datavorous/entrance-exam-dataset huggingface dataset:
What are the common types of secondary structure of proteins?

Options:
A The common types of secondary structure of proteins include primary and tertiary structures.
B The common types of secondary structure of proteins are alpha helix and beta sheets.
C The common types of secondary structure of proteins are globular and fibrous.
D The common types of secondary structure of proteins are hydrophobic and hydrophilic regions.

The correct answer is:
The common types of secondary structure of proteins are alpha helix and beta sheets.

Solution:
Correct Option is : (B)
The common types of secondary structure of proteins are alpha helix and beta sheets.

Topic: Chemistry, Biomolecules, NEET
What are the common types of secondary structure of proteins?

Options:

None

Solution:
The conformation which the polypeptide chains assume as a result of hydrogen bonding is called secondary structure of the proteins. The two types of secondary structures are \(\alpha\)-helix and \(\beta\)-pleated sheet structure.

Topic: Chemistry, Biomolecules
What are the co-ordinates of the foot of the perpendicular drawn from the point $(3,5,4)$ on the plane $\mathrm{z}=0$ ?

Options:
A $\quad(0,5,4)$
B $\quad(3,5,0) \quad$
C $\quad(3,0,4)$
D $(0,0,4)$

The correct answer is:
$\quad(3,5,0) \quad$

Solution:
Plane z $=0$ is simply xy plane, so z quadrant value will be zero.
So, options (b) is correct option.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 2)
What are the co-ordinates of the foot of the perpendicular from the point $(2,3)$ on the line $x+y-11=0 ?$

Options:
A $(2,9)$
B $(5,6)$
C $(-5,6)$
D $(6,5)$

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

Solution:
Let B be the foot of perpendicular $\mathrm{AB}$.
Now, $x+y-11=0$
$\Rightarrow \mathrm{y}=-\mathrm{x}+11$ ...(1)
$\Rightarrow$ Slope $=-1$ ...(2)
Since, $A B$ is perpendicular to $x+y-11=0$ product of their slopes $=-1$ $\Rightarrow-1+$ Slope of $A B=-1$
$\Rightarrow$ Slope of $\mathrm{AB}=1$
Now, equation of $\mathrm{AB}$ is given as $y-3=1(x-2) \quad$ (using slope point form) $\Rightarrow y-x=1$ ...(3)
Now, foot of perpendicular
$=$ Point of intersection of line $A B$ and $x+y-11=0$ So, on solving equation (1) and (2) we get $x=5, y=6$.
Hence, $B=(5,6)$.

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 2)
What are the coordinates of the point equidistant from the four points $(0,0,0),(2,0,0),(0,4,0),(0,0,6) ?$

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

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

Solution:
The equation of sphere passing through the given points is
$x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z+d=0$
$\therefore(0)^{2}+(0)^{2}+(0)^{2}+2 u(0)+2 v(0)+2 w(0)+d=0$
$\Rightarrow d=0$
$(2)^{2}+(0)^{2}+(0)^{2}+2 u(2)+2 v(0)+2 u(0)+0=0$
$\Rightarrow u=-1$
$(0)^{2}+(4)^{2}+(0)^{2}+2 u(0)+2 v(4)+2 u(0)+0=0$
$\Rightarrow v=-2$
$(0)^{2}+(0)^{2}+(6)^{2}+2 u(0)+2 v(0)+2 w(6)+0=0$
$\Rightarrow w=-3$
$\therefore$ The centre of sphere $=(1,2,3)$
It will be the point equidistant from the four points $(0,0,0),(2,0,0),(0,4,0)$ and $(0,0,6)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 2)
What are the degree and order respectively of differential equation of the family of rectangular hyperbolas whose axis of symmetry are the coordinate axis?

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

The correct answer is:
1,1

Solution:
The equation of family of rectangular hyperbola is $\mathrm{xy}=\mathrm{c}^{2}$
On differentiating w.r.t. $\mathrm{x}$, we get
$$
y+x \frac{d y}{d x}=0
$$
Thus, the order and degree of differential equation are 1 and 1 respectively.

Topic: Mathematics, Differential Equations, NDA, NDA 2009 (Phase 1)
What are the degree and order respectively of the differential
equation satisfying $\mathrm{e}^{\mathrm{y} \sqrt{1-\mathrm{x}^{2}}+\mathrm{x} \sqrt{1-\mathrm{y}^{2}}}=\mathrm{ce}^{\mathrm{x}}$,
$($ where $c>0,|x| < 1,|y| < 1) ? \quad$

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

The correct answer is:
1,1

Solution:
$\begin{aligned} & \mathrm{e}^{\mathrm{y}\left(\sqrt{1-\mathrm{x}^{2}}+\mathrm{x} \sqrt{1-\mathrm{y}^{2}}\right)-\mathrm{s}-\mathrm{c}} \\ & \Rightarrow \mathrm{y} \sqrt{1-\mathrm{x}^{2}}+\mathrm{x} \sqrt{1-\mathrm{y}^{2}}-\mathrm{x}=\log \mathrm{c} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}} \sqrt{1-\mathrm{x}^{2}}+\mathrm{y} \cdot \frac{1}{\not 2 \sqrt{1-\mathrm{x}^{2}}}(-\not{2} \mathrm{x})+\sqrt{1-\mathrm{y}^{2}} \\ & & \\+\mathrm{x} \cdot \frac{1}{\not{2} \sqrt{1-\mathrm{y}^{2}}}(-\not{2} \mathrm{y}) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}-1=0 \\ \Rightarrow & \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}} \sqrt{1-\mathrm{x}^{2}}-\frac{\mathrm{x} \mathrm{y}}{\sqrt{1-\mathrm{x}^{2}}}+\sqrt{1-\mathrm{y}^{2}}-\frac{\mathrm{xy}}{\sqrt{1-\mathrm{y}^{2}}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=1 \\ & \text { Degree }=1, \text { order }=1 \end{aligned}$

Topic: Mathematics, Differential Equations, NDA, NDA 2016 (Phase 2)
What are the degree and order respectively of the
differential equation $\mathrm{y}=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)^{2} ? \quad$

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

The correct answer is:
4,1

Solution:
$y=x\left(\frac{d y}{d x}\right)^{2}+\left(\frac{d x}{d y}\right)^{2}$
$=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-2}$
$\Rightarrow \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=\mathrm{x} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{4}+1$
So, degree $=4$, order $=1$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 1)
What are the difficulties that one would face in classification of animals, if common fundamental features are not taken into account?

Options:
A It is impossible to add new species detected every day without common features
B No difficulties are faced
C Cannot be defined
D None of the above

The correct answer is:
It is impossible to add new species detected every day without common features

Solution:
Without common features, it is very difficult to treat each organism individually. It is impossible to add new species detected every day without common features. To study diversity in animals, classification must be based on common basic features.

Topic: Biology, Animal Kingdom, NEET
What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?

Options:

None

Solution:
If fundamental characters are not taken into account then the classification of animals is difficult due to difference in characters in different animals which are as following :
(i) Grade of organisation
(ii) Symmetry
(iii) Coelom
(iv) Segmentation
(v) Notochord
Common fundamental features help in grouping animals in certain categories or sub-categories. For examples, all birds have fore-limbs modified to assist in flying. This gives us one clue to categorise a particular animals among aves.

Topic: Biology, Animal Kingdom
What are the direction cosines of a line which is equally inclined to the positive directions of the axes?

Options:
A $\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$
B $\left\langle-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$
C $\left\langle-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$
D $\left\langle\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right\rangle$

The correct answer is:
$\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$

Solution:
Let $\ell, \mathrm{m}, \mathrm{n}$ are the dc's of a line that is inclined equally at $\alpha$ to the $+$ ve direction of axes.
Now, $\begin{aligned} \ell &=\cos \alpha \\ \mathrm{m} &=\cos \alpha \\ \mathrm{n} &=\cos \alpha . \end{aligned}$
Also, $\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=1$
$3 \cos ^{2} \alpha=1$
$\cos \alpha=\frac{1}{\sqrt{3}}$
$\therefore \quad$ dc's of the line are : $\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
What are the direction cosines of the line represented
by $3 x+y+2 z=7, x+2 y+3 z=5 ?$

Options:
A $(-1,-7,5)$
B $(-1,7,5)$
C $\left(-\frac{1}{\sqrt{75}},-\frac{7}{\sqrt{75}}, \frac{5}{\sqrt{75}}\right)$
D $\left(-\frac{1}{\sqrt{75}}, \frac{7}{\sqrt{75}}, \frac{5}{\sqrt{75}}\right)$

The correct answer is:
$\left(-\frac{1}{\sqrt{75}},-\frac{7}{\sqrt{75}}, \frac{5}{\sqrt{75}}\right)$

Solution:
Let the direction ratio of the line be, $\mathrm{a}, \mathrm{b}, \mathrm{c}$
This line is contained by both plane,
$3 x+y+2 z=7$ and $x+2 y+3 z=5$
$\Rightarrow 3 \mathrm{a}+\mathrm{b}+2 \mathrm{c}=0$
and $a+2 b+3 c=0$
Solving these two equations
$\frac{a}{\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|}=\frac{-b}{\left|\begin{array}{ll}3 & 2 \\ 1 & 3\end{array}\right|}=\frac{c}{\left|\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right|}=k$
Let
$\Rightarrow \frac{\mathrm{a}}{-1}=\frac{-\mathrm{b}}{7}=\frac{\mathrm{c}}{5}=\mathrm{k}$
$\mathrm{a}=-\mathrm{k}, \mathrm{b}=-7 \mathrm{k}, \mathrm{c}=5 \mathrm{k}$
Direction cosines are
$\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$\sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{(-k)^{2}+(-7 k)^{2}+(5 k)^{2}}$
$=\sqrt{\mathrm{k}^{2}+49 \mathrm{k}^{2}+25^{2}}=\mathrm{k} \sqrt{75}$
So, Direction cosines are $\left(\frac{-\mathrm{k}}{\mathrm{k} \sqrt{75}}, \frac{-7 \mathrm{k}}{\mathrm{k} \sqrt{75}}, \frac{5 \mathrm{k}}{\mathrm{k} \sqrt{75}}\right)$
$=\left(-\frac{1}{\sqrt{75}},-\frac{7}{\sqrt{75}}, \frac{5}{\sqrt{75}}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 2)
What are the direction cosines of $z$ -axis?

Options:
A $ < 1,1,1>$
B $ < 1, \underline{0,0}>$
C $ < 0,1,0>$
D $ < 0,0,1>$

The correct answer is:
$ < 0,0,1>$

Solution:
No Solution

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2019 (Phase 1)
What are the direction ratios of normal to the plane $2 \mathrm{x}-\mathrm{y}+2 \mathrm{z}+1=0 \quad$

Options:
A $\langle 2,1,2\rangle$
B $\left\langle 1,-\frac{1}{2}, 1\right\rangle$
C $\langle 1,-2,1\rangle$
D None of the above

The correct answer is:
$\left\langle 1,-\frac{1}{2}, 1\right\rangle$

Solution:
Given equation of plane is $2 \mathrm{x}-\mathrm{y}+2 \mathrm{z}+1=0$ $\Rightarrow a=2, b=-1, c=2$
Hence d.R $\langle 2,-1,2\rangle$ i.e., $\left\langle 1,-\frac{1}{2}, 1\right\rangle$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 1)
What are the direction ratios of the line determined by the planes $x-y+2 z=1$ and $x+y-z=3 ? \quad$

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

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

Solution:
The intersection of given plane is $x-y+2 z-1+\lambda(x+y-z-3)=0$
$\Rightarrow x(1+\lambda)+y(\lambda-1)+z(2-\lambda)-3 \lambda-1=0$
DR's of normal to the above plane is $(1+\lambda, \lambda-1,2-\lambda)$
By taking option (a) $-1(1+\lambda)+3(\lambda-1)+2(2-\lambda)=0$
$\Rightarrow-1-\lambda+3 \lambda-3+4-2 \lambda=0$
$\Rightarrow 0=0$ which is true.
Hence, option (a) is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 2)
What are the direction ratios of the line of intersection of the planes $x=3 z+4$ and $y=2 z-3 ?$

Options:
A $\langle 1,2,3\rangle$
B $\langle 2,1,3\rangle$
C $\langle 3,2,1\rangle$
D $\langle 1,3,2\rangle$

The correct answer is:
$\langle 3,2,1\rangle$

Solution:
Let $P_{1}: x-3 z-4=0$ and $P_{2}: y-2 z+3=0$ be two planes Let $a x+b y+c z=d$ be the equation of line. Since, the line of intersection will be perpendicular to the normal of both the planes $\therefore \quad a(1)+b(0)+c(-3)=0$
$\Rightarrow \mathrm{a}-3 \mathrm{c}=0$
and $a(0)+b(1)+c(-2)=0$
$\Rightarrow \mathrm{b}-2 \mathrm{c}=0$
(ii) From (1) and (2) we have
$$
\frac{\mathrm{a}}{3}=\frac{\mathrm{b}}{2}=\frac{\mathrm{c}}{1}
$$
Hence, $\mathrm{d} \cdot \mathrm{Rs}= < 3,2,1>$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
What are the effects of androgens on male individuals?

Options:
A Stimulation of female secondary sex organs
B Stimulation of muscular growth
C Stimulation of spermatozoa formation
D Stimulation of mammary gland development

The correct answer is:
Stimulation of muscular growth

Solution:
Androgens regulate the development, maturation and functions of the male accessory sex organs and stimulate muscular growth, growth of facial and axillary hair, aggressiveness, low pitch of voice, and influence the male sexual behaviour (libido). They also produce anabolic effects on protein and carbohydrate metabolism.

Topic: Biology, Chemical Coordination and Integration, NEET
What are the equations of the directrices of the ellipse $25 x^{2}+16 y^{2}=400 ?$

Options:
A $\quad 3 x \pm 25=0$
B $3 y \pm 25=0$
C $x \pm 15=0$
D $y \pm 25=0$

The correct answer is:
$3 y \pm 25=0$

Solution:
Given equation of ellipse is $25 x^{2}+16 y^{2}=400$ which can be rewritten as
$\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$
We know standard equation of ellipse is
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
On comparing given equation with standard equation, we get $a=4$ and $b=5(b>a)$ Equations of the directrices are
$y=\pm \frac{b}{e}=\pm \frac{5}{\sqrt{1-\frac{16}{25}}}=\pm \frac{25}{3} \quad\left(\because e=\sqrt{1-\frac{a^{2}}{b^{2}}}\right)$
$\Rightarrow 3 y \pm 25=0$

Topic: Mathematics, Ellipse, NDA, NDA 2010 (Phase 1)
What are the events that take place in the ovary an uterus during follicular phase of the menstrual cycle.

Options:

None

Solution:
The events involved in the menstrual cycle are menstrual phase, follilcular phase, ovulatory phase and luteal phase. In follicular please there is development of follicles in ovary This phase is also called as proliferative phase as there is proliferation of endometrium in uterus. It last for about 14 days. Lining of the uterus and fallopian tubes proliferates.
A graafian follicle grows, matures and secretes oestrogen. It ruptures to release its egg (secondary oocyte) after about 14 days. The secretion of gonadotropins (LH and FSH) increases gradually during the follicular phase, and stimulates follicular development as well as secretion of estrogens by the growing follicles.
Both LH and FSH attain a peak level in the middle of cycle (about 14th day). Rapid secretion of LH leading to its maximum level during the mid-cycle called LH surge that induces rupture of Graafian follicle and thereby the release of ovum (ovulation). The follicular phase ends with ovulation.

Topic: Biology, Human Reproduction
What are the individual numbers of primary, secondary and tertiary bromide across all the structural isomers of C5H11Br?

Options:
A 4 primary , 3 secondary and 3 tertiary bromides
B 4 primary , 3 secondary and 1 tertiary bromides
C 4 primary , 2 secondary and 2 tertiary bromides
D 5 primary , 4 secondary and 2 tertiary bromides

The correct answer is:
4 primary , 3 secondary and 1 tertiary bromides

Solution:
Correct Option is : (B)
4 primary , 3 secondary and 1 tertiary bromides

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
What are the last two digits of the number $9^{200}$ ?

Options:
A 19
B 21
C 41
D 1

The correct answer is:
1

Solution:
Using binomial theorem $9^{200}=(1+8)^{200}$
$=1+8.200+\frac{200 \times 199}{2 !} \times 8^{2}+\ldots$
$=1+1600+1273600+\ldots$
From above, it is clear that the last two digits of the number $9^{200}$ are 01 .

Topic: Mathematics, Binomial Theorem, NDA, NDA 2006 (Phase 2)
What are the order and degree respectively of the differential
equation
$\left.\left(d^{4} y / d x^{4}\right)^{3}\right\}^{2 / 3}-7 x\left(d^{3} y / d x^{3}\right)^{2}=8 ? \quad$

Options:
A 3,2
B 4,3
C 4,2
D 3,3

The correct answer is:
4,2

Solution:
$\left\{\left(\frac{d^{4} y}{d x^{4}}\right)^{3}\right\}^{\frac{2}{3}}-7 x\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=8$
$\Rightarrow\left(\frac{d^{4} y}{d x^{4}}\right)^{2}-7 x\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=8$
- The order and degree of the given differential equation are 4 and 2 respectively.

Topic: Mathematics, Differential Equations, NDA, NDA 2008 (Phase 2)
What are the order and degree, respectively of the differential
equation $\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{5 / 6}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{1 / 3} ?$3

Options:
A 2,1
B 2,5
C $2, \frac{5}{6} \quad$
D $1, \frac{1}{3} \quad$

The correct answer is:
2,5

Solution:
Given differential equation is:
$\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\right)^{5 / 6}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{1 / 3}$
Raising both the side to power of 6 , to make it a polynomial of derivatives.
$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{5}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{6 / 3} \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\right)^{5}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}$
Highest derivative has power of $5 .$ So, the order and degree of given differential equation are 2 and 5 respectively.

Topic: Mathematics, Differential Equations, NDA, NDA 2008 (Phase 1)
What are the order and degree, respectively, of the differential
equation $\left(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\right)^{2}=\mathrm{y}^{4}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{5} ?$

Options:
A 4,5
B 2,3
C 3,2
D 5,4

The correct answer is:
3,2

Solution:
$\left(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\right)^{2}=\mathrm{y}^{4}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{5}$
order $=3$, degree $=2$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 1)
What are the order and degree respectively of the differential equation whose solution is $\mathrm{y}=\mathrm{cx}+\mathrm{c}^{2}-3 \mathrm{c}^{3 / 2}+2$, where $\mathrm{c}$ is a parameter?

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

The correct answer is:
1,4

Solution:
Given:
Solution of differential equation is
$$
y=c x+c^{2}-3 c^{\frac{3}{2}}+2
$$
To find order and degree of differential equation, we will find differential equation first. Now differentiating equation (1) w.r.t. $x$ and putting value of $c$ to remove it, we get
$$
\begin{array}{l}
\frac{d y}{d x}=c \\
y=x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{2}-3\left(\frac{d y}{d x}\right)^{\frac{3}{2}}+2 \\
\Rightarrow(y-2)^{2}+\left(\frac{d y}{d x}\right)^{4}+(2 x-9)\left(\frac{d y}{d x}\right)^{3}
\end{array}
$$
$$
+\left(x^{2}-2 y+4\right)\left(\frac{d y}{d x}\right)^{2}+(-2 x y+4 x) \frac{d y}{d x}=0
$$
Hence order of differential equation is 1 and degree is $4 .$

Topic: Mathematics, Differential Equations, NDA, NDA 2016 (Phase 1)
What are the order and degree respectively of the
differential equation $\mathrm{y}=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{dx}}{\mathrm{dy}} ?$

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

The correct answer is:
1,2

Solution:
The given differential equation is
$\mathrm{y}=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{dx}}{\mathrm{dy}}$
Multiplying both the sides by $\frac{\mathrm{dy}}{\mathrm{dx}}$
We get $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \mathrm{y}=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+1$
$\Rightarrow \quad x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)+1=0$
Hence, order and degree of differential equation are 1 and 2.

Topic: Mathematics, Differential Equations, NDA, NDA 2006 (Phase 2)
What are the paired appendages of Vertebrates?

Options:
A Fins
B Limbs
C Both A) and B)
D None of the above

The correct answer is:
Both A) and B)

Solution:
Vertebrates have paired appendages which may be fins or limbs, Fish have fins and Tetrapods have limbs.

Topic: Biology, Animal Kingdom, NEET
What are the points of intersection of the curve $4 \mathrm{x}^{2}-9 y^{2}=1$ with its conjugate axis?

Options:
A $(1 / 2,0)$ and $(-1 / 2,0)$
B $(0,2)$ and $(0,-2)$
C $(0,3)$ and $(0,-3)$
D No such point exists

The correct answer is:
No such point exists

Solution:
The given equation of curve is $4 x^{2}-9 y^{2}=1$
$\Rightarrow \quad \frac{x^{2}}{1 / 4}-\frac{y^{2}}{1 / 9}=1$
This is an equation of a hyperbola which does not
inter sect with conjugate Hence, no point of inter section exists

Topic: Mathematics, Hyperbola, NDA, NDA 2011 (Phase 1)
What are the points on the curve $x^{2}+y^{2}-2 x-3=0$ where the tangents are parallel tox-axis?

Options:
A $(1,2)$ and $(1,-2)$
B $(0, \sqrt{3})$ and $(0,-\sqrt{3})$
C $(3,0)$ and $(-3,0)$
D $(2,1)$ and $(2,-1)$

The correct answer is:
$(1,2)$ and $(1,-2)$

Solution:
Given equation curve is $x^{2}+y^{2}-2 x-3=0$
On differentiating we get
$2 x+2 y \frac{d y}{d x}-2=0$
$\Rightarrow y \frac{d y}{d x}=-x+1$
$\Rightarrow \frac{d y}{d x}=-\frac{x+1}{y}$
Since, tangent is parallel to $x$ -axis
$\therefore \frac{d y}{d x}=0 \Rightarrow-\frac{x+1}{y}=0 \Rightarrow x=1$
From equation (1), we have $1+y^{2}-2-3=0 \Rightarrow y=\pm 2$
Hence, required points are $(1,2)$ and $(1,-2)$.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
What are the products of the reduction of aldehydes and ketones by sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4)?

Options:
A Alkenes
B Primary and secondary alcohols
C Ketones and aldehydes
D Hydrocarbons

The correct answer is:
Primary and secondary alcohols

Solution:
Aldehydes are reduced to primary alcohols, and ketones are reduced to secondary alcohols by sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4).

Topic: Chemistry, Aldehydes, Ketones and Carboxylic Acids, NEET
What are the roots of the equation $2(\mathrm{y}+2)^{2}-5(\mathrm{y}+2)=12 ?$

Options:
A $-7 / 2,2$
B $-3 / 2,4$
C $-5 / 3$
D $3 / 2,4$

The correct answer is:
$-7 / 2,2$

Solution:
Given equation is $2(y+2)^{2}-5(y+2)=12$ Let $y+2=a$ So, quadratic equation can be rewritten as $2 a^{2}-5 a-12=0$
$\Rightarrow 2 a^{2}-8 a+3 a-12=0$
$\Rightarrow 2 a(a-4)+3(a-4)=0$
$\Rightarrow(2 a+3)(a-4)=0$
$\Rightarrow 2 a+3=0$ or $a-4=0$
$\Rightarrow a=\frac{-3}{2}$ or $a=4$
$\Rightarrow y+2=\frac{-3}{2}$ or $y+2=4$
$\Rightarrow y=\frac{-3}{2}-2$ or $y=2$
$\Rightarrow y=\frac{-7}{2}$ or 2 (Required roots)

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
What are the roots of the equation $\left|x^{2}-x-6\right|=x+2 ?$

Options:
A $-2,1,4$
B $0,2,4$
C $0,1,4$
D $-2,2,4$

The correct answer is:
$-2,2,4$

Solution:
Given, $\left|x^{2}-x-6\right|=x+2$ $\therefore \mathrm{x}^{2}-\mathrm{x}-6=\mathrm{x}+2$ and $\mathrm{x}^{2}-\mathrm{x}-6=-(\mathrm{x}+2)$
$\Rightarrow \mathrm{x}^{2}-2 \mathrm{x}-8=0$
$\Rightarrow \mathrm{x}^{2}=\mathrm{X}=6=-\mathrm{X}=2$
$\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}+2 \mathrm{x}-8=0$
$\Rightarrow \mathrm{x}^{2}=4$
$\Rightarrow \mathrm{x}(\mathrm{x}-4)+2(\mathrm{x}-4)=0$
$\Rightarrow \mathrm{x}-+2,-2$
$\Rightarrow \mathrm{x}=4,-2$
$\therefore \mathrm{x}=-2,2,4$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2019 (Phase 1)
What are the square roots of -2 i ? $(i=\sqrt{-1})$

Options:
A $\pm(1+i)$
B $\pm(1-i)$
C $\pm i$
D $\pm 1$

The correct answer is:
$\pm(1-i)$

Solution:
Let $Z=x+i y=-2 i$
Let square root of $z$ be $a+i b$.
Then $\sqrt{x+i y}=a+i b$
$\Rightarrow x+i y=(a+i b)^{2}=\left(a^{2}-b^{2}\right)+(2 a b) i$
$\Rightarrow-2 i=\left(a^{2}-b^{2}\right)+i(2 a b)$
Equating real and imaginary part, we get $a^{2}-b^{2}=0$ and $2 a b=-2 \quad \Rightarrow a b=-1$
Since, $a b < 0$ therefore
$\sqrt{x+i y}=\pm\left[\sqrt{\frac{\sqrt{x^{2}+y^{2}}+x}{2}}-i \sqrt{\frac{\sqrt{x^{2}+y^{2}}-x}{2}}\right]$
$=\pm\left[\sqrt{\frac{\sqrt{4}+0}{2}}-i \sqrt{\frac{\sqrt{4}-0}{2}}\right]$
$=\pm\left[\sqrt{\frac{2}{2}}-i \sqrt{\frac{2}{2}}\right]=\pm[1-i]$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
What are the structures that subtend the sporangia in pteridophytes?

Options:
A Stems
B Roots
C Sporophylls
D Rhizomes

The correct answer is:
Sporophylls

Solution:
The sporangia in pteridophytes are subtended by leaf-like appendages called sporophylls.

Topic: Biology, Plant Kingdom, NEET
What are the two broad categories into which unit cells can be divided?

Options:
A Primary and secondary unit cells
B Simple and complex unit cells
C Primitive and centred unit cells
D Hexagonal and cubic unit cells

The correct answer is:
Primitive and centred unit cells

Solution:
Unit cells can be broadly divided into two categories: primitive and centred unit cells, as mentioned in the paragraph.

Topic: Chemistry, Solid State, NEET