datavorous/entrance-exam-dataset huggingface dataset:
What is the equation to the straight line joining the origin
to the point of intersection of the lines $\frac{x}{a}+\frac{y}{b}=1$ and
$\frac{x}{b}+\frac{y}{a}=1 ?$

Options:
A $x+y=0$
B $x+y+1=0$
C $x-y=0$
D $x+y+2=0$

The correct answer is:
$x-y=0$

Solution:
We know that the equation of straight line passing
through the intersection point of two lines $\frac{x}{a}+\frac{y}{b}=1$
and $\frac{x}{b}+\frac{y}{a}=1$, is
$$
\left(\frac{x}{a}+\frac{y}{b}-1\right)+\lambda\left(\frac{x}{b}+\frac{y}{a}-1\right)=0 ...(i)
$$
This line passes through the origin. $\therefore \quad(0-1)+\lambda(0-1)=0 \Rightarrow \lambda=-1$
On putting the value of $\lambda$ in Eq. (i), we get
$$
\begin{aligned}
&\left(\frac{x}{a}+\frac{y}{b}-1\right)-1\left(\frac{x}{b}+\frac{y}{a}-1\right)=0 \\
\Rightarrow & \frac{x}{a}+\frac{y}{b}-1-\frac{x}{b}-\frac{y}{a}+1=0 \\
\Rightarrow & x\left(\frac{1}{a}-\frac{1}{b}\right)-y\left(\frac{1}{a}-\frac{1}{b}\right)=0 \\
\Rightarrow & x-y=0
\end{aligned}
$$

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 2)
What is the equation to the straight line passing through $(5,-2)$ and $(-4,7) ?$

Options:
A $\quad 5 x-2 y=4$
B $-4 x+7 y=9$
C $x+y=3$
D $x-y=-1$

The correct answer is:
$x+y=3$

Solution:
Equation of line
$\mathrm{y}+2=\frac{7+2}{-4-5}(\mathrm{x}-5)$
$\Rightarrow \mathrm{y}+2=-\mathrm{x}+5$
$\Rightarrow x+y=3$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
What is the equation to the straight line passing through $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ and parallel to z-axis?

Options:
A $\frac{x-a}{1}=\frac{y-b}{0}=\frac{z-c}{0}$
B $\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$
C $\frac{x-a}{0}=\frac{y-b}{1}=\frac{z-c}{0}$
D $\frac{x-a}{0}=\frac{y-b}{1}=\frac{z-c}{1}$

The correct answer is:
$\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$

Solution:
We know that dr's of z-axis are $(0,0,1)$ So, dr's of the required line are 0,0 and 1
Now, equation of the line passing through $(a, b, c)$ and having dr's 0,0 and 1 is
$\frac{\mathrm{x}-\mathrm{a}}{0}=\frac{\mathrm{y}-\mathrm{b}}{0}=\frac{\mathrm{z}-\mathrm{c}}{1}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
What is the equilibrium constant (Kc) for the reaction: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) if the standard cell potential (Eo(cell) is 0.46 V)?

Options:
A 0.46
B 3.19
C 1015 × 2
D 3.92 × 1015

The correct answer is:
3.92 × 1015

Solution:
Eo(cell) = 0.059V ÷ 2 log KC = 0.46 or log KC= (0.46V x 2) ÷ 0.059V = 15.6 KC = 3.92 x 1015

Topic: Chemistry, Electrochemistry, NEET
What is the equivalent binary number of the decimal number $13.625$ ?

Options:
A $1101.111$
B $1111.101$
C $1101.101$
D $1111.111$

The correct answer is:
$1101.101$

Solution:
$13.625$
\begin{array}{c|c|c}
2 & 13 & \\
\hline 2 & 6 & 1 \\
\hline 2 & 3 & 0 \\
\hline & 1 & 1
\end{array}
$0.625 \times 2=1.250$
ot $0.5 \times 2=1$
$0.250 \times 2=0.5$
1
0
1 $13.625=1101.101$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
What is the equivalent definition of the function given by
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2 \mathrm{x}, \mathrm{x} \geq 0 \\ 0, \mathrm{x} < 0\end{array} ?\right.$

Options:
A $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$
B $\mathrm{f}(\mathrm{x})=2 \mathrm{x}$
C $\mathrm{f}(\mathrm{x})=|\mathrm{x}|+\mathrm{x}$
D $\mathrm{f}(\mathrm{x})=2|\mathrm{x}|$

The correct answer is:
$\mathrm{f}(\mathrm{x})=|\mathrm{x}|+\mathrm{x}$

Solution:
The given function is $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2 \mathrm{x}, \mathrm{x} \geq 0 \\ 0, \mathrm{x} < 0\end{array}\right.$
The equation can be re-written as
$f(x)=\left\{\begin{array}{ll}x+x, & x \geq 0 \\ -x+x & x < 0\end{array}\right.$
Hence, equivalent definition of given function is $\mathrm{f}(\mathrm{x})=|\mathrm{x}|+\mathrm{x}$

Topic: Mathematics, Functions, NDA, NDA 2006 (Phase 2)
What is the fate of primary xylem in a dicot root showing extensive secondary growth?

Options:
A
It is retained in the centre of the axis

B
It gets crushed

C
May or may not get crushed

D
It gets surrounded by primary phloem

The correct answer is:
It is retained in the centre of the axis

Solution:
No Solution Available

Topic: Biology, Anatomy of Flowering Plants
What is the focal distance of any point $P\left(x_{1}, y_{1}\right)$ on the parabola $y^{2}=4 a x ?$

Options:
A $x_{1}+y_{1}$
B $x_{1} y_{1}$
C $a x_{1}$
D $a+x_{1}$

The correct answer is:
$a+x_{1}$

Solution:
Focal-Distans The distance between a point on a parabola and its focus is called its focal distance Let $F(a, 0)$ be a focus on parabola $y^{2}=4 a x$. Since, $P\left(x_{1}, y_{1}\right)$ on $y^{2}=4 a x$
$y_{1}^{2}=4 a x_{1}$
Now, Focal distance
Hence, focal distance $=a+x_{1}$.

Topic: Mathematics, Parabola, NDA, NDA 2011 (Phase 2)
What is the foot of the perpendicular from the point $(2,3)$ on the line $x+y-11=0 ?$

Options:
A $(1,10)$
B $(5,6)$
C $(6,5)$
D $(7,4)$

The correct answer is:
$(5,6)$

Solution:
We know a line perpendicular to a given line $a x-b y+c=0$ is $b x+a y+k=0$
The equation of line perpendicular to given line
$$
x+y-11=0 ...(i)
$$
is $-x+y+\lambda=0$ ...(ii)
Since, this equation passes through $(2,3)$. Therefore $(2,3)$ satisfies the equation of line $\therefore \quad-2+3+\lambda=0$
$\Rightarrow \quad \lambda=-1$
From Eq. (ii), $-x+y-1=0$
$\Rightarrow \quad y=x+1$
And from eq. (i), $x+x+1-11=0$
$\Rightarrow \quad 2 x=10$
$\Rightarrow \quad x=5$
Hence, coordinates of foot of perpendicular from $(2,3)$ to given line is $(5,6)$

Topic: Mathematics, Straight Lines, NDA, NDA 2009 (Phase 1)
What is the formula for finding coefficient of variation, given \(\sigma=\) standard deviation and \(\bar{x}=\) mean \(\neq 0\) ?

Options:
A \(\frac{\bar{x}}{\sigma} \times 100\)
B \(\frac{\bar{x}}{\sigma}\)
C \(\frac{\sigma}{\bar{x}}\)
D \(\frac{\sigma}{\bar{x}} \times 100\)

The correct answer is:
\(\frac{\sigma}{\bar{x}} \times 100\)

Solution:
Coefficient of variation, having standard deviation \(\sigma\) and mean \(\bar{x} \neq 0\), is
\(\frac{\sigma}{\bar{x}} \times 100\)
Hence, option (d) is correct.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 1)
What is the foundation of collision theory?

Options:
A Quantum theory
B Thermodynamics
C Kinetic theory of gases
D Chemical equilibrium

The correct answer is:
Kinetic theory of gases

Solution:
Collision theory is based on the kinetic theory of gases.

Topic: Chemistry, Chemical Kinetics, NEET
What is the Fourth term of an AP of $n$ terms whose sum is $n$ $(\mathrm{n}+1) ?$

Options:
A 6
B 8
C 12
D 20

The correct answer is:
8

Solution:
Given, sum of n terms $\left(\mathrm{S}_{\mathrm{n}}\right)=\mathrm{n}(\mathrm{n}+1)$
We know, $\mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$
$=[\mathrm{n}(\mathrm{n}+1)]-[(\mathrm{n}-1)(\mathrm{n}-1+1)]$
$=\mathrm{n}(\mathrm{n}+1)-(\mathrm{n}-1) \mathrm{n}$
$=\mathrm{n}(\mathrm{n}+1-\mathrm{n}+1)$
$=2 \mathrm{n}$
Fourth term, $\mathrm{T}_{4}=2(4)=8$

Topic: Mathematics, Sequences and Series, NDA, NDA 2019 (Phase 1)
What is the full form of EFB?

Options:
A Enzyme Federation of Biochemistry
B Ecological Foundation of Biodiversity
C European Federation of Biotechnology
D European Foundation of Botany

The correct answer is:
European Federation of Biotechnology

Solution:
The European Federation of Biotechnology was established by European scientists in 1978. It is a non-profit federation of national biotechnology associations, learned societies, universities, scientific institutes, biotechnology companies and individual biotechnologists working to promote biotechnology throughout Europe and beyond.

Topic: Biology, Biotechnology - Principles and Processes, NEET
What is the function of neutrophils in the body?

Options:
A Transport of respiratory gases
B Secrete histamine, serotonin and heparin
C Destroy foreign organisms entering the body
D Resisting infections and associated with allergic reactions

The correct answer is:
Destroy foreign organisms entering the body

Solution:
Neutrophils are the most abundant cells (60-65%) of the total WBCs and they are phagocytic cells which destroy foreign organisms entering the body.

Topic: Biology, Body Fluids and Circulation, NEET
What is the function of starch in plants?

Options:
A To form helical secondary structures
B To hold I2 molecules
C To act as a storehouse of energy
D To act as a reducing end

The correct answer is:
To act as a storehouse of energy

Solution:
Starch is described as present in plant tissues as a storehouse of energy.

Topic: Biology, Biomolecules (B), NEET
What is the function of the polar nuclei in an embryo-sac?

Options:
A Produce an egg cell
B Enclose ovules
C Assist in fertilization
D Produce a diploid secondary nucleus

The correct answer is:
Produce a diploid secondary nucleus

Solution:
The polar nuclei in an embryo-sac eventually fuse to produce a diploid secondary nucleus, which is necessary for the development of the endosperm, the tissue that nourishes the developing embryo.

Topic: Biology, Plant Kingdom, NEET
What is the function of the synergids in an embryo-sac?

Options:
A Produce an egg cell
B Enclose ovules
C Assist in fertilization
D Produce a diploid secondary nucleus

The correct answer is:
Assist in fertilization

Solution:
Synergids are cells present within an embryo-sac that assist in fertilization by releasing chemical signals that attract pollen tubes to the egg cell.

Topic: Biology, Plant Kingdom, NEET
What is the fundamental characteristic of animals belonging to the phylum Chordata?

Options:
A presence of a notochord
B bilaterality
C triploblasty
D coelomate body

The correct answer is:
presence of a notochord

Solution:
The fundamental characteristic of animals belonging to the phylum Chordata is the presence of a notochord, which is a flexible rod-like structure located between the gut and the nerve cord.

Topic: Biology, Animal Kingdom, NEET
What is the fundamental idea behind the Crystal Field Theory (CFT)?

Options:
A It assumes that metal-ligand bonds are purely covalent.
B It considers the metal-ligand bond as purely ionic due to electrostatic interactions.
C It assumes that the d orbitals in metal atoms are non-degenerate.
D It explains the color exhibited by coordination compounds.

The correct answer is:
It considers the metal-ligand bond as purely ionic due to electrostatic interactions.

Solution:
The Crystal Field Theory (CFT) is based on the assumption that the metal-ligand bond is purely ionic due to electrostatic interactions between the metal ion and ligands.

Topic: Chemistry, Coordination Compounds, NEET
What is the fundamental principle behind the solubility rule "like dissolves like" in relation to the intermolecular forces present in solutions?

Options:
A Solubility is determined by the similarity of intermolecular forces between the solute and solvent.
B Solubility is solely dependent on the chemical composition of the solute.
C Solubility is determined by the temperature of the solvent.
D Solubility is unrelated to intermolecular forces.

The correct answer is:
Solubility is determined by the similarity of intermolecular forces between the solute and solvent.

Solution:
Correct Option is : (A)
Solubility is determined by the similarity of intermolecular forces between the solute and solvent.

Topic: Chemistry, Solutions, NEET
What is the fundamental structural and functional unit of all living organisms?

Options:
A Tissue
B Organ
C Cell
D Organism

The correct answer is:
Cell

Solution:
A cell is the fundamental structural and functional unit of all living organisms. It is capable of independent existence and performing the essential functions of life.

Topic: Biology, Cell - The Unit of Life, NEET
What is the general solution of $\left(1+e^{x}\right) y d y=e^{x} d x ? \quad$

Options:
A $y^{2}=\ln \left[c^{2}\left(e^{x}+1\right)^{2}\right]$
B $y=\ln \left[c\left(e^{x}+1\right)\right]$
C $y^{2}=\ln \left[c\left(e^{x}+1\right)\right]$
D None of these Where 'c' is a constant of integration

The correct answer is:
$y^{2}=\ln \left[c^{2}\left(e^{x}+1\right)^{2}\right]$

Solution:
The given differential equation is
$$
\left(1+e^{x}\right) y d y=e^{x} d x
$$
By separating the variable, we get
$$
y d y=\frac{e^{x}}{1+e^{x}} d x
$$
Integrating on both the sides,
$$
\begin{array}{l}
\Rightarrow \quad \int y d y=\int\left(\frac{e^{x}}{1+e^{x}}\right) d x \\
\Rightarrow \quad \frac{y^{2}}{2}=\log \left(1+e^{x}\right)+\log c \\
\Rightarrow \quad y^{2}=2 \log \left[c\left(1+e^{x}\right)\right] \\
(\because \log m+\log n=\log m n)
\end{array}
$$
$\Rightarrow \quad y^{2}=\log \left[c^{2}\left(1+e^{x}\right)^{2}\right]$

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 2)
What is the general solution of the differential equation
$\frac{d y}{d x}+\frac{x}{y}=0 ?$

Options:
A $x^{2}+y^{2}=c$
B $x^{2}-y^{2}=c$
C $x^{2}+y^{2}=c x y$
D $x+y=c$

The correct answer is:
$x^{2}+y^{2}=c$

Solution:
$\begin{aligned} & \frac{d y}{d x}+\frac{x}{y}=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{-x}{y} \\ & \Rightarrow y d y+x d x=0 \\ & \Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}+c=0 \\ & \Rightarrow \mathrm{y}^{2}+\mathrm{x}^{2}+2 \mathrm{c}=0 \\ & \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c} . \end{aligned}$

Topic: Mathematics, Differential Equations, NDA, NDA 2019 (Phase 1)
What is the general solution of the differential equation $\mathrm{e}^{\mathrm{x}} \tan \mathrm{y} \mathrm{d} \mathrm{x}+\left(1-\mathrm{e}^{\mathrm{x}}\right) \sec ^{2} \mathrm{y} \mathrm{dy}=0 ?$

Options:
A $\sin \mathrm{y}=\mathrm{c}\left(1-\mathrm{e}^{\mathrm{x}}\right)$
B $\cos \mathrm{y}=\mathrm{c}\left(1-\mathrm{e}^{\mathrm{x}}\right)$
C $\cot \mathrm{y}=\mathrm{c}\left(1-\mathrm{e}^{\mathrm{x}}\right)$
D None of the above where $c$ is the constant of integration

The correct answer is:
None of the above where $c$ is the constant of integration

Solution:
Given diff equation can be written as
$\frac{e^{x}}{1-e^{x}} d x=-\frac{\sec ^{2} y}{\tan y} d y$
On integrating both the sides, we get
$\int \frac{\mathrm{e}^{\mathrm{x}}}{1-\mathrm{e}^{\mathrm{x}}} \mathrm{dx}=-\int \frac{\mathrm{sec}^{2} \mathrm{y}}{\tan \mathrm{y}} \mathrm{dy}$
$\Rightarrow-\log \left(1-\mathrm{e}^{\mathrm{x}}\right)=-\log (\tan \mathrm{y})+\log \mathrm{c}$
$$
\begin{array}{l}
\Rightarrow \log (\tan y)=\log c+\log \left(1-e^{x}\right) \\
\Rightarrow \quad \log (\tan y)=\log \left[c\left(1-e^{x}\right)\right] \\
\Rightarrow \quad \tan y=c\left(1-e^{x}\right)
\end{array}
$$
Where ' $\mathrm{c}$ ' is the constant of integration.

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 2)
What is the general solution of the differential equation $\mathrm{x}^{2} \mathrm{dy}+\mathrm{y}^{2} \mathrm{dx}=0 ?$

Options:
A $x+y=c$
B $x y=c$
C $c(x+y)=x y$
D None of the above where $c$ is the constant of integration.

The correct answer is:
$c(x+y)=x y$

Solution:
Given differential equation is $\mathrm{x}^{2} \mathrm{dy}+\mathrm{y}^{2} \mathrm{dx}=0$ $\Rightarrow \quad x^{2} d y=-y^{2} d x$
$\Rightarrow \quad \frac{d y}{y^{2}}+\frac{d x}{x^{2}}=0$
$\Rightarrow \quad \int \mathrm{y}^{-2} \mathrm{dy}+\int \mathrm{x}^{-2} \mathrm{dx}=0$
$\Rightarrow \frac{\mathrm{y}^{-2+1}}{-2+1}+\frac{\mathrm{x}^{-2+1}}{-2+1}=\mathrm{a}$ where 'a' is a constant of
integration
$\Rightarrow \quad-\frac{1}{y}-\frac{1}{x}=a$
$-(x+y)=a x y \Rightarrow c(x+y)=x y$
where $\mathrm{c}=-\frac{1}{\mathrm{a}}$ is a constant of integration.

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 2)
What is the general solution of the differential equation $x$ $d y-y d x=y^{2} ? \quad$

Options:
A $x=c y$
B $y^{2}=c x$
C $x+x y-c y=0$
D None of these

The correct answer is:
$x=c y$

Solution:
Differential equation $x \mathrm{dy}-\mathrm{y} \mathrm{dx}=\mathrm{y}^{2}$ $=(\mathrm{y} \mathrm{dx}-\mathrm{x} \mathrm{dy})=\mathrm{y}^{2}$
$\therefore \mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=0$
$\frac{x}{y}=C \therefore x=C y$
$\left(\mathrm{x}^{2}+\mathrm{x}+1\right) \mathrm{dy}+\left(\mathrm{y}^{2}+\mathrm{y}+1\right) \mathrm{dx}=0$
$\left(\mathrm{x}^{2}+\mathrm{x}+1\right) \mathrm{d} \mathrm{y}=-\left(\mathrm{y}^{2}+\mathrm{y}+1\right) \mathrm{d} \mathrm{x}$
$\quad \frac{\mathrm{d} \mathrm{x}}{\left(1+\mathrm{x}+\mathrm{x}^{2}\right)}+\frac{\mathrm{dy}}{\left(1+\mathrm{y}+\mathrm{y}^{2}\right)}=0$
$\Rightarrow \int \frac{\mathrm{dx}}{\left(\mathrm{x}+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\int \frac{\mathrm{dy}}{\left(\mathrm{y}+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=0$
$\Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{y}+1}{\sqrt{3}}\right)$
$=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\mathrm{C}_{1}\right)$
$\Rightarrow \tan ^{-1}\left\{\frac{\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)+\left(\frac{2 \mathrm{y}+1}{\sqrt{3}}\right)}{1-\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)\left(\frac{2 \mathrm{y}+1}{\sqrt{3}}\right)}\right\}=\tan ^{-1} \mathrm{C}_{1}$
$\left[\because \tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]$
$\Rightarrow \frac{\sqrt{3}[(2 \mathrm{x}+1)+(2 \mathrm{y}+1)]}{3-(2 \mathrm{x}+1)(2 \mathrm{y}+1)}=\mathrm{C}_{1}$
$\Rightarrow \frac{2 \sqrt{3}(\mathrm{x}+\mathrm{y}+1)}{-4 \mathrm{xy}-2 \mathrm{y}-2 \mathrm{x}+2}=\mathrm{C}_{1}$
$\Rightarrow 2 \sqrt{3}(\mathrm{x}+\mathrm{y}+1)=\mathrm{C}_{1}(2-2 \mathrm{x}-2 \mathrm{y}-4 \mathrm{xy})$
$\Rightarrow 2 \sqrt{3}(\mathrm{x}+\mathrm{y}+1)=2 \mathrm{C}_{1}(1-\mathrm{x}-\mathrm{y}-2 \mathrm{xy})$
$\Rightarrow(\mathrm{x}+\mathrm{y}+1)=\frac{\mathrm{C}_{1}}{\sqrt{3}}(1-\mathrm{x}-\mathrm{y}-2 \mathrm{xy})$
$(x+y+1)=A(1+B x+C y+D x y)$

Topic: Mathematics, Differential Equations, NDA, NDA 2014 (Phase 1)
What is the general solution of the differential equation $\mathrm{ydx}-\left(\mathrm{x}+2 \mathrm{y}^{2}\right) \mathrm{dy}=0 ? \quad$

Options:
A $x=y^{2}+c y$
B $x=2 \mathrm{cy}^{2}$
C $x=2 y^{2}+c y$
D None of the above

The correct answer is:
$x=2 y^{2}+c y$

Solution:
ydx $-\left(x+2 y^{2}\right) d y=0$
$\quad \Rightarrow y d x-x d y-2 y^{2} d y=0$
$\Rightarrow y-x \cdot \frac{d y}{d x}=2 y^{2} \cdot \frac{d y}{d x}$
$\Rightarrow \frac{y-x \cdot \frac{d y}{d x}}{y^{2}}=2 \cdot \frac{d y}{d x}$
Integrating
$\Rightarrow \frac{x}{y}=2 y+c$
$\Rightarrow \mathrm{x}=2 \mathrm{y}^{2}+\mathrm{cy}$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 1)
What is the geometric interpretation of the identity
$(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$ ?
$1.$ If the diagonals of a given parallelogram are used as sides of a second parallelogram, then the area of the second parallelogram is twice that of the given parallelogram.
$2.$ If the semi-diagonals of a given parallelogram are used as sides of a second parallelogram, then the area of the second parallelogram is half that of the given parallelogram. 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:
Both 1 and 2

Solution:
Both statements (1) and (2) are correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 1)
What is the geometric mean of 10,40 and 60 (appox)?

Options:
A 10
B 28
C $29.6$
D 70

The correct answer is:
28

Solution:
G.M= $(10 \times 40 \times 60)^{1 / 3}=28.84 \simeq 28$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 2)
What is the geometric mean of the data $2,4,8,16,32 ?$

Options:
A 2
B 4
C 8
D 16

The correct answer is:
8

Solution:
Required geometric mean $=\sqrt[5]{2.4 .8 .16 .32}$ $=\left(32^{3}\right)^{1 / 5}=\left(2^{15}\right)^{1 / 5}=2^{3}=8$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 1)
What is the geometric mean of the ratio of corresponding terms of two series where $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are geometric means of the two series?

Options:
A $\log \mathrm{G}_{1}-\log \mathrm{G}_{2}$
B $\log \mathrm{G}_{1}+\log \mathrm{G}_{2}$
C $\frac{\mathrm{G}_{1}}{\mathrm{G}_{2}}$
D $\mathrm{G}_{1} \mathrm{G}_{2}$

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

Solution:
Let series be a, $\mathrm{G}_{1}$, b and $\mathrm{a}^{\prime}, \mathrm{G}_{2}, \mathrm{~b}^{\prime} \mathrm{so}, \mathrm{G}_{1}=\sqrt{\mathrm{ab}}$ and
$\mathrm{G}_{2}=\sqrt{\mathrm{a}^{\prime} \mathrm{b}^{\prime}}$
Series formed by ratio of the corresponding terms are :
$\frac{\mathrm{a}}{\mathrm{a}^{\prime}}, \frac{\mathrm{G}_{1}}{\mathrm{G}_{2}}, \frac{\mathrm{b}}{\mathrm{b}^{\prime}}$
Geometric means of this series $=\sqrt{\frac{a}{a^{\prime}} \cdot \frac{b}{b^{\prime}}}$
$=\sqrt{\frac{\mathrm{ab}}{\mathrm{a}^{\prime} \mathrm{b}^{\prime}}}=\frac{\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{a}^{\prime} \mathrm{b}^{\prime}}}=\frac{\mathrm{G}_{1}}{\mathrm{G}_{2}}$
So, geometric mean of the ratio of corresponding term of two series where $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are geometric means of
two series is $\frac{\mathrm{G}_{1}}{\mathrm{G}_{2}}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
What is the geometric mean of the sequence $1,2,4,8, \ldots . .$ $2^{\mathrm{n}} ?$

Options:
A $2^{\mathrm{n} / 2}$
B $2^{(\mathrm{n}+1) / 2}$
C $2^{(\mathrm{n}+1)}-1$
D $2^{(\mathrm{n}-1)}$

The correct answer is:
$2^{\mathrm{n} / 2}$

Solution:
Geometric mean $=\sqrt[n+1]{1.2 .4 .8 \ldots 2^{n}}$
$\left(\because\right.$ there are $(n+1)$ multiples from $2^{\circ}$ to $\left.2^{n}\right)$
$=\sqrt[n+1]{2^{0} \cdot 2^{1} \cdot 2^{2} \cdot 2^{3} \ldots 2^{n}}$
$=n+1]{2^{1+2+3+\ldots+n}}=\sqrt[n+1]{2 \frac{n(n+1)}{2}}$
$=\left[2 \frac{n(n+1)}{2}\right]^{\frac{1}{n+1}}=2^{\frac{n}{2}}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 2)
What is the Graafian follicle ?

Options:
A The mature follicle
B The primary follicle
C The secondary follicle
D The tertiary follicle

The correct answer is:
The mature follicle

Solution:
The Graafian follicle now transforms into the mature follicle.

Topic: Biology, Human Reproduction, NEET
What is the greatest integer among the following by which the number $5^{5}+7^{5}$ is divisible?

Options:
A 6
B 8
C 11
D 12

The correct answer is:
12

Solution:
We know, $a^{n}+b^{n}$ is divisible by $(a+b)$, if $n$ is odd. Here, $\mathrm{n}=5$ is odd $\therefore 5^{5}+7^{5}$ is divisible by $5+7=12$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
What is the greatest value of the positive integer $\mathrm{n}$ satisfying
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots+\frac{1}{2^{n-1}} < 2-\frac{1}{1000} ?$

Options:
A 8
B 1
C 10
D 11

The correct answer is:
10

Solution:
$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots \ldots . .+\frac{1}{2^{\mathrm{n}-1}} < 2-\frac{1}{1000}$
$\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}} < 2-\frac{1}{1000}$
$\Rightarrow \quad 2-\frac{1}{2^{D-1}} < 2-\frac{1}{1000}$
$2^{n-1} < 1000$
Now, $(2)^{9}=512 \&(2)^{10}=1024$
$\therefore \quad n-1=9$
$\Rightarrow \quad n=10 .$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
What is the image of the point $(1,2)$ on the line $3 x+4 y-1=0$ ?

Options:
A $\left(-\frac{7}{5},-\frac{6}{5}\right)$
B $\left(\frac{7}{8}, \frac{1}{2}\right)$
C $\left(\frac{7}{8},-\frac{1}{2}\right)$
D $\left(-\frac{7}{5}, \frac{1}{2}\right)$

The correct answer is:
$\left(-\frac{7}{5},-\frac{6}{5}\right)$

Solution:
Let $(a, b)$ be the image of point $(1,2)$ w.r.t. line $3 x+4 y-1=0$
$\therefore\left(\frac{a+1}{2}, \frac{b+2}{2}\right)$ will be on the line $3 x+4 y-1=0$
$\Rightarrow 3\left(\frac{a+1}{2}\right)+4\left(\frac{b+2}{2}\right)-1=0$
$\Rightarrow \quad 3 a+3+4 b+8-2=0$
$\Rightarrow \quad 3 a+4 b+9=0$
Now, co-ordinates given in option 'a' satisfies the equation $3 a+4 b+9=0$ Thus, the image of point $(1,2)$ is $\left(-\frac{7}{5},-\frac{6}{5}\right)$.

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 1)