datavorous/entrance-exam-dataset huggingface dataset:
If \( \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \), then \( A A^{\prime}= \)

Options:
A \( A \)
B zero matrix
C A
D \( 11 \)

The correct answer is:
\( 11 \)

Solution:
Given that \( A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \)
So, \( A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
Then,
\[
\begin{array}{l}
A \cdot A=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
=\left[\begin{array}{cc}
\cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\
-\cos \alpha \sin \alpha+\cos \alpha \sin \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I
\end{array}
\]

Topic: Mathematics, Matrices, KCET, KCET 2018
If \( \mathrm{A}=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right] \) and \( \left|\mathrm{A}^{3}\right|=27 \), then \( \alpha= \)

Options:
A \( \pm 1 \)
B \( \pm 2 \)
C \( \pm \sqrt{7} \)
D \( 0 \pm \sqrt{5} \)

The correct answer is:
\( \pm \sqrt{7} \)

Solution:
Given that $A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right]$
And $\left|A^{3}\right|=27=|A|^{3}$
$\Rightarrow|A|=3$
So, $|A|=\alpha^{2}-4=3$
$\Rightarrow \alpha^{2}=7 \Rightarrow \alpha=\pm \sqrt{7}$

Topic: Mathematics, Continuity and Differentiability, KCET, KCET 2015
If \( \mathrm{P}=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right| \) and \( \mathrm{Q}=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right| \),then \( \frac{d Q}{d x}= \)

Options:
A \( 3 P+1 \)
B \( 1-3 P \)
C \( -3 P \)
D \( 3 \mathrm{P} \)

The correct answer is:
\( 3 \mathrm{P} \)

Solution:
Given that
\( P=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|, Q=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right| \)
So, \( P=x^{2}-1 \)
And \( Q=x\left(x^{2}-1\right)-(x-1)+(1-x) \)
\( =x^{3}-3 x+2 \)
Now, \( \frac{d Q}{d x}=3 x^{2}-3 \)
\( =3\left(x^{2}-1\right)=3 P \)

Topic: Mathematics, Differentiation, KCET, KCET 2015
If mean and variance of a Binomial variate $\mathrm{X}$ are 2 and 1 respectively, then the probability that $\mathrm{X}$ takes a value greater than 1 is

Options:
A $\frac{2}{3}$
B $\frac{4}{5}$
C $\frac{7}{8}$
D $\frac{11}{16}$

The correct answer is:
$\frac{11}{16}$

Solution:
We have; $\mathrm{np}=2=$ mean $\mathrm{npq}=1=$ variance
$\Rightarrow \mathrm{p}=\frac{1}{2} ; \mathrm{q}=\frac{1}{2} \& \mathrm{n}=4$
$\begin{aligned} \text { Required probability } &=\mathrm{P}(\mathrm{x}>1) \\ &=1-\mathrm{P}(\mathrm{x} \leq 1) \\ &=1-[\mathrm{P}(\mathrm{x}=0)+\mathrm{P}(\mathrm{x}=1)] \\ &=1-\left[{ }^{4} \mathrm{C}_{0} \mathrm{q}^{4}+{ }^{4} \mathrm{C}_{1} \mathrm{q}^{3} \mathrm{p}^{1}\right] \\ &=1-\frac{5}{16}=\frac{11}{16} \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
If median of the $\Delta \mathrm{ABC}$ through $\mathrm{A}$ is perpendicular to $\mathrm{BC}$, then which one of the following is correct?

Options:
A $\tan A+\tan B=0$
B $\tan B-\tan C=0$
C $\tan \mathrm{C}+2 \tan \mathrm{A}=0$
D $\tan B+\tan C=0$

The correct answer is:
$\tan B-\tan C=0$

Solution:
In the given $\Delta \mathrm{ABC}$ let $\mathrm{BC}=\mathrm{a}$
$\therefore \quad \mathrm{BD}=\mathrm{CD}=\frac{\mathrm{a}}{2}$
In $\Delta \mathrm{ADB}$,
$\tan B=\frac{A D}{B D}=\frac{A D}{a / 2}$
$\Rightarrow \quad \tan B=\frac{2 \mathrm{AD}}{\mathrm{a}}$ ...(1)
In $\Delta \mathrm{ADC}$
$\tan \mathrm{C}=\frac{\mathrm{AD}}{\mathrm{CD}}=\frac{\mathrm{AD}}{\mathrm{a} / 2}$
$\tan C=\frac{2 A D}{a}$ ...(2)
From eqs. (1) and (2), we get
$\tan B=\tan C$
$\Rightarrow \tan B-\tan C=0$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2007 (Phase 2)
If momentum $(p)$, area $(A)$ and time $(T)$ are taken to be fundamental quantities, then energy has the dimensional formula

Options:
A
$\left[p \mathrm{~A}^{-1} \mathrm{~T}^1\right]$

B
$\left[p^2 \mathrm{AT}\right]$

C
$\left[p \mathrm{~A}^{-1 / 2} \mathrm{~T}\right]$

D
$\left[p \mathrm{~A}^{1 / 2} \mathrm{~T}\right]$

The correct answer is:
$\left[p \mathrm{~A}^{1 / 2} \mathrm{~T}\right]$

Solution:
Given that fundamental quantities are momentum $(p)$, area $(A)$ and time $(T)$.
Let us consider the dimensional formula for
$$
\begin{aligned}
&E \propto\left[p^a A^b T^c\right] \\
&\left.E=k p^a A^b T^c\right]
\end{aligned}
$$
where $k$ is dimensionless constant of proportionality.
Dimensions of energy $[E]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$ and Dimension of momentum $p=m v=\left[\mathrm{MLT}^{-1}\right]$
Dimension of Area $[A]=\left[\mathrm{L}^2\right]$
Dimension of Time $[T]=[\mathrm{T}]$
Dimension of energy $[E]=[K][p]^a[A]^b[T]^c$
Putting all the dimensions, value
$\mathrm{ML}^2 \mathrm{~T}^{-2}=\left[\mathrm{MLT}^{-1}\right]^a\left[\mathrm{~L}^2\right]^b\left[\mathrm{~T}^c\right.$
$=\mathrm{M}^a \mathrm{~L}^{2 b+a} \mathrm{~T}^{-a+c}$
By principle of homogeneity of dimensions,
$$
\begin{aligned}
&a=1,2 b+a=2 \Rightarrow 2 b+1=2 \Rightarrow b=\frac{1}{2} \\
&-a+c=-2 \\
&c=-2+a=-2+1=-1
\end{aligned}
$$
So, Dimensional formula (of energy)
$$
E=\left[p A^{1 / 2} T^{-1}\right]
$$

Topic: Physics, Units and Dimensions
If $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ are the sizes, $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ the geometric means of two series respectively, then which one of the following expresses the geometric mean (G) of the combined series?

Options:
A $\log \mathrm{G}=\frac{\mathrm{n}_{1} \mathrm{G}_{1}+\mathrm{n}_{2} \mathrm{G}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
B $\log \mathrm{G}=\frac{\mathrm{n}_{2} \log \mathrm{G}_{1}+\mathrm{n}_{1} \log \mathrm{G}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
C $\mathrm{G}=\frac{\mathrm{n}_{1} \log \mathrm{G}_{1}+\mathrm{n}_{2} \log \mathrm{G}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
D None of the above

The correct answer is:
$\log \mathrm{G}=\frac{\mathrm{n}_{2} \log \mathrm{G}_{1}+\mathrm{n}_{1} \log \mathrm{G}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

Solution:
Geometric Mean of combined series is given by the expression
$\log \mathrm{G}=\frac{\mathrm{n}_{2} \log \mathrm{G}_{1}+\mathrm{n}_{1} \log \mathrm{G}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 1)
If n1,n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by

Options:
A 1n=1n1+1n2+1n3
B 1n=1n1+1n2+1n3
C n=n1+n2+n3
D n=n1+n2+n3

The correct answer is:
1n=1n1+1n2+1n3

Solution:
In stretched string for transverse wave, Frequency is inversely proportional to lengthAs nâ1l and l=l1+l2+l3
â´ 1n=1n1+1n2+1n3

Topic: Physics, Waves and Sound, NEET
If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by:

Options:
A 1n=1n1+1n2+1n3
B 1n=1n1+1n2+1n3
C n=n1+n2+n3
D n=n1+n2+n3

The correct answer is:
1n=1n1+1n2+1n3

Solution:
For first segment we have,

n1=12l1 Tμ

For second segment we have,
n2=12l2 Tμ

For third segment we have,
n3=12l3Tμ

For the whole string we have,
n=12l Tμ

Total length of the string is,
l=l1+l2+l3

Thus, the fundamental frequency for the whole string is,
1n=1n1+1n2+1n3

Topic: Physics, Waves and Sound, NEET, NEET 2014
If $n=(2017) !$, then what is $[2018-1]$
$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots .+\frac{1}{\log _{2017} n}$
equal to?

Options:
A 0
B 1
C $\frac{\mathrm{n}}{2}$
D $\mathrm{n}$

The correct answer is:
1

Solution:
$n=(2017) !$
$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots .+\frac{1}{\log _{2017} n}$
$=\frac{1}{\frac{\log _{n}}{\log _{2}}}+\frac{1}{\frac{\log _{n}}{\log _{3}}}+\frac{1}{\frac{\log _{n}}{\log _{4}}}+\ldots .+\frac{1}{\frac{\log _{n}}{\log _{2017}}}$
$\left(\because \log _{\mathrm{a}} \mathrm{b}=\frac{\log _{\mathrm{b}}}{\log _{\mathrm{a}}}\right)$
$=\frac{\log _{2}}{\log _{n}}+\frac{\log _{3}}{\log _{n}}+\frac{\log _{4}}{\log _{n}}+\ldots .+\frac{\log _{2017}}{\log _{n}}$
$=\frac{\log _{2}+\log _{3}+\log _{4}+\ldots .+\log _{2017}}{\log _{n}}$
$=\frac{\log (2.3 .4 \ldots .2017)}{\log _{\mathrm{n}}}$
$(\because \log a+\log b+\log c+\ldots=\log a . b . c \ldots \ldots)$
$=\frac{\log (2017 !)}{\log _{\mathrm{n}}}=\frac{\log _{\mathrm{n}}}{\log _{\mathrm{n}}}=1$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
If $n(A)=115, n(B)=326, n(A-B)=47$, then what is
$\begin{array}{ll}n(A \cup B) \text { equal to? } &\end{array}$

Options:
A 373
B 165
C 370
D 394

The correct answer is:
373

Solution:
We know, for two sets $\mathrm{A}$ and $\mathrm{B}$ $\mathrm{A}-\mathrm{B}=\mathrm{A}-(\mathrm{A} \cap \mathrm{B})$
$n(A-B)=n(A)-n(A \cap B)$
Given, $n(A)=115, n(B)=326$ and $n(A-B)=47$.
$\Rightarrow 47=115-n(A \cap B)$
$\Rightarrow n(A \cap B)=68$
Consider $n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=115+326-68=373$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $N_{a}=\{a x \mid x \in N\}$, then what is $N_{12} \cap N_{8}$ equal to?

Options:
A $N_{12}$
B $N_{24}$
C $N_{20}$
D $N_{48}$

The correct answer is:
$N_{20}$

Solution:
Given, $N_{a}=\{a x \mid x \in N$ $N_{12}=\{12,24,36,48, \ldots$
and $N_{8}=\{8,16,24, .$
$N_{12} \cap N_{8}=\{24,48, \ldots\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $n \in \mathrm{N}$, then $121^{\mathrm{n}}-25^{\mathrm{n}}+1900^{\mathrm{n}}-(-4)^{\mathrm{n}}$ is divisible by which one of the following?

Options:
A 1904
B 2000
C 2002
D 2006

The correct answer is:
2000

Solution:
So, $z=1+\sqrt{5}$
$121^{n}-25^{n}+1900^{n}-\left(-4^{n}\right)$
Let us substitute $\mathrm{n}=1$ We get,$(121)^{1}-(25)^{1}+(1900)^{1}-\left(-4^{1}\right)$
$=121-25+1900+4$
$=2025-25$
$=2000$
So, given expression is divisible by 2000

Topic: Mathematics, Mathematical Induction, NDA, NDA 2018 (Phase 1)
If $\mathrm{O}$ be the origin and $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ are two points,
then what is $(\mathrm{OA})(\mathrm{OB}) \cos \angle \mathrm{AOB}$ ?

Options:
A $\mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}$
B $\mathrm{y}_{1}^{2}+\mathrm{y}_{2}^{2}$
C $\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}$
D $\mathrm{x}_{1} \mathrm{y}_{1}+\mathrm{x}_{2} \mathrm{y}_{2}$

The correct answer is:
$\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}$

Solution:
Let $\mathrm{O}(0,0), \mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ and $\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ be three points
$\mathrm{OA}=\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}}, \mathrm{OB}=\sqrt{\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}}$
$\mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
In $\Delta \mathrm{AOB}$
$\cos \angle \mathrm{AOB}=\frac{\mathrm{OA}^{2}+\mathrm{OB}^{2}-\mathrm{AB}^{2}}{2 . \mathrm{OA} \cdot \mathrm{OB}}$
$\Rightarrow \quad \mathrm{OA} \cdot \mathrm{OB} \cos \angle \mathrm{AOB}=\frac{\mathrm{OA}^{2}+\mathrm{OB}^{2}-\mathrm{AB}^{2}}{2}$
$=\frac{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}-\left\{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}\right\}}{2}$
$\Rightarrow \quad \mathrm{OA} \cdot \mathrm{OB} \cdot \cos \angle \mathrm{AOB}$
$=\frac{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}-\left\{\mathrm{x}_{2}^{2}+\mathrm{x}_{1}^{2}-2 \mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{2}^{2}+\mathrm{y}_{1}^{2}-2 \mathrm{y}_{1} \mathrm{y}_{2}\right\}}{2}$
$=\frac{2\left(\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}\right)}{2}=\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2008 (Phase 1)
If one debarks a tree, what parts of the plant is being removed?

Options:

None

Solution:
Debarking refers to removal of bark, i.e., all tissues exterior to the vascular cambium, including secondary phloem. Bark includes periderm (phellogen, phellem and phelloderm) and secondary phloem.

Topic: Biology, Anatomy of Flowering Plants
If one of the cube roots of 1 be \(\omega\), then
\(\left|\begin{array}{ccc}1 & 1+\omega^2 & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega & -1\end{array}\right|=\)

Options:
A \(\omega\)
B 1
C 1
D 0

The correct answer is:
0

Solution:
Hints: \(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_3\)
\(\begin{aligned}
& \mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_2 \\
& \mathrm{C}_3 \rightarrow \mathrm{C}_3+\omega \mathrm{C}_1 \\
& \mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1
\end{aligned}\)

Topic: Mathematics, Determinants, WBJEE, WBJEE 2011
If one of the roots of the equation $a(b-c) x^{2}+b(c-a) x+$ $c(a-b)=0$ is 1, what is the second root?

Options:
A $-\frac{b(c-a)}{a(b-c)}$
B $\frac{b(c-a)}{a(b-c)}$
C $\frac{c(a-b)}{a(b-c)}$
D $-\frac{c(a-b)}{a(b-c)}$

The correct answer is:
$\frac{c(a-b)}{a(b-c)}$

Solution:
Givenequation is
$a(b-c) x^{2}+b(c-a) x+o(a-b)=0$
Let $\alpha$ be the second root.
$\mathrm{So},(\alpha)(1)=\frac{c(a-b)}{a(b-c)}$
Hence, $\alpha=$ second root $=\frac{c(a-b)}{a(b-c)}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
If one of the roots of the equation $x^{2}+a x-6=0$ is 1 , then what is $(\mathrm{a}-6)$ equal to?

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

The correct answer is:
-1

Solution:
Since 1 is the root of given equation $\therefore$ it satisfies the equation. $\therefore(1)^{2}+a(1)-6=0$
$\Rightarrow a-6=-1$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 1)
If one root of the equation $(1-\mathrm{m}) \mathrm{x}^{2}+1 \mathrm{x}+1=0$ is double the other and 1 is real, then what is the greatest value of $\mathrm{m}$ ?

Options:
A $-\frac{9}{8}$
B $\frac{9}{8}$
C $-\frac{8}{9}$
D $\frac{8}{9}$

The correct answer is:
$\frac{9}{8}$

Solution:
Given equation is
$(\ell-m) x^{2}+\ell x+1=0$
Roots are $\alpha, \beta$. One root is double the other. $\beta=2 \alpha$
Sum of roots $=\alpha+\beta$
$\begin{array}{ll}3 \alpha=\frac{-\ell}{\ell-m} & \alpha(2 \alpha)=\frac{1}{(\ell-m)} \\ \Rightarrow \alpha^{2}=\frac{\ell^{2}}{9(\ell-m)^{2}} \quad 2 \alpha^{2}=\frac{1}{\ell-m} \\ \Rightarrow 2 \frac{\ell^{2}}{9(\ell-m)^{2}}=\frac{1}{(\ell-m)} \\ \Rightarrow \frac{2 \ell^{2}}{9(\ell-m)}=1 \\ \Rightarrow 2 \ell^{2}=9(\ell-m) \Rightarrow 2 \ell^{2}-9 \ell+9 m=0\end{array}$
For $\ell$ to be real discriminant should be $b^{2}-4 a c \geq 0$ $81-4 \times 2 \times 9 m \geq 0$
$m \leq \frac{9}{8}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 1)
If one root of the equation $a x^{2}+b x+c=0, a \neq 0$ is reciprocal ofthe other root, then which one of the following is correct?

Options:
A $a=c$
B $b=c$
C $a=-c$
D $b=0$

The correct answer is:
$a=c$

Solution:
Let the roots of $\alpha x^{2}+b x+c=0, a \neq 0$ be $\alpha$ and $\frac{1}{\alpha}$
eroduct of roots $=\alpha \cdot \frac{1}{\alpha}=\frac{c}{a}$
$\Rightarrow c=a$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 1)
If $\overrightarrow{\mathrm{p}} \neq \overrightarrow{0}$ and the conditions $\overrightarrow{\mathrm{p}} \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}}$ hold
simultaneously, then which one of the following is correct?

Options:
A $\overrightarrow{\mathrm{q}} \neq \overrightarrow{\mathrm{r}}$
B $\overrightarrow{\mathrm{q}}=\overrightarrow{-\mathrm{r}}$
C $\overrightarrow{\text { q. } \cdot \mathrm{r}}=0 \quad$
D $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}$

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

Solution:
Given that $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}$
$\Rightarrow \overrightarrow{\mathrm{p}} \cdot(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$
$\Rightarrow \overrightarrow{\mathrm{p}}$ is perpendicular to $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$
Also, $\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}}$ (given).
$\Rightarrow \overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$
$\Rightarrow \overrightarrow{\mathrm{p}}$ is parallel to $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$
Which is not possible simultaneously unless either $\mathrm{p}$ or $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$ is zero, since $\overrightarrow{\mathrm{p}} \neq 0, \Rightarrow \overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}=0$
Thus, the given conditions hold simultaneously if $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
If $\left|\begin{array}{ccc}p & -q & 0 \\ 0 & p & q \\ q & 0 & p\end{array}\right|=0$, then which one of the following is correct?

Options:
A $p$ is one of the cube roots of unity
B $q$ is one of the cube roots of unity
C $\frac{p}{q}$ is one of the cube roots of unity
D None of the above

The correct answer is:
$\frac{p}{q}$ is one of the cube roots of unity

Solution:
Let $\left|\begin{array}{ccc}p & -q & 0 \\ 0 & p & q \\ q & 0 & p\end{array}\right|=0$
$\Rightarrow p\left(p^{2}\right)+q\left(-q^{2}\right)+0=0$
$\Rightarrow p^{3}-q^{3}=0$
$\Rightarrow p^{3}=q^{3}$
$\Rightarrow \frac{p^{3}}{q^{3}}=1 \Rightarrow\left(\frac{p}{q}\right)^{3}=1$
$\Rightarrow \frac{p}{q}$ is one of the cube roots of unity.

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 2)
If $P(32,6)=k C(32,6)$, then what is the value of $k$ ?

Options:
A 6
B 32
C 120
D 720

The correct answer is:
720

Solution:
Since ${ }^{32} \mathrm{P}_{6}=\mathrm{k}^{32} \mathrm{C}_{6}$
$\Rightarrow \frac{32 !}{(32-6) !}=k \cdot \frac{32 !}{6 !(32-6) !}$
$\Rightarrow \mathrm{k}=6 !=720$

Topic: Mathematics, Permutation Combination, NDA, NDA 2009 (Phase 1)
If $\mathrm{P}(77,31)=\mathrm{x}$ and $\mathrm{C}(77,31)=\mathrm{y}$, then which one of the following is correct?

Options:
A $x=y$
B $2 x=y$
C $77 x=31 y$
D $x>y$

The correct answer is:
$x>y$

Solution:
As we know $\mathrm{P}(\mathrm{n}, \mathrm{r})=\mathrm{r} ! \mathrm{C}(\mathrm{n}, \mathrm{r})$
$\therefore \quad$ From the question, we have $\mathrm{x}=\mathrm{r} !(\mathrm{y})$
Here $\mathrm{r}=31$
$\therefore \quad x=(31) ! . y \rightarrow x>y$

Topic: Mathematics, Permutation Combination, NDA, NDA 2013 (Phase 1)
If $P(A)=\frac{2}{3}, P(B)=\frac{2}{5}$ and $P(A \cup B)-P(A \cap B)=\frac{2}{5}$, then
what is $P(A \cap B)$ equal to ?

Options:
A $\frac{3}{5}$
B $\frac{5}{11}$
C $\frac{1}{3}$
D None of these

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

Solution:
$\quad P(A \cup B)-P(A \cap B)=\frac{2}{5}$
We know that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$\frac{2}{3}+\frac{2}{5}-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}-\frac{2}{3}-\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{6-16}{15} \Rightarrow \frac{-10}{15}=-\frac{2}{3} \times \frac{-1}{2}=\frac{1}{3}$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
If $P(A)=0.8, P(B)=0.9, P(A B)=p$, which one ofthe following is correct?

Options:
A $0.72 \leq p \leq 0.8$
B $\quad 0.7 \leq p \leq 0.8$
C $0.72 < p < 0.8$
D $0.7 < p < 0.8$

The correct answer is:
$\quad 0.7 \leq p \leq 0.8$

Solution:
We know,
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 0.8+0.9-p \leq 1$
$\Rightarrow \quad 1.7-p \leq 1$
$\Rightarrow \quad 0.7 \leq p$
Now, $\quad P(A) < P(B)$
$\therefore \quad P(A \cap B) \leq P(A)$
$\Rightarrow \quad p \leq 0.8$
Hence, $0.7 \leq p \leq 0.8$

Topic: Mathematics, Probability, NDA, NDA 2008 (Phase 2)
If $P(A)$ denotes the power set of $A$ and $A$ is the void set, then what is number of elements in $P\{P\{P\{P(A)\}\}\}$ ?

Options:
A 0
B 1
C 4
D 16

The correct answer is:
16

Solution:
Since, $\mathrm{A}$ is void set therefore the number of elements in power set of $A$ is 1 . $P\{P(A)\}=2^{1}=2$
because If set $A$ has $n$ elements then $P(A)$ has $2^{n}$ elements.
$\Rightarrow P\{P\{P(A)\}\}=2^{2}=4$
$\Rightarrow P\left\{P\{P\{P(A)\}\}=2^{4}=16\right.$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $p$ and $q$ are positive integers, then which one of the
following equations has $p-\sqrt{q}$ as one of its roots?

Options:
A $x^{2}-2 p x-\left(p^{2}-q\right)=0$
B $x^{2}-2 p x+\left(p^{2}-q\right)=0$
C $x^{2}+2 p x-\left(p^{2}-q\right)=0$
D $x^{2}+2 p x+\left(p^{2}-q\right)=0$

The correct answer is:
$x^{2}-2 p x+\left(p^{2}-q\right)=0$

Solution:
If any equation has $p-\sqrt{q}$ as a root, then another root will be $p+\sqrt{q}$.
So, sum of roots $=p-\sqrt{q}+p+\sqrt{q}=2 p$
and product of roots $=(p-\sqrt{q})(p+\sqrt{q})=p^{2}-q$
Now, required equation is $x^{2}-($ sum of roots $) x+($ product of roots $)=0$ $\Rightarrow x^{2}-2 p x+\left(p^{2}-q\right)=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 1)
If $p$ and $q$ are the roots of the equation $x^{2}-p x+q=0$, then what are the values of $p$ and $q$ respectively?

Options:
A 1,0
B $\overline{0}, 1$
C $-2,0$
D $-2,1$

The correct answer is:
1,0

Solution:
Given quadratic equation is $x^{2}-p x+q=0$
$p$ and $q$ are the roots of $x^{2}-p x+q=0$ : Sum of roots $=p+q=p \Rightarrow q=p-p=$ and product of roots $=p q=q \Rightarrow p=q / q=$
$\Rightarrow q(p-1)=0 \Rightarrow p=1, q=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 1)
If $P$ and $Q$ are two complex numbers, then the modulus of the quotient of $P$ and $Q$ is:

Options:
A Greater than the quotient of their moduli
B Less than the quotient of their moduli
C Less than or equal to the quotient of their moduli
D Equal to the quotient of their moduli

The correct answer is:
Equal to the quotient of their moduli

Solution:
Thetwo complex numbers are $\mathrm{P}=\mathrm{x}+$ iy and $\mathrm{Q}=\alpha+\mathrm{i} \beta$
Quotient $=\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{x}+\mathrm{iy}}{\alpha+\mathrm{i} \beta},\left|\frac{\mathrm{P}}{\mathrm{Q}}\right|=\left|\frac{\mathrm{x}+\mathrm{iy}}{\alpha+\mathrm{i} \beta}\right|$
$=\frac{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}{\sqrt{\alpha^{2}+\beta^{2}}}=\sqrt{\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{\alpha^{2}+\beta^{2}}}=\frac{\left|\mathrm{Z}_{1}\right|}{\left|\mathrm{Z}_{2}\right|}=\left|\frac{\mathrm{Z}_{1}}{\mathrm{Z}_{2}}\right|$
Hence, the quotient of their modulus is equal to the quotient of their moduli.

Topic: Mathematics, Complex Number, NDA, NDA 2014 (Phase 1)
If $\mathrm{P}(\mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \overline{\mathrm{C}})=\frac{1}{3}$ and $\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B} \cap \overline{\mathrm{C}})=\frac{1}{3}$,
then what is $\mathrm{P}(\mathrm{B} \cap \mathrm{C})$ equal to?

Options:
A $\frac{1}{12}$
B $\frac{3}{4}$
C $\frac{1}{15}$
D $\frac{1}{9}$

The correct answer is:
$\frac{1}{12}$

Solution:
$\mathrm{P}(\mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \overline{\mathrm{C}})=\frac{1}{3}, \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B} \cap \overline{\mathrm{C}})=\frac{1}{3}$.
We know. $\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \overline{\mathrm{C}})+\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B} \cap \overline{\mathrm{C}})$
$=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$
$\mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{C}})$
$\therefore \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{C}})$
$=\frac{3}{4}-\frac{2}{3}=\frac{9-8}{12}=\frac{1}{12} .$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)