datavorous/entrance-exam-dataset huggingface dataset:
If p be the length of the perpendicular from the orgin on the straight line $\mathrm{x}+2 \mathrm{by}+2 \mathrm{p}=0$, then what is the value of b?

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

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

Solution:
Length of perpendicular from the origin on the straight line $x+2 b y-2 p=0$ is
$\left|\frac{0+2 b \times 0-2 p}{\sqrt{1^{2}+(2 b)^{2}}}\right|=p$
or $\mathrm{p}=\left|\frac{-2 \mathrm{p}}{\sqrt{1^{2}+4 \mathrm{~b}^{2}}}\right|$
or $p^{2}=\frac{4 p^{2}}{1+4 b^{2}}$
$\frac{4}{1+4 b^{2}}=1$
$\Rightarrow 1+4 b^{2}=4$ or $4 b^{2}=3$
$\Rightarrow \quad b^{2}=\frac{3}{4} \Rightarrow b=\pm \frac{\sqrt{3}}{2} \Rightarrow b=\frac{\sqrt{3}}{2}$ matches with the
given option.

Topic: Mathematics, Straight Lines, NDA, NDA 2007 (Phase 1)
If p be the length of the perpendicular from the origin on the
straight line $\mathrm{ax}+\mathrm{by}=\mathrm{p}$ and $\mathrm{b}=\frac{\sqrt{3}}{2}$, then what is the angle between the perpendicular and the positive direction of $\begin{array}{ll}\text { x-axis? } & {}\end{array}$

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

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

Solution:
Equation of line is $a x+b y-p=0$, then length of perpendicular, from the origin.
$p=\left|\frac{a \times 0+b \times 0-p}{\sqrt{a^{2}+b^{2}}}\right|$ or $p=\left|\frac{-p}{\sqrt{a^{2}+b^{2}}}\right|$
or $\left|\frac{1}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|=1$ or $\mathrm{a}^{2}+\mathrm{b}^{2}=1$
$\mathrm{b}=\frac{\sqrt{3}}{2}$ or $\mathrm{b}^{2}=\frac{3}{4}$
$a^{2}+\frac{3}{4}=1$
$\mathrm{a}^{2}=\frac{1}{4} \Rightarrow \mathrm{a}=\frac{1}{2} \quad\left[\mathrm{a}=-\frac{1}{2}\right.$ not taken since
angle is with $+$ ve direction to $\mathrm{x}$ -axis. $]$ Equation is $\frac{1}{2} x+\frac{\sqrt{3}}{2} y=p$ or $x \cos 60^{\circ}+y \sin 60^{\circ}=p$
Angle $=60^{\circ}$

Topic: Mathematics, Straight Lines, NDA, NDA 2007 (Phase 1)
If $\mathrm{p}=\operatorname{cosec} \theta-\cot \theta$ and $\mathrm{q}=(\operatorname{cosec} \theta+\cot \theta)^{-1}$, then which
one of the following is correct?

Options:
A $\mathrm{pq}=1$
B $\mathrm{p}=\mathrm{q}$
C $p+q=1$
D $\mathrm{p}+\mathrm{q}=0$

The correct answer is:
$\mathrm{p}=\mathrm{q}$

Solution:
& \mathrm{p}=\operatorname{cosec} \theta-\cot \theta \\
& \mathrm{q}=(\operatorname{cosec} \theta+\cot \theta)^{-1} \\
& \Rightarrow \frac{1}{q}=\operatorname{cosec} \theta+\cot \theta \\
& \text { We know, } \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1 \\
& \Rightarrow(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)=1 \\
& \Rightarrow\left(\frac{1}{q}\right)(p)=1 \\
& \Rightarrow \mathrm{p}=\mathrm{q}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $p$ is the length of the perpendicular drawn from the
origin to the line $\frac{x}{a}+\frac{y}{b}=1$, then which one of the following
is correct?

Options:
A $\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}$
B $\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}$
C $\frac{1}{p}=\frac{1}{a}+\frac{1}{b}$
D $\frac{1}{p}=\frac{1}{a}-\frac{1}{b}$

The correct answer is:
$\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}$

Solution:
Given equation of line is $\frac{x}{a}+\frac{y}{b}=1$
$\Rightarrow b x+a y-a b=0$
$p=\frac{|b \cdot 0+a \cdot 0-a b|}{\sqrt{b^{2}+a^{2}}}$
$p=\frac{a b}{\sqrt{b^{2}+a^{2}}}$
on squaring both side, we get
$p^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \Rightarrow \frac{1}{p^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{1}{b^{2}}+\frac{1}{a^{2}}$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 2)
If $p, q$ and $r$ are rational numbers, then the roots of the equation $x^{2}-2 p x+p^{2}-q^{2}+2 q r-r^{2}=0$ are

Options:
A complex
B pure imaginary
C irrational
D rational

The correct answer is:
rational

Solution:
The given equation is $x^{2}-2 p x+p^{2}-q^{2}+2 q r-r^{2}=0$ Now, discriminant $=(-2 p)^{2}-4(1)\left(p^{2}-q^{2}+2 q r-r^{2}\right)$
$=4 p^{2}-4 p^{2}+4 q^{2}-8 q r+4 r^{2}$
$=(2 q-2 r)^{2}=4(q-r)^{2}$
which is always greater than zero. Therefore, the roots of given equation are rational

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 1)
If $P, Q$ and $R$ are three non-collinear points, then what is
$P Q \cap P R$ equal to?

Options:
A Null set
B $\{P\}$
C $\{P, Q, R\}$
D $\{Q, R\}$

The correct answer is:
$\{P\}$

Solution:
Since, $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are three non-collinear points.
We have $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are like as

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
If $(p, q)$ is the point on the $x$ -axis equidistant from the points $(1,2)$ and $(2,3)$, then which one of the following is correct?

Options:
A $p=0, q=4$
B $p=4, q=0$
C $p=3 / 2, q=0$
D $p=1, q=0$

The correct answer is:
$p=4, q=0$

Solution:
Since $(p, q)$ is the point on the $\mathrm{x}$ -axis $\therefore q=0$
Let $P=(p, 0)$
$A=(1,2)$ and $B=(2,3)$
Given: $P A=P B$ $\Rightarrow P A^{2}=P B^{2}$
$\Rightarrow(1-p)^{2}+4=(2-p)^{2}+9$
$\Rightarrow 1+p^{2}-2 p-4-p^{2}+4 p=5$
$\Rightarrow 2 p=8$
$\Rightarrow p=4$
Hence, $p=4, q=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 2)
If $\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}=0$, then the determinant $\left|\begin{array}{lll}p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a\end{array}\right|$ equals

Options:
A 0
B 1
C $\mathrm{pa}+\mathrm{qb}+\mathrm{rc}$
D $\mathrm{pa}+\mathrm{qb}+\mathrm{rc}+\mathrm{a}+\mathrm{b}+\mathrm{c}$

The correct answer is:
0

Solution:
$\left|\begin{array}{lll}\mathrm{pa} & \mathrm{qb} & \mathrm{rc} \\ \mathrm{qc} & \mathrm{ra} & \mathrm{pb} \\ \mathrm{rb} & \mathrm{pc} & \mathrm{qa}\end{array}\right|=\mathrm{pa}\left(\mathrm{rqa}^{2}-\mathrm{p}^{2} \mathrm{bc}\right)-\mathrm{qb}\left(\mathrm{q}^{2} \mathrm{ac}-\mathrm{prb}^{2}\right)+$
$\mathrm{rc}\left(\mathrm{qpc}^{2}-\mathrm{r}^{2} \mathrm{ab}\right)$
$=\mathrm{pqra}^{3}-\mathrm{p}^{3} \mathrm{abc}-\mathrm{q}^{3} \mathrm{abc}+\mathrm{pqrb}^{3}+\mathrm{pqrc}^{3}-\mathrm{r}^{3} \mathrm{abc}$
$=\mathrm{pqr}\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)-\mathrm{abc}\left(\mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}\right)$
Given, $\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}=0$
$\Rightarrow \mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}=3$ pqr and $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=3 \mathrm{abc} .$
$\therefore \mathrm{pqr}\left(\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}\right)-\mathrm{abc}\left(\mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}\right)=\mathrm{pqr}(3 \mathrm{abc})-$
$\mathrm{abc}(3 \mathrm{pqr})=0$

Topic: Mathematics, Determinants, NDA, NDA 2018 (Phase 1)
If $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in AP as well as G.P., then which one of the following is correct?

Options:
A $\quad \mathrm{p}=\mathrm{q} \neq \mathrm{r}$
B $\quad p \neq q \neq r$
C $\quad \mathrm{p} \neq \mathrm{q}=\mathrm{r}$
D $\mathrm{p}=\mathrm{q}=\mathrm{r}$

The correct answer is:
$\mathrm{p}=\mathrm{q}=\mathrm{r}$

Solution:
Since $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in A.P.
$\therefore 2 \mathrm{q}=\mathrm{p}+\mathrm{r}$ ...(i)
$\mathrm{q}^{2}=\mathrm{pr} \Rightarrow \mathrm{q}=\sqrt{\mathrm{pr}}$
$\therefore 2 \sqrt{\mathrm{pr}}=\mathrm{p}+\mathrm{r}$
$(\sqrt{\mathrm{p}})^{2}+(\sqrt{\mathrm{r}})^{2}-2 \sqrt{\mathrm{p}} \cdot \sqrt{\mathrm{r}}=0$
$\Rightarrow(\sqrt{\mathrm{p}}-\sqrt{\mathrm{r}})^{2}=0$
$\Rightarrow \sqrt{\mathrm{p}}=\sqrt{\mathrm{r}}$
$\Rightarrow \mathrm{p}=\mathrm{r}$ ...(ii)
$2 q=2 p$
$\Rightarrow q=p$...(iii)
from (2) and (3)
$p=q=r$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 1)
If $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in one geometric progression and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in another geometric progression, then ap, bq, cr are in

Options:
A Arithmetic progression
B Geometric progression
C Harmonic progression
D None of the above

The correct answer is:
Geometric progression

Solution:
Let the common ratio be $\mathrm{K}_{1}$ for $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$. $\therefore \quad \mathrm{q}=\mathrm{K}_{1} \mathrm{p}$
$\mathrm{\&} \mathrm{r}=\left(\mathrm{K}_{1}\right)^{2} \mathrm{p}$
Let the common ratio be $\mathrm{K}_{2}$ for $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$
$\begin{array}{ll}\therefore & \mathrm{b}=\mathrm{K}_{2} \mathrm{a} \\ \ & \mathrm{c}=\left(\mathrm{K}_{2}\right)^{2} \mathrm{a} \\ \therefore & \mathrm{bq}=\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) \mathrm{ap} \\ \ & \mathrm{Cr}=\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)^{2} \text { ap } \\ \text { So } \mathrm{ap}, \mathrm{bq}, \mathrm{cr} \text { are in G.P. }\end{array}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2015 (Phase 2)
If $p, q, r$ are positive integers and $\omega$ is the cube root of unity and $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3 \mathrm{p}}+\mathrm{x}^{3 \mathrm{q}}+1+\mathrm{x}^{3 \mathrm{r}+2}$, then what is $\mathrm{f}(\omega)$ equal to?

Options:
A $\omega$
B $-\omega^{2}$
C $-\omega$
D 0

The correct answer is:
0

Solution:
Since $\omega$ is a cube root of unity $\therefore \omega^{3}=1$ and $1+\omega+\omega^{2}=0$
Let $\mathrm{f}(\mathrm{x})=\left(x^{3}\right) p+\left(x^{3}\right)^{\mathrm{q}} \cdot x+\left(x^{3}\right)^{r} \cdot x^{2}$
Now, put $x=\omega$ $\mathrm{f}(\omega)=\left(\omega^{3}\right)^{p}+\left(\omega^{3}\right)^{q} . \omega+\left(\omega^{3}\right)^{r} \cdot \omega^{2}$
$=1^{p}+1^{q} \cdot \omega+1^{r} . \omega^{2}=1+\omega+\omega^{2}=0$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
If $P, Q, R$ are the mid points of the sides $A B, B C, C A$, respectively of a triangle $A B C$ and if $\vec{a}, \vec{p}, \vec{q}$ are the position vector of $A, P, Q$ respectively, then what is the position vector of $R ?$

Options:
A $2 \vec{a}-(\vec{p}-\vec{q})$
B $(\vec{p}-\vec{q})-2 \vec{a}$
C $\quad \vec{a}-(\vec{p}-\vec{q})$
D $\vec{a} / 2-(\vec{p}-\vec{q}) / 2$

The correct answer is:
$\quad \vec{a}-(\vec{p}-\vec{q})$

Solution:
Let the position vectors of $B, C$ and $R$ are $\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ and
$\overrightarrow{\mathbf{r}}$ respectively.
$\therefore \overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}}{2} \Rightarrow \overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}}$
$\overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{2}$
$\Rightarrow \vec{c}=2 \vec{q}-\vec{b}$
$\Rightarrow \overrightarrow{\mathrm{c}}=2 \overrightarrow{\mathrm{q}}-(2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}})=2 \overrightarrow{\mathrm{q}}-2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}$
and $\overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}}{2}$
$=\frac{2 \overrightarrow{\mathrm{q}}-2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}}}{2}=\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
If $p=\sin \left(989^{\circ}\right) \cos \left(991^{\circ}\right)$, then which one of the following is correct?

Options:
A $p$ is finite and positive
B $p$ is finite and negative
C $p=0$
D $p$ is undefined

The correct answer is:
$p$ is finite and negative

Solution:
Given, $p=\sin \left(989^{\circ}\right) \cos \left(991^{\circ}\right)$
Which can be written as $=\sin \left(1080^{\circ}-91^{\circ}\right) \cos \left(1080^{\circ}-89^{\circ}\right)$
$=-\sin 91^{\circ} \cos 89^{\circ}$
$=-\sin \left(90^{\circ}+1^{\circ}\right) \cos 89^{\circ}$
$=-\cos 1^{\circ} \cos 89^{\circ}$
As $\cos 1^{\circ}$ and $\cos 89^{\circ}$ are positive. therefore their product is also $+\mathrm{ve}$ Hence, $p$ is finite and negative.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
If $\mathrm{p}=\tan \left(-\frac{11 \pi}{6}\right), \mathrm{q}=\tan \left(\frac{21 \pi}{4}\right)$ and $\mathrm{r}=\cot \left(\frac{283 \pi}{6}\right)$,
then which of the following is/are correct?
1. The value of $p \times r$ is 2 .
2. $\mathrm{p}, \mathrm{q}$ andr are in G.P. 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:
2 only

Solution:
\begin{array}{l}
\mathrm{p}=\tan \left(-\frac{11 \pi}{6}\right) \\
\mathrm{p}=-\tan \left(2 \pi-\frac{\pi}{6}\right) \\
\mathrm{p}=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \\
\mathrm{q}=\tan \left(\frac{21 \pi}{4}\right)=\tan \left(6 \pi-\frac{3 \pi}{4}\right) \\
\mathrm{q}=-\tan \frac{3 \pi}{4}=+\tan \frac{\pi}{4}=1 \\
\mathrm{r}=\cot \left(\frac{283 \pi}{6}\right)=\cot \left(46 \pi+\frac{7 \pi}{6}\right)=\cot \left(\pi+\frac{\pi}{6}\right) \\
\mathrm{r}=\cot \frac{\pi}{6}=\sqrt{3} \\
\mathrm{p} \times \mathrm{r}=\frac{1}{\sqrt{3}} \times \sqrt{3}=1 \\
\therefore \text { Statement(1) is incorrect. }
\end{array}
also
$\frac{\mathrm{p}}{\mathrm{q}}=\frac{\mathrm{q}}{\mathrm{r}}=\frac{1}{\sqrt{3}}$
So p, q, rare in G.P. $\therefore$ Statement $(2)$ is correct.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 2)
If $\mathrm{p}^{\text {th }}$ term of an $\mathrm{AP}$ is $\mathrm{q}$, and its $\mathrm{q}^{\text {th }}$ term is $\mathrm{p}$, then what is the common difference?

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

The correct answer is:
$-1$

Solution:
Let first term and common difference of an AP are a and d respectively. Its $\mathrm{P}^{\text {th }}$ term $=\mathrm{a}+(\mathrm{p}-1) \mathrm{d}=\mathrm{q}$ ...(i)
and $q^{\text {th }}$ term $=a+(q-1) d=p$ ...(ii)
Solving Eqs. (i) and (ii), we find
$a=p+q-1, d=-1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
If $p$ times the $p$ th term of an AP is $q$ times the $q$ th term, then what is the $(p+q)$ th term equal to? $[2010-I]$

Options:
A $p+q$
B $p q$
C 1
D 0

The correct answer is:
0

Solution:
Let a and d be the first term and common difference of an AP respectively $\mathrm{p}^{\text {th }}$ term $=\mathrm{a}+(\mathrm{p}-1) \mathrm{d}$
and $q^{\text {th }}$ term $=\mathrm{a}+(\mathrm{q}-1) \mathrm{d}$
According to question.
$p[a+(p-1) d]=q[a+(q-1) d]$
$\Rightarrow p a+\left(p^{2}-p\right) d=q a+\left(q^{2}-q\right) d$
$\Rightarrow(p-q) a=\left(q^{2}-p^{2}+p-q\right) d$
$\Rightarrow(p-q) a=(p-q)(-p-q+1) d$
$\Rightarrow a=-(p+q-1) d$
Now, $(p+q)^{t h}$ term $=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) d$
$\quad=-(p+q-1) d+(p+q-1) d=0$

Topic: Mathematics, Sequences and Series, NDA, NDA 2010 (Phase 1)
If $\mathrm{p}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta, \mathrm{q}=\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta$ and $\mathrm{p}^{2}+4 \mathrm{pq}+\mathrm{q}^{2}=$
$\mathrm{A} \mathrm{X}^{2}+\mathrm{BY}^{2}, 0 \leq 0 \leq \frac{\pi}{2}$
What is the value of A?

Options:
A 4
B 3
C 2
D 1

The correct answer is:
3

Solution:
\quad \begin{array}{l}\mathrm{p}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta \\ \mathrm{q}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta \\ \text { Given, } \mathrm{p}^{2}+4 \mathrm{pq}+\mathrm{q}^{2}=\mathrm{Ax}^{2}+\mathrm{By}^{2}\end{array} \end{aligned}$
Let us take $\theta=\frac{\pi}{4}$.
$p=x \cos \frac{\pi}{4}-y \cos \frac{\pi}{4}=\frac{x-y}{\sqrt{2}}$
$$
\begin{array}{l}
q=x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}=\frac{x+y}{\sqrt{2}} \\
p q=\frac{x^{2}-y^{2}}{2} \Rightarrow 2 p q=x^{2}-y^{2} \\
\Rightarrow 4 \mathrm{pq}=2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
\text { Now, } \mathrm{p}^{2}+\mathrm{q}^{2}=\mathrm{x}^{2} \cos ^{2} \theta+\mathrm{y}^{2} \sin ^{2} \theta-2 \mathrm{xy} \cos \theta \sin \theta \\
+\mathrm{x}^{2} \sin ^{2} \theta+\mathrm{y}^{2} \cos ^{2} \theta+2 \mathrm{x} \mathrm{sin} \theta \cos \theta=\mathrm{x}^{2}+\mathrm{y}^{2} \\
\text { From }(1),(2), \mathrm{p}^{2}+\mathrm{q}^{2}+4 \mathrm{pq}=\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
=3 \mathrm{x}^{2}-\mathrm{y}^{2}
\end{array}
$$
Comparing this with the given form, we get
$$
\theta=\frac{\pi}{4}, A=3, B=-1
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\mathrm{p}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta, \mathrm{q}=\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta$ and $\mathrm{p}^{2}+4 \mathrm{pq}+\mathrm{q}^{2}=$
$\mathrm{A} \mathrm{X}^{2}+\mathrm{BY}^{2}, 0 \leq 0 \leq \frac{\pi}{2}$
What is the value of B ?

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

The correct answer is:
$-1$

Solution:
\quad \begin{array}{l}\mathrm{p}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta \\ \mathrm{q}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta \\ \text { Given, } \mathrm{p}^{2}+4 \mathrm{pq}+\mathrm{q}^{2}=\mathrm{Ax}^{2}+\mathrm{By}^{2}\end{array} \end{aligned}$
Let us take $\theta=\frac{\pi}{4}$.
$p=x \cos \frac{\pi}{4}-y \cos \frac{\pi}{4}=\frac{x-y}{\sqrt{2}}$
$$
\begin{array}{l}
q=x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}=\frac{x+y}{\sqrt{2}} \\
p q=\frac{x^{2}-y^{2}}{2} \Rightarrow 2 p q=x^{2}-y^{2} \\
\Rightarrow 4 \mathrm{pq}=2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
\text { Now, } \mathrm{p}^{2}+\mathrm{q}^{2}=\mathrm{x}^{2} \cos ^{2} \theta+\mathrm{y}^{2} \sin ^{2} \theta-2 \mathrm{xy} \cos \theta \sin \theta \\
+\mathrm{x}^{2} \sin ^{2} \theta+\mathrm{y}^{2} \cos ^{2} \theta+2 \mathrm{x} \mathrm{sin} \theta \cos \theta=\mathrm{x}^{2}+\mathrm{y}^{2} \\
\text { From }(1),(2), \mathrm{p}^{2}+\mathrm{q}^{2}+4 \mathrm{pq}=\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
=3 \mathrm{x}^{2}-\mathrm{y}^{2}
\end{array}
$$
Comparing this with the given form, we get
$$
\theta=\frac{\pi}{4}, A=3, B=-1
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\mathrm{p}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta, \mathrm{q}=\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta$ and $\mathrm{p}^{2}+4 \mathrm{pq}+\mathrm{q}^{2}=$
$\mathrm{A} \mathrm{X}^{2}+\mathrm{BY}^{2}, 0 \leq 0 \leq \frac{\pi}{2}$
What is the value of \theta

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{4}$
D $\frac{\pi}{6}$

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

Solution:
\quad \begin{array}{l}\mathrm{p}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta \\ \mathrm{q}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta \\ \text { Given, } \mathrm{p}^{2}+4 \mathrm{pq}+\mathrm{q}^{2}=\mathrm{Ax}^{2}+\mathrm{By}^{2}\end{array} \end{aligned}$
Let us take $\theta=\frac{\pi}{4}$.
$p=x \cos \frac{\pi}{4}-y \cos \frac{\pi}{4}=\frac{x-y}{\sqrt{2}}$
$$
\begin{array}{l}
q=x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}=\frac{x+y}{\sqrt{2}} \\
p q=\frac{x^{2}-y^{2}}{2} \Rightarrow 2 p q=x^{2}-y^{2} \\
\Rightarrow 4 \mathrm{pq}=2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
\text { Now, } \mathrm{p}^{2}+\mathrm{q}^{2}=\mathrm{x}^{2} \cos ^{2} \theta+\mathrm{y}^{2} \sin ^{2} \theta-2 \mathrm{xy} \cos \theta \sin \theta \\
+\mathrm{x}^{2} \sin ^{2} \theta+\mathrm{y}^{2} \cos ^{2} \theta+2 \mathrm{x} \mathrm{sin} \theta \cos \theta=\mathrm{x}^{2}+\mathrm{y}^{2} \\
\text { From }(1),(2), \mathrm{p}^{2}+\mathrm{q}^{2}+4 \mathrm{pq}=\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
=3 \mathrm{x}^{2}-\mathrm{y}^{2}
\end{array}
$$
Comparing this with the given form, we get
$$
\theta=\frac{\pi}{4}, A=3, B=-1
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If Planck's constant $(h)$ and speed of light in vacuum $(c)$ are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?

Options:
A
Mass of electron $\left(m_e\right)$

B
Universal gravitational constant $(G)$

C
Charge of electron $(e)$

D
Mass of proton $\left(m_p\right)$

The correct answers are:
Mass of electron $\left(m_e\right)$
,
Universal gravitational constant $(G)$
,
Mass of proton $\left(m_p\right)$

Solution:
As we know that dimension of $[h]$
$$
=\frac{E}{v}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
$$
Dimension of $[c]=\frac{s}{t}=\frac{[\mathrm{L}]}{[\mathrm{T}]}$ $=\left[\mathrm{LT}^{-1}\right]$
Dimension of mass of electron $\left[m_e\right]=\mathrm{M}$
$$
\begin{aligned}
&{[G]=\frac{F r^2}{M_1 M_2}=\frac{\left[\mathrm{ML}^3 \mathrm{~T}^{-2}\right]}{[\mathrm{M}][\mathrm{M}]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]} \\
&{[e]=[A T]}
\end{aligned}
$$
Dimension of mass of proton $\left[m_p\right]=[\mathbf{M}]$
$$
\begin{aligned}
&{\left[\frac{h c}{G}\right]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\left[\mathrm{LT}^{-1}\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]}=\left[\mathrm{M}^2\right]} \\
&\Rightarrow M=\sqrt{\frac{h c}{G}}
\end{aligned}
$$
Similarly,
$$
\frac{h}{c}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]}{\left[\mathrm{LT}^{-1}\right]}=[\mathrm{ML}]
$$
$$
\begin{aligned}
&L=\frac{h}{c M}=\frac{h}{c} \sqrt{\frac{G}{h c}}=\frac{\sqrt{G h}}{c^{3 / 2}} \\
&\text { As, } c=\left[\mathrm{LT}^{-1}\right] \\
&\Rightarrow[T]=\frac{[L]}{[c]}=\frac{\sqrt{G h}}{c^{3 / 2} \cdot c}=\frac{\sqrt{G h}}{c^{5 / 2}}
\end{aligned}
$$
Hence physical quantities (a), (b) or (d) any can be used to represent L, M and T in terms of three chosen fundamental quantities.

Topic: Physics, Units and Dimensions
If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensional formula of force is

Options:
A PV2T2
B P-1V2T-2
C PVT2
D P-1VT2

The correct answer is:
PV2T2

Solution:
Let F∝PxVyTz
by substituting the following dimensions :
[P]=[ML-1T-2], [V]=[LT-1], [T]=[T]
and comparing the dimension of both sides
x=1, y=2, z=2 F=PV2T2

Topic: Physics, Units and Dimensions, NEET
If $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=\mathrm{p}(\mathrm{x}-\alpha)(\mathrm{x}-\beta)$, and $\mathrm{p}^{3}+\mathrm{pq}+\mathrm{r}=0 ; \mathrm{p}, \mathrm{q}$ and r being real numbers, then which of the following is not possible?

Options:
A $\alpha=\beta=\mathrm{p}$
B $\alpha \neq \beta=\mathrm{p}$
C $\alpha=\beta \neq \mathrm{p}$
D $\beta \neq \alpha=\mathrm{p}$

The correct answer is:
$\alpha=\beta=\mathrm{p}$

Solution:
Given equation is $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=\mathrm{p}(\mathrm{x}-\alpha)(\mathrm{x}-\beta)$
$\Rightarrow p x^{2}+q x+r=p x^{2}-p(\alpha+\beta) x+\alpha \beta p$
$\Rightarrow \alpha \beta \mathrm{p}=\mathrm{r}$ and $\mathrm{q}=-(\alpha+\beta) \mathrm{p}$...(1)
Also given that $\mathrm{p}^{3}+\mathrm{pq}+\mathrm{r}=0$
Putting value of $q$ and $r$ from $(1)$ $\Rightarrow p^{3}-p^{2}(\alpha+\beta)+\alpha \beta p=0$
$\Rightarrow \mathrm{p}^{2}-\mathrm{p}(\alpha+\beta)+\alpha \beta=0$
$\Rightarrow(p-\alpha)(p-\beta)=0 \Rightarrow \alpha=\beta=p$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 2)
If \(Q\) is the inverse of \(A\), when \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\) and \(10 \times Q=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3\end{array}\right]\), find \(x=\)

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
Given, \(Q=A^{-1}\)
\(\begin{aligned}
A & =\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right] \\
|A| & =1(1+3)+1(2+3)+1(2-1)=4+5+1 \\
|A| & =10
\end{aligned}\)
\(\operatorname{Adj} A=\left[\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^{\mathrm{T}}\)
Adj \(A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\)
\(\begin{aligned}
& A^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{Adj} A \\
& A^{-1}=\frac{1}{10}\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right] \\
& 10 Q=\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & x \\
1 & -2 & 3
\end{array}\right] \\
& 10 A^{-1}=\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & x \\
1 & -2 & 3
\end{array}\right] \\
& {\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]=\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & x \\
1 & -2 & 3
\end{array}\right]} \\
& \therefore \quad x=5 \\
\end{aligned}\)
Hence, option (d) is correct.

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)