datavorous/entrance-exam-dataset huggingface dataset:
If the foci of the conics $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{7}=1$ and $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$
were to coincide, then what is the value of a ?

Options:
A 2
B 3
C 4
D 16

The correct answer is:
4

Solution:
The equation of ellipse is given as
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{7}=1$
Eccentricity is given by
$e=\sqrt{1-\frac{7}{a^{2}}}$
Therefore, foci of ellipse are $(\pm \mathrm{ae}, 0) \mathrm{ie}$,
$\left(\pm a \sqrt{1-\frac{7}{a^{2}}}, 0\right)$
Now, the equation of given hyper bola is
$\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25} \Rightarrow \frac{x^{2}}{\frac{144}{25}}-\frac{y^{2}}{\frac{81}{25}}=1$
So, a $=\frac{12}{5}$ and $b=\frac{9}{5}$
$\therefore \quad e^{\prime}=\sqrt{1+\frac{81 / 25}{144 / 25}}=\sqrt{\frac{144+81}{144}}=\sqrt{\frac{225}{144}}$
$\quad=\frac{15}{12}$
Foci of hyper bola are $\left(\pm \frac{12}{5}, \frac{15}{12}, 0\right)$ ie, $(\pm 3,0)$.
Since these foci coincides
$\Rightarrow 3=a \sqrt{1-\frac{7}{a^{2}}}$
$\Rightarrow \frac{3}{\mathrm{a}}=\sqrt{1-\frac{7}{\mathrm{a}^{2}}}$
$\Rightarrow \frac{9}{\mathrm{a}^{2}}=1-\frac{7}{\mathrm{a}^{2}}$
$\Rightarrow \frac{16}{a^{2}}=1 \Rightarrow a=4$

Topic: Mathematics, Ellipse, NDA, NDA 2007 (Phase 2)
If the free energy change of a reaction is greater than zero, then the reaction is

Options:
A spontaneous
B non-spontaneous
C at equilibrium
D endothermic

The correct answer is:
non-spontaneous

Solution:
An endergonic reaction (also called a non-spontaneous reaction) is a chemical reaction in which the standard change in free energy is positive and energy is absorbed.

Topic: Biology, Biomolecules (B), JIPMER, JIPMER 2015
If the function
$$
f(x)=\frac{x(x-2)}{x^{2}-4}, x \neq \pm 2
$$
is continuous at $x=2$, then what is $f(2)$ equal to?

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

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

Solution:
Let $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}(\mathrm{x}-2)}{\mathrm{x}^{2}-4}=\frac{\mathrm{x}(\mathrm{x}-2)}{(\mathrm{x}-2)(\mathrm{x}+2)}=\frac{x}{x+2}$
Since $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$
$\therefore \quad \lim _{x \rightarrow 2} f(x)=f(2)$
$\Rightarrow \quad \lim _{x \rightarrow 2} \frac{x}{x+2}=f(2)$
$\Rightarrow \quad \mathrm{f}(2)=\frac{2}{4}=\frac{1}{2}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2010 (Phase 2)
If the function \(f:[-1,1] \rightarrow R\) defined by \(f(x)=\left\{\begin{array}{cc}2^x+1, & \text { for } x \in[-1,0) \\ 1, & \text { for } x=0 \\ 2^x-1, & \text { for } x \in(0,1]\end{array}\right.\) then, in \([-1,1], f(x)\) has

Options:
A a maximum
B a minimum
C both maximum and minimum
D neither maximum nor minimum

The correct answer is:
neither maximum nor minimum

Solution:
The given function \(f:[-1,1] \rightarrow R\)
defined by \(f(x)=\left[\begin{array}{cc}2^x+1, & \text { for } x \in[-1,0) \\ 1, & \text { for } x=0 \\ 2^x-1, & \text { for } x \in(0,1]\end{array}\right.\)
The given function \(f(x)\) in \(x \in(-1,0)\) strictly increasing and \(f\left(0^{-}\right) \rightarrow 2\)
but \(f(0)=1\)
and in interval \(x \in(0,1)\), again it is strictly increasing, but \(f\left(0^{+}\right)=0\)
So, function has neither maximum nor minimum. Hence, option (4) is correct.

Topic: Mathematics, Application of Derivatives, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
If the graph of a quadratic polynomial lies entirely above $\mathrm{x}$ -axis, then which one of the following is correct?

Options:
A Both the roots are real
B One root is real and the other is complex
C Both the roots are complex
D Cannot say

The correct answer is:
Both the roots are complex

Solution:
Since the graph is not meeting the $\mathrm{x}$ -axis at all, roots are Complex numbers.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2017 (Phase 1)
If the inverse of $\left[\begin{array}{lll}1 & \mathrm{p} & \mathrm{q} \\ 0 & \mathrm{x} & 0 \\ 0 & 0 & 1\end{array}\right]$ is $\left[\begin{array}{ccc}1 & -\mathrm{p} & -\mathrm{q} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ then what is the value of $\mathrm{x}$?

Options:
A 1
B Zero
C $-1$
D $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}$

The correct answer is:
1

Solution:
Let $A=\left[\begin{array}{lll}1 & p & q \\ 0 & x & 0 \\ 0 & 0 & 1\end{array}\right]$ and $A^{-1}=\left[\begin{array}{ccc}1 & -p & -q \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\mathrm{AA}^{-1}=\mathrm{I}$
$\Rightarrow\left[\begin{array}{lll}1 & \mathrm{p} & \mathrm{q} \\ 0 & \mathrm{x} & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -\mathrm{p} & -\mathrm{q} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \mathrm{x} & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow \mathrm{x}=1$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 2)
If the latus rectum of an ellipse is equal to half of the minor axis, then what is its eccentricity?

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

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

Solution:
Length of minor axis $=2 \mathrm{~b}$ and latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$
According to given condition $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\mathrm{b}$
$\Rightarrow 2 \mathrm{~b}=\mathrm{a}$
Now, $\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\sqrt{1-\frac{\mathrm{b}^{2}}{4 \mathrm{~b}^{2}}}=\sqrt{\frac{3}{4}}$
$\Rightarrow \mathrm{e}=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Ellipse, NDA, NDA 2012 (Phase 1)
If the latus rectum of an ellipse is equal to one half its minor axis, what is the eccentricity of the ellipse?

Options:
A $\frac{1}{2}$
B $\frac{\sqrt{3}}{2}$
C $\frac{3}{4}$
D $\frac{\sqrt{15}}{4}$

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

Solution:
Length of latus rectum of an ellipse is $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$ where bis
semi minor axis and a is semi-major axis. As given, $\frac{2 b^{2}}{a}=b$
$\Rightarrow 2 \mathrm{~b}=\mathrm{a} \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{2}$
Weknow that eccentricity $\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Ellipse, NDA, NDA 2006 (Phase 1)
If the least number of zeroes in a lower triangular matrix is 10, then what is the order of the matrix?

Options:
A $3 \times 3$
B $4 \times 4$
C $5 \times 5$
D $10 \times 10$

The correct answer is:
$4 \times 4$

Solution:
Number of zeroes in a lower triangular matrix of order $\mathrm{n} \times \mathrm{n}$ is
$1+2+3+\ldots .+\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
Number of zeros $=10$
$\Rightarrow \frac{\mathrm{n}(\mathrm{n}+1)}{2}=10 \quad \Rightarrow \mathrm{n}^{2}+\mathrm{n}-20=0$
$\Rightarrow(n+5)(n-4)=0 \quad \Rightarrow n=4$ or $-5$
$(-5$ is meaningless $)$ $\Rightarrow \mathrm{n}=4 .$
$\Rightarrow$ order of the matrix is $4 \times 4$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 2)
If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$, then the ratio of the final and initial fundamental frequencies is :

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

The correct answer is:
$2: 1$

Solution:
Fundamental frequency, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
Initial frequency, $n_1=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
New length, $l^{\prime}=l-\frac{l \times 40}{100}=0.6 l$
New tension, $T^{\prime}=T+\frac{T \times 44}{100}$
$=1.44 \mathrm{~T}$
New frequency $n_2=\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}$
$\therefore \quad \frac{n_2}{n_1}=\frac{\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}$
$\frac{n_2}{n_1}=\frac{l}{l^{\prime}} \sqrt{\left(\frac{T^{\prime}}{T}\right)}$
$=\frac{l}{0.6 l} \sqrt{\frac{1.44 T}{T}}$
$=\frac{1}{0.6} \times 1.2$
$=2$
$\therefore \quad n_2: n_1=2: 1$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2003
If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$, then the ratio of the final and initial fundamental frequencies is :

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

The correct answer is:
$2: 1$

Solution:
Fundamental frequency, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
Initial frequency, $n_1=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
New length, $l^{\prime}=l-\frac{l \times 40}{100}=0.6 l$
New tension, $T^{\prime}=T+\frac{T \times 44}{100}$
$=1.44 \mathrm{~T}$
New frequency $n_2=\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}$
$\therefore \quad \frac{n_2}{n_1}=\frac{\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}$
$\frac{n_2}{n_1}=\frac{l}{l^{\prime}} \sqrt{\left(\frac{T^{\prime}}{T}\right)}$
$=\frac{l}{0.6 l} \sqrt{\frac{1.44 T}{T}}$
$=\frac{1}{0.6} \times 1.2$
$=2$
$\therefore \quad n_2: n_1=2: 1$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2003
If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$. Then the ratio of the final and initial fundamental frequencies is $1: 2$, then the value of $x$ is

Options:
A 20
B 30
C 40
D 60

The correct answer is:
40

Solution:
Fundamental frequency for string, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
So, $n \propto \sqrt{T} / l$
Now, $\quad T^{\prime}=T+0.44 T=\frac{144 T}{100}$
and $\quad l^{\prime}=l-\frac{x}{100} l=\left(\frac{100-x}{100}\right) l$
Also, $\quad \frac{n^{\prime}}{n}=\frac{2}{1}$
$\frac{2}{1}=\sqrt{\frac{144 T}{100 T}} \times \frac{l}{\left(\frac{100-x}{100}\right) l}=\frac{12}{10} \times \frac{100}{100-x}$
or $(100-x)=60$ or $x=40$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 1)
If the length of a stretched string is shortened by x% and the tension is increased by 44% then the ratio of the initial and final fundamental frequencies is 1:2, the value of 'x' is

Options:
A 20
B 30
C 40
D 60

The correct answer is:
40

Solution:
Recall the formula of fundamental frequency in terms of tension and mass per unit length,

f=12lTμ

it means,

f1f2=12l1T1μf=12l2T2μ=l2l1×T1T2

Put given data,

12=l1-x100l×TT1+44100⇒12=100-x10×112⇒60=100-x⇒x=40

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
If the length of stretched string is reduced by $40 \%$ and tension is increased by $44 \%$ then the ratio of final to initial frequencies of stretched string is

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

The correct answer is:
$2: 1$

Solution:
Let the initial length and tension be $l$ and $\mathrm{T}$ respectively.
After shortening,
The new length $l_{\text {new }}=l-\frac{40}{100} l=\frac{3}{5} l$
After increase in tension, the new tension $\mathrm{T}_{\text {new }}=\mathrm{T}+\frac{44}{100} \mathrm{~T}=\frac{144 \mathrm{~T}}{100}$
Fundamental frequency of a vibrating string is given by
$\begin{aligned}
\mathrm{n} & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
\therefore \quad \mathrm{n}_1 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
\mathrm{n}_2 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{\mathrm{new}}}{\mathrm{m}}} \\
\therefore \quad \frac{\mathrm{n}_1}{\mathrm{n}_2} & =\frac{l \text { aew }}{l} \times \frac{\sqrt{\mathrm{T}}}{\sqrt{\mathrm{T}_{\mathrm{new}}}} \\
& =\frac{\frac{3}{5} l}{l} \times \sqrt{\frac{100 \mathrm{~T}}{144 \mathrm{~T}}} \\
& =\frac{3}{5} \times \frac{10}{12}=\frac{1}{2} \\
\therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1} & =\frac{2}{1}
\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (10 May Shift 1)
If the letters of the word BAZAR are arranged in dictionary order, then what is the 50th word?

Options:
A $\quad$ ZAABR
B ZBAAR
C $\quad$ ZBRAA
D $\quad$ ZAARB

The correct answer is:
$\quad$ ZAARB

Solution:
With Aat first place, rest 4 places will be arranged in $4 !$ ways so, Number of words begin with $\mathrm{A}=4 !=24$ Similarly with $\mathrm{B}$ at first place,
Number of words begin with $\mathrm{B}=\frac{4 !}{2 !}=12$
[As there are two $\left.\mathrm{A}_{\mathrm{s}}\right]$
Number of words begin with $\mathrm{R}=\frac{4 !}{2 !}=12$
Thus, 48 words have starting letter $\mathrm{A}, \mathrm{B}$ and $\mathrm{R}$. So, 49 th word will be ZAABR and 50 th word will be ZAARB.

Topic: Mathematics, Permutation Combination, NDA, NDA 2006 (Phase 2)
If the line through the points $A(k, 1,-1)$ and $B(2 k, 0,2)$ is perpendicular to the line through the points $B$ and $C(2+2 k, k, 1)$, then what is the value of $k ? \quad$

Options:
A $-1$
B 1
C $-3$
D 3

The correct answer is:
3

Solution:
Givenpoints are $A(k, 1,-1), B(2 k, 0,2)$ and $C(2+2 k, k, 1)$
Let $r_{1}=$ length of line
$A B=\sqrt{(2 k-k)^{2}+(0-1)^{2}+(2+1)^{2}}=\sqrt{k^{2}+10}$
and $r_{2}=$ length of line $B C=\sqrt{(2)^{2}+k^{2}+(-1)^{2}}$
$=\sqrt{k^{2}+5}$
Now, let $\ell_{1}, m_{1}, n_{1}$ be direction-cosines of line $A B$ and
$\ell_{2}, m_{2}, n_{2}$ be the direction cosines of $B C$.
Since $\mathrm{AB}$ is perpendicular to $\mathrm{BC}$
$\ell_{1} \ell_{2}+m_{1} m_{2}+n_{1} n_{2}=0$
Now, $\ell_{1}=\frac{k}{\sqrt{k^{2}+10}}, m_{1}=\frac{-1}{\sqrt{k^{2}+10}}, n_{1}=\frac{3}{\sqrt{k^{2}+10}}$
and $\ell_{2}=\frac{2}{\sqrt{k^{2}+5}}, m_{2}=\frac{k}{\sqrt{k^{2}+5}}, n_{2}=\frac{-1}{\sqrt{k^{2}+5}}$
So, $\ell_{1} \ell_{2}+m_{1} m_{2}+n_{1} n_{2}=0$
$\Rightarrow \frac{2 k}{\sqrt{k^{2}+10} \sqrt{k^{2}+5}}-\frac{k}{\sqrt{k^{2}+10} \sqrt{k^{2}+5}}-\frac{3}{\sqrt{k^{2}+10} \sqrt{k^{2}+5}}=0$
$\Rightarrow 2 k-k-3=0$
$\Rightarrow k=3$
For $k=3, A B$ is perpendicular to $B C$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 1)
If the lines $3 y+4 x=1, y=x+5$ and $5 y+b x=3$ are ooncurrent, then what is the value of $b ?$

Options:
A 1
B 3
C 6
D 0

The correct answer is:
6

Solution:
The equation of given lines are
$3 y+4 x=1$ ...(i)
$y=x+5$ ...(ii)
and $5 y+b x=3$ ...(iii)
On solving Eqs. (i) and (ii), we get $x=-2$ and $y=3$
If these lines are concurrent, then these values must satisfy the third equation $15-2 b=3 \Rightarrow 2 b=12 \Rightarrow b=6$
$ALTERNATESOLUTION$:
Given equation of lines are $3 y+4 x-1=0$
$x+5-y=0$ and
$b x+5 y-3=0$
Since, the given lines are concurrent $\therefore$ The value of determinant made by coeff of equations is $0 .$
ie, $\left|\begin{array}{ccc}4 & 3 & -1 \\ 1 & -1 & 5 \\ b & 5 & -3\end{array}\right|=0$
$\Rightarrow 4(3-25)-3(-3-5 b)-1(5+b)=0$
$\Rightarrow-88+4+14 b=0$
$\Rightarrow-84=-14 b$
$\Rightarrow b=6 .$

Topic: Mathematics, Determinants, NDA, NDA 2010 (Phase 1)
If the lines $3 \mathrm{y}+4 \mathrm{x}=1, \mathrm{y}=\mathrm{x}+5$ and $5 \mathrm{y}+\mathrm{bx}=3$ are concurrent, then what is the value of b?

Options:
A 1
B 3
C 6
D $\frac{1}{2}$

The correct answer is:
6

Solution:
Given lines, $3 \mathrm{y}+4 \mathrm{x}=1 \Rightarrow 4 \mathrm{x}+3 \mathrm{y}-1=0$
$y=x+5 \Rightarrow x-y+5=0$
$5 y+b x=3 \Rightarrow b x+5 y-3=0$
Since, these lines are concurrent,
$\left|\begin{array}{ccc}4 & 3 & -1 \\ 1 & -1 & 5 \\ b & 5 & -3\end{array}\right|=0$
$\Rightarrow 4(3-25)-3(-3-5 b)-1(5+b)=0$
$\Rightarrow 4(-22)+9+15 b-5-b=0$
$\Rightarrow-88+4+14 b=0$
$\Rightarrow-84+14 b=0$
$\Rightarrow \mathrm{b}=6$

Topic: Mathematics, Straight Lines, NDA, NDA 2019 (Phase 1)
If the magnitude of $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ equals to $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}$, then which one of the following is correct?

Options:
A $\quad \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}$
B The angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is $45^{\circ}$
C $\vec{a}$ is parallel to $\vec{b}$
D $\vec{a}$ is perpendicular to $\overrightarrow{\mathrm{b}}$

The correct answer is:
The angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is $45^{\circ}$

Solution:
Since magnitude of $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=$ magnitude of $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}$
$\therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|$
$\Rightarrow|\vec{a}||\vec{b}| \sin \theta=|\vec{a}||\vec{b}| \cos \theta$ (By Definition)
$\Rightarrow \tan \theta=1$
$\Rightarrow \theta=\frac{\pi}{4}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 1)
If the magnitude of difference of two unit vectors is $\sqrt{3}$, then the magnitude of sum of the two vectors is

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

The correct answer is:
1 unit

Solution:
$|\hat{a}-\hat{b}|=\sqrt{3}$
$\Rightarrow|\hat{a}-\hat{b}|^{2}=(\sqrt{3})^{2}$
$\Rightarrow \hat{a} . \hat{a}+\hat{b} \cdot \hat{b}-2 \hat{a} \cdot \hat{b}=3$
$\Rightarrow 2 \hat{a} \cdot \hat{b}=-1$
Now; $|\hat{a}+\hat{b}|^{2}=\hat{a} \cdot \hat{a}+\hat{b} \cdot \hat{b}+2 \hat{a} \cdot \hat{b}=1+1-1$
$\Rightarrow|\hat{a}+\hat{b}|^{2}=1$
$\Rightarrow|\hat{a}+\hat{b}|=1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
If the magnitudes of two vectors a and $\mathrm{b}$ are equal then which one of the following is correct?

Options:
A $(\vec{a}+\vec{b})$ is parallel to $(\vec{a}-\vec{b})$
B $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=1$
C $(\vec{a}+\vec{b})$ is perpendicular to $(\vec{a}-\vec{b})$
D None of the above

The correct answer is:
$(\vec{a}+\vec{b})$ is perpendicular to $(\vec{a}-\vec{b})$

Solution:
Given $|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|$
Consider $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})$
$=\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}$
$=|\overrightarrow{\mathrm{a}}|^{2}-|\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}-|\overrightarrow{\mathrm{a}}|^{2}=0$
Hence $(\vec{a}+\vec{b})$ is perpendicular to $(\vec{a}-\vec{b})$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 1)
If the matrix $\left[\begin{array}{ccc}\alpha & 2 & 2 \\ -3 & 0 & 4 \\ 1 & -1 & 1\end{array}\right]$ is not invertible, then:

Options:
A $\alpha=-5$
B $\alpha=5$
C $\alpha=0$
D $\quad \alpha=1$

The correct answer is:
$\alpha=-5$

Solution:
Let $\mathrm{A}=\left[\begin{array}{ccc}\alpha & 2 & 2 \\ -3 & 0 & 4 \\ 1 & -1 & 1\end{array}\right]$
$|\mathrm{A}|=\left|\begin{array}{ccc}\alpha & 2 & 2 \\ -3 & 0 & 4 \\ 1 & -1 & 1\end{array}\right|$
$|A|=\alpha(0+4)-2(-3-4)+2(3-0)=4 \alpha+20$
Since $\mathrm{A}^{-1}$ does not exist, $\therefore|\mathrm{A}|=0$
$4 \alpha+20=0$
$4 \alpha=-20$
$\alpha=-5$

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 2)
If the matrix
$A=\left[\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right]$
is singular, then what is the solution set $S ?$

Options:
A $\mathrm{S}=\{0,2,3\}$
B $\mathrm{S}=\{-1,2,3\}$
C $\mathrm{S}=\{1,2,3\}$
D $\mathrm{S}=\{2,3\}$

The correct answer is:
$\mathrm{S}=\{0,2,3\}$

Solution:
Let $A=\left[\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right]$
Since, this matrix is singular.
$\therefore |A|=0$
$\Rightarrow\left|\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right|=0$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3}$
$\Rightarrow\left|\begin{array}{ccc}2-x & 1 & 1 \\ 0 & -x & -x \\ -1 & -3 & -x\end{array}\right|=0$
$\Rightarrow(2-x)\left(x^{2}-3 x\right)-1(-x)+1(-x)=0$
$\Rightarrow(2-x)(x)(x-3)=0$
$\Rightarrow x=2,0,3$
Hence, solution set $S=\{0,2,3\}$

Topic: Mathematics, Matrices, NDA, NDA 2011 (Phase 1)
If the matrix $A$ is such that $\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right) A=\left(\begin{array}{cc}1 & 1 \\ 0 & -1\end{array}\right)$, then what is $A$ equal to?

Options:
A $\left(\begin{array}{cc}1 & 4 \\ 0 & -1\end{array}\right)$
B $\left(\begin{array}{ll}1 & 4 \\ 0 & 1\end{array}\right)$
C $\left(\begin{array}{cc}-1 & 4 \\ 0 & -1\end{array}\right)$
D $\left(\begin{array}{ll}1 & -4 \\ 0 & -1\end{array}\right)$

The correct answer is:
$\left(\begin{array}{cc}1 & 4 \\ 0 & -1\end{array}\right)$

Solution:
$\because\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right) \mathrm{A}=\left(\begin{array}{rr}1 & 1 \\ 0 & -1\end{array}\right)$
Let $\mathrm{B} \quad=\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right)$ and $|\mathrm{B}|=1$
$\therefore \mathrm{B}^{-1}=\left(\begin{array}{rr}1 & -3 \\ 0 & 1\end{array}\right)$ $\quad\left(\because \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\right)$
$\therefore \mathrm{A}=\left(\begin{array}{rr}1 & -3 \\ 0 & 1\end{array}\right)\left(\begin{array}{rr}1 & 1 \\ 0 & -1\end{array}\right)=\left(\begin{array}{rr}1 & 4 \\ 0 & -1\end{array}\right)$

Topic: Mathematics, Determinants, NDA, NDA 2014 (Phase 2)
If the matrix $\mathrm{AB}$ is a zero matrix, then which one of the following is correct?

Options:
A A must be equal to zero matrix or B must be equal to zeromatrix.
B A must be equal to zero matrix and $\mathrm{B}$ must be equal to zeromatrix.
C It is not necessary that either $\mathrm{A}$ is zero matrix or $\mathrm{B}$ is zeromatrix.
D None of the above

The correct answer is:
It is not necessary that either $\mathrm{A}$ is zero matrix or $\mathrm{B}$ is zeromatrix.

Solution:
If matrix $\mathrm{AB}=0$ then it is not necessary that either A is zero matrix or $\mathrm{B}$ is zero matrix.

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 2)
If the matrix $\mathrm{B}$ is the adjoint of the square matrix A and $\alpha$ is the value of the determinant of $\mathrm{A}$, then what is $\mathrm{AB}$ equal to ?

Options:
A $\alpha$
B $\left(\frac{1}{\alpha}\right)$ I
C I I
D $\alpha$

The correct answer is:
$\alpha$

Solution:
Since, adjoint of the square matrix A is $\mathrm{B} \Rightarrow \frac{\mathrm{B}}{|\mathrm{A}|}=\mathrm{A}^{-1}$
$\Rightarrow \frac{\mathrm{AB}}{|\mathrm{A}|}=\mathrm{AA}^{-1}=\mathrm{I}$
$\Rightarrow \mathrm{AB}=|\mathrm{A}| \mathrm{I} \Rightarrow \mathrm{AB}=\alpha \mathrm{I}$

Topic: Mathematics, Matrices, NDA, NDA 2008 (Phase 1)
If the matrix $\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$ is singular, then what is one of the values of $\theta$?

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{2}$
C $\pi$
D 0

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

Solution:
A matrix is singular if value of its determinant is zero.
Given that matrix $\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$ is singular,
$\left|\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ll}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=0 \Rightarrow \cos ^{2} \theta-\sin ^{2} \theta=0=\cos \frac{\pi}{2}$
$\Rightarrow \cos 2 \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
If the matrix \(M_r\) is given by \(M_r=\left(\begin{array}{cc}r & r-1 \\ r-1 & r\end{array}\right)\) for \(r=1,2,3, \ldots\) then \(\operatorname{det}\left(\mathrm{M}_1\right)+\operatorname{det}\left(\mathrm{M}_2\right)+\ldots . .+\operatorname{det}\left(\mathrm{M}_{2008}\right)=\)

Options:
A 2007
B 2008
C \((2008)^2\)
D \((2007)^2\)

The correct answer is:
\((2008)^2\)

Solution:
\(\begin{aligned}
& \text {Hint : } \operatorname{det}\left(M_r\right)=r^2-(r-1)^2=r^2-\left(r^2-2 r+1\right)=2 r-1 \\
& \begin{aligned}
\therefore \sum_{r=1}^{2008} \operatorname{det}\left(M_r\right)=\sum_{r=1}^{2008}(2 r-1)= & +3+5+\cdots+4015 \\
= & (2008)^2
\end{aligned}
\end{aligned}\)

Topic: Mathematics, Determinants, WBJEE, WBJEE 2023
If the mean and standard deviation of a binomial distribution are 20 and 4, respectively, then the number of trials is ________

Options:
A 25
B 50
C 200
D 100

The correct answer is:
25

Solution:
Given, mean of a binomial distribution =20

⇒ np=20 …i

and standard deviation of a binomial distribution =4

⇒ np1-p=4 …ii

Put np=20 from equation i to equation ii, we get

⇒ 201-p=4

⇒ p=45

Put the value of p in the equation i, we get

⇒ n45=20

⇒ n=25

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2020 (23 Sep Shift 1)