datavorous/entrance-exam-dataset huggingface dataset:
If $\hat{a}$ and $\hat{b}$ are the unit vectors along $\vec{a}$ and $\vec{b}$ respectively, then what is the projection of $\vec{b}$ on $\vec{a}$ ?

Options:
A $\vec{a} \cdot \vec{b}$
B $\hat{a} \cdot \hat{b}$
C $\hat{a} \cdot \vec{b}$
D $|\vec{a} \times \vec{b}|$

The correct answer is:
$\vec{a} \cdot \vec{b}$

Solution:
The projection of $\vec{b}$ on $\vec{a}=\frac{\vec{a} \vec{b}}{|\vec{a}|}$
Since, $\vec{a}$ is the unit vector
$\therefore|\vec{a}|=1$
Hence, projection of $\vec{b}$ on $\vec{a}=\frac{\vec{a} \cdot \vec{b}}{1}=\vec{a} \cdot \vec{b}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 2)
If $A$ and $B$ are two disjoint sets, then which one of the
following is correct?

Options:
A $A-B=A-(A \cap B)$
B $B-A^{\prime}=(A \cap B)$
C $A \cap B=(A-B) \cap B$
D All of these

The correct answer is:
$A-B=A-(A \cap B)$

Solution:
Since, Aand $B$ are two disjoints therefore $A \cap B=\phi$
$A \quad A-B=A-(A \cap B)$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $2 \mathrm{P}(\mathrm{A})=3 \mathrm{P}(\mathrm{B})$, where 0 $ < \mathrm{P}(\mathrm{A}) < \mathrm{P}(\mathrm{B}) < 1$, then which one of the following is correct?

Options:
A $\quad \mathrm{P}(\mathrm{A} \mid \mathrm{B}) < \mathrm{P}(\mathrm{B} \mid \mathrm{A}) < \mathrm{P}(\mathrm{A} \cap \mathrm{B})$
B $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) < \mathrm{P}(\mathrm{B} \mid \mathrm{A}) < \mathrm{P}(\mathrm{A} \mid \mathrm{B})$
C $\quad \mathrm{P}(\mathrm{B} \mid \mathrm{A}) < \mathrm{P}(\mathrm{A} \mid \mathrm{B}) < \mathrm{P}(\mathrm{A} \cap \mathrm{B})$
D $\quad P(A \cap B) < P(A \mid B) < P(B \mid A)$

The correct answer is:
$\mathrm{P}(\mathrm{A} \cap \mathrm{B}) < \mathrm{P}(\mathrm{B} \mid \mathrm{A}) < \mathrm{P}(\mathrm{A} \mid \mathrm{B})$

Solution:
Given, $2 . \mathrm{P}(\mathrm{A})=3 . \mathrm{P}(\mathrm{B})$
$\Rightarrow \frac{2 \mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}=\frac{3 \mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}$
$\Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{2 \cdot \mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{3 \mathrm{P}(\mathrm{B})}$
$\Rightarrow \frac{1}{2} \cdot \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{1}{3} \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)$
$\Rightarrow \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right) < \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right) .$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)
If $A$ and $B$ are two events such that $P(A)=0.5, P(B)=0.6$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.4$, then what is $P(\overline{A \cup B})$ equal to ?

Options:
A $0.9$
B 7
C $0.5$
D $0.3$

The correct answer is:
$0.3$

Solution:
$\begin{aligned} &=\frac{{ }^{4} C_{1}}{52} C_{1} \\ & P(\mathrm{~A})=0.5, \mathrm{P}(\mathrm{B})=0.6, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.4 \\ & P(\overline{A \cup B})=1-P(A \cup B) \\ &=1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \\ &=1-[0.5+0.6-0.4] \\ &=1-0.7=0.3 \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
If $A$ and $B$ are two events such that $P(A \cup B)=\frac{3}{4}$,
$P(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{2}{3}$, then what is $P(B)$ equal to?

Options:
A $\frac{1}{3}$
B $\frac{2}{3}$
C $\frac{1}{8}$
D $\frac{2}{9}$

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

Solution:
$\begin{array}{ll} & \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4} \\ & \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4} \\ & \mathrm{P}(\overline{\mathrm{A}})=\frac{2}{3}, \mathrm{P}(\mathrm{A})=\frac{1}{3} \\ \therefore \quad & \mathrm{P}(\mathrm{B})=\frac{3}{4}-\frac{1}{3}+\frac{1}{4}=\frac{9-4+3}{12}=\frac{\not 8}{12}=\frac{2}{3} \\ \therefore & \text { Option (b) is correct. }\end{array}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 1)
If $A$ and $B$ are two events such that $P(A \cup B)=\frac{3}{4}$, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}, \mathrm{P}(\overline{\mathrm{A}})=\frac{2}{3}$ where $\overline{\mathrm{A}}$ is the complement of
A, then what is $\mathrm{P}(\mathrm{B})$ equal to ?

Options:
A $1 / 3$
B $2 / 3$
C $1 / 9$
D $2 / 9$

The correct answer is:
$2 / 3$

Solution:
Given $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$,
$\mathrm{P}(\overline{\mathrm{A}})=\frac{2}{3} \Rightarrow \mathrm{P}(\mathrm{A})=\frac{1}{3}$
As we know
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\therefore \quad \frac{3}{4}=\frac{1}{3}+\mathrm{P}(\mathrm{B})-\frac{1}{4}$
$\Rightarrow \quad \frac{3}{4}+\frac{1}{4}-\frac{1}{3}=\mathrm{P}(\mathrm{B})$
$\Rightarrow \mathrm{P}(\mathrm{B})=\frac{2}{3}$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
If $\mathrm{A}$ and $\mathrm{B}$ are two events, then what is the probability of occurrence of either event A or event B?

Options:
A $\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$
B $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
C $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
D $\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$

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

Solution:
Probability of occurence of either event A or event $\mathrm{B}=$ $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
If $\mathrm{A}$ and $\mathrm{B}$ are two invertible square matrices of same order, then what is $(\mathrm{AB})^{-1}$ equal to?

Options:
A $\mathrm{B}^{-1} \mathrm{~A}^{-1}$
B $\mathrm{A}^{-1} \mathrm{~B}^{-1}$
C $\mathrm{B}^{-1} \mathrm{~A}$
D $\mathrm{A}^{-1} \mathrm{~B}$

The correct answer is:
$\mathrm{B}^{-1} \mathrm{~A}^{-1}$

Solution:
$(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
If $A$ and $B$ are two matrices such that $A B=A$ and $B A=B$, then which one of the following is correct?

Options:
A $\left(A^{T}\right)^{2}=A^{T}$
B $\left(A^{T}\right)^{2}=B^{T}$
C $\left(A^{T}\right)^{2}=\left(A^{-1}\right)^{-1}$
D None of the above

The correct answer is:
$\left(A^{T}\right)^{2}=A^{T}$

Solution:
Let $A$ and $B$ be two matrices such that $A B=A$ and $B A$ $=B$
Now, consider $A B=A$ Take Transpose on both side $(A B)^{T}=A^{T}$
$\Rightarrow A^{T}=B^{T} . A^{T}$ ...(1)
Now, $\mathrm{BA}=\mathrm{B}$
Take, Transpose on both side $(B A)^{T}=\mathrm{B}^{\mathrm{T}}$
$\Rightarrow \mathrm{B}^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}} \cdot \mathrm{B}^{\mathrm{T}}$ ...(2)
Now, from equation (1) and (2). we have $A^{T}=\left(A^{T} \cdot B^{T}\right) A^{T}$
$A^{T}=A^{T}\left(B^{T} A^{T}\right)$
$=A^{T}(A B)^{T} \quad\left(\because(A B)^{T}=B^{T}=B^{T} A^{T}\right)$
$=A^{T} \cdot A^{T}$
Thus, $A^{T}=\left(A^{T}\right)^{2}$

Topic: Mathematics, Matrices, NDA, NDA 2008 (Phase 2)
If $A$ and $B$ are two mutually exclusive and exhaustive events
with $P(B)=3 P(A)$, then what is the value of $P(\bar{B})$ ?

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

The correct answer is:
$1 / 4$

Solution:
A and B are mutually exclusive and exhaustive events with $P(A \cap B)=0, P(A \cup B)=1$
we know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 1=P(A)+3 P(A)$
$\Rightarrow \quad P(A)=\frac{1}{4}$
$\therefore P(B)=\frac{3}{4}$
Hence, $P(\bar{B})=1-P(B)=1-\frac{3}{4}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 1)
If $A$ and $B$ are two mutually exclusive events, then what is P(A B) equal to?

Options:
A 0
B $P(A)+P(B)$
C $P(A) P(B)$
D $\mathrm{P}(\mathrm{A}) P\left(\frac{B}{A}\right)$

The correct answer is:
0

Solution:
Since, A and B are mutually exclusive events. $\therefore \quad \mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 1)
If A and B are two non-empty sets having $n$ elements in common, then what is the number of common elements in
the sets $\mathrm{A} \times \mathrm{B}$ and $\mathrm{B} \times \mathrm{A}$

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

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

Solution:
Since A and B have $n$ elements in common so, $A \times B$ and $\mathrm{B} \times \mathrm{A}$ have $\mathrm{n}^{2}$ elements in common.

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
If A and B are two non-singular square matrices such that $\mathrm{AB}=\mathrm{A}$, then which one of the following is correct?

Options:
A $\mathrm{B}$ is an identity matrix
B $\quad \mathrm{B}=\mathrm{A}^{-1}$
C $\mathrm{B}=\mathrm{A}^{2}$
D Determinant of $\mathrm{B}$ is zero

The correct answer is:
$\mathrm{B}$ is an identity matrix

Solution:
Since, A and B are twonon-singular matrices therefore their determinant is non-zero. $\therefore \mathrm{A}^{-1}$ and $\mathrm{B}^{-1}$ defined.
Consider $\mathrm{AB}=\mathrm{A} \Rightarrow \mathrm{A}^{-1} \mathrm{AB}=\mathrm{A}^{-1} \mathrm{~A} \Rightarrow \mathrm{B}=\mathrm{I}$

Topic: Mathematics, Matrices, NDA, NDA 2013 (Phase 1)
If $\mathrm{A}$ and $\mathrm{B}$ are two subsets of a set $\mathrm{X}$, then what is $A \cap(A \cup B)^{\prime}$ equal to?

Options:
A $\mathrm{A}$
B $\mathrm{B}$
C $\phi$
D $A^{\prime}$

The correct answer is:
$\phi$

Solution:
On applying Demorgan's law, we get
On applying associative law, we get,
$=\left(\mathrm{A} \cap \mathrm{A}^{\prime}\right) \cap \mathrm{B}^{\prime}=\phi \cap \mathrm{B}^{\prime}=\phi$

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $60^{\circ}$ to each other, then which one of the following is correct?

Options:
A $|\vec{a}+\vec{b}| < 1$
B $|\vec{a}+\vec{b}|>1$
C $|\vec{a}-\vec{b}| < 1$
D $|\vec{a}-\vec{b}|>1$

The correct answer is:
$|\vec{a}+\vec{b}|>1$

Solution:
$\begin{aligned}|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}| &=\sqrt{|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|^{2}} \\ &=\sqrt{|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+2|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}| \cos 60^{\circ}} \\ &=\sqrt{1^{2}+1^{2}+2 \cdot 1 \cdot 1 \cdot \frac{1}{2}}=\sqrt{3} \\ \therefore|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|>1 \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
If $\hat{a}$ and $\hat{b}$ are two unit vectors, then the vector
$(\hat{a}+\hat{b}) \times(\hat{a} \times \hat{b})$ is parallel to

Options:
A $\quad(\hat{a}-\hat{b})$
B $(\hat{a}+\hat{b})$
C $(2 \hat{\mathrm{a}}-\hat{\mathrm{b}})$
D $(2 \hat{a}+\hat{b})$

The correct answer is:
$\quad(\hat{a}-\hat{b})$

Solution:
$\begin{aligned} & \\ \text { } &(\hat{a}+\hat{b}) \times(\hat{a} \times \hat{b}) \\ &=\hat{a} \times(\hat{a} \times \hat{b})+\hat{b} \times(\hat{a} \times \hat{b}) \\ &=(\hat{a} \cdot \hat{b}) \hat{a}-(\hat{a} \cdot \hat{a}) \hat{b}+(\hat{b} \cdot \hat{b}) \hat{a}-(\hat{b} \cdot \hat{a}) \hat{b} \\ &=\lambda \hat{a}-\hat{b}+\hat{a}-\lambda \hat{b} \\ &=\lambda(\hat{a}-\hat{b})+1(\hat{a}-\hat{b}) . \\ &=(\lambda+1)(\hat{a}-\hat{b}) \end{aligned}$
So, it is parallel to $\hat{a}-\hat{b}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 2)
If $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a} \bullet \vec{b}=0$ and
$\vec{a} \times \vec{b}=0$, then which one of the following is correct?

Options:
A $\vec{a} \text { is parallel to } \vec{b}$
B $\vec{a}$ is perpendicular to $\vec{b}$
C Either $\vec{a}$ or $\vec{b}$ is a null vector
D None of the above

The correct answer is:
Either $\vec{a}$ or $\vec{b}$ is a null vector

Solution:
Let $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a} \cdot \vec{b}=0$ and
$\vec{a} \times \vec{b}=0$ then either $\vec{a}$ or $\vec{b}$ is a null vector.

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 2)
If $A$ and $B$ are twosets satisfying $A-B=B-A$, then which one of the following is correct?

Options:
A $\quad A=\phi$
B $A \cap B=\phi$
C $A=B$
D None of these

The correct answer is:
$A=B$

Solution:
We drawthe Venn diagram,
$A-B=A-(A \cap B)$ and $B-A=B-(A \cap B)$
$\mathrm{A}-\mathrm{B}=\mathrm{B}-\mathrm{A}$
$\Rightarrow \quad A-(A \cap B)=B-(A \cap B)$
$\Rightarrow A=B$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
$\mathrm{If} \overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are unit vectors inclined at an angle of $30^{\circ}$ to each other, then which one of the following is correct?

Options:
A $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|>1$
B $\quad 1 < |\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}| < 2$
C $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=2$
D $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|>2$

The correct answer is:
$\quad 1 < |\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}| < 2$

Solution:
$\begin{aligned} &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{b}} \cos \theta \\ \Rightarrow &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=1+1+2|\overrightarrow{\mathrm{a}} \| \overrightarrow{\mathrm{b}}| \cos 30^{\circ} \\ &=1+1+2 \times \frac{\sqrt{3}}{2} \\ \Rightarrow &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=2+\sqrt{3} \\ \Rightarrow &|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{2+\sqrt{3}} \\ & 1 < \sqrt{2+\sqrt{3}} < 2 \\ \Rightarrow & 1 < |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}| < 2 \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
If $\vec{a}$ and $\vec{b}$ are unit vectors, then what is the value of
$|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}$ ?

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

The correct answer is:
1

Solution:
$\quad|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^{2}+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^{2}$
$=(|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}| \cdot \sin \theta)^{2}+(|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}| \cdot \cos \theta)^{2}$
$=(1.1 \cdot \sin \theta)^{2}+(1.1 \cdot \cos \theta)^{2}$
$=\sin ^{2} \theta+\cos ^{2} \theta=1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
If $\vec{a}$ and $\overrightarrow{\mathrm{b}}$ are vectors such that $|\vec{a}|=2,|\overrightarrow{\mathrm{b}}|=7$ and
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$, then what is the acute angle between
$\vec{a}$ and $\overrightarrow{\mathrm{b}}$ ?

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$\begin{aligned} &|\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=7 \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \end{aligned}$
We know, $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta \cdot \hat{\mathrm{n}}$ where $\hat{\mathrm{n}}$ is unit
vector.
$\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta$
$\Rightarrow|3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}|=(2)(7) \sin \theta$
$\Rightarrow \sqrt{9+4+36}=(2)(7) \sin \theta .$
$\Rightarrow \pm 7=(2)(7) \sin \theta .$
$\therefore \sin \theta=\pm \frac{1}{2} .$
$\sin \theta$ is acute angle, $\theta=30^{\circ}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 1)
If $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$ and $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0}$ then which one of the following is
correct ?

Options:
A $\overrightarrow{\mathrm{a}}$ is parallel to $\overrightarrow{\mathrm{b}}$
B $\overrightarrow{\mathrm{a}}$ is perpendicular to $\overrightarrow{\mathrm{b}}$
C $\overrightarrow{\mathrm{a}}=\overrightarrow{0}$ or $\overrightarrow{\mathrm{b}}=\overrightarrow{0}$
D None of the above

The correct answer is:
$\overrightarrow{\mathrm{a}}=\overrightarrow{0}$ or $\overrightarrow{\mathrm{b}}=\overrightarrow{0}$

Solution:
$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \Rightarrow \overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{b}}$
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=0 \Rightarrow \overrightarrow{\mathrm{a}} \| \overrightarrow{\mathrm{b}}$
(ii) it is clear that $\overrightarrow{\mathrm{a}}=0$ or $\overrightarrow{\mathrm{b}}=0$ From (i) and

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
If $\mathrm{A}+\mathrm{B}=90^{\circ}$, then what is $\sqrt{\sin \mathrm{A} \sec \mathrm{B}-\sin \mathrm{A} \cos \mathrm{B}}$
$\begin{array}{ll}\text { equal to ? }

Options:
A $\sin \mathrm{A}$
B $\cos A$
C $\tan \mathrm{A}$
D 0

The correct answer is:
$\cos A$

Solution:
Let $\mathrm{A}+\mathrm{B}=90^{\circ}$
Consider $\sqrt{\sin A \sec B-\sin A \cos B}$
$\begin{aligned} &=\sqrt{\sin \mathrm{A} \sec \left(90^{\circ}-\mathrm{A}\right)-\sin \mathrm{A} \cos \left(90^{\circ}-\mathrm{A}\right)} \\ &=\sqrt{\sin \mathrm{A} \operatorname{cosec} \mathrm{A}-\sin \mathrm{A} \sin \mathrm{A}}=\sqrt{1-\sin ^{2} \mathrm{~A}}=\cos \mathrm{A} . \\ \text { 140. (a) } & \text { Consider } \tan ^{4} \mathrm{~A}-\sec ^{4} \mathrm{~A}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} \\ &=\left(\tan ^{2} \mathrm{~A}\right)^{2}-\left(\sec ^{2} \mathrm{~A}\right)^{2}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} . \\ &=\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)\left(\tan ^{2} \mathrm{~A}-\sec ^{2} \mathrm{~A}\right) \\ &=\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} \\ &=-\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}=0 . \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
If $|\vec{a}+\vec{b}|=\vec{a}-\vec{b} \mid$, then which one of the following is correct?

Options:
A $\vec{a}=\lambda \vec{b}$ for some scalar $\lambda$
B $\vec{a}$ is parallel to $\vec{b}$
C $\vec{a}$ is perpendicular to $\vec{b}$
D $\vec{a}=\vec{b}=\overrightarrow{0}$

The correct answer is:
$\vec{a}$ is perpendicular to $\vec{b}$

Solution:
If $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|$
We know that when $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0$, then $\vec{a}$ is
perpendicular to $\overrightarrow{\mathrm{b}}$. $=\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}=0$
$\vec{a}$ is perpendicular to $\overrightarrow{\mathrm{b}}$
$\therefore \quad$ Option (c) is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 1)
If $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$,then which one of the following is correct ?

Options:
A $|\vec{a}|=|\vec{b}|$.
B $\vec{a}$ is parrallel to $\vec{b}$.
C $\vec{a}$ is perpendicular to $\vec{b}$.
D $\vec{a}$ is a unit vector .

The correct answer is:
$\vec{a}$ is perpendicular to $\vec{b}$.

Solution:
Since, $|a+b|=|a-b|$ $\Rightarrow[|a+b|]^{2}=[a-b]^{2}$
$\Rightarrow a \cdot a+b \cdot b+a \cdot b+b \cdot a=a \cdot a+b \cdot b-a \cdot b-b \cdot a$
$\Rightarrow 4 a \cdot b=0 \quad(\because a \cdot b=b . a)$
$\Rightarrow a \cdot b=0$
Hence, a is perpendicular to $\mathrm{b}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
If $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$, then which one of the following is correct?

Options:
A $\vec{a}$ is parallel to $\vec{b}$
B $\vec{a}$ is perpendicular to $\vec{b}$
C $\vec{a}=\vec{b}$
D Both $\vec{a}$ and $\vec{b}$ are unit vectors

The correct answer is:
$\vec{a}$ is perpendicular to $\vec{b}$

Solution:
Given, $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|$
$\Rightarrow|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|^{2}=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}$
$\Rightarrow|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+2|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}-2|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|$
$\Rightarrow 4|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|=0$
$\Rightarrow \overrightarrow{\mathbf{a}}$ is perpendicular to $\overrightarrow{\mathbf{b}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 1)
If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are angles of a triangle such that $\tan \mathrm{A}=1$, tan $B=2$, then what is the value of $\tan C$ ?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
In any triangle $\mathrm{ABC}, \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$
or, $\mathrm{A}+\mathrm{B}=\pi-\mathrm{C}$
so, $\tan (A+B)=\tan (\pi-C)$
or $-\tan C=\tan (A+B)$
so, $\tan C=-\tan (A+B)$
$=-\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}$
As given, $\tan A=1$, and $\tan B=2$
putting these values,
$\tan C=-\frac{1+3}{1-1 \times 3}=\frac{4}{2}=2$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2006 (Phase 2)
If A. B and $\mathrm{C}$ are any three arbitraryevents then which one of the following expressions shows that both $\mathrm{A}$ and $\mathrm{B}$ occur but not C ?

Options:
A $\mathrm{A} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}}$
B $\mathrm{A} \cap \mathrm{B} \cap \overline{\mathrm{C}}$
C $\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}}$
D $(\mathrm{A} \cup \mathrm{B}) \cap \overline{\mathrm{C}}$

The correct answer is:
$\mathrm{A} \cap \mathrm{B} \cap \overline{\mathrm{C}}$

Solution:
If A, B and C are any three arbitrary events occurrence of both A and B is given by A $\cap$ Band non-occurrence
of $\mathrm{C} a s \overline{\mathrm{C}}$ then the event both $\mathrm{A}$ and $\mathrm{B}$ occur but not $\mathrm{C}$
is represented by $\mathrm{A} \cap \mathrm{B} \cap \overline{\mathrm{C}}$.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
If $A, B$ and $C$ are in AP and $b: c=\sqrt{3}: \sqrt{2}$, then what is the value of sin C?

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

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

Solution:
Let $\mathrm{a}-\mathrm{d}$, a and $\mathrm{a}+\mathrm{d}$ be three numbers which are in
A.P. since $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in $\mathrm{A} . \mathrm{P}$. $\therefore \quad A=a-d, B=a, C=a+d$
$\Rightarrow \quad a-d+a+a+d=180^{\circ}$
$(\because \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are angles of a triangle $)$ $\Rightarrow \mathrm{a}=60^{\circ}$
$\Rightarrow A=60^{\circ}-d, B=60^{\circ}, C=60^{\circ}+d$
Now by sine rule,
$\frac{b}{c}=\frac{\sin B}{\sin C} \Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\sin 60^{\circ}}{\sin C}$
$\Rightarrow \quad \sin C=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}}=\frac{1}{\sqrt{2}}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 1)
If \( a, b \) and \( c \) are in A.P., then the value of \( \left|\begin{array}{lll}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c\end{array}\right| \) is

Options:
A \( x-(a+b+c) \)
B \( 9 x^{2}+a+b+c \)
C \( 00 \)
D \( a+b+c \)

The correct answer is:
\( 00 \)

Solution:
Given \( \left|\begin{array}{ccc}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+6\end{array}\right| \) \( R_{1} \rightarrow R_{1}-R_{2} ; R_{2} \rightarrow R_{2}-R_{3} \) \( \Delta=\left|\begin{array}{ccc}-2 & -2 & a-b \\ -2 & -2 & b-c \\ x+6 & x+7 & x+c\end{array}\right| \) Since, a, b, c are in A.P. then \( a-b=b-c \) So, \( R_{1} \equiv R_{2} \) Therefore, \( \Delta=0 \)

Topic: Mathematics, Mathematical Reasoning, KCET, KCET 2014
If $A, B$ and $C$ are non-empty sets such that
$A \cap C=\phi$, then what is $(A \times B) \cap(C \times B)$ equal to?

Options:
A $\mathrm{A} \times \mathrm{C}$
B $\mathrm{A} \times \mathrm{B}$
C $\mathrm{B} \times \mathrm{C}$
D $\phi$

The correct answer is:
$\phi$

Solution:
Let $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are non-empty sets such that $\mathrm{A} \cap \mathrm{C}=\phi$
Consider, $(A \times B) \cap(C \times B)=(A \cap C) \times(B \cap B)$
$=(\mathrm{A} \cap \mathrm{C}) \times \mathrm{B}=\phi \times \mathrm{B}=\phi$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)