datavorous/entrance-exam-dataset huggingface dataset:
Companion cells in plants are associated with

Options:
A vessels
B sperms
C sieve elements
D guard cells.

The correct answer is:
sieve elements

Solution:
The companion cells occur only in angiosperms where they accompany most sieve tube elements. Companion cell is a speciliased type of parenchyma cell which is closely associated in origin, position and function with sieve tube elements. These cells are living having abundant granular cytoplasm and a prominent elongated nucleus. Usually, the nuclei of the companion cell serve for the nuclei of the sieve tubes as they lack them. The companion cells do not contain starch. They live only so long as the sieve tube element with which they are associated and they are crushed with those cells. Longitudinal division of the mother cell of the sieve tube element forms the companion cells. One daughter cell becomes a companion cell and other sieve tube element.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2004
Compare the following on the basis of acidity

Options:
A p-cresol=o-cresol>m-cresol
B p-cresol
C p-cresol>o-cresol>m-cresol
D p-cresol=o-cresol

The correct answer is:
p-cresol=o-cresol

Solution:
Correct Option is : 4
p-cresol=o-cresol

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Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Compound A, C8H10O , is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell.
A and Yare respectively

Options:
A
B
C
D

The correct answer is:

Solution:
Option (a) is secondary alcohol which on oxidation gives phenylmethyl ketone (Acetophenone). This on reaction with I2 and NaOH form iodoform and sodium benzoate. (Iodoform Test).

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Compressed Natural Gas (CNG) is

Options:
A
propane

B
methane

C
ethane

D
butane

The correct answer is:
methane

Solution:
Compressed Natural Gas $(\mathrm{CNG})$ is the best, cheap fuel, that has replaced petrol and diesel fuel and is considered as clean fuel. it consists of around $90 \%$ methane, by compressing it to less than $1 \%$ of volume it occupies at standard atmospheric pressure. Propane and butane together form LPG or liquified petroleum gas while ethane is used in chemical industry to produce ethene.

Topic: Biology, Environmental Issues
Consider a cell given below
$\mathrm{Cu}\left|\mathrm{Cu}^{2+} \| \mathrm{Cl}^{-}\right| \mathrm{Cl}_2, \mathrm{Pt}$
Write the reactions that occur at anode and cathode.

Options:

None

Solution:
Standard Hydrogen Electrode (SHE) is the reference electrode whose electrode potential is taken to be zero. The electrode potential of other electrodes are measured with respect to it.

Topic: Chemistry, Electrochemistry
Consider a circle of radius $\mathrm{R}$. What is the length of a chord which subtends an angle $\theta$ at the centre?

Options:
A $2 \mathrm{R} \sin \left(\frac{\theta}{2}\right)$
B $2 \mathrm{R} \sin \theta$
C $2 \mathrm{R} \tan \left(\frac{\theta}{2}\right)$
D $2 \mathrm{R} \tan \theta$

The correct answer is:
$2 \mathrm{R} \sin \left(\frac{\theta}{2}\right)$

Solution:
Let there be a circle of radius $\mathrm{R}$ and chord $\mathrm{AB}$.
$\mathrm{OD} \perp \mathrm{AB}$ and $\mathrm{AD}=\mathrm{DB}$.
and $\mathrm{AD}=2 \mathrm{AD}$
$\angle \mathrm{AOB}=\theta$
$\Rightarrow \quad \angle \mathrm{AOD}=\frac{\theta}{2}$
In $\triangle \mathrm{AOD}$,
$\sin \frac{\theta}{2}=\frac{\mathrm{AD}}{\mathrm{OA}}$
$\sin \frac{\theta}{2}=\frac{\mathrm{AD}}{\mathrm{R}}$
$\mathrm{AD}=\mathrm{R} \sin \frac{\theta}{2}$
$\therefore \quad$ Length of chord $\mathrm{AB}=2 \mathrm{~A} \mathrm{D}=2 \mathrm{R} \sin \frac{\theta}{2}$.

Topic: Mathematics, Circle, NDA, NDA 2007 (Phase 2)
Consider a circle passing through the origin and the points $(\mathrm{a}, \mathrm{b})$ and $(-b,-a)$.
On which line does the centre of the circle lie?

Options:
A $x+y=0$
B $x-y=0$
C $x+y=a+b$
D $x-y=a^{2}-b^{2}$

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

Solution:
Suppose; $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ is the eq. of the
circle. Since; it passes through $(0,0) ;(a, b) \&(-b,-a)$
$\therefore \mathrm{C}=0$
$a^{2}+b^{2}+2 g a+2 f b=0$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{gb}-2 \mathrm{fb}=0$
on solving:
$\mathrm{g}=-\mathrm{f}$
$\therefore$ centre $=(-\mathrm{g},-\mathrm{f})$ or $(+\mathrm{f},-\mathrm{f})$
$\therefore$ from options:
$\mathrm{x}+\mathrm{y}=0$ is the line which passes through $(f,-f)$

Topic: Mathematics, Circle, NDA, NDA 2016 (Phase 1)
Consider a circle passing through the origin and the points $(\mathrm{a}, \mathrm{b})$ and $(-b,-a)$.
What is the sum of the squares of the intercepts cut off by the circle on the axes?

Options:
A $\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right)^{2}$
B $2\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}$
C $4\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}$
D None of the above

The correct answer is:
$2\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}$

Solution:
The two intercepts are : $-2 \mathrm{~g} \&-2 \mathrm{f}$ $\therefore$ from eq $(1) \&(2)$ we get;
$\mathrm{g}=\frac{-1}{2}\left(\frac{a^{2}+b^{2}}{a-b}\right) \& f=\frac{1}{2}\left(\frac{a^{2}+b^{2}}{a-b}\right)$
is sum of squares of intercepts
$=\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}+\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}$
$=2\left[\frac{a^{2}+b^{2}}{a-b}\right]^{2}$

Topic: Mathematics, Circle, NDA, NDA 2016 (Phase 1)
Consider a differential equation of order $\mathrm{m}$ and degree $\mathrm{n}$. Which one of the following pairs is not feasible?

Options:
A $(3,2)$
B $(2,3 / 2)$
C $(2,4)$
D $(2,2)$

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

Solution:
Degree of differential equation is always a positive integer.
$\therefore \quad\left(2, \frac{3}{2}\right)$ can not be the feasible.

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 2)
Consider a gas with molar mass M. If the sound at frequency f is introduced to a tube of this gas at temperature T then an internal acoustic standing wave is set up with nodes separated by L. The adiabatic constant γ=CpCv is

Options:
A Mf2L2RT
B Mf2L24RT
C 4Mf2L2RT
D 3Mf2L22RT

The correct answer is:
4Mf2L2RT

Solution:
Recall the formula of the fundamental frequency of the standing wave, f=v2L, and the formula of sound waves in terms of absolute temperature and molecular mass,

v=3RTM,
⇒f=γRTM2L⇒f2×4L2=γRTM⇒γ=f2×4L2MRT

Topic: Physics, Kinetic Theory of Gases, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 2)
Consider a matrix $M=\left[\begin{array}{lll}3 & 4 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & k\end{array}\right]$ and the following
statements
Statement $\mathbf{A}:$ Inverse of $\mathrm{M}$ exists.
Statement $\mathbf{B}: k \neq 0$ Which one of the following in respect of the above matrix and statement is correct?

Options:
A A implies $\mathrm{B}$, but B does not imply A
B $\mathrm{B}$ implies $\mathrm{A}$, but A does not imply $\mathrm{B}$
C Neither A implies B nor B implies A
D A implies B as well as B implies A

The correct answer is:
A implies B as well as B implies A

Solution:
Given $M=\left[\begin{array}{lll}3 & 4 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & k\end{array}\right]$
Now $|M|=\left|\begin{array}{lll}3 & 4 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & k\end{array}\right|=k(3-8)=-5 k$
From statement II, $k \neq 0$ then inverse of $\mathrm{M}$ exist (statement I). Thus, statement Aimplies B as well as B implies A.

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 1)
Consider a pair of insulating blocks with thermal resistances $R_{1}$ and $R_{2}$ as shown in the figure. The temperature $\theta$ at the boundary between the two blocks is

Options:
A $\left(\theta_{1} \theta_{2} \sqrt{R}_{1} R_{2}\right) /\left(\theta_{1}+\theta_{2}\right)\left(R_{1}+R_{2}\right)$
B $\left(\theta_{1} R_{1}+\theta_{2} R_{2}\right) /\left(R_{1}+R_{2}\right)$
C $\left[\left(\theta_{1}+\theta_{2}\right) R_{1} R_{2}\right] /\left(R_{1}{ }^{2}+R_{2}{ }^{2}\right)$
D $\left(\theta_{1} R_{2}+\theta_{2} R_{1}\right) /\left(R_{1}+R_{2}\right)$

The correct answer is:
$\left(\theta_{1} R_{2}+\theta_{2} R_{1}\right) /\left(R_{1}+R_{2}\right)$

Solution:
Rate of transmission of heat
$$
\begin{array}{l}
=\frac{\text { Temperature difference }}{\text { Thermal Resistance }} \\
\therefore \quad \frac{d Q}{d t}=\frac{d \theta}{R}
\end{array}
$$
Here, $$\frac{d Q}{d t}=\frac{\left(\theta-\theta_{2}\right)}{R_{2}}=\frac{\theta_{1}-\theta}{R_{1}}$$
$$
\begin{array}{l}
\Rightarrow \frac{\theta-\theta_{2}}{R_{2}}=\frac{\theta_{1}-\theta}{R_{1}} \\
\Rightarrow \quad R_{1} \theta-R_{1} \theta_{2}=R_{2} \theta_{1}-R_{2} \theta \\
\Rightarrow \quad \theta\left(R_{1}+R_{2}\right)=R_{2} \theta_{1}+R_{1} \theta_{2} \\
\therefore \quad \theta=\frac{\left(R_{2} \theta_{1}+R_{1} \theta_{2}\right)}{\left(R_{1}+R_{2}\right)}
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, VITEEE, VITEEE 2019
Consider a parallelogram whose vertices are $\mathrm{A}(1,2), \mathrm{B}(4, \mathrm{y}), \mathrm{C}(\mathrm{x}, 6)$ and $\mathrm{D}(3,5)$ taken in order.
What is the area of the parallelogram?

Options:
A $\frac{7}{2}$ square units
B 4 square units
C $\frac{11}{2}$ square units
D 7 square units

The correct answer is:
7 square units

Solution:
Area of parallelogram $=2$ area of $\Delta \mathrm{ADB}$ $\vec{a}=\overline{\mathrm{AB}}=(4-1) \hat{i}+(3-2) \hat{j}$
$\vec{b}=\overrightarrow{\mathrm{AD}}=(3-1) \hat{i}+(5-2) \hat{j}$
$\therefore$ Area of parallelogram $=2\left[\frac{1}{2}|\vec{a} \times \vec{b}|\right]=|\vec{a} \times \vec{b}|$
Now; $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 2 & 3 & 0\end{array}\right|=7 \hat{k}$.
Area $=|\vec{a} \times \vec{b}|=|7 \hat{\mathrm{k}}|=\sqrt{49}=7$ square unit

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)
Consider a parallelogram whose vertices are $\mathrm{A}(1,2), \mathrm{B}(4, \mathrm{y}), \mathrm{C}(\mathrm{x}, 6)$ and $\mathrm{D}(3,5)$ taken in order.
What is the point of intersection of the diagonals?

Options:
A $\left(\frac{7}{2}, 4\right)$
B $(3,4)$
C $\left(\frac{7}{2}, 5\right)$
D $(3,5)$

The correct answer is:
$\left(\frac{7}{2}, 4\right)$

Solution:
Point of intersection $(a, b)$ is $\left(\frac{7}{2}, 4\right)$.

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider a parallelogram whose vertices are $\mathrm{A}(1,2), \mathrm{B}(4, \mathrm{y}), \mathrm{C}(\mathrm{x}, 6)$ and $\mathrm{D}(3,5)$ taken in order.
What is the value of $\mathrm{AC}^{2}-\mathrm{BD}^{2}$ ?

Options:
A 25
B 30
C 36
D 40

The correct answer is:
36

Solution:
Suppose Mid point of $A C$ and $B D$ is $M(a, b)$.
$a=\frac{1+x}{2}=\frac{3+4}{2}$
$\Rightarrow x=6$
$b=\frac{5+y}{2}=\frac{2+6}{2}$
$\Rightarrow y=3$
$a=\frac{7}{2}, b=4$
$A C^{2}=(1-x)^{2}+(2-6)^{2}=(1-6)^{2}+(-4)^{2}=41$
$B D^{2}=(3-4)^{2}+(5-3)^{2}=1+4=5$
$A C^{2}-B D^{2}=41-5$
$A C^{2}-B D^{2}=36$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider a random experiment of throwing together a die and two coins. The associated sample space has

Options:
A 8 points
B 12 points
C 24 points
D 36 points

The correct answer is:
24 points

Solution:
$6 \times 2 \times 2=24$ sample points $(\because$ no. of points in sample space of a die $=6$ and no. of points in sample space of a $\operatorname{coin}=2$ )

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)
Consider a sphere passing through the origin and the points $(2,1,-1),(1,5,-4),(-2,4,-6)$
Consider the following statements:
1 . The sphere passes through the point $(0,4,0)$.
2 . The point $(1,1,1)$ is at a distance of 5 unit from the centre of the sphere. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
1 only

Solution:
(1) Equation of sphere is $x^{2}+y^{2}+z^{2}+2 x-4 y+6 z=0$
Put the value $(0,4,0)$, we get $0+16+0+0-16+0=0$
So, the sphere passes through the point $(0,4,0)$. Hence, Statement 1 is correct.
2 . Distance between the point $(1,1,1)$ and centre of sphere $(-1,2,-3)$
$=\sqrt{(1+1)^{2}+(1-2)^{2}+(1+3)^{2}}$
$=\sqrt{4+1+16}=\sqrt{21} \neq 5$
Hence, statement 2 is not correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 2)
Consider a sphere passing through the origin and the points $(2,1,-1),(1,5,-4),(-2,4,-6)$
What is the radius of the sphere?

Options:
A $\sqrt{12}$
B $\sqrt{14}$
C 12
D 14

The correct answer is:
$\sqrt{14}$

Solution:
Equation of sphere passing through origin is $x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z=0$
which passes through the points $(2,1,-1),(1,5,-4)$, and $(-2,4,-6)$
$\therefore 4 u+2 v-2 w=-6$
$2 \mathrm{u}+10 \mathrm{v}-8 \mathrm{w}=-42$
and $-4 u+8 v-12 w=-56$
Erom eqns (i),
(ii) and (iii), we get $\mathrm{u}=1, \mathrm{v}=-2$ and $\mathrm{w}=3$
$\therefore$ Radius of sphere $=\sqrt{u^{2}+v^{2}+w^{2}}$
$=\sqrt{1+4+9}=\sqrt{14}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 2)
Consider a sphere passing through the origin and the points (2,1,-1),(1,5,-4),(-2,4,-6) What is the centre of the sphere?

Options:
A $\quad(-1,2,-3)$
B $(1,-2,3)$
C $(1,2,-3)$
D $(-1,-2,-3)$

The correct answer is:
$\quad(-1,2,-3)$

Solution:
$\quad$ From explanation 54 Centre of sphere, $(-\mathrm{u},-\mathrm{v},-\mathrm{w})=(-1,2,-3)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 2)
Consider a tiangle $\mathrm{ABC}$ in which
$\cos A+\cos B+\cos C=\sqrt{3} \sin \frac{\pi}{3}$
What is the value of
$$
\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{C}+\mathrm{A}}{2}\right) ?
$$

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{1}{16}$
D None of the above

The correct answer is:
None of the above

Solution:
As we know that
$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{1}{8}$
$\because\left(\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}\right)$
$\therefore \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{C+A}{2}\right)=\frac{1}{8}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2016 (Phase 1)
Consider a tiangle $\mathrm{ABC}$ in which
$\cos A+\cos B+\cos C=\sqrt{3} \sin \frac{\pi}{3}$
What is the value of $\sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}$ ?

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{8}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{8}$

Solution:
Given $\cos A+\cos B+\cos C=\sqrt{3} \sin \frac{\pi}{3}$
$\cos A+\cos B+\cos C$
$=2 \cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}+\left(1-2 \sin ^{2} \frac{C}{2}\right)$
$=2 \sin \frac{C}{2} \cos \frac{A-B}{2}+1-2 \sin ^{2} \frac{C}{2}$
$\left(\because \frac{A+B}{2}=90^{\circ}-\frac{C}{2}\right)$
$=1+2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}-\sin \frac{C}{2}\right]$
$=1+2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right]$
$=1+4 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2}$
$\cos A+\cos B+\cos C=\sqrt{3} \sin \frac{\pi}{3}$
$1+4 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2}=\sqrt{3} \sin \frac{\pi}{3}$
$\Rightarrow 1+4 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2}=\sqrt{3} \times \frac{\sqrt{3}}{2}$
$\Rightarrow 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{3}{2}-1$
$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{1}{8}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2016 (Phase 1)
Consider a triangle $\mathrm{ABC}$ satisfying $2 a \sin ^{2}\left(\frac{C}{2}\right)+2 c \sin ^{2}\left(\frac{A}{2}\right)=2 a+2 c-3 b$
$\sin \mathrm{A}, \sin \mathrm{B}_{,} \sin \mathrm{C}$ are in

Options:
A G.P.
B A.P.
C HP.
D Neither in G.P. nor in A.Pnor in H.P.

The correct answer is:
A.P.

Solution:
As we have already proven $2 \mathrm{~b}=\mathrm{a}+\mathrm{c}$
as-
$\mathrm{a}=\mathrm{RsinA}$
$\mathrm{b}=\mathrm{RsinB}$
$\mathrm{c}=\mathrm{RsinC}$
$\Rightarrow 2(\mathrm{RsinB})=\mathrm{Rsin} \mathrm{A}+\mathrm{RsinC}$
$\Rightarrow 2 \mathrm{R}(\sin \mathrm{B})=\mathrm{R}(\sin \mathrm{A}+\sin \mathrm{C})$
$2 \sin \mathrm{B}=\sin \mathrm{A}+\sin \mathrm{C}$
$\therefore$ Statement (1) is incorrect.
also $\frac{\mathrm{p}}{\mathrm{q}}=\frac{\mathrm{q}}{\mathrm{r}}=\frac{1}{\sqrt{3}}$
So $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in G.P.
$\therefore$ Statement $(2)$ is correct.
$\begin{array}{lll}\text { 184. (d) } x^{2}+b x+c=0 & b \neq 0\end{array}$
$\because \tan \alpha$ and $\tan \beta$ are roots of equation.
$\therefore \tan \alpha+\tan \beta=\frac{-b}{1}=-\mathrm{b}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 2)
Consider a triangle $\mathrm{ABC}$ satisfying $2 a \sin ^{2}\left(\frac{C}{2}\right)+2 c \sin ^{2}\left(\frac{A}{2}\right)=2 a+2 c-3 b$
The sides of the triangle are in

Options:
A G.P.
B A.P.
C HP.
D Neither in G.P. nor in A.Pnor in H.P.

The correct answer is:
A.P.

Solution:
$\quad 2 \mathrm{a} \sin ^{2}\left(\frac{\mathrm{C}}{2}\right)+2 \mathrm{c} \sin ^{2}\left(\frac{\mathrm{A}}{2}\right)=2 \mathrm{a}+2 \mathrm{c}-3 \mathrm{~b}$
$\Rightarrow 2 \mathrm{a} \times \frac{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}{\mathrm{ab}}+2 \mathrm{c} \times \frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{bc}}=2 \mathrm{a}+2 \mathrm{c}-3 \mathrm{~b}$
$\Rightarrow \frac{2}{\mathrm{~b}}(\mathrm{~s}-\mathrm{b})[\mathrm{s}-\mathrm{a}+\mathrm{s}-\mathrm{c}]=2 \mathrm{a}+2 \mathrm{c}-3 \mathrm{~b}$
$\Rightarrow \frac{2}{\mathrm{~b}}(\mathrm{~s}-\mathrm{b}) \mathrm{b}=2 \mathrm{a}+2 \mathrm{c}-3 \mathrm{~b} \quad[\because 2 \mathrm{~s}-\mathrm{a}-\mathrm{c}=\mathrm{b}]$
$\Rightarrow \mathrm{a}+\mathrm{c}=2 \mathrm{~b}$
So $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2015 (Phase 2)
Consider an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ .
What is the area included between the ellipseand the greatest rectangle inscribed in the ellipse?

Options:
A $a b(\pi-1)$
B $2 a b(\pi-1)$
C $a b(\pi-2)$
D None of these

The correct answer is:
$a b(\pi-2)$

Solution:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Area of ellipse is $\pi a b$ Area ofshaded region $=$ Area of ellipse- Area of rectangle $=\pi a b-2 a b=a b(\pi-2)$

Topic: Mathematics, Ellipse, NDA, NDA 2014 (Phase 1)
Consider an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ .
What is the area of the greatest rectangle that can be inscribed in the ellipse?

Options:
A $a b$
B $2 a b$
C $a b / 2$
D $\sqrt{a b}$

The correct answer is:
$2 a b$

Solution:
Given equation of ellipse, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Let $\mathrm{A}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$ be any point on ellipse(1st quadrant)
Coordinate of $\mathrm{B}=[\mathrm{a} \cos (\pi-\theta), \mathrm{b} \sin (\pi-\theta)]$
$=(-a \cos \theta, b \sin \theta)$(2nd quadrant)
Coordinate of $C=[a \cos (\pi+\theta), b \sin (\pi+\theta)]$(3rd quadrant)
Coordinate of $\mathrm{D}=[\mathrm{a} \cos (2 \pi-\theta), \mathrm{b} \sin (2 \pi-\theta)]$
$=(a \cos \theta,-b \sin \theta) \quad$ (4th quadrant)
Area of the rectangle $\mathrm{ABCD}$ $=(a \cos \theta+a \cos \theta)(b \sin \theta+b \sin \theta)$
$=2 a \cos \theta \times 2 b \sin \theta=2 a b \sin 2 \theta$
$=2 a b \times 1=2 a b$

Topic: Mathematics, Ellipse, NDA, NDA 2014 (Phase 1)
Consider any point $P$ on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ in the first quadrant. Let $\mathrm{r}$ and s represent its distances from $(4,0)$ and
$(-4,0)$ respectively, then $(\mathrm{r}+\mathrm{s})$ is equal to

Options:
A 10 unit
B 9 unit
C 8 unit
D 6 unit

The correct answer is:
10 unit

Solution:
$\frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{9}=1$
Put $\mathrm{x}=3$
$\frac{9}{25}+\frac{y^{2}}{9}=1$
$y=\frac{12}{5}$
$\mathrm{P}=(3,12 / 5)$
$r=P O=\sqrt{(4-3)^{2}+\left(0-\frac{12}{5}\right)^{2}}$
$=13 / 5$
$\mathrm{S}=\mathrm{PO}^{\prime}=\sqrt{(-4-3)^{2}+\left(0-\frac{12}{5}\right)^{2}}$
$=37 / 5$

Topic: Mathematics, Ellipse, NDA, NDA 2015 (Phase 2)
Consider events $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ of the sample space $\mathrm{S}=\{\mathrm{n}: \mathrm{n}$ is an
integer such that $10 \leq \mathrm{n} \leq 20\}$ given by:
A is the set of all even numbers.
B is the set of all prime numbers.
$\mathrm{C}=(15)$
$\mathrm{D}$ is the set of all integers $\leq 16$.
E is the set of all double digit numbers expressible as a power of 2.
$\mathrm{B}$ and $\mathrm{C}$ are

Options:
A Mutually exclusive events but not exhaustive events
B Compound events
C Mutually exclusive and exhaustive events
D Elementary events

The correct answer is:
Mutually exclusive events but not exhaustive events

Solution:
$\mathrm{B}$ and $\mathrm{C}$ are mutually exclusive events but not exhaustive events.

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
Consider events $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ of the sample space $\mathrm{S}=\{\mathrm{n}: \mathrm{n}$ is an
integer such that $10 \leq \mathrm{n} \leq 20\}$ given by:
A is the set of all even numbers.
B is the set of all prime numbers.
$\mathrm{C}=(15)$
$\mathrm{D}$ is the set of all integers $\leq 16$.
E is the set of all double digit numbers expressible as a power of 2.
A, B and $C$ are

Options:
A Mutually exclusive events but not exhaustive events
B Exhaustive events but not mutually exclusive events
C Mutually exclusive and exhaustive events
D Elementary events

The correct answer is:
Mutually exclusive and exhaustive events

Solution:
A, B and C are exhaustive events and mutually exclusive events.

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
Consider events $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ of the sample space $\mathrm{S}=\{\mathrm{n}: \mathrm{n}$ is an
integer such that $10 \leq \mathrm{n} \leq 20\}$ given by:
A is the set of all even numbers.
B is the set of all prime numbers.
$\mathrm{C}=(15)$
$\mathrm{D}$ is the set of all integers $\leq 16$.
E is the set of all double digit numbers expressible as a power of 2.
A, $\mathrm{B}$ and $\mathrm{D}$ are

Options:
A Mutually exclusive events but not exhaustive events
B Exhaustive events but not mutually exclusive events
C Mutually exclusive and exhaustive events
D Elementary events

The correct answer is:
Exhaustive events but not mutually exclusive events

Solution:
A, B and D are exhaustive events but not mutually
exclusive events.

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
Consider events $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ of the sample space $\mathrm{S}=\{\mathrm{n}: \mathrm{n}$ is an
integer such that $10 \leq \mathrm{n} \leq 20\}$ given by:
A is the set of all even numbers.
B is the set of all prime numbers.
$\mathrm{C}=(15)$
$\mathrm{D}$ is the set of all integers $\leq 16$.
E is the set of all double digit numbers expressible as a power of 2.
$\mathrm{C}$ and $\mathrm{E}$ are

Options:
A Mutually exclusive events but not elementary events
B Exhaustive events but not mutually exclusive events
C Mutually exclusive and exhaustive events
D Elementary and mutually exclusive events

The correct answer is:
Elementary and mutually exclusive events

Solution:
$\mathrm{C}$ and $\mathrm{E}$ are mutually exclusive and elementary events.

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
Consider following features:

Organ system level of organisation

Bilateral symmetry

True coelomates with segmentation of body.

Select the correct option of animal group which possess all the above characteristics:

Options:
A Annelida, Arthropoda and Chordata
B Annelida, Arthropoda and Mollusca
C Arthropoda, Mollusca and Chordata
D Annelida, Mollusca and Chordata

The correct answer is:
Annelida, Arthropoda and Chordata

Solution:
Organ system level organization, bilateral symmetry, true coelomates with the segmentation of body are seen in Annelida, Arthropoda, and Chordata.

Annelids possess metameric segmentation.

Arthropods have segmented bodies, like the annelids. These segments have become specialized, however, with one pair of jointed appendages added to each segment. Among living arthropods, the millipedes most closely suggest what the ancestral arthropod might have looked like.

Chordates use heteromeric segmentation in which the link between segmented parts appears more subtly.

Topic: Biology, Animal Kingdom, NEET, NEET 2019 (All India)