datavorous/entrance-exam-dataset huggingface dataset:
Let $-1 \leq x \leq 1$. If $\cos \left(\sin ^{-1} x\right)=\frac{1}{2}$, then how many value does $\tan \left(\cos ^{-1} \mathrm{x}\right)$ assume?

Options:
A One
B Two
C Four
D Infinite

The correct answer is:
Two

Solution:
As given :
$\cos \left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{2}$
$\Rightarrow \sin ^{-1} x=\cos ^{-1}\left(\frac{1}{2}\right)$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{3} \Rightarrow x=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
$\therefore \tan \left(\cos ^{-1} x\right)=\tan \left(\cos ^{-1} \frac{\sqrt{3}}{2}\right)$
$=\tan \left(\pm \frac{\pi}{6}\right)=\pm \frac{1}{\sqrt{3}}$
Hence, $\tan \left(\cos ^{-1} \mathrm{x}\right)$ have two values.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2008 (Phase 1)
Let $45^{\circ} \leq \theta < 90^{\circ}$. If $\tan \theta+\cot \theta=(\tan \theta)^{\mathrm{i}}+(\cot \theta)^{\mathrm{j}}$ for some
$\mathrm{i} \geq 2$, then what is the value of $\sin \theta+\cos \theta$ ?

Options:
A $\sqrt{2}$
B $\frac{1}{\sqrt{2}}$
C $\frac{(\sqrt{3}+1)}{2}$
D $\frac{2}{(\sqrt{3}+1)}

The correct answer is:
$\sqrt{2}$

Solution:
As given, $\tan \theta+\cot \theta=(\tan \theta)^{\mathrm{i}}+(\cot \theta)^{\mathrm{i}}$
Also, $45^{\circ} \leq \theta < 90^{\circ}$ and $\mathrm{i} \geq 2$
which is possible only when $\theta=45^{\circ}$ Since, $\tan 45^{\circ}+\cot 45^{\circ}=1+1=2$
and $\left(\tan 45^{\circ}\right)^{\mathrm{i}}+\left(\cot 45^{\circ}\right)^{\mathrm{i}}=1+1=2$
Thus, $\sin \theta+\cos \theta=\sin 45^{\circ}+\cos 45^{\circ}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{1+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 2)
Let $16 x^{2}-3 y^{2}-32 x-12 y=44$ represents a hyperbola. Then,

Options:
A length of the transverse axis is $2 \sqrt{3}$
B length of each latusrecturn is $32 / \sqrt{3}$
C eccentricity is $\sqrt{19 / 3}$
D equation of a directrix is $x=\frac{\sqrt{19}}{3}$

The correct answers are:
length of the transverse axis is $2 \sqrt{3}$, length of each latusrecturn is $32 / \sqrt{3}$, eccentricity is $\sqrt{19 / 3}$

Solution:
Given equation of hyperbola is $16 x^{2}-3 y^{2}-32 x-12 y=44$
$\Rightarrow 16 x^{2}+16-32 x-3 y^{2}-12-12 y$
$=44+4$
$\Rightarrow 16(x-1)^{2}-3(y+2)^{2}=48$
$\Rightarrow \quad \frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1$
On comparing the equation with standard equation of hyperbola, we get $a=\sqrt{3}, b=4$
Now, length of transverse axis $=2 a=2 \sqrt{3}$
and length of latusrectum $=\frac{2 b^{2}}{a}=\frac{2 \times 16}{\sqrt{3}}=\frac{32}{\sqrt{3}}$
$\therefore$ Eccentricity $(e)=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$=\sqrt{1+\frac{16}{3}}=\sqrt{\frac{19}{3}}$
Equation of directrix is $x=\pm \frac{a}{e}$ $\Rightarrow x-1=\pm \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{19}}$
$\Rightarrow \quad x=1 \pm \frac{3}{\sqrt{19}}$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2015
Let $|\overrightarrow{\mathrm{a}}| \neq 0,|\overrightarrow{\mathrm{b}}| \neq 0$
$(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$ holds if and only if

Options:
A $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are perpendicular
B $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are parallel
C $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are inclined at an angle of $45^{\circ}$
D $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are anti-parallel

The correct answer is:
$\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are perpendicular

Solution:
$(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=a^{2}+2 \vec{a} \cdot \vec{b}+b^{2}=a^{2}+b^{2}$
$\Rightarrow \quad \vec{a} \cdot \vec{b}=0 \Rightarrow \vec{a} \perp \vec{b}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)
Let $\mathrm{A}=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right]$
If $\mathrm{AB}=\mathrm{BA}$, then what is the value of $\mathrm{x}$?

Options:
A $-1$
B 0
C 1
D Any real number

The correct answer is:
0

Solution:
As given $\mathrm{A}=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ and $\mathrm{B}=\left|\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right|$
$\mathrm{AB}=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & \mathrm{x} \\ 0 & -1\end{array}\right]$
and $\mathrm{BA}=\left[\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]=\left[\begin{array}{cc}1 & -\mathrm{x} \\ 0 & -1\end{array}\right]$
Given that $\mathrm{AB}=\mathrm{BA}$
We have, $\left[\begin{array}{cc}1 & \mathrm{x} \\ 0 & -1\end{array}\right]=\left[\begin{array}{cc}1 & -\mathrm{x} \\ 0 & -1\end{array}\right]$
$\Rightarrow \mathrm{x}=-\mathrm{x} \Rightarrow 2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
Let $\mathrm{A}=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$ be a square matrix of order 3 . Then for any positive integer $\mathrm{n}$, what is $\mathrm{A}^{\mathrm{n}}$ equal to?

Options:
A A
B $3^{\mathrm{n}} \mathrm{A}$
C $\left(3^{n-1}\right) \mathrm{A}$
D $3 \mathrm{~A}$

The correct answer is:
$\left(3^{n-1}\right) \mathrm{A}$

Solution:
Given matrix is $\mathrm{A}=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
So, $\mathrm{A}^{2}=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
$=\left[\begin{array}{lll}3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3\end{array}\right]=3\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
and $\mathrm{A}^{3}=3\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
$=3\left[\begin{array}{lll}3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3\end{array}\right]=9\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
$=3^{2} \mathrm{~A}$
Similarly $\mathrm{A}^{4}=3^{3} \mathrm{~A} .$ Hence, $A^{\mathrm{n}}=3^{\mathrm{n}-1} \mathrm{~A}$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
Let $\mathrm{A}=\{1,2,3,4,5,6,7,8,9,10\}$. Then the number of subsets of A containing exactly two elements is

Options:
A 20
B 40
C 45
D 90

The correct answer is:
45

Solution:
$\mathrm{A}=\{1,2,3,4,5,6,7,8,9,10\}$
Number of subsets of A containing two elements $=9+8+7+6+5+4+3+2+1$
$\frac{9(9+1)}{2}=\frac{90}{2}=45$
$\therefore$ Option $(\mathrm{c})$ is correct Alternate Method The number of subsets of A containing exactly two elements is:
${ }^{10} \mathrm{C}_{2}=\frac{10 \times 9}{2 \times 1}=45$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
Let $\mathrm{A}=\{1,2,3,4,5,6,7,8,9,10\}$. Then the number of subsets of A containing two or three elements is

Options:
A 45
B 120
C 165
D 32

The correct answer is:
165

Solution:
Given, A $\{1,2,3,4,5,6,7,8,9,10\}$ Set $\mathrm{A}$ has 10 elements. Number of sub sets Containing 2 and 3 elements is
${ }^{10} \mathrm{c}_{\mathrm{c}}+10_{\mathrm{c}_{3}}$
$10_{\mathrm{c}_{2}}+10_{\mathrm{c}_{3}}=\frac{10 \times 9}{2}+\frac{10 \times 9 \times 8}{3 \times 2}$
$=45+120=165$

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 1)
Let $\vec{a}=(1,-2,3)$ and $\vec{b}=(3,1,2)$ be two vectors and $\vec{c}$ be a
vector of length $l$ and parallel to $(\vec{a}+\vec{b})$. What is $\vec{c}$ equal
to?

Options:
A $\frac{1}{\sqrt{4}}(-2,-3,1)$
B $\frac{1}{\sqrt{2}}(1,0,1)$
C $\frac{1}{\sqrt{42}}(-5,-4,-1)$
D None of these

The correct answer is:
None of these

Solution:
Given $\overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
and $\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$\therefore \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-\mathbf{j}+5 \hat{\mathbf{k}}$
Then, $\overrightarrow{\mathbf{c}}=\lambda(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$
$=\lambda(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}})$
$\Rightarrow l=\sqrt{16 \lambda^{2}+\lambda^{2}+25 \lambda^{2}}$
$\Rightarrow l=\sqrt{42} \lambda$
$\lambda=\frac{l}{\sqrt{42}}$
$\therefore \quad \overrightarrow{\mathbf{c}}=\frac{l}{\sqrt{42}}(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}})=\frac{1}{\sqrt{42}}(4,-1,5)$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
Let $A=\{-1,2,5,8\}, B=\{0,1,3,6,7\}$ and $R$ be the relation 'is
one less than' from $A$ to $B$, then how many elements will $R$ contain?

Options:
A 2
B 3
C 5
D 9

The correct answer is:
3

Solution:
Given, $\mathrm{A}=\{-1,2,5,8\}$ and $\mathrm{B}=\{0,1,3,6,7\}$
Since, $\mathrm{R}$ be the relation 'is one less than' from $\mathrm{A}$ to $\mathrm{B}$ $\mathrm{R}=\{(-1,0),(2,3),(5,6)\}$
Hence, R contains 3 elements.

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[a_{i j}\right]$, where $i, j=1,2$, If its inverse matrix is $\left[b_{i j}\right]$, what is $b_{22} ?$

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

The correct answer is:
$-\frac{1}{2}$

Solution:
Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
$|A|=4 \times 1-2 \times 3=4-6=-2$
$\therefore \quad A^{-1}=-\frac{1}{2}\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]$
$\Rightarrow\left[b_{i j}\right]=-\frac{1}{2}\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]=\left[\begin{array}{lc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$
$\Rightarrow b_{22}=\frac{1}{2}$

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 1)
Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]$ where $a, b$ are natural numbers, then which one of the following is correct?

Options:
A There exist more than one but finite number of $B$ 's such that $A B=B A$
B There exists exactly one $\mathrm{B}$ such that $A B=B A$
C There exist infinitely many $B$ 's such that $A B=B A$
D There cannot exist any $B$ such that $A B=B A$

The correct answer is:
There exist infinitely many $B$ 's such that $A B=B A$

Solution:
$A B=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]=\left[\begin{array}{cc}a & 2 b \\ 3 a & 4 b\end{array}\right]$
and $B A=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[\begin{array}{cc}a & 2 a \\ 3 b & 4 b\end{array}\right]$
If $A B=B A$
$\Rightarrow\left[\begin{array}{cc}a & 2 b \\ 3 a & 4 b\end{array}\right]=\left[\begin{array}{cc}a & 2 a \\ 3 b & 4 b\end{array}\right] \Rightarrow a=b$
From the above it is clear that there exist infinitely many B's such that $\mathrm{AB}=\mathrm{BA}$.

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 1)
$\quad$ Let $\overrightarrow{\mathbf{a}}=2 \overrightarrow{\mathbf{j}}-3 \overrightarrow{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=-3 \overrightarrow{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$. Let $\hat{\mathbf{n}}$ be
a unit vector such that $\overrightarrow{\mathbf{a}} \hat{\mathbf{n}}=\overrightarrow{\mathbf{b}} \hat{\mathbf{n}}=0$. What is the value of $\overrightarrow{\mathbf{c}} . \hat{\mathbf{n}} ?$

Options:
A 1
B $\sqrt{19}$
C 3
D $-3$

The correct answer is:
$-3$

Solution:
As given $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
and $\overrightarrow{\mathrm{c}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Let $\overrightarrow{\mathrm{n}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
Since $\vec{a}$ and $\hat{n}$ are perpendicular to each other. $\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{n}}=0 \Rightarrow(2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=0$
$\Rightarrow 2 \mathrm{y}-3 \mathrm{z}=0$
and $\overrightarrow{\mathrm{b}} \cdot \hat{\mathrm{n}}=0$
$\Rightarrow(\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=0$
$\Rightarrow \mathrm{y}+3 \mathrm{z}=0$
$\ldots(2)$
On solving Eqs. (1) and (2) $y=z=0$
Since $\hat{n}$ is a unit vector, $\sqrt{x^{2}+y^{2}+z^{2}}=1 \Rightarrow x=1 \quad[$ since, $y=z=0]$
hence, $\hat{n}=\hat{i}$
This gives, $\overrightarrow{\text { c. }} \hat{\text { n }}=(-3 \hat{i}+3 \hat{j}+\hat{k}) \cdot(\hat{\mathrm{i}})=-3$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
Let $\mathrm{A}=\left[\begin{array}{ccc}5 & 6 & 1 \\ 2 & -1 & 5\end{array}\right]$. Let there exist a matrix $B$ such that
$A B=\left[\begin{array}{ll}35 & 49 \\ 29 & 13\end{array}\right] .$ What is $B$ equal to?

Options:
A $\left[\begin{array}{lll}5 & 1 & 4 \\ 2 & 6 & 3\end{array}\right]$
B $\left[\begin{array}{lll}2 & 6 & 3 \\ 5 & 1 & 4\end{array}\right]$
C $\left[\begin{array}{ll}5 & 2 \\ 1 & 6 \\ 4 & 3\end{array}\right]$
D $\left[\begin{array}{ll}2 & 5 \\ 6 & 1 \\ 3 & 4\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ll}5 & 2 \\ 1 & 6 \\ 4 & 3\end{array}\right]$

Solution:
$A=\left[\begin{array}{ccc}5 & 6 & 1 \\ 2 & -1 & 5\end{array}\right]$ and let $B=\left[\begin{array}{ll}5 & 2 \\ 1 & 6 \\ 4 & 3\end{array}\right]$
$\therefore \quad A B=\left[\begin{array}{lrl}5 & 6 & 1 \\ 2 & -1 & 5\end{array}\right]\left[\begin{array}{ll}5 & 2 \\ 1 & 6 \\ 4 & 3\end{array}\right]$
$=\left[\begin{array}{ll}25+6+4 & 10+36+3 \\ 10-1+20 & 4-6+15\end{array}\right]$
$=\left[\begin{array}{ll}35 & 49 \\ 29 & 13\end{array}\right]$
Hence, option (c) is correct.
$ALTERNATE SOLUTION$
Given $A=\left[\begin{array}{lll}5 & 6 & 1 \\ 2 & -1 & 5\end{array}\right]_{2 \times 3}$ and $A B=\left[\begin{array}{cc}35 & 49 \\ 29 & 13\end{array}\right]_{2 \times 2}$
Since, order of $\mathrm{A}=2 \times 3$ and order of $\mathrm{AB}=2 \times 2$ $\therefore$ order of $\mathrm{B}=3 \times 2$
Let $B=\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right]$
$\therefore \quad A B=\left[\begin{array}{ccc}5 & 6 & 1 \\ 2 & -1 & 5\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right]$
$\left[\begin{array}{ll}35 & 49 \\ 29 & 13\end{array}\right]=\left[\begin{array}{cc}5 a+6 c+e & 5 b+6 d+f \\ 2 a-c+5 e & 2 b-d+5 f\end{array}\right]$
$\Rightarrow 5 a+6 c+e=35$
$5 b+6 d+f=49$
$2 a-c+5 e=29$
$2 b-d+5 f=13$
On solving above four equations we get $\mathrm{a}=5, \mathrm{~b}=2, \mathrm{c}=1, \mathrm{~d}=6, \mathrm{e}=4$ and $\mathrm{f}=3$
Hence $B=\left[\begin{array}{ll}5 & 2 \\ 1 & 6 \\ 4 & 3\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 1)
Let $|\bar{a}|=7,|\bar{b}|=11,|\bar{a}+\bar{b}|=10 \sqrt{3}$
What is $|\bar{a}-\bar{b}|$ equal to ?

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

The correct answer is:
$2 \sqrt{10}$

Solution:
Let angle between $\vec{a}$ and $\overrightarrow{\mathrm{b}}$ be $\theta$.
$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2|\overrightarrow{\mathrm{a}}| \overrightarrow{\mathrm{b}}|\cos \theta|}$
$\sqrt[10]{3}=\sqrt{49+121+2 \times 7 \times 11 \cos \theta}$
$300=170+154 \cos \theta$
$154 \cos \theta=130$
$|\vec{a}-\vec{b}|=\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a} \| \vec{b}| \cos \theta}$
$|\vec{a}-\vec{b}|=\sqrt{170-154 \cos \theta}$
$|\vec{a}-\vec{b}|=\sqrt{170-130}=\sqrt{40}$ or $2 \sqrt{10}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
Let $|\bar{a}|=7,|\bar{b}|=11,|\bar{a}+\bar{b}|=10 \sqrt{3}$
What is the angle between $(\bar{a}+\bar{b})$ and $(\bar{a}-\bar{b})$ ?

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{6}$
D None of these

The correct answer is:
None of these

Solution:
Let angle between $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$ be $\alpha$
$\cos \alpha=\frac{(\vec{a}+\vec{b})(\vec{a}-\vec{b})}{|\vec{a}+\vec{b} \| \vec{a}-\vec{b}|}$
$=\frac{(7)^{2}-(11)^{2}}{10 \sqrt{3} \times 2 \sqrt{10}}=\frac{(7+11)(7-11)}{20 \sqrt{3} \times \sqrt{10}}=\frac{-18}{5 \sqrt{30}}$
$=\frac{-6 \times 3}{5 \sqrt{30}} \times \frac{\sqrt{30}}{\sqrt{30}}=-\frac{3 \sqrt{30}}{25}$
$\alpha=\cos ^{-1}\left(\frac{-3}{5} \sqrt{\frac{6}{5}}\right)$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
Let $A=\{a, b, c, d\}$ and $B=\{x, y, z\} .$ What is the number of element in $A \times B$?

Options:
A 6
B 7
C 12
D 64

The correct answer is:
12

Solution:
$\mathrm{n}(\mathrm{A})=4$ and $\mathrm{n}(\mathrm{B})=3$
Number of elements in $n(A \times B)=4 \times 3=12$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{m}}$ be a matrix and $\mathrm{C}=\left[\mathrm{c}_{\mathrm{iij}}\right]_{\mathrm{m} \times \mathrm{m}}$ be another matrix where $\mathrm{c}_{\mathrm{ij}}$ is the cofactor of $\mathrm{a}_{\mathrm{ij}}$. Then, what is the value of $|\mathrm{AC}|$?

Options:
A $|\mathrm{A}|^{\mathrm{m}-1}$
B $|\mathrm{A}|^{\mathrm{m}}$
C $|\mathrm{A}|^{\mathrm{m}+1}$
D Zero

The correct answer is:
$|\mathrm{A}|^{\mathrm{m}+1}$

Solution:
Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{i} j}\right]_{\mathrm{m} \times \mathrm{m}}$ be a matrix and $\mathrm{C}=\left[\mathrm{c}_{\mathrm{i} j}\right]_{\mathrm{m} \times \mathrm{m}}$ be
another matrix where cij is the cofactor of $\mathrm{a}_{\mathrm{ij}}$
$\therefore \quad$ The value of $|\mathrm{AC}|=|\mathrm{A}|^{\mathrm{m}+1}$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 2)
Let $\mathrm{A}=\left(\mathrm{a}_{\mathrm{ij}}\right)_{\mathrm{n} \times \mathrm{n}}$ and adj $\mathrm{A}=\left(\alpha_{\mathrm{ij}}\right)$
If $\mathrm{A}=\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 4 \\ 2 & 3 & -1\end{array}\right]$, what is the value of $\alpha_{23}$?

Options:
A 1
B $-1$
C 8
D -8

The correct answer is:
8

Solution:
$\because \quad A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 4 \\ 2 & 3 & -1\end{array}\right]$
Since, adj $\mathrm{A}=\left(\alpha_{\mathrm{ij}}\right)$ so, $\alpha_{\mathrm{ij}}=\mathrm{a}_{\mathrm{ji}}$ of $\mathrm{A}$.
$\therefore \alpha_{23}=\mathrm{a}_{32}=(-1)^{2+3}\left|\begin{array}{ll}1 & 3 \\ 4 & 4\end{array}\right|=-(4-12)=8$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
Let $A$ and $B$ are two mutually exclusive events with $\mathrm{P}(\mathrm{A})=$ $\frac{1}{3} \mathrm{P}(\mathrm{B})=\frac{1}{4} .$ What is the value of $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) ?$

Options:
A $\frac{1}{6}$
B $\frac{1}{4}$
C $\frac{1}{3}$
D $\frac{5}{12}$

The correct answer is:
$\frac{5}{12}$

Solution:
$A \& B$ are mutualy exclusive events. i.e., $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$
We know, $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})$
$=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]$
$=1-\left[\frac{1}{3}+\frac{1}{4}-0\right]$
$=1-\left[\frac{7}{12}\right]=\frac{5}{12} .$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 1)
Let $A$ and $B$ be $(3 \times 3)$ matrices with $\operatorname{det} A=4$ and det $B=3$.
What is det ( $2 \mathrm{AB}$ ) equal to ?

Options:
A 96
B 72
C 48
D 36

The correct answer is:
96

Solution:
A and B are $(3 \times 3)$ matrices det $\mathrm{A}=4$
$\operatorname{det} B=3$
$\therefore \operatorname{det}(2 \mathrm{AB})=(2)^{3}|\mathrm{~A}||\mathrm{B}|$
$=2^{3}(4)(3)=8(4)(3)=96$

Topic: Mathematics, Determinants, NDA, NDA 2019 (Phase 1)
Let $A$ and $B$ be $(3 \times 3)$ matrices with $\operatorname{det} A=4$ and det $B=3$.
What is det $\left(3 \mathrm{AB}^{-1}\right)$ equal to ?

Options:
A 12
B 18
C 36
D 48

The correct answer is:
36

Solution:
$\operatorname{det}\left(3 \mathrm{AB}^{-1}\right)=(3)^{3}|\mathrm{~A}|\left|\mathrm{B}^{-1}\right|$

Topic: Mathematics, Determinants, NDA, NDA 2019 (Phase 1)
Let $A$ and $B$ be matrices of order $3 \times 3$. If $A B=0$, then which of the following can be concluded?

Options:
A $A=0$ and $B=0$
B $|A|=0$ and $|B|=0$
C Either $|A|=0$ or $|B|=0$
D Either $A=0$ or $B=0$

The correct answer is:
Either $A=0$ or $B=0$

Solution:
If AB = 0 , then it may be concluded that either $\mathrm{A}=0$ or $\mathrm{B}=0$
But, it should be noticed that it is not necessary that either $\mathrm{A}=0$ or $\mathrm{B}=0$.

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 2)
Let $\mathrm{A}$ and $\mathrm{B}$ be obtuse angles such that $\sin \mathrm{A}=\frac{4}{5}$ and
$\cos B=-\frac{12}{13}$. What is the value of $\sin (A+B) ?$

Options:
A $-\frac{63}{65}$
B $-\frac{33}{65}$
C $-\frac{33}{65}$
D $-\frac{63}{65}$

The correct answer is:
$-\frac{63}{65}$

Solution:
$\sin \mathrm{A}=\frac{4}{5}$ and $\cos \mathrm{B}=-\frac{12}{13}$
It is given that $\mathrm{A}$ and $\mathrm{B}$ are obtuse angle
$\Rightarrow \cos \mathrm{A}-\pm \sqrt{1-\sin ^{2} \mathrm{~A}}-\pm \sqrt{1-\frac{16}{25}}--\frac{3}{5}$
Negative sign is taken for cos A since A being obtuse lies in second quadrant. $\sin B=\pm \sqrt{1-\cos ^{2} B}=\pm \sqrt{1-\left(\frac{-12}{13}\right)^{2}}$
$=\sqrt{\frac{169-144}{169}}=\frac{5}{13}$
Positive sign is taken since, $\sin B$ is positive in second quadrant.
$\Rightarrow \cos \mathrm{A}=\frac{-3}{5}$ and $\sin \mathrm{B}=\frac{3}{13}$
$\therefore \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$
$=\frac{4}{5} \times\left(\frac{-12}{13}\right)+\frac{-3}{5} \times\left(\frac{5}{13}\right)=-\frac{48}{65}-\frac{15}{65}$
$=\frac{-48-15}{65}=\frac{-63}{65}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 2)
Let A and B be subsets of $\mathrm{X}$ and $\mathrm{C}=\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right) \cup\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$,
where $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ are complements of A and $\mathrm{B}$ respectively in
X. What is $C$ equal to?

Options:
A $\left(\mathrm{A} \cup \mathrm{B}^{\prime}\right)-\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)$
B $\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)-\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$
C $(\mathrm{A} \cup \mathrm{B})-(\mathrm{A} \cap \mathrm{B})$
D $\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)-\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)$

The correct answer is:
$(\mathrm{A} \cup \mathrm{B})-(\mathrm{A} \cap \mathrm{B})$

Solution:
$C=\left(A \cap B^{\prime}\right) \cup\left(A^{\prime} \cap B\right)$
Let usdraw venn diagram and compare it with options.
This also represents $(\mathrm{A} \cup \mathrm{B})-(\mathrm{A} \cap \mathrm{B})$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)