datavorous/entrance-exam-dataset huggingface dataset:
Consider the two circles $(x-1)^{2}+(y-3)^{2}=r^{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}-8 \mathrm{x}+2 \mathrm{y}+8=0$
What is the distance between the centres of the two circles?

Options:
A 5 units
B 6 units
C 8 units
D 10 units

The correct answer is:
5 units

Solution:
Given equation of circles $(x-1)^{2}+(y-3)^{2}=r^{2}$
$\left(h_{1}, k_{1}\right) \equiv$ coordinates of centre $\equiv(1,3)$
$$
\begin{array}{l}
\therefore x^{2}+y^{2}-8 x+2 y+8=0 \\
\Rightarrow(x-4)^{2}+(y+1)^{2}=(3)^{2}
\end{array}
$$
$\left(h_{2}, k_{2}\right) \equiv$ coordinates of centre $\equiv(4,-1)$
Distance between centres of two circles
$d=\sqrt{\left(h_{1}-h_{2}\right)^{2}+\left(k_{1}-k_{2}\right)^{2}}$
$\Rightarrow d=\sqrt{(1-4)^{2}+(3+1)^{2}}$
$\Rightarrow d=\sqrt{25}$
$d=5$ units

Topic: Mathematics, Circle, NDA, NDA 2016 (Phase 1)
Consider the two idealised systems (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid $\quad$ of $\quad$ length $\mathrm{L}>\mathrm{R}$, radius of cross-section. In (i) $\mathrm{E}$ is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below

Options:
A
case (i) contradicts Gauss' law for electrostatic fields

B
case (ii) contradicts Gauss' law for magnetic fields

C
case (i) agrees with $\oint \mathrm{E} . \mathrm{dl}=0$.

D
case (ii) contradicts $\oint \mathrm{H} . \mathrm{dl}=\mathrm{I}_{\mathrm{en}}$

The correct answer is:
case (ii) contradicts Gauss' law for magnetic fields

Solution:
The electric field lines, do not form a continuous path while the magnetic field lines form the closed paths.
Gauss's law states that, $\oint_{\mathrm{s}}$ E.ds $=\frac{\mathrm{q}}{\varepsilon_0}$ for electrostatic field. So, it does not contradict for electrostatic fields as the electric field lines do not form closed continuous path.
According to Gauss' law in magnetic field, $\oint E \mathrm{ds}=0$
It contradicts for magnetic field, because there is a magnetic field inside the solenoid and no field outside the solenoid carrying current but the magnetic field lines form the closed path.
According to Gauss' law in magnetic field,
$$
\oint_{\mathrm{s}} \mathrm{E} . \mathrm{ds}=0
$$
It contradicts for magnetic field, because there is a magnetic field inside the solenoid and no field outside the solenoid carrying current but the magnetic field lines form the closed path.

Topic: Physics, Magnetic Properties of Matter
Consider the two lines $x+y+1=0$ and $3 x+2 y+1=0$
What is the equation of the line passing through the point of intersection of the given lines and parallel to y-axis? $[2016-I]$

Options:
A $x+1=0$
B $\mathrm{x}-1=0$
C $x-2=0$
D $x+2=0$

The correct answer is:
$\mathrm{x}-1=0$

Solution:
Equation of $y$ -axis $x=0$
Equation of line parallel to $y$ -axis is $x=k$
If this line passes through $(1,-2$ then $)$ $x=1$
Hence equation of line which passes through point of intersection of given line $(1,-2)$ and parallel to $y$ -axis $x=1$
$\Rightarrow x-1=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider the two lines $x+y+1=0$ and $3 x+2 y+1=0$
What is the equation of the line passing through the point of intersection of the given lines and parallel to $\mathrm{x}$ -axis?

Options:
A $\mathrm{y}+1=0$
B $\mathrm{y}-1=0$
C $\mathrm{y}-2=0$
D $\mathrm{y}+2=0$

The correct answer is:
$\mathrm{y}+2=0$

Solution:
Equation of $x$ -axi $y=0$
Line parallel to $x$ axis is $y=k$
If this line passes through $(1,-2)$ then $k=-2$
$\Rightarrow y=-2$
$\Rightarrow y+2=0$
Equation of line passing through $(1,-2)$ and parallel to $x$ -axis is $y+2=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider the two series of observations Aand B as follows:
\begin{array}{l|l|l|l|l|l}
\hline Series A & 1019 & 1008 & 1015 & 1006 & 1002 \\
\hline Series B & 1.9 & 0.8 & 1.5 & 0.6 & 0.2 \\
\hline
\end{array}
If the standard deviation of the Series A is $\sqrt{38,}$ then what is the standard deviation of the Series $\mathrm{B}$ ?

Options:
A $3.8$
B $\sqrt{0.38}$
C $0.38$
D $\sqrt{38}$

The correct answer is:
$\sqrt{0.38}$

Solution:
Standard deviation (series B)
$=\sqrt{\left(\frac{1.9+0.8+1.5+0.6+0.2}{5}\right)^{2}}$
$=\sqrt{\frac{6.9}{5}-1}=\sqrt{1.38-1}$
$=\sqrt{0.38}$

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 1)
Consider the vectors $\bar{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $\bar{b}=4 \hat{i}-4 \hat{j}+7 \hat{k}$.
What is the scalar projection of $\bar{a}$ on $\bar{b}$ ?

Options:
A 1
B $19 / 9$
C $17 / 9$
D $23 / 9$

The correct answer is:
$19 / 9$

Solution:
Projection of $\vec{a}$ on $\overrightarrow{\mathrm{b}}=\frac{\overrightarrow{\mathrm{a}} \overrightarrow{\vec{b}}}{|\overrightarrow{\mathrm{b}}|}=\frac{(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})}{\left|\sqrt{4^{2}+(-4)^{2}+7^{2}}\right|}$
$=\frac{19}{9}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
Consider the vectors $\bar{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $\bar{b}=4 \hat{i}-4 \hat{j}+7 \hat{k}$.
What is the vector perpendicular to both the vectors?

Options:
A $\quad-10 \hat{i}-3 \hat{j}+4 \hat{k}$
B $-10 \hat{i}+3 \hat{j}+4 \hat{k}$
C $10 \hat{i}-3 \hat{j}+4 \hat{k}$
D None of these

The correct answer is:
$\quad-10 \hat{i}-3 \hat{j}+4 \hat{k}$

Solution:
Vector perpendicular to $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 4 & -4 & 7\end{array}\right|=\hat{i}(-14+4)-j(7-4)+\hat{k}(-4+8)$
$=-10 \hat{i}-3 \hat{j}+4 \hat{k}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
Consider two identical tubes A and B. Tube A has both ends open while tube B has one end closed. The ratio of the fundamental frequency of tube A and B is

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

The correct answer is:
2 : 1

Solution:
In tube A, Î»A=2l
In tube B, Î»B=4l
â´ ÏA=Î½Î»A=Î½2l
â´ ÏB=Î½Î»B=Î½4l
â´ ÏAÏB=21.

Topic: Physics, Waves and Sound, NEET
Consider two matrices $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -4 \\ 2 & 1 & -4\end{array}\right]$.
Which one of the following is correct?

Options:
A $B$ is the right inverse of $A$
B $B$ is the left inverse of $A$
C $B$ is the both sided inverse of $A$
D None of the above

The correct answer is:
$B$ is the left inverse of $A$

Solution:
I.
$\mathrm{AB}=\left|\begin{array}{cc}1 & 2 \\ 2 & 1 \\ 1 & 1\end{array}\right|_{3 \times 2}\left|\begin{array}{ccc}1 & 2 & -4 \\ 2 & 1 & -4\end{array}\right|_{2 \times 3}$
$=\left|\begin{array}{rrr}5 & 4 & -12 \\ 4 & 5 & -12 \\ 3 & 3 & -8\end{array}\right|_{3 \times 3}$
II.
$\begin{aligned} \mathrm{BA} &=\left|\begin{array}{lll}1 & 2 & -4 \\ 2 & 1 & -4\end{array}\right|_{2 \times 3}\left|\begin{array}{cc}1 & 2 \\ 2 & 1 \\ 1 & 1\end{array}\right|_{3 \times 2} \\ &=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|_{2 \times 2} \end{aligned}$
Here, $\mathrm{B}$ is not the right inverse of $\mathrm{A}$ but $\mathrm{B}$ is the left inverse of $\mathrm{A}$.

Topic: Mathematics, Matrices, NDA, NDA 2014 (Phase 1)
Consider $x=4 \tan ^{-1}\left(\frac{1}{5}\right), y=\tan ^{-1}\left(\frac{1}{70}\right)$ and $z=\tan ^{-1}\left(\frac{1}{99}\right)$.
What is $x$ equal to?

Options:
A $\tan ^{-1}\left(\frac{60}{119}\right)$
B $\tan ^{-1}\left(\frac{120}{119}\right)$
C $\tan ^{-1}\left(\frac{90}{169}\right)$
D $\tan ^{-1}\left(\frac{170}{169}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{120}{119}\right)$

Solution:
$\mathrm{x}=4 \tan ^{-1}\left(\frac{1}{5}\right)$
$=2\left[2 \tan ^{-1} \frac{1}{5}\right]=2\left[\tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right]$
$=2 \tan ^{-1}\left(\frac{\frac{2}{5} \times 25}{24}\right)=2 \tan ^{-1} \frac{10}{24}=2 \tan ^{-1} \frac{5}{12}$
$=\tan ^{-1}\left(\frac{2 \times \frac{5}{12}}{1-\frac{25}{144}}\right)=\tan ^{-1} \frac{120}{119} .$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2015 (Phase 1)
Consider $x=4 \tan ^{-1}\left(\frac{1}{5}\right), y=\tan ^{-1}\left(\frac{1}{70}\right)$ and $z=\tan ^{-1}\left(\frac{1}{99}\right)$.
What is $x-y$ equal to?

Options:
A $\tan ^{-1}\left(\frac{828}{845}\right)$
B $\tan ^{-1}\left(\frac{8287}{8450}\right)$
C $\tan ^{-1}\left(\frac{8281}{8450}\right)$
D $\tan ^{-1}\left(\frac{8287}{8471}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{8281}{8450}\right)$

Solution:
$x-\mathrm{y}=\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1} \frac{1}{70}$
$=\tan ^{-1} \mid \frac{\frac{120}{119}-\frac{1}{70}}{1+\left(\frac{120}{119} \times \frac{1}{70}\right) \mid}$
$$
=\tan ^{-1}\left|\frac{\frac{8400-119}{8330}}{1+\frac{120}{8330}}\right|
$$
$\therefore \quad$ Option (c) is correct.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2015 (Phase 1)
Consider $x=4 \tan ^{-1}\left(\frac{1}{5}\right), y=\tan ^{-1}\left(\frac{1}{70}\right)$ and $z=\tan ^{-1}\left(\frac{1}{99}\right)$.
What is $x-y+z$ equal to?

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{6}$
D $\frac{\pi}{4}$

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

Solution:
$x-y=\tan ^{-1} \frac{8281}{8450} \Rightarrow \tan ^{-1}\left|\frac{8281}{8450}\right|+\tan ^{-1}\left(\frac{1}{99}\right)$
$=\tan ^{-1}\left|\frac{\frac{8281}{8450}+\frac{1}{99}}{1-\frac{8281}{8450} \times \frac{1}{99}}\right|=\tan ^{-1} \mid \frac{\frac{828269}{836550}}{\frac{828269}{836550} \mid=1}$
$\tan ^{-1}=(1)=\frac{\pi}{4}$
$\therefore \quad$ Option (d) is correct.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2015 (Phase 1)
Consider $\int x \tan ^{-1} x d x=A\left(x^{2}+1\right) \tan ^{-1} x+B x+C$,
where $\mathrm{C}$ is the constant of integration.
What is the value of $A$?

Options:
A 1
B $1 / 2$
C $-1 / 2$
D $1 / 4$

The correct answer is:
$1 / 2$

Solution:
Given, $\int x \tan ^{-1} x d x=A\left(x^{2}+1\right) \tan ^{-1} x+B x+C$
where, $\mathrm{C}$ is the constant of integration
Consider, $\int_{I I}{x \tan ^{-1} x d x}$
$\quad=\tan ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{d}{d x}\left(\tan ^{-1} x\right) \cdot \frac{x^{2}}{2} d x$
(using integration by parts)
$=\frac{x^{2} \cdot \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(\int\left(\frac{1+x^{2}-1}{1+x^{2}}\right) d x\right)$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(\int d x-\int \frac{d x}{1+x^{2}}\right)$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+C$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{x}{2}+\frac{\tan ^{-1} x}{2}+C$
$=\frac{1}{2}\left(x^{2}+1\right) \tan ^{-1} x-\frac{x}{2}+C$
$\mathrm{A}=\frac{1}{2}$, hence option (b) is correct.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2014 (Phase 2)
Consider $\int x \tan ^{-1} x d x=A\left(x^{2}+1\right) \tan ^{-1} x+B x+C$,
where $\mathrm{C}$ is the constant of integration.
What is the value of $B$?

Options:
A 1
B $1 / 2$
C $-1 / 2$
D $1 / 4$

The correct answer is:
$-1 / 2$

Solution:
Given, $\int x \tan ^{-1} x d x=A\left(x^{2}+1\right) \tan ^{-1} x+B x+C$
where, $\mathrm{C}$ is the constant of integration
Consider, $\int_{I I}{x \tan ^{-1} x d x}$
$\quad=\tan ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{d}{d x}\left(\tan ^{-1} x\right) \cdot \frac{x^{2}}{2} d x$
(using integration by parts)
$=\frac{x^{2} \cdot \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(\int\left(\frac{1+x^{2}-1}{1+x^{2}}\right) d x\right)$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(\int d x-\int \frac{d x}{1+x^{2}}\right)$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+C$
$=\frac{x^{2} \tan ^{-1} x}{2}-\frac{x}{2}+\frac{\tan ^{-1} x}{2}+C$
$=\frac{1}{2}\left(x^{2}+1\right) \tan ^{-1} x-\frac{x}{2}+C$
$\mathrm{B}=-\frac{1}{2}$, hence option (c) is correct.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2014 (Phase 2)
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Options:

None

Solution:
Oxidation of \(\mathrm{Fe}^{2+}\) converts it to \(\mathrm{Fe}^{3+}\), i.e.,
\(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} ; \mathrm{E}_{\mathrm{ox}}^{\mathrm{o}}=-0.77 \mathrm{~V}\)
Only those substances can oxidise \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) which are stronger oxidizing agents and have positive reduction potentials greater than \(0.77 \mathrm{~V}\), so that EMF of the cell reaction is positive. This is so for elements lying below \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) in the series ex: \(\mathrm{BO}_2, \mathrm{Cl}_2\) and \(\mathrm{F}_2\).

Topic: Chemistry, Electrochemistry
Consult the table of standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.

Options:
A Silver, Platinum, and Gold
B Zinc, Aluminium, and Iron
C Copper, Mercury, and Lead
D Nickel, Tin, and Chromium

The correct answer is:
Copper, Mercury, and Lead

Solution:
Correct Option is : (C)
Copper, Mercury, and Lead

Topic: Chemistry, Electrochemistry, NEET
Coordinates of the points $\mathrm{O}, \mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are respectively $(0,0,0),(4,6,2 \mathrm{~m}),(2,0,2 \mathrm{n})$ and $(2,4,6)$. Let $\mathrm{L}, \mathrm{M}, \mathrm{N}$ and $\mathrm{K}$ be
points on the sides OR, OP, PQ and QR respectively such that LMNK is a parallelogram whose two adjacent sides
side L.K are each of length $\sqrt{2}$. What are the values of $\mathrm{m}$ and
n respectively?

Options:
A 6,2
B 1,3
C 3,1
D None of the above

The correct answer is:
3,1

Solution:
Go through the option (c)

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 2)
Coordination number of nickel in $\left[\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{4-}$ is

Options:
A 3
B 12
C 6
D 4

The correct answer is:
6

Solution:
Oxalate $\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right)$ is an example of bidendate ligand. Three oxalate ligands form six coordinate bond around $\mathrm{Ni}^{2+}$ ion. Hence, coordination number is 6 .

Topic: Chemistry, Coordination Compounds, COMEDK, COMEDK 2012
Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.

Options:

None

Solution:
Cork cambium : As the stem continues to increase in girth due to the activity of vascular cambium, the outer cortical and epidermal layers get broken and need to be replaced to provide new protective cell layers. Hence, sooner or later, another meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled and nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. This is clear how cork cambium forms tissues that form the cork.

Topic: Biology, Anatomy of Flowering Plants
Cork cambium is developed from

Options:
A Apical meristem
B Intercalary meristem
C Primitive meristem
D Lateral meristem

The correct answer is:
Lateral meristem

Solution:
Cork cambium (phellogen) is an outermost lateral meristem in woody plants, that develops from permanent tissues in the region of epidermis hypodermis, cortex, and even in outer layers of phloem. It is also known as a phellogen that forms a layer of cells that produces a secondary protective layer of the stem called the periderm.

Topic: Biology, Anatomy of Flowering Plants, NEET
Cork cambium results in the formation of cork which becomes impermeable to water due to the accumulation of

Options:
A resins
B suberin
C lignins
D tannins.

The correct answer is:
suberin

Solution:
Cork cambium is a secondary lateral meristem that may arise from permanent living cells of hypodermis or outer cortex. It is composed of a single layer of meristematic cells. Its cells divide tangentially and periclinally towards the outer face and produce cork cells. These cells are compactly arranged and have thin cellulose walls in the beginning. As they mature there is a gradual loss of living matter and cells get elongated radially, vertically or tangentially. The cell walls become thick because of development of fatty substance called suberin, which is impervious to water.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2004
Cork is commonly used for sealing bottles. It is produced by

Options:
A dermatogen
B phellogen
C xylem
D vascular cambium

The correct answer is:
phellogen

Solution:
Cork cambium or phellogen is a type of cambium arising within the outer layer of the stems of woody plants, usually as a complete ring surrounding the inner tissues. The cells of the cork cambium divide to produce an outer corky tissue (cork or phellem) and an innerly secondary cortex (phelloderm). The common bottle cork produced from Quercus suber is a product of phellogen.

Topic: Biology, Anatomy of Flowering Plants, NEET
Correct diagram for the determination of internal resistance of a primary cell by potentiometer is

Options:
A
B
C
D

The correct answer is:

Solution:
The polarity of primary cell and secondary cell should be in same sense. Also, a key is kept in series with resistance box to measure emf when the key is open.

Topic: Physics, Current Electricity, NEET
Correlation between two variable is said to be perfect if

Options:
A one variable increases, the other also increases
B one variable increases, the other decreases
C one variable increases, the other also increases proportionally
D one variable increases, the other decreases proportionally

The correct answer is:
one variable increases, the other also increases proportionally

Solution:
Correlation between two variables is said to be perfect if one variable increases, the other also increase:
proportionally.

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 2)
$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

Options:
A $\frac{\sin 2^n A}{2^n \sin A}$
B $\frac{2^n \sin 2^n A}{\sin A}$
C $\frac{2^n \sin A}{\sin 2^n A}$
D $\frac{\sin A}{2^n \sin 2^n A}$

The correct answer is:
$\frac{\sin 2^n A}{2^n \sin A}$

Solution:
It is a standard result.
$\begin{gathered}
\cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A \\
=\frac{\sin 2^n A}{2^n \sin A}
\end{gathered}$

Topic: Mathematics, Trigonometric Ratios & Identities, AP EAMCET, AP EAMCET 2009
$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

Options:
A $\frac{\sin 2^n A}{2^n \sin A}$
B $\frac{2^n \sin 2^n A}{\sin A}$
C $\frac{2^n \sin A}{\sin 2^n A}$
D $\frac{\sin A}{2^n \sin 2^n A}$

The correct answer is:
$\frac{\sin 2^n A}{2^n \sin A}$

Solution:
It is a standard result.
$\begin{gathered}
\cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A \\
=\frac{\sin 2^n A}{2^n \sin A}
\end{gathered}$

Topic: Mathematics, Trigonometric Ratios & Identities, TS EAMCET, TS EAMCET 2009
Cryopreservation, seed bank, gene bank and wild safari parks are the methods of

Options:
A ex-situ conservation of biodiversity
B in-situ conservation of biodiversity
C collection of different types of soil
D taxonomical aids

The correct answer is:
ex-situ conservation of biodiversity

Solution:
Ex-situ conservation literally means 'off-site conservation'. It is the process of protecting an endangered species, variety or breed of plant or animal outside its natural habitat. Threatened plant or animal is taken out from its natural habitat and placed in special setting, where they can be protected and given special care.

Topic: Biology, Biodiversity and its Conservation, NEET
Culturing of shoot apex as explant on nutrient medium in tissue culture gives to disease-free plants because:

Options:
A Such an explant contains secondary metabolites
B It contains toxins that prevent growth of other organism
C Such explant is pathogen free
D Sterilization kills all pathogens

The correct answer is:
Such explant is pathogen free

Solution:
Culturing of shoot apex as an explant on nutrient medium in tissue gives rise to disease-free plants because the shoot tips regenerated are free of infections or pathogen such as bacteria, fungi, virus, etc. The infection cannot invade into the shoot tip because the shoot tip cells show a high rate of cell division.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Curve of intersection of two spheres is

Options:
A an ellipse
B a circle
C a parabola
D None of these

The correct answer is:
a circle

Solution:
a circle (obviously)

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 2)
$d B$ is a standard abbreviation used for the quantitative expression of

Options:
A the density of bacteria in a medium
B a particular pollutant
C the dominant Bacillus in a culture
D a certain pesticide

The correct answer is:
a particular pollutant

Solution:
Noise pollution is a physical form of pollution that affects the receiver directly. $\mathrm{dB}$ (decibel) is a standard abbreviation used for the quantitative expression of noise. Noise or pollutant sound has a value of $80 \mathrm{~dB}$ and above, it harms hearing system. The WHO has fixed $45 \mathrm{~dB}$ as the safe noise level for a city.

Topic: Biology, Environmental Issues, NEET, NEET 2010 (Screening)
$\beta-\mathrm{D}-(+)-$ Glucopyranose is

Options:
A

B

C

D

The correct answer is:

Solution:
No solution. Refer to answer key.

Topic: Chemistry, Biomolecules, AP EAMCET, AP EAMCET 2017 (24 Apr Shift 1)
$\frac{d}{d x}\left(\sin ^{-1}\left(3 x-4 x^3\right)\right), \frac{1}{2}

MathematicsDifferentiationMHT CETMHT CET 2022 (05 Aug Shift 1)

Options:

A $\frac{1}{3 \sqrt{1-x^2}}$
B $\frac{-3}{\sqrt{1-x^2}}$
C $\frac{-1}{3 \sqrt{1-x^2}}$
D $\frac{3}{\sqrt{1-x^2}}$

Solution:

2930 Upvotes

Verified Answer

The correct answer is:
$\frac{-3}{\sqrt{1-x^2}}$

$\begin{aligned} & \frac{\mathrm{d}\left(\sin ^{-1}\left(3 \mathrm{x}-4 \mathrm{x}^3\right)\right)}{\mathrm{dx}} \\ & =\frac{\mathrm{d}\left(\pi-3 \sin ^{-1} \mathrm{x}\right)}{\mathrm{dx}}\left[\text { as } \frac{1}{2}<\mathrm{x}<1\right]=\frac{-3}{\sqrt{1-\mathrm{x}^2}}\end{aligned}$

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

The correct answer is:
$\frac{-3}{\sqrt{1-x^2}}$

Solution:
$\begin{aligned} & \frac{\mathrm{d}\left(\sin ^{-1}\left(3 \mathrm{x}-4 \mathrm{x}^3\right)\right)}{\mathrm{dx}} \\ & =\frac{\mathrm{d}\left(\pi-3 \sin ^{-1} \mathrm{x}\right)}{\mathrm{dx}}\left[\text { as } \frac{1}{2}<\mathrm{x}<1\right]=\frac{-3}{\sqrt{1-\mathrm{x}^2}}\end{aligned}$

Topic: Mathematics, Differentiation, MHT CET, MHT CET 2022 (05 Aug Shift 1)
Danny ate spinach, salad and mutton curry for lunch. At which of the following trophic levels did Danny function in this food chain?

(i) Producer

(ii) Primary Consumer

(iii) Secondary Consumer

(iv) Tertiary Consumer

Options:
A i and ii
B ii only
C ii and iii
D iv only

The correct answer is:
ii and iii

Solution:
Plants are the producers in the food chain. The organism consuming plants is the Primary consumer. The organism consuming animals which in turn derive energy from plants is a Secondary consumer. Thus in the above example Danny is consuming plants is a primary consumer as well as he had mutton which comes from goats which are dependent on plants for their food and acts as the secondary consumer also.

Topic: Biology, Ecosystem, NEET
Data can be represented in which of the following forms? Textual form Tabular form 3 . Graphical form Select the correct answer using the code given below

Options:
A 1 and 2 only
B 2 and 3 onlv
C 1 and 3 only
D 1,2 and

The correct answer is:
1,2 and

Solution:
Data can be represented in all three forms

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)