datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are three events, then what is the probability that at least two of these events occur together?

Options:
A $\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{C} \cap \mathrm{A})$
B $\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{C} \cap \mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$
C $\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{C} \cap \mathrm{A})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$
D $P(A \cap B)+P(B \cap C)+P(C \cap A)-3 P(A \cap B \cap C)$

The correct answer is:
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{C} \cap \mathrm{A})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$

Solution:
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
If $A, B, C$ are three sets such that $A \cup B=A \cup C$ and $A \cap B=A \cap C$, then which one of the following is correct?

Options:
A $A=\underline{B}$ only
B $B=C$ only
C $A=C$ only
D $A=B=C$

The correct answer is:
$B=C$ only

Solution:
Let $A, B, C$ be three sets such that $A \cup B=A \cup C$ and $A \cap B=A \cap C$
Let $x \in A \cup B \Rightarrow x \in A$ or $x \in B$
(1) Since, $A \cup B=A \cup C$ there fore
$x \in A$ or $x \in C$
Also, Given $A \cap B=A \cap C$ $x \in A \cap B \Rightarrow x \in A$ and $x \in B$
and $x \in A$ and $x \in C(\because A \cap B=A \cap C)$
Thus, from $(1),(2),(3)$, and $(4)$, we have $B=C$ only

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are three sets, then what is $\mathrm{A}-(\mathrm{B}-\mathrm{C})$ equal to?

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

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

Solution:
Following venn diagram shows the relation $\mathrm{A}-(\mathrm{B}-\mathrm{C})$
In the above venn diagram, horizontal lines shows
(A-B) and vertical lines show $(\mathrm{A} \cap \mathrm{C})$
$(A-B) \cup(A \cap C)=A-(B-C)$

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to
the plane of $\vec{b}, \vec{c}$; and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$
Then, what is $|\vec{a}+\vec{b}+\vec{c}| ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
As given: a is perpendicular to $\overrightarrow{\mathrm{b}}$ and $\mathrm{c}$ $\Rightarrow \vec{a} \cdot \vec{b}=0 \& \overrightarrow{a \cdot c}=0$
and angle between $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}=\frac{\pi}{3}$
$\therefore \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \frac{\pi}{3}=1.1 \cdot \frac{1}{2}$
$=\frac{1}{2} \quad(\because \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are unit vectors)
Now, $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+|\vec{c}|^{2}$
$+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a})$
$=1+1+1+2 \cdot\left(0+\frac{1}{2}+0\right)$
$=1+1+1+1=4$
$\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|=2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
If $(a, b),(c, d)$ and $(a-c, b-d)$ are collinear, then which one of the following is correct?

Options:
A $\mathrm{bc}-\mathrm{ad}=0$
B $a b-c d=0$
C $\mathrm{bc}+\mathrm{ad}=0$
D $a b+c d=0$

The correct answer is:
$\mathrm{bc}-\mathrm{ad}=0$

Solution:
Let $A, B$ and $C$ having co-ordinates $(a, b),(c, d)$ and $\{(a-c),(b-d)\}$ respectively be the points If these poins are collinear then
$\left|\begin{array}{ccc}a & b & 1 \\ c & d & 1 \\ a-c & b-d & 1\end{array}\right|=0$
$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ gives
$\left|\begin{array}{ccc}a & b & 1 \\ c-a & d-b & 0 \\ a-c & b-d & 1\end{array}\right|=0$
$\mathrm{R}_{3} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3}$ gives
$\left|\begin{array}{ccc}a & b & 1 \\ c-a & d-b & 0 \\ 0 & 0 & 1\end{array}\right|=0$
$\Rightarrow 1 .\{\mathrm{a}(\mathrm{d}-\mathrm{b})-\mathrm{b}(\mathrm{c}-\mathrm{a})\}=0$
$\Rightarrow \mathrm{ad}-\mathrm{ab}-\mathrm{bc}+\mathrm{ab}=0$
$\Rightarrow b c-a d=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2008 (Phase 1)
If $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in harmonical progression such that $\mathrm{a}>\mathrm{d}$, then which one of the following is correct?

Options:
A $a+c=b+d$
B $a+c>b+d$
C $\mathrm{ac}=\mathrm{bd}$
D $a b=c d$

The correct answer is:
$a+c>b+d$

Solution:
Since, $\mathrm{a}>\mathrm{d}$ and $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in $\mathrm{HP}$.
$\Rightarrow \mathrm{a}>\mathrm{b}>\mathrm{c}>\mathrm{d}$.
$\mathrm{b}=\frac{2 \mathrm{a} \mathrm{c}}{\mathrm{a}+\mathrm{c}} \Rightarrow \mathrm{b}^{2} \mathrm{~d}=\frac{2 \mathrm{a} \mathrm{b} \mathrm{cd}}{\mathrm{a}+\mathrm{c}}$
and $c=\frac{2 b d}{b+d} \Rightarrow c^{2} a=\frac{2 a b c d}{b+d}$
$\frac{c^{2} a}{b^{2} d}=\frac{a+c}{b+d}$
$\Rightarrow \quad \frac{a+c}{b+d}=\left(\frac{a}{b}\right) \cdot\left(\frac{c}{d}\right) \cdot\left(\frac{c}{b}\right)>1$
$\Rightarrow a+c>b+d$

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
If $a, b, c, d, e, f$ are in AP, then $(e-c)$ is equal to which one of the following?

Options:
A $2(c-a)$
B $2(d-c)$
C $2(f-d)$
D $(d-c)$

The correct answer is:
$2(d-c)$

Solution:
Given, $a, b, c, d, e, f$ are in $A . P$
$\therefore \quad 2 \mathrm{~d}=\mathrm{e}+\mathrm{c}$...(1)
Consider $\mathrm{e}-\mathrm{c}=2 \mathrm{~d}-\mathrm{c}-\mathrm{c} \quad$ (from 1$)$
$=2 \mathrm{~d}-2 \mathrm{c}=2(\mathrm{~d}-\mathrm{c})$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 2)
If $a \neq b \neq c$, then one value of $x$ which satisfies the equation $\left|\begin{array}{ccc}0 & \mathrm{x}-\mathrm{a} & \mathrm{x}-\mathrm{b} \\ \mathrm{x}+\mathrm{a} & 0 & \mathrm{x}-\mathrm{c} \\ \mathrm{x}+\mathrm{b} & \mathrm{x}+\mathrm{c} & 0\end{array}\right|=0$ is given by

Options:
A a
B $\mathrm{b}$
C $\mathrm{c}$
D 0

The correct answer is:
0

Solution:
$\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0$
We know, the value of symmetric matrix's determinant is $0\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|=0$
In the given matrix, determinant is 0, if $x=0$

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 1)
If $A+B+C=\pi$, then what is $\cos (A+B)+\cos C$ equal to ?

Options:
A 0
B $2 \cos C$
C $\cos C-\sin C$
D $2 \sin C$

The correct answer is:
0

Solution:
$A+B+C=\pi$
$\quad A+B=\pi-C$
$\quad \cos (A+B)=\cos (\pi-C)$
$\quad \cos (A+B)=-\cos C$
$\quad$ or $\cos (A+B)+\cos C=0$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2014 (Phase 1)
If a, b, care in geometric progression and a, $2 \mathrm{~b}, 3 \mathrm{c}$ are in arithmetic progression, then what is the common ratio $\mathrm{r}$ such that $0 < \mathrm{r} < 1 ?$

Options:
A $\frac{1}{3}$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{3}$

Solution:
Given that a, b, c, are in GP. Let r be common ratio of GP. So, $a=a, b=a r$ and $c=a r^{2}$ ...(i)
Also, given that $a, 2 b, 3 c$ are in AP.
$\Rightarrow 2 b=\frac{a+3 c}{2}$
$\Rightarrow \quad 4 \mathrm{~b}=\mathrm{a}+3 \mathrm{c}$
From Eq. (1) $4 a r=a+3 a r^{2}$
$\Rightarrow \quad 3 \mathrm{ar}^{2}-4 \mathrm{ar}+\mathrm{a}=0$
$\Rightarrow \quad 3 r^{2}-4 r+1=0$
$\Rightarrow \quad 3 r^{2}-3 r-r+1=0$
$\Rightarrow 3 \mathrm{r}(\mathrm{r}-1)-1(\mathrm{r}-1)=0$
$\Rightarrow(\mathrm{r}-1)(3 \mathrm{r}-1)=0$
$\Rightarrow \mathrm{r}=1$ or $\mathrm{r}=\frac{1}{3}, \mathrm{r}=\frac{1}{3}$ is in the choice.

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
If $(\mathrm{a}, \mathrm{b})$ is at unit distance from the line $8 \mathrm{x}+6 \mathrm{y}+1=0$, then which of the following conditions are correct?
1) $\quad 3 \mathrm{a}-4 \mathrm{~b}-4=0$
2) $8 \mathrm{a}+6 \mathrm{~b}+11=0$
3) $\quad 8 a+6 b-9=0$
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:
2 and 3 only

Solution:
Here $\frac{|8 a+6 b+1|}{\sqrt{8^{2}+6^{2}}}=1 \Rightarrow|8 a+6 b+1|=10$
$\Rightarrow 8 a+6 b+1=\pm 10$
$\Rightarrow 8 \mathrm{a}+6 \mathrm{~b}+1=10 \& 8 \mathrm{a}+6 \mathrm{~b}+1=-10$
$\Rightarrow 8 a+6 b-9=0 \& 8 a+6 b+11=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)
If $A \subseteq B$, then which one of the following is not correct?

Options:
A $\quad P(A \cap \bar{B})=0$
B $\quad P(A \mid B)=\frac{P(A)}{P(B)}$
C $\quad P(B \mid A)=\frac{P(B)}{P(A)}$
D $P(A \mid(A \cup B))=\frac{P(A)}{P(B)}$

The correct answer is:
$\quad P(B \mid A)=\frac{P(B)}{P(A)}$

Solution:
No Solution

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 1)
If $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{x}^{3} \mathrm{dx}=0$ and $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{x}^{2} \mathrm{dx}=\frac{2}{3}$, then what are the values of a and b respectively?

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

The correct answer is:
$-1,1$

Solution:
$\begin{aligned} & \int_{a}^{b} x^{3} \cdot d x=0 \\ \Rightarrow &\left(\frac{x^{4}}{4}\right)_{a}^{b}=0 \\ \Rightarrow & b^{4}-a^{4}=0 \\ \Rightarrow &\left(b^{2}-a^{2}\right)\left(b^{2}+a^{2}\right)=0 \\ \Rightarrow & b^{2}-a^{2}=0(\text { or }) b^{2}+a^{2}=0 \\ \Rightarrow &(b+a)(b-a)=0(\text { or }) b^{2}+a^{2}=0 \\ \Rightarrow & b=-a(\text { or }) a=b(\text { or }) b^{2}+a^{2}=0 \end{aligned}$
$\int_{a}^{b} x^{2} \cdot d x=\frac{2}{3}$
$\left(\frac{x^{3}}{3}\right)_{a}^{b}=\frac{2}{3}$
$\Rightarrow b^{3}-a^{3}=2$
Since, $b=-a$
$\Rightarrow b^{3}-\left(-b^{3}\right)=2$
$\Rightarrow 2 b^{3}=2$
$\Rightarrow b^{3}=1$
$\Rightarrow b=1, a=-1$

Topic: Mathematics, Definite Integration, NDA, NDA 2018 (Phase 1)
If $A$ be a real skew- symmetric matrix of order $\mathrm{n}$ such that $A^{2}+I=0$, I being the identity matrix of the same order as that of $A$, then what is the order of $A$?

Options:
A 3
B Odd
C Prime number
D Even

The correct answer is:
3

Solution:
We know if $A$ is a real skew-symmetric matrix of order $n$ such that $A^{2}+I=0$, then order of $A$ is 3.

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 1)
If a circle of radius b units with centre at $(0, \mathrm{~b})$ touches the
line $\mathrm{y}=\mathrm{x}-\sqrt{2}$, then what is the value of b?

Options:
A $2+\sqrt{2}$
B $2-\sqrt{2}$
C $2 \sqrt{2}$
D $\sqrt{2}$

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

Solution:
Distance from the centre to the point of line which touches circle is $\mathrm{OM}=$ radius
$$
y=x-\sqrt{2} \text { or } x-y-\sqrt{2}=0
$$
$r=\frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}}$
$\because x_{0}=0$
$\& Y_{0}=b$
$b=\frac{|1(0)+(-1)(b)-\sqrt{2}|}{\sqrt{(1)^{2}+(-1)^{2}}}$
$b=\frac{b+\sqrt{2}}{\sqrt{2}}$
$(\sqrt{2}-1) b=\sqrt{2}$
$\Rightarrow b=2+\sqrt{2}$

Topic: Mathematics, Circle, NDA, NDA 2016 (Phase 1)
If $\mathrm{A}=\left(\cos 12^{\circ}-\cos 36^{\circ}\right)\left(\sin 96^{\circ}+\sin 24^{\circ}\right)$
and $\mathrm{B}=\left(\sin 60^{\circ}-\sin 12^{\circ}\right)\left(\cos 48^{\circ}-\cos 72^{\circ}\right)$, then what is
$\frac{\mathrm{A}}{\mathrm{B}}$ equal to?

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

The correct answer is:
0

Solution:
Given A=\left(\cos 12^{\circ}-\cos 36^{\circ}\right)\left(\sin 96^{\circ}+\sin 24^{\circ}\right) \\
B=\left(\sin 60^{\circ}-\sin 12^{\circ}\right)\left(\cos 48^{\circ}-\cos 72^{\circ}\right) \\
& \frac{A}{B}=\frac{\left[2 \sin 24^{\circ} \sin 12^{\circ}\right]\left[2 \sin 60^{\circ} \cos 36^{\circ}\right]}{\left[2 \cos 36^{\circ} \sin 24^{\circ}\right]\left[2 \sin 60^{\circ} \sin 12^{\circ}\right]} \\
& \Rightarrow \frac{A}{B}=1 \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{ll}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then what is $\mathbf{A}^{3}$ equal to?

Options:
A $\left[\begin{array}{ll}\cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta\end{array}\right]$
B $\left[\begin{array}{cc}\cos ^{3} \theta & \sin ^{3} \theta \\ -\sin ^{3} \theta & \cos ^{3} \theta\end{array}\right]$
C $\left[\begin{array}{cc}\cos 3 \theta & -\sin 3 \theta \\ \sin 3 \theta & \cos 3 \theta\end{array}\right]$
D $\left[\begin{array}{cc}\cos ^{3} \theta & -\sin ^{3} \theta \\ \sin ^{3} \theta & \cos ^{3} \theta\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ll}\cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta\end{array}\right]$

Solution:
$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
We know, $A^{n}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$
$\therefore A^{3}=\left[\begin{array}{cc}\cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ then what is $\mathrm{AA}^{\mathrm{T}}$ equal to (where $\begin{array}{ll} \left.\mathrm{A}^{\mathrm{T}} \text { is the transpose of } \mathrm{A}\right) ?\end{array}$

Options:
A Null matrix
B Identify matrix
C $\mathrm{A}$
D $-\mathrm{A}$

The correct answer is:
Identify matrix

Solution:
$A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$A A^{T}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$=\left[\begin{array}{cc}\cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha+\sin \alpha \cos \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 1)
If | a | denotes the absolutevalue of an integer, then which of the following are correct?
1.$|\mathrm{ab}|=|\mathrm{a}||\mathrm{b}|$
$2.|\mathrm{a}+\mathrm{b}| \leq|\mathrm{a}|+|\mathrm{b}|$
$3.\quad|\mathrm{a}-\mathrm{b}| \geq|| \mathrm{a}|-| \mathrm{b} \mid$
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:
All the given three options are correct.

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2017 (Phase 2)
If a differentiable function $\mathrm{f}$ defined for $\mathrm{x}>0$ satisfies the relation $\mathrm{f}\left(\mathrm{x}^{2}\right)=\mathrm{x}^{3}, \mathrm{x}>0$, then what is the value of $\mathrm{f}^{\prime}(4) ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
According to given relation.
$\because \mathrm{f}\left(\mathrm{x}^{2}\right)=\mathrm{x}^{3}$
Putting $\mathrm{x}=\sqrt{\mathrm{x}}$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}^{3 / 2}$
Differentiating both the sides,
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{3}{2} \mathrm{x}^{1 / 2}$
$\Rightarrow \quad f^{\prime}(4)=\frac{3}{2} \cdot 4^{1 / 2}=\frac{3}{2}(2)=3$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
If a flag-staff of $6 \mathrm{~m}$ height placed on the top of a tower
throws a shadow of $2 \sqrt{3} \mathrm{~m}$ along the ground, then what is the angle that the sun makes with the ground?

Options:
A $60^{\circ}$
B $45^{\circ}$
C $30^{\circ}$
D $15^{\circ}$

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

Solution:
$\mathrm{AB}$ is the tower. $\mathrm{BC}$ is flag staff. The angles made by the shadows of tower and flag staff are same.
In $\Delta \mathrm{ABE}, \tan \theta=\frac{\mathrm{h}}{\mathrm{x}}$
In $\Delta \mathrm{ACD}, \tan \theta=\frac{\mathrm{h}+6}{\mathrm{x}+2 \sqrt{3}}$
(2), $\frac{\mathrm{h}}{\mathrm{x}}=\frac{\mathrm{h}+6}{\mathrm{x}+2 \sqrt{3}} \Rightarrow \mathrm{hx}+2 \sqrt{3} \mathrm{~h}=\mathrm{hx}+6 \mathrm{x}$
from(1),
$\Rightarrow \frac{\mathrm{h}}{\mathrm{x}}=\frac{6}{2 \sqrt{3}}=\frac{3}{\sqrt{3}}=\frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}}=\sqrt{3}$
from(1), $\tan \theta=\frac{\mathrm{h}}{\mathrm{x}}=\sqrt{3} \Rightarrow \theta=60^{\circ}$.

Topic: Mathematics, Heights and Distances, NDA, NDA 2018 (Phase 1)
If a function \(f\) is defined by :
\(\begin{aligned}
f(x) & =0, \text { when } x=1, \\
& =x^3-1, \text { when } 1 < x < \infty,
\end{aligned}\)
\(=x-1\), when \(-\infty < x < 1\), then at \(x=1, f\) is

Options:
A continuous and differentiable
B continuous but not differentiable
C discontinuous and differentiable
D discontinuous and not differentiable

The correct answer is:
continuous but not differentiable

Solution:
We have,
\(f(x)=\left\{\begin{array}{rc}
x-1, & -\infty < x < 1 \\
0, & x=1 \\
x^3-1, & 1 < x < \infty
\end{array}\right.\)
Now, (LHL at \(x=1\) )
\(\begin{aligned}
& =\lim _{x \rightarrow 1}(x-1) \\
& =\mathbf{l}-\mathbf{l}=\mathbf{0}
\end{aligned}\)
(RHL at \(x=1\) ) \(=\lim _{x \rightarrow 1} x^3-1\)
\(=(\mathrm{l})^3-\mathbf{1}=0 \text { and } f(\mathrm{l})=0\)
\(\therefore \quad \text { LHL }=\text { RHL }=f(\mathrm{l})\)
So, \(f(x)\) is continuous at \(x=1\)
Now, (LHD at \(x=1\) )
\(=\lim _{x \rightarrow 1} \frac{(x-1)-0}{x-1}=1\)
and (RHD at \(x=1\) )
\(\begin{aligned}
& =\lim _{x \rightarrow 1} \frac{\left(x^3-1\right)-0}{x-1} \\
& =\lim _{x \rightarrow 1}\left(x^2+x+1\right)=3
\end{aligned}\)
\(\therefore\) LHD \(\neq\) RHD
So, \(f(x)\) is not differentiable at \(x=1\)

Topic: Mathematics, Continuity and Differentiability, AP EAMCET, AP EAMCET 2019 (23 Apr Shift 1)
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as:

Options:
A Gene mapping
B Expressed sequence tags
C Bioinformatics
D Sequence annotation

The correct answer is:
Sequence annotation

Solution:
In sequence annotation, the whole set of genome containing all coding and non-coding sequences is sequenced and the functions are assigned to different segments.

It could be used for marking specific features in a DNA, RNA or protein sequence with descriptive information about structure or function.

It helps in describing regions or sites of interest in the protein sequence, such as enzyme active sites, secondary structure or other characteristics reported in the cited references.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2022 (Phase 1)
If $A=\left(\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right), B=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right), C=\left(\begin{array}{ll}0 & i \\ i & 0\end{array}\right)$
where $i=\sqrt{-1}$, then which one of the following is correct?

Options:
A $A B=-C$
B $A B=C$
C $A^{2}=B^{2}=C^{2}=I$, where $I$ is the identity matrix
D $B A \neq C$

The correct answer is:
$A B=-C$

Solution:
$\mathrm{AB}=\left(\begin{array}{cc}\mathrm{i} & 0 \\ 0 & -1\end{array}\right)\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)=\left(\begin{array}{cc}0 & -1 \\ -\mathrm{i} & 0\end{array}\right)$
$\mathrm{AB}=-\left(\begin{array}{ll}0 & \mathrm{i} \\ \mathrm{i} & 0\end{array}\right)=-\mathrm{C}$

Topic: Mathematics, Matrices, NDA, NDA 2013 (Phase 2)
If $\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=3 \overrightarrow{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}, \quad$ and $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})$ is
perpendicular to $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$, then what is the value of $\lambda ?$

Options:
A $-2$ only
B $\pm 2$
C 3 only
D $\pm 3$

The correct answer is:
$\pm 2$

Solution:
As given:
$\begin{array}{ll} & \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}} \\ & =4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+(\lambda-3) \hat{\mathrm{k}} \\ \text { and } & \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}} \\ & =-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-(3+\lambda) \hat{\mathrm{k}}\end{array}$
$(\vec{a}+\vec{b})$ is perpendicular to $(\vec{a}-\vec{b})$
$\Rightarrow \quad(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=0$
$\Rightarrow \quad\{\hat{\mathrm{i}}+\hat{\mathrm{j}}+(\lambda-3) \hat{\mathrm{k}}\}\{-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-(3-\lambda) \hat{\mathrm{k}}\}=0$
$\Rightarrow-8+3+\left(3^{2}-\lambda^{2}\right)=0$
$\Rightarrow 4-\lambda^{2}=0$
$\Rightarrow \lambda=\pm 2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 2)
If $\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$ then what is
$(\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})$ equal to?

Options:
A 106
B $-106$
C 53
D $-53$

The correct answer is:
$-106$

Solution:
$\begin{aligned} \vec{a} &=\hat{i}-2 \hat{j}+5 \hat{k} \\ \vec{b} &=2 \hat{i}+\hat{j}-3 \hat{k} \\ & \vec{b}-\vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}-(\hat{i}-2 \hat{j}+5 \hat{k}) \\ &=\hat{i}(2-1)+\hat{j}(1+2)+\hat{k}(-3-5)=\hat{i}+3 \hat{j}-8 \hat{k} \\ & 3 \vec{a}+\vec{b}=3(\hat{i}-2 \hat{j}+5 \hat{k})+(2 \hat{i}+\hat{j}-3 \hat{k}) \\ &=3 \hat{i}-6 \hat{j}+15 \hat{k}+2 \hat{i}+\hat{j}-3 \hat{k} \\ &=\hat{i}(3+2)+\hat{j}(-6+1)+\hat{k}(15-3) \\ &=5 \hat{i}-5 \hat{j}+12 \hat{k} \\ &(\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})=(\hat{i}+3 \hat{j}-8 \hat{k}) \cdot(5 \hat{i}-5 \hat{j}+12 \hat{k}) \\ &=(1)(5)+(3)(-5)+(-8)(12) \\ &=5-15-96=-106 \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2019 (Phase 1)
If $\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}, \quad \vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$, then what is
$(\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})$ equal to?

Options:
A 106
B $-106$
C 53
D $-53$

The correct answer is:
$-106$

Solution:
Given, $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
and $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
$\therefore \quad \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}$
and $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=(3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+15 \hat{\mathbf{k}})+(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})$
$\begin{aligned}=5 \hat{i}-5 \hat{j}+12 \hat{k} & \\ \text { Hence, }(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) &=(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}) \cdot(5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ &=5-15-96 \\ &=-106 \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 1)