+s_2 \\ 1+s_1 & 1+s_2 & 1+s_3 \\ 1+s_2 & 1+s_3 & 1+s_4\end{array}\right|\)
\(=\left|\begin{array}{ccc}
1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \\
1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \\
1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4
\end{array}\right|\)
\(\begin{aligned} & =\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2\end{array}\right| \times\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2\end{array}\right| \\ & =\{(1-\alpha)(\alpha-\beta)(1-\beta)\}^2 \\ & =(1-(\alpha+\beta)+\alpha \beta)^2(\alpha-\beta)^2 \\ & =\left(1+\frac{b}{a}+\frac{c}{a}\right)^2\left(\frac{b^2}{a^2}-\frac{4 c}{a}\right) \\ & =\frac{(a+b+c)^2}{a^2} \times \frac{b^2-4 a c}{a^2}=\left(b^2-4 a c\right) \frac{(a+b+c)^2}{a^4} \\ & \therefore k=b^2-4 a c\end{aligned}\)

Topic: Mathematics, Determinants, WBJEE, WBJEE 2023
Let $\alpha$ and $\beta$ be real numbers and $z$ be a complex number. If $z^{2}+\alpha z+\beta=0$ has two distinct non-real roots with $\operatorname{Re}(\mathrm{z})=1$, then it is necessary that

Options:
A $\beta \in(-1,0)$
B $|\beta|=1$
C $\beta \in(1, \infty)$
D $\beta \in(0,1)$

The correct answer is:
$\beta \in(1, \infty)$

Solution:
Let $z=x+$ iy $\therefore z^{2}+\alpha z+\beta=(x+i y)^{2}+\alpha(x+i y)+\beta$
$=\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{ixy}+\alpha \mathrm{x}+\mathrm{i} \alpha \mathrm{y}+\beta$
Given, $z^{2}+\alpha z+\beta=0$
$\therefore x^{2}-y^{2}+2 i x y+\alpha x+i \alpha y+\beta=0$
$\Rightarrow x^{2}-y^{2}+\alpha x+\beta+i(2 x y+\alpha y)=0+i .0$
Comparing real and imaginary parts, we get $\mathrm{x}^{2}-\mathrm{y}^{2}+\alpha \mathrm{x}+\beta=0$
$\mathrm{y}(2 \mathrm{x}+\alpha)=0$
$(2) \Rightarrow 2 \mathrm{x}+\alpha=0 \quad(\because \mathrm{y} \neq 0)$
$\Rightarrow 2(1)+\alpha=0 \quad(\because$ Given $\operatorname{Re}(z)=1)$
$\Rightarrow \alpha=-2$
Now, (1) $\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}+\alpha \mathrm{x}+\beta=0$
$\Rightarrow(1)^{2}-\mathrm{y}^{2}+(-2)(1)+\beta=0$
$\Rightarrow 1-\mathrm{y}^{2}-2+\beta=0$
$\Rightarrow-1-\mathrm{y}^{2}+\beta=0$
$\Rightarrow \beta=1+\mathrm{y}^{2}$
Since, $\mathrm{y} \in \mathrm{R}$ and $\mathrm{y} \neq 0$,
$\beta$ is always greater than $1 .$ So, $\beta \in(1, \infty)$

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 1)
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-\left(1-2 a^{2}\right) x+\left(1-2 a^{2}\right)=0$
Under what condition does the above equation have real roots?

Options:
A $\mathrm{a}^{2} < \frac{1}{2}$
B $\mathrm{a}^{2}>\frac{1}{2}$
C $\mathrm{a}^{2} \leq \frac{1}{2}$
D $a^{2} \geq \frac{1}{2}$

The correct answer is:
$a^{2} \geq \frac{1}{2}$

Solution:
Using $a x^{2}+b x+c=0$
$a=1, b=-\left(1-2 a^{2}\right) \& c=\left(1-2 a^{2}\right)$
For roots to be real,
$b^{2}-4 a c>0$
$\Rightarrow\left[-\left(1-2 a^{2}\right)\right]^{2}-4(1)\left(1-2 a^{2}\right) \geq 0$
$\Rightarrow 4 a^{4}+4 a^{2}-3 \geq 0$
$\Rightarrow\left(2 a^{2}-1\right)\left(2 a^{2}+3\right) \geq 0$
$\Rightarrow a^{2} \geq \frac{1}{2}$ or $a^{2} \leq-\frac{3}{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 2)