datavorous/entrance-exam-dataset huggingface dataset:
What is the image of the point $(2,3)$ in the line $y=-x$ ?

Options:
A $(-3,-2)$
B $(-3,2)$
C $(-2,-3)$
D $(3,2)$

The correct answer is:
$(-3,-2)$

Solution:
Let there be a point $\mathrm{P}(2,3)$ on cartesian plane. Image of this point in the line $\mathrm{y}=-\mathrm{x}$ will lie on a line which is perpendicular to this line and distance of this point from $\mathrm{y}=-\mathrm{x}$ will be equal to distance of the image from this line.
Let $\mathrm{Q}$ be the image of $\mathrm{p}$ and let the co-ordinate of $\mathrm{Q}$ be $(\mathrm{h}, \mathrm{k})$
Slope of line $\mathrm{y}=-\mathrm{x}$ is $-1$ Line joining $P, Q$ will be perpendicular to $\mathrm{y}=-\mathrm{x} \mathrm{so}$, its slope $=1$ Let the equation of the line be $\mathrm{y}=\mathrm{x}+\mathrm{c}$ since this passes through point $(2,3)$ $3=2+c \Rightarrow c=1$
and the equation $\mathrm{y}=\mathrm{x}+1$ The point of intersection $\mathrm{R}$ lies in the middle of $\mathrm{P} \& \mathrm{Q}$. Point of intersection of line $\mathrm{y}=-\mathrm{x}$ and $\mathrm{y}=\mathrm{x}+1$ is $2 \mathrm{y}=1, \Rightarrow \mathrm{y}=\frac{1}{2}$ and $\mathrm{x}=-\frac{1}{2}$
Hence, $\frac{\mathrm{h}+2}{2}=-\frac{1}{2}$ and $\frac{\mathrm{k}+3}{2}=\frac{1}{2}$
$\Rightarrow \quad \mathrm{h}=-3$ and $\mathrm{k}=-2$
So, the image of the point $(2,3)$ in the line $\mathrm{y}=-\mathrm{x}$ is $(-3,-2)$

Topic: Mathematics, Straight Lines, NDA, NDA 2006 (Phase 2)
What is the importance of record keeping in dairy farming?

Options:
A It helps to identify and rectify problems early
B It is mandatory for regular inspections
C It is not important
D It helps to reduce the chance of direct contact between the produce and the handler

The correct answer is:
It helps to identify and rectify problems early

Solution:
Proper record keeping in dairy farming is important to help identify and rectify problems early.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
What is the inclination of the line $\sqrt{3} \mathrm{x}-\mathrm{y}-1=0 ?$

Options:
A $30^{\circ}$
B $60^{\circ}$
C $135^{\circ}$
D $150^{\circ}$

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

Solution:
Given equation can be written as $\mathrm{y}=\sqrt{3} \mathrm{x}-1$ on comparing with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
We get $\tan \theta=\sqrt{3} \Rightarrow \theta=60^{\circ}$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 1)
What is the initial reason a clear boundary is visible between the liquid and vapor phases when heating a liquid in a closed vessel?

Options:
A The vapor becomes denser than the liquid
B The liquid becomes less dense
C The temperature decreases
D The density of the vapor decreases

The correct answer is:
The liquid becomes less dense

Solution:
Initially, a clear boundary is visible because the liquid becomes less dense as it is heated.

Topic: Chemistry, States of Matter, NEET
What is the interior acute angle of the parallelogram whose
sides are represented by the vectors $\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and
$\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\hat{\mathrm{k}} ?$

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

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

Solution:
Let $\overrightarrow{\mathrm{a}}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\hat{k}$
and $\overrightarrow{\mathrm{b}}=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\therefore \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$=\frac{\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\hat{k}\right) \cdot\left(\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{j}+\hat{k}\right)}{\sqrt{\frac{1}{2}+\frac{1}{2}+1} \sqrt{\frac{1}{2}+\frac{1}{2}+1}}$
$=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{2}+1\right]=\frac{1}{2}=\cos 60^{\circ}$
$\therefore \theta=60^{\circ}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
What is the interval in which the function $\mathrm{f}(\mathrm{x})=\sqrt{9-\mathrm{x}^{2}}$ is increasing? $(f(x)>0)$

Options:
A $0 < x < 3$
B $-3 < x < 0$
C $\quad 0 < \mathrm{x} < 9$
D $-3 < x < 3$

The correct answer is:
$-3 < x < 0$

Solution:
$f(x)=\sqrt{9-x^{2}}$
$f^{\prime}(x)=\frac{1}{2 \sqrt{9-x^{2}}} \times(-2 x)=-\frac{x}{\sqrt{9-x^{2}}}$
For function to be increasing
$-\frac{x}{\sqrt{9-x^{2}}}>0$
or $-x>0$ or $x < 0$
but $\sqrt{9-\mathrm{x}^{2}}$ is defined only when $9-x^{2}>0$ or $x^{2}-9 < 0$
$(x+3)(x-3) < 0$
ie $-3 < x < 3$
$-3 < x < 3 \cap x < 0$
$\Rightarrow-3 < x < 0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 1)
What is the interval over which the function $f(x)=6 x-x^{2}, x>0$ is increasing $? \quad$

Options:
A $(0,3)$
B $(3,6)$
C $(6,9)$
D None of these

The correct answer is:
$(0,3)$

Solution:
Given $f(x)=6 x-x^{2}, x>0$ Differentiate w.r.t. $x$ both side we get $f^{\prime}(x)=6-2 x$
As we know $f(x)$ will be increasing function, if $f^{\prime}(x)>0 \quad \therefore 6-2 x>0 \Rightarrow x < 3$
Thus, required interval is $(0,3)$.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
What is the inverse of $\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$?

Options:
A $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
B $\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$
C $\left[\begin{array}{ccc}-1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
D $\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]$

The correct answer is:
$\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$

Solution:
$A=\left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$
$\therefore \quad|A|=[0(0)-0(0)+1(-1)]=-1$
and $\operatorname{adj} A=\left[\begin{array}{rrr}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]$
Hence, $A^{-1}=\frac{1}{|A|} \operatorname{adj} A$
$=-\frac{1}{1}\left[\begin{array}{rrr}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 1)
What is the inverse of $\mathrm{A}=\left[\begin{array}{cc}1+\mathrm{i} & 1+\mathrm{i} \\ -1+\mathrm{i} & 1-\mathrm{i}\end{array}\right]$?

Options:
A $\frac{1}{4}\left[\begin{array}{cc}1-\mathrm{i} & -1-\mathrm{i} \\ 1-\mathrm{i} & 1+\mathrm{i}\end{array}\right]$
B $\frac{1}{4}\left[\begin{array}{cc}1+i & -1+i \\ 1+i & -1-i\end{array}\right]$
C $\frac{1}{4}\left[\begin{array}{cc}1+i & 1-i \\ -1-i & 1+i\end{array}\right]$
D $\frac{1}{4}\left[\begin{array}{cc}1+i & 1-i \\ -1-i & -1+i\end{array}\right]$

The correct answer is:
$\frac{1}{4}\left[\begin{array}{cc}1-\mathrm{i} & -1-\mathrm{i} \\ 1-\mathrm{i} & 1+\mathrm{i}\end{array}\right]$

Solution:
We know that
If $\mathrm{A}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right], \operatorname{adj} \mathrm{A}=\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]$
and $|\mathrm{A}|=\mathrm{ad}-\mathrm{bc}$
$\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}$ if $|\mathrm{A}| \neq 0$
So, $\mathrm{A}^{-1}=\frac{1}{\mathrm{ad}-\mathrm{bc}}\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]=\frac{1}{|\mathrm{~A}|}\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]$
Given that $\mathrm{A}=\left[\begin{array}{cc}1+\mathrm{i} & 1+\mathrm{i} \\ -1+\mathrm{i} & 1-\mathrm{i}\end{array}\right]$
Then adj $\mathrm{A}=\left[\begin{array}{cc}1-\mathrm{i} & -1-\mathrm{i} \\ 1-\mathrm{i} & 1+\mathrm{i}\end{array}\right]$
and, $|\mathrm{A}|=\left|\begin{array}{cc}1+\mathrm{i} & 1+\mathrm{i} \\ -1+\mathrm{i} & 1-\mathrm{i}\end{array}\right|$
$=\left(1-i^{2}\right)-\left(i^{2}-1\right)=\left(1-i^{2}-i^{2}+1\right)=1+1+1+1=4$
$\Rightarrow \quad \mathrm{A}^{-1}=\frac{1}{4}\left[\begin{array}{cc}1-\mathrm{i} & -1-\mathrm{i} \\ 1-\mathrm{i} & 1+\mathrm{i}\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 2)
What is the inverse of the function $\mathrm{y}=5^{\log \mathrm{x}}$ ?

Options:
A $x=5^{1 / \log y}$
B $x=y^{1 / \log 5}$
C $x=5^{\log y}$
D $x=y^{\log 5}$

The correct answer is:
$x=y^{1 / \log 5}$

Solution:
Given equation is $\mathrm{y}=5 \log \mathrm{x}$ To get the inverse, we express $\mathrm{x}$ in terms of $\mathrm{y}$ Taking log on both the sides, $\begin{aligned} & \log y=\log x \cdot \log 5 \\ \Rightarrow & \log x=\frac{\log y}{\log 5}=\log y^{\frac{1}{\log 5}} \\ \Rightarrow & x=y^{\frac{1}{\log 5}} \end{aligned}$

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 1)
What is the inverse of the matrix
$A=\left(\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right) ?$

Options:
A $\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right)$
B $\left(\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right)$
C $\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta\end{array}\right)$
D $\left(\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right)$

The correct answer is:
$\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right)$

Solution:
$A=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$
We know, $\mathrm{AA}^{-1}=\mathrm{I}$
Let us take first option (a) as $\overline{\mathrm{A}}^{1}$.
$\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}\cos ^{2} \theta+\sin ^{2} \theta & -\sin \theta \cos \theta+\cos \theta \sin \theta & 0 \\ -\sin \theta \cos \theta+\cos \theta \sin \theta & \sin ^{2} \theta+\cos ^{2} \theta & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\mathrm{I}$

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 1)
What is the IUPAC name of CH3CHCH2CH3CHO?

Options:
A Butan-2-al
B 2-methylbutanal
C 3-methylisobutyraldehyde
D 2-ethylpropanal

The correct answer is:
2-methylbutanal

Solution:
The given compound is:

C4H3C3H2C2H(CH3)C1HO

There are four carbon in the longest chain, hence, the root word is but. All carbon in the longest chain are connected with single bonds, hence, an is the primary suffix. Aldehyde is the major functional group, hence, al is the secondary suffix. Methyl substituent is attached to the second carbon so 2-methyl will be used as prefix.

Hence, by combing all of these things IUPAC name of this compound is 2-methylbutanal.

Topic: Chemistry, General Organic Chemistry, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
What is the key difference between primary and secondary sewage treatment?

Options:
A Primary sewage treatment occurs first followed by secondary treatment
B In primary sewage treatment, there is mechanical or physical removal of small and large substances while the secondary sewage treatment includes the removal of organic matter by microbes.
C Cannot be defined
D None of the above

The correct answer is:
In primary sewage treatment, there is mechanical or physical removal of small and large substances while the secondary sewage treatment includes the removal of organic matter by microbes.

Solution:
The key difference between primary and secondary sewage treatment is In primary sewage treatment, there is mechanical or physical removal of small and large substances while the secondary sewage treatment includes the removal of organic matter by microbes.

Topic: Biology, Microbes in Human Welfare, NEET
What is the key difference between primary and secondary sewage treatment?

Options:

None

Solution:
Primary $\left(1^{\circ}\right)$ treatment is a physical process that involves for removal of particulate as settelable particle.
Secondary $\left(2^{\circ}\right)$ treatment is purely a biological treatment involving microbial oxidation.

Topic: Biology, Microbes in Human Welfare
What is the largest value of a third order determinant whose elements are 0 or 1?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
In the third order determinant $\left|\begin{array}{lll}1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|$ which
satisfied these conditions. Its value is 2 , which is largest if the elements of a determinant are 0 or 1.

Topic: Mathematics, Determinants, NDA, NDA 2006 (Phase 2)
What is the last digit of $3^{4^{4 \mathrm{n}}}+1$, where $n$ is a natural number?

Options:
A 2
B 7
C 8
D None of these

The correct answer is:
None of these

Solution:
In $3^{\text {n }}$, last digit is 3 , if $n=1,9$ if $n=2,7$ if $n=3$ and 1 if $\mathrm{n}=4$ and it is repeated after than
Given expression is $3^{3^{4 \mathrm{n}}}+1$
Let $x=3^{3^{4 n}}+1=3^{81 n}+1$
$\Rightarrow \mathrm{x}=3^{80 \mathrm{n}} \cdot 3^{\mathrm{n}}+1$
Last digit of $x$ will be decided by $3^{\text {n }}$ since $3^{80 \text { n }}$ has power multiple of 4 . $\operatorname{If} n=1$ last digit is $3+1=4$ $\mathrm{n}=2$ last digit is $3^{2}+1=9+1=10$ So, last digit is zero. $\mathrm{n}=3$ last digit is $3^{3}+1=27+1=28$
last digit is 8 If $n=4$ last digit is $3^{4}+1=81+1=82$
last digit is 2 . So, there is no definite value of last digit.

Topic: Mathematics, Binomial Theorem, NDA, NDA 2008 (Phase 1)
What is the least positive integer $n$ for which $\left(\frac{1+i}{1-i}\right)^{n}=1$ ?

Options:
A 16
B 12
C 8
D 4

The correct answer is:
4

Solution:
Consider $\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{(1+i)^{2}}{1-i^{2}}=\frac{(1+i)^{2}}{1+1}$
$=\frac{1+i^{2}+2 i}{2}=\frac{2 i}{2}=i$
$\therefore\left(\frac{1+i}{1-i}\right)^{n}=i^{n}$
Now, $i^{n}=1$ is possible for $n=4$.

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 2)
What is the least value of $25 \operatorname{cosec}^{2} x+36 \sec ^{2} x$ ?

Options:
A $\overline{1}$
B 11
C 120
D 121

The correct answer is:
121

Solution:
$\begin{aligned} & 25 \operatorname{cosec}^{2} x+36 \sec ^{2} x . \\ & \text { Minimum value }=(\sqrt{25}+\sqrt{36})^{2} \\ &=(5+6)^{2}=(11)^{2}=121 \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
What is the least value of the standard deviation of 5 integers no two of which are equal?

Options:
A $\sqrt{5}$
B 2
C $\sqrt{2}$
D No such least value can be computed

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

Solution:
Let us consider any five integers which are $3,4,5,6,7$.
mean $=\frac{3+4+5+6+7}{5}=\frac{25}{5}=5$
$S D=\sqrt{\frac{(5-3)^{2}+(5-4)^{2}+(5-5)^{2}+(5-6)^{2}+(5-7)^{2}}{5}}$
$\quad=\sqrt{\frac{4+1+0+1+4}{5}=\sqrt{2}}$
Hence, the least value of the standard deviation of $^{5}$ integers is $\sqrt{2}$

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 2)
What is the least vlaue of $f(x)=2 x^{3}-3 x^{2}-12 x+1$ on $[-2,2.5] ?$

Options:
A $-3$
B 8
C $-19$
D $-16.5$

The correct answer is:
$-19$

Solution:
Let $f(x)=2 x^{3}-3 x^{2}-12 x+1$ $\Rightarrow f^{\prime}(x)=6 x^{2}-6 x-12$
Put $f^{\prime}(x)=0$ for maxima or minima
$\Rightarrow 6 x^{2}-6 x-12=0 \Rightarrow x^{2}-x-2=0$
$\Rightarrow(x-2)(x+1)=0$
$\Rightarrow \quad x=2, x=-1$
Now, $f^{\prime \prime}(x)=12 x-6$
$\Rightarrow f^{\prime \prime}(2)=24-6=18>0$
Since, $f^{\prime \prime}(x)$ is $+v e$ at $x=2$
$\therefore f(x)$ is minimum at $x=2$. Hence, minimum value is $f(2)=2(2)^{3}-3(2)^{2}-12 \times 2+1$
$=16-12-24+1=-19$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
What is the length of arc of a circle of radius $5 \mathrm{~cm}$ subtending a central angle measuring $15^{\circ} ?$

Options:
A $5 \pi / 12 \mathrm{~cm}$
B $7 \pi / 12 \mathrm{~cm}$
C $\pi / 12 \mathrm{~cm}$
D $\pi / 5 \mathrm{~cm}$

The correct answer is:
$5 \pi / 12 \mathrm{~cm}$

Solution:
We know length of arc of a circle $=2 \pi r \frac{\theta}{360}$ where ' $r$ ' is the radius and $\theta$ is the central angle. So, $r=5, \theta=15^{\circ}$
$\therefore$ Length $=2 \pi \times 5 \times \frac{15^{\circ}}{360^{\circ}}=\frac{5 \pi}{12} \mathrm{~cm}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
What is the length of the intercept made on the $x$ -axis by the circle $x^{2}+y^{2}+2 g x+2 f y+c=0 ?

Options:
A $\frac{\sqrt{\left(\mathrm{g}^{2}-\mathrm{c}\right)}}{2}$
B $\frac{\sqrt{\left(\mathrm{g}^{2}-4 \mathrm{c}\right)}}{2}$
C $2 \sqrt{\left(\mathrm{g}^{2}-4 \mathrm{c}\right)}$
D $2 \sqrt{\left(\mathrm{g}^{2}-\mathrm{c}\right)}$

The correct answer is:
$2 \sqrt{\left(\mathrm{g}^{2}-\mathrm{c}\right)}$

Solution:
Let the circle $\mathrm{S} \equiv \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ cut $\mathrm{x}$ -axis at
two points $\mathrm{A}$, and $\mathrm{B}$ at $\mathrm{x}=\mathrm{a}$ and $\mathrm{x}=\mathrm{b}$. Let $C$ be the centre and $\mathrm{AB}$ the chord.
Length of chord $\mathrm{AB}=\mathrm{b}-\mathrm{a}$. Perpendicular from centre c on chord bisect this chord at point M. Radius $r=\sqrt{g^{2}+f^{2}-c}$ also radius $A C=$
$r=\sqrt{A M^{2}+C M^{2}}$
$\mathrm{AM}=\frac{\mathrm{AB}}{2}=\frac{\mathrm{b}-\mathrm{a}}{2}$
So, $\mathrm{r}=\sqrt{\left(\frac{\mathrm{b}-\mathrm{a}}{2}\right)^{2}+(-\mathrm{f})^{2}}$
Thus, $\sqrt{g^{2}+f^{2}-c}=\sqrt{\left(\frac{b-a}{2}\right)^{2}+f^{2}}$
or $\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}=\left(\frac{\mathrm{b}-\mathrm{a}}{2}\right)^{2}+\mathrm{f}^{2}$
or $\left(\frac{\mathrm{b}-\mathrm{a}}{2}\right)^{2}=\mathrm{g}^{2}-\mathrm{c}$
$\frac{\mathrm{b}-\mathrm{a}}{2}=\sqrt{\mathrm{g}^{2}-\mathrm{c}}$
Hence, $\mathrm{b}-\mathrm{a}=\mathrm{AB}=2 \sqrt{\mathrm{g}^{2}-\mathrm{c}}$
The length of the intercept made on the $x$ -axis by the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ is $2 \sqrt{g^{2}-c}$.

Topic: Mathematics, Circle, NDA, NDA 2006 (Phase 2)
What is the length of the latus rectum of the ellipse $25 x^{2}+16 y^{2}=400 ?$

Options:
A $25 / 2$
B 25/4
C $16 / 5$
D $32/5$

The correct answer is:
$32/5$

Solution:
Equation of ellipse is $25 x^{2}+16 y^{2}=400$ $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$

Topic: Mathematics, Ellipse, NDA, NDA 2014 (Phase 2)
What is the length of the longest interval in which the function $f(\mathrm{x})=3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}$ is increasing?

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

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

Solution:
$\mathrm{f}(\mathrm{x})=3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}=\sin 3 \mathrm{x}$
$\sin 3 x$ increases in $\left[\frac{-\pi}{6}, \frac{\pi}{6}\right]$
So, interval length $=\frac{\pi}{3}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 1)
What is the length of the perpendicular from the origin to
the plane $a x+b y+\sqrt{2 a b} z=1 ?$

Options:
A $1 /($ ab $)$
B $1 /(a+b)$
C $\mathrm{a}+\mathrm{b}$
D $\mathrm{ab}$

The correct answer is:
$1 /(a+b)$

Solution:
Length of perpendicular from the origin to the plane $a x+b y+\sqrt{2 a b} z-1=0$ is
$=\left|\frac{0+0+0-1}{\sqrt{a^{2}+b^{2}+2 a b}}\right|=\frac{1}{\sqrt{(a+b)^{2}}}=\frac{1}{(a+b)}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 2)
What is the length of the vector $(1,1)$ ?

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

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

Solution:
Given, vector is $(1,1)$.
$\therefore \quad$ Length of vector $=\sqrt{1^{2}+1^{2}+0^{2}}=\sqrt{2}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
What is the locus of a point which is equidistant from the point $(m+n, n-m)$ and the point $(m-n, n+m)$ ?

Options:
A $m x=n y$
B $n x=-m y$
C $n x=m y$
D $\quad m x=-n y$

The correct answer is:
$n x=m y$

Solution:
Let the locus of a point be $(h, k)$ Let the given points be $P(m+n, n-m)$ and $\mathrm{Q}(m-n, n+m)$
$\therefore$ By distance formula, we have $\sqrt{[h-(m+n)]^{2}+[k-(n-m)]^{2}}$
$\quad=\sqrt{[h-(m-n)]^{2}+[k-(n+m)]^{2}}$
$\Rightarrow h^{2}+(m+n)^{2}-2 h(m+n)+k^{2}$
$+h^{2}+(m-n)^{2}-2 h(m-n)+k^{2}+(n-m)^{2}-2 k(n-m)$
$\Rightarrow 2 h(m-n-m-n)+2 k(n+m-n+m)=0$
$\Rightarrow-4 h n+4 k m=0 \Rightarrow m k=n h$
Hence, locus of a point is $n x=m y$.

Topic: Mathematics, Straight Lines, NDA, NDA 2009 (Phase 2)
What is the locus of a point which is equidistant from the points $(a+b, a-b)$ and $(b-a, a+b) ?$

Options:
A $b x-a y=0$
B $b x+a y=0$
C $-a x+b y=0$
D $a x+b y=0$

The correct answer is:
$-a x+b y=0$

Solution:
Let the coordinates of the point are $(x, y)$ $\therefore \sqrt{\{x-(a+b)\}^{2}+\{y-(a-b)\}^{2}}$
$=\sqrt{\{x-(b-a)\}^{2}+\{y-(a+b)\}^{2}}$
Squaring both sides, we get $\Rightarrow x^{2}+(a+b)^{2}-2 x(a+b)+y^{2}+(a-b)^{2}-2 y(a-b)$
$=x^{2}+(b-a)^{2}-2 x(b-a)+y^{2}+(a+b)^{2}-2 y(a+b)$
$\Rightarrow 2 x(a+b)+2 y(a-b)=2 x(b-a)+2 y(a+b)$
$\Rightarrow x\{(a+b)-(b-a)\}+y\{(a-b)-(a+b)\}=0$
$\Rightarrow 2 a x+(-2 b y)=0$
$\Rightarrow a x-b y=0$
$\Rightarrow-a x+b y=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2008 (Phase 2)
What is the locus of a point which moves equidistant from the coordinate axes?

Options:
A $x \pm y=0$
B $x+2 y=0$
C $2 x+y=0$
D None of these

The correct answer is:
$x \pm y=0$

Solution:
We have, $y=x$ ...(i)
$y=-x$ ...(ii)
$-x=-y$ ...(iii)
$x=-y$ ...(iv)
Thus, the required locus of a point which moves equidistant from the co-ordinate axes is $x \pm v=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
What is the locus of points of intersection of a sphere and a plane?

Options:
A Circle
B Ellipse
C Parabola
D Hyperbola

The correct answer is:
Circle

Solution:
Locus of points of intersection of a sphere and a plane is circle.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 2)
What is the locus of points, the difference of whose distances from two points being constant?

Options:
A Pair of straight lines
B An ellipse
C A hyperbola
D A parabola

The correct answer is:
A hyperbola

Solution:
We know that the locus of the difference of whose distances from two points being constant, is a hyperbola.

Topic: Mathematics, Hyperbola, NDA, NDA 2010 (Phase 1)
What is the locus of the point of intersection of the straight lines $(\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})=\mathrm{m}$ and $(\mathrm{x} / \mathrm{a})-(\mathrm{y} / \mathrm{b})=1 / \mathrm{m}$ ?

Options:
A Circle
B Parabola
C Ellipse
D Hyperbola

The correct answer is:
Hyperbola

Solution:
The given equation of straight lines are
$\frac{x}{a}+\frac{y}{b}=m$...(i)
and $\frac{x}{a}-\frac{y}{b}=\frac{1}{m} \quad \ldots$ (ii)
From Eqs. (1) and (2), we get $\left(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}\right)\left(\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}\right)=\mathrm{m} \times \frac{1}{\mathrm{~m}}=1$
$\Rightarrow \quad \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
which represents a Hyperbola and is the locus of point of intersection the given straight lines.

Topic: Mathematics, Pair of Lines, NDA, NDA 2006 (Phase 2)
What is the locus of the point which is at a distance 8 units to the left of $Y$ -axis?

Options:
A $X=8$
B $Y=8$
C $\quad X=-8$
D $Y=-8$

The correct answer is:
$\quad X=-8$

Solution:
Required locus is $X=-8$ which is at a distance of 8 units to the left of $Y$ -axis.

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
What is the locus of the point $(\mathrm{x}, \mathrm{y})$ for which the vectors $(\hat{i}-x \hat{j}-2 \hat{k})$ and $(2 \hat{i}+\hat{j}+y \hat{k})$ are orthogonal ?

Options:
A A circle
B An ellipse
C A parabola
D A straight line

The correct answer is:
A straight line

Solution:
As given: $(\hat{i}-x \hat{j}-2 \hat{k})$ and $(2 \hat{i}+\hat{j}+y \hat{k})$ areorthogonal.
So, there dot product $=0$
$\Rightarrow(\hat{i}-x \bar{j}-2 \hat{k}) \cdot(2 \hat{i}+\hat{j}+y \hat{k})=0$
$\Rightarrow 2-x-2 y=0$
$\Rightarrow x+2 y=2$
Which is an equation of straight line. Thus. the locus of the point $(\mathrm{x}, \mathrm{v})$ is a straight line.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)