datavorous/entrance-exam-dataset huggingface dataset:
The correct statement about dialysis is:

Options:
A It acts as a permanent treatment solution to treat people with end-stage kidney disease.
B The dialysing unit into which the blood is drained contains heparin to prevent coagulation.
C The cellophane tube is surrounded by a dialysing fluid, which has the same concentration as plasma including the nitrogenous waste.
D Dialysis replaces all of the functions of the kidney through diffusion, ultrafiltration and tubular secretion.

The correct answer is:
The dialysing unit into which the blood is drained contains heparin to prevent coagulation.

Solution:
Dialysis is used to provide an artificial replacement for lost kidney function in people with renal failure. Dialysis may be necessary in patients with acute kidney failure to bridge the time gap required for treating the fundamental causes and as a holding measure in patients with irreversible end stage chronic kidney disease (CKD) stage 5 until transplantation is performed. Sometimes Dialysis is the only supportive measure in patients with end stage CKD when transplantation is not possible.
In this process, the blood is drained from a convenient artery and an anti-coagulant like heparin is added to prevent clotting of blood and the cleared blood is pumped back to the body through a vein after adding anti-heparin to it. Dialysis treatment replaces only some of the kidney functions like diffusion (waste removal) and ultrafiltration (fluid removal). Thus Dialysis is an imperfect treatment to replace kidney function because it does not correct the endocrine functions of the kidney.

Topic: Biology, Excretory Products and their Elimination, NEET
The correlation coefficient between two variables $X$ and $y$ found to be $0.6 .$ All the observations on $X$ and $Y$ are transformed using the transformations $U=2-3 X$ and $V=$. $Y+1$. The correlation coefficient between the transformed variables $U$ anc

Options:
A $-0 \cdot 5$
B $+0 \cdot 5$
C $-0 \cdot f$
D $\pm 0$.

The correct answer is:
$-0 \cdot f$

Solution:
No Solution

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 1)
The correlation coefficient computed from a set of 30 observations is $0.8 .$ Then the percentage of variation not explained by linear regression is $\quad$

Options:
A $80 \%$
B $20 \%$
C $64 \%$
D $36 \%$

The correct answer is:
$36 \%$

Solution:
No Solution

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 2)
The cumulative frequency of the largest observed value must always be : $\quad$

Options:
A Less than the total number of observations
B Greater than the total number of observations
C Equal to total number of observations
D Equal to mid point of the last class interval

The correct answer is:
Equal to total number of observations

Solution:
The cumulative frequency of the largest observed value must always be equal to the total number of observations

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
The curve $y^{2}=-4 a x(a>0)$ lies in

Options:
A First and fourth quadrants
B First and second quadrants
C Second and third quadrants
D Third and fourth quadrants

The correct answer is:
Second and third quadrants

Solution:
Given curve $y^{2}=-4 a x$ which is one of the form of parabola
It is clear from the figure that curve lies in the second and third quadrants.

Topic: Mathematics, Parabola, NDA, NDA 2009 (Phase 2)
The curve $\mathrm{y}=\mathrm{xe}^{\mathrm{x}}$ has minimum value equal to

Options:
A $-\frac{1}{\mathrm{e}}$
B $\frac{1}{\mathrm{e}} \quad$
C -e
D e

The correct answer is:
$-\frac{1}{\mathrm{e}}$

Solution:
Let $\mathrm{y}=\mathrm{xe}^{\mathrm{x}}$. Differentiate both side w.r.t. ' $\mathrm{x}$ '.
$\Rightarrow \frac{d y}{d x}=e^{x}+x e^{x}=e^{x}(1+x)$
Put $\frac{d y}{d x}=0$
$\Rightarrow e^{x}(1+x)=0 \quad \Rightarrow x=-1$
Now, $\frac{d^{2} y}{d x^{2}}=e^{x}+e^{x}(1+x)=e^{x}(x+2)$
$\left(\frac{d^{2} y}{d x^{2}}\right)_{(x=-1)}=\frac{1}{e}+0>0$
Hence, $\mathrm{y}=\mathrm{xe}^{\mathrm{x}}$ is minimum function and
$\mathrm{y}_{\min }=-\frac{1}{\mathrm{e}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)
The data below record the itemwise quarterly expenditure of a private organization $\begin{array}{ll}\text { Item of expenditure } & \text { Amount (in lakh rupees) }\end{array}$
1 Salaries $6.0$
$\begin{array}{lll}\text { 2. } & \text { TA \& DA } & \text { 4.9 }\end{array}$
3 . House rent and postage $3.6$
4 . All other expenses
Total The data is represented by a pie diagram. What is the sectorial angle of the sector with largest area?

Options:
A $120^{4}$
B 108
C $100^{2}$
D $90^{\circ}$

The correct answer is:
108

Solution:
In a pie chart largest amount occupies largest area. So, the salaries occupies largest area
$\Rightarrow$ Sectorial angle $=\frac{6}{20} \times 360^{\circ}=108^{\circ}$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 2)
The decimal number ( $127.25)_{10}$, when converted to binary nunber, takes the form

Options:
A $(1111111.11)_{3}$
B $(1111110.01)$.
C $(1110111.11)_{2}^{2}$
D $(1111111.01)_{2}^{2}$

The correct answer is:
$(1111111.01)_{2}^{2}$

Solution:
$127=(1111111)_{2}$
Now,
\begin{array}{l|l}
2 & 127 \\
\hline 2 & 63-1 \\
\hline 2 & 31-1 \\
\hline 2 & 15-1 \\
\hline 2 & 7-1 \\
\hline 2 & 3-1 \\
\hline & 1-1
\end{array}
$127=(1111111)_{2}$
Now,
$\therefore$ Option (d)

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
The decimal number $(57.375)_{10}$ when converted to binary number takes the form:

Options:
A $(111001.011)_{2}$
B $(100111.110)_{2}$
C $(110011.101)$
D $(111011.011)$.

The correct answer is:
$(111001.011)_{2}$

Solution:
$(111001.011)_{2}=\left[2^{5} \times 1+2^{4} \times 1+2^{3} \times 1+2^{2} \times 0+2^{1} \times 0\right.$
$\left.+2^{0} \times 1\right] \cdot\left[2^{-1} \times 0+2^{-2} \times 1+2^{-3} \times 1\right]=(57.375)_{10}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 2)
The decimal representation of the number ( 1011 ), in binary system is:

Options:
A 5
B 7
C 9
D 11

The correct answer is:
11

Solution:
$(1011)_{2}=2^{3} \times 1+2^{2} \times 0+2^{1} \times 1+2^{0} \times 1$
$=8+0+2+1=(11)_{10}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 2)
The definition of Mode fails if:

Options:
A the maximum frequency is repeated
B the maximum frequency is not repeated
C the maximum frequency occurs in the middle
D the curve drawn with the help of given data is $\begin{array}{ll}\text { symmetrical } & {[2006-I I]}\end{array}$

The correct answer is:
the curve drawn with the help of given data is $\begin{array}{ll}\text { symmetrical } & {[2006-I I]}\end{array}$

Solution:
The definition of Mode fails if the curve drawn with the help of given data is symmetrical.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 2)
The degree and order respectively of the differential
equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}+\mathrm{y}+1}$ are $\quad$

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

The correct answer is:
1,1

Solution:
Since order of the highest derivative in the given diff. equation is 1 and exponent of the derivative is also 1 therefore degree and order is $(1,1)$.

Topic: Mathematics, Differential Equations, NDA, NDA 2013 (Phase 1)
The degree of the differential equation $\quad$ $\frac{d y}{d x}-x=\left(y-x \frac{d y}{d x}\right)^{-4}$ is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
$\begin{aligned} & \because \frac{d y}{d x}-x=\left(y-x \frac{d y}{d x}\right)^{-4} \\ & \Rightarrow\left(\frac{d y}{d x}-x\right)\left(y-x \cdot \frac{d y}{d x}\right)^{4}=1 \end{aligned}$
$\therefore$ Order of the above differential equation $=1$ $\&$ degree $=5$

Topic: Mathematics, Differential Equations, NDA, NDA 2015 (Phase 2)
The densities of graphite and diamond at $298 \mathrm{~K}$ are 2.25 and $3.31 \mathrm{gcm}^{-3}$, respectively. If the standard free energy difference $\left(\Delta \mathrm{G}^{\circ}\right)$ is equal to $1895 \mathrm{~J} \mathrm{~mol}^{-1}$, the pressure at which graphite will be transformed into diamond at $298 \mathrm{~K}$ is:

Options:
A $9.92 \times 10^8 \mathrm{~Pa}$
B $9.92 \times 10^7 \mathrm{~Pa}$
C $9.92 \times 10^6 \mathrm{~Pa}$
D none

The correct answer is:
none

Solution:
$\Delta \mathrm{G}=-\mathrm{P} \Delta \mathrm{V}$
$$
\begin{aligned}
& 1895=P\left[\left(\frac{12}{2.25}-\frac{12}{3.31}\right) \times 10^{-6} \mathrm{~m}^3\right] \\
& P=\frac{1895 \times 10^6}{1.71}=1108.187 \times 10^6 \mathrm{Bar} \\
& =1108.1871 \times 10^{11} \mathrm{~Pa}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2003
The derivative of $\ln (x+\sin x)$ with respect to $(x+\cos x)$ is

Options:
A $\frac{1+\cos x}{(x+\sin x)(1-\sin x)}$
B $\frac{1-\cos x}{(x+\sin x)(1+\sin x)}$
C $\frac{1-\cos x}{(x-\sin x)(1+\cos x)}$
D $\frac{1+\cos x}{(x-\sin x)(1-\cos x)}$

The correct answer is:
$\frac{1+\cos x}{(x+\sin x)(1-\sin x)}$

Solution:
$\ln (x+\sin x)=y$ (say)
$\frac{d y}{d x}=\frac{1}{(x+\sin x)}(1+\cos x)$
$=\frac{(1+\cos x)}{(x+\sin x)}$
$x+\cos x=z$ (say)
$\quad \frac{d z}{d x}=(1-\sin x)$
derivative of $\ln (x+\sin x)$ w.r.t $(x+\cos x)$ is
$\frac{\mathrm{dy}}{\mathrm{dz}}=\frac{(1+\cos \mathrm{x})}{(\mathrm{x}+\sin \mathrm{x})(1-\sin \mathrm{x})}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2015 (Phase 2)
The derivative of $\sec ^{2} x$ with respect to $\tan ^{2} x$ is

Options:
A 1
B 2
C $2 \sec x \tan x$
D $2 \sec ^{2} x \tan x$

The correct answer is:
1

Solution:
Let $u=\sec ^{2} x, v=\tan ^{2} x$
To find : $\frac{\mathrm{du}}{\mathrm{dv}}$.
Now, $\frac{\mathrm{du}}{\mathrm{dx}}=2 \sec \mathrm{x} \cdot \sec \mathrm{x} \cdot \tan \mathrm{x}$
and $\frac{d v}{d x}=2 \tan x \cdot \sec ^{2} x$
Thus, $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\mathrm{du}}{\mathrm{dx}} \times \frac{\mathrm{dx}}{\mathrm{dv}}=\frac{2 \sec \mathrm{x} \cdot \sec \mathrm{x} \cdot \tan \mathrm{x}}{2 \tan \mathrm{x} \mathrm{sec}^{2} \mathrm{x}}=1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)
The derivative of $|x|$ at $x=0$

Options:
A is 1
B is $-1$
C is 0
D does not exist

The correct answer is:
does not exist

Solution:
Since $|\mathrm{x}|=\left\{\begin{array}{l}\mathrm{x} \text { if } \mathrm{x} \geq 0 \\ -\mathrm{x} \text { if } \mathrm{x} < 0\end{array}\right.$
$\therefore \mathrm{R} . \mathrm{H} \cdot \mathrm{D}$ of $|\mathrm{x}|=1$ at $\mathrm{x}=0$
and $\mathrm{L} . \mathrm{H} .$ Dof $|\mathrm{x}|=-1$ at $\mathrm{x}=0$
Now, R. H. $\mathrm{D} \neq \mathrm{L} . \mathrm{H} . \mathrm{D}$ at $\mathrm{x}=0$
Hence, the derivative of $|\mathrm{x}|$ at $\mathrm{x}=0$ does not exist.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)
The determinant
$\left|\begin{array}{lll}a+b+c & a+b & a \\ 4 a+3 b+2 c & 3 a+2 b & 2 a \\ 10 a+6 b+3 c & 6 a+3 b & 3 a\end{array}\right|$
is independent of which one of the following?

Options:
A $\mathrm{a}$ and $\mathrm{b}$
B $\mathrm{b}$ and $\mathrm{c}$
C a and c
D All of these

The correct answer is:
$\mathrm{b}$ and $\mathrm{c}$

Solution:
Let $\mathrm{D}=\left|\begin{array}{ccc}\mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b} & \mathrm{a} \\ 4 \mathrm{a}+3 \mathrm{~b}+2 \mathrm{c} & 3 \mathrm{a}+2 \mathrm{~b} & 2 \mathrm{a} \\ 10 \mathrm{a}+6 \mathrm{~b}+3 \mathrm{c} & 6 \mathrm{a}+3 \mathrm{~b} & 3 \mathrm{a}\end{array}\right|$
$\mathrm{By} \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$, we get
$\Rightarrow\left|\begin{array}{ccc}a+b+c & a+b & a \\ 2 a+b & a & 0 \\ 7 a+3 b & 3 a & 0\end{array}\right|$
$\mathrm{By} \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}$ gives:
$\Rightarrow\left|\begin{array}{ccc}\mathrm{c} & \mathrm{a}+\mathrm{b} & \mathrm{a} \\ \mathrm{a}+\mathrm{b} & \mathrm{a} & 0 \\ 4 \mathrm{a}+3 \mathrm{~b} & 3 \mathrm{a} & 0\end{array}\right|$
Again by $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{2}$, we get :
$D=\left|\begin{array}{ccc}c & a+b & a \\ a+b & a & 0 \\ a & 0 & 0\end{array}\right|$
$=\mathrm{a}\{0 .(\mathrm{a}+\mathrm{b})-\mathrm{a} \cdot \mathrm{a}\}$
$=-\mathrm{a}^{3}$ which is independent of $\mathrm{b}$ and $\mathrm{c}$.

Topic: Mathematics, Determinants, NDA, NDA 2007 (Phase 2)
The determinant of a orthogonal matrix is:

Options:
A $\pm 1$
B 2
C 0
D $\pm 2$

The correct answer is:
$\pm 1$

Solution:
Let A be orthogonal matrix, therefore $\mathrm{AA}^{\mathrm{T}}=1$ $\Rightarrow\left|\mathrm{AA}^{\mathrm{T}}\right|=1 \Rightarrow|\mathrm{A}| \cdot\left|\mathrm{A}^{\mathrm{T}}\right|=1$
$\Rightarrow|\mathrm{A}|^{2}=1$
$\Rightarrow|\mathrm{A}|=\pm 1$

Topic: Mathematics, Matrices, NDA, NDA 2013 (Phase 1)
The determinant of an odd order skew symmetric matrix is always:

Options:
A Zero
B One
C Negative
D Depends on the matrix

The correct answer is:
Zero

Solution:
We know that, elements of principal diagonals of a skew-symmetric matrix are all zero.
$A=\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & -c \\ -b & c & 0\end{array}\right|_{3 \times 3} \Rightarrow|A|\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & -c \\ -b & c & 0\end{array}\right|$
$=a b c-a b c=0$

Topic: Mathematics, Matrices, NDA, NDA 2014 (Phase 1)
The diagram given here is the standard ECG of a normal person. The P-wave represents the

Options:
A contraction of both the atria
B initiation of the ventricular contraction
C beginning of the systole
D end of systole

The correct answer is:
contraction of both the atria

Solution:
In ECG, P-wave represents the depolarisation of atria, which leads to the contraction of both atria. T-wave represents the return of ventricles from excited to normal state. The QRS complex represents the depolarisation of the ventricles, which initiates ventricular contraction. The contraction starts shortly after $Q$ and marks the beginning of systole.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2013 (All India)
The diagram shows three features found in the tissues of the human respiratory system. Which tubes of the gas exchange system could be represented at position P in the diagram?

Options:
A Bronchus, Bronchiole and Trachea
B Bronchus and Bronchiole
C Bronchus and Trachea
D Trachea and Bronchiole

The correct answer is:
Bronchus and Trachea

Solution:
The mucosal surface of the trachea is lined by ciliated, pseudostratified columnar epithelium (i.e., respiratory epithelium) interspersed with goblet cells. The trachea divides into the left and right bronchi and they are histologically similar to the trachea. Bronchi are lined by ciliated, pseudostratified columnar epithelium (respiratory epithelium) and interspersed with goblet cells. Bronchi give way to smaller conducting airways, bronchioles which are thin-walled airways lined by variably ciliated or non-ciliated columnar epithelium but lack glandular cells. The tracheae, primary, secondary and tertiary bronchi and initial bronchioles are supported by incomplete cartilaginous rings.

Topic: Biology, Breathing and Exchange of Gases, NEET
The diameter of circles (in mm) drawn in a design are given below :
Calculate the standard deviation and mean diameter of the circles.
[Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,40.5-44.5,44.5$ $-48.5,48.5-52.5$ and then proceed]

Options:

None

Solution:
Making the distribution overlapping type, we get
$\begin{aligned}
&\text { Mean }=A+\frac{\Sigma f_i d_i}{N} \times h \\
&=42.5+\frac{25 \times 4}{100}=42.5+1=43.5
\end{aligned}$
Standard Deviation $=h \sqrt{\frac{\Sigma f_i d_i^2}{N}-\left(\frac{\Sigma f_i d_i}{N}\right)^2}$
$\begin{aligned}
&=4 \sqrt{\frac{199}{100}-\left(\frac{25}{100}\right)^2} \\
&=4 \sqrt{1.99-0.0625}=4 \sqrt{1.9275}=4 \times 1.388 \\
&=5.552=5.55
\end{aligned}$
Hence $\bar{x}=43.5$ and S.D. $=5.55$.

Topic: Mathematics, Statistics
The difference between the nth term and $(\mathrm{n}-1)$ th term of a sequence is independent of $\mathrm{n}$. Then the sequence follows which one of the following?

Options:
A AP
B GP
C $\mathrm{HP}$
D None of these

The correct answer is:
AP

Solution:
If the sequence is in AP with first term, a and common difference, $\mathrm{d}$. $\Rightarrow \mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
Also $T_{n-1}=a+(n-2) d$
So, the sequence is in AP for which difference between the nth term and $(\mathrm{n}-1)$ th term is independent of $\mathrm{n}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 1)
The difference between the Red List and Red Data Book is

Options:
A The Red Data Book contains three colored pages - Red, pink and green and The Red List has the names typed with brown, blue and orange.
B The Red List contains names of endangered species and Red Data Book contains information of endangered species.
C Both (a) and (b).
D None of the above.

The correct answer is:
The Red List contains names of endangered species and Red Data Book contains information of endangered species.

Solution:
IUCN is the International Union of Conservation of Nature and Natural Resources which is now called the World Conservation Union (WCU). The (IUCN) Red List also known as the Red List or Red Data List was founded in 1964. This book gives the global conservation status of biological species. It uses a set of criteria to evaluate the Threatened species and subspecies under the extinction risk.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2014
The difference between the Red List and Red Data Book is

Options:
A The Red Data Book contains three colored pages - Red, pink and green and The Red List has the animals typed with brown, blue and orange.
B The Red List contains names of endangered species and Red Data Book contains information of endangered species.
C Both (a) and (b).
D None of the above.

The correct answer is:
The Red List contains names of endangered species and Red Data Book contains information of endangered species.

Solution:
IUCN is the International Union of Conservation of Nature and Natural Resources which is now called the World Conservation Union (WCU). The (IUCN) Red List also known as the Red List or Red Data List was founded in 1964. This book gives the global conservation status of biological species. It uses a set of criteria to evaluate the Threatened species and subspecies under the extinction risk.

Topic: Biology, Biodiversity and its Conservation, NEET
The difference of focal distances of any point on a hyperbola is equal to

Options:
A latus rectum
B semi-transverse axis
C transverse axis
D semi-latus rectum

The correct answer is:
transverse axis

Solution:
Transverse axis Proof: The focal distance of any point $(x, y)$ on
hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is
$e|x|-a$ from the nearer focus
$e|x|+a$ from the farther focus Difference $=(e|x|+a)-(e|x|-a)=2 a$
$=$ length of transverse axes.

Topic: Mathematics, Hyperbola, NDA, NDA 2013 (Phase 1)
The difference of two angles is $1^{\circ}$; the circular measure of their sum is $1 .$ What is the smaller angle in circular measure?

Options:
A $\left[\frac{180}{\pi}-1\right]$
B $\left[1-\frac{\pi}{180}\right]$
C $\frac{1}{2}\left[1-\frac{\pi}{180}\right]$
D $\frac{1}{2}\left[\frac{180}{\pi}-1\right]

The correct answer is:
$\frac{1}{2}\left[1-\frac{\pi}{180}\right]$

Solution:
Let the angles are $\alpha$ and $\beta$, then $\alpha-\beta=1^{\circ}$
$\Rightarrow \alpha-\beta=\frac{\pi}{180^{\circ}}$ is circular measure
As given, $\quad \alpha+\beta=1$
On solving Eqs. (i) and (ii), we get, $\alpha=\frac{1}{2}\left[1+\frac{\pi}{180^{\circ}}\right]$ and $\beta=\frac{1}{2}\left[1-\frac{\pi}{180^{\circ}}\right]$
$\beta$ is the smaller angle. Hence, smaller angle $=\frac{1}{2}\left[1-\frac{\pi}{180^{\circ}}\right]$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 1)
The difference of two numbers 10001100 and 1101101 in binary system is expressed in decimal system by which one of the following?

Options:
A 27
B 29
C 31
D 33

The correct answer is:
31

Solution:
The given binary number $10001100=1 \times 2^{7}+1 \times 2^{3}+1$ $\times 2^{2}$
$=128+8+4=140$ (decimal numbers)
and $1101101=1 \times 2^{6}+1 \times 2^{5}+1 \times 2^{3}+1 \times 2^{2}+1 \times 2^{0}$
$=64+32+8+4+1=109$
their difference $=140-109=31$

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 1)
The differential equation $(3 x+4 y+1) d x+(4 x+5 y+1) d y=0$ represents a family of

Options:
A circles
B parabolas
C ellipses
D hyperbolas

The correct answer is:
hyperbolas

Solution:
The given differential equation is
$(3 x+4 y+1) d x+(4 x+5 y+1) d y=0$
Comparing eq. (i) with $\mathrm{Mdx}+\mathrm{Ndy}=0$, we get
$M=3 x+4 y+1$
and $\mathrm{N}=4 \mathrm{x}+5 \mathrm{y}+1$
Here, $\frac{\partial \mathrm{M}}{\partial \mathrm{y}}=\frac{\partial \mathrm{N}}{\partial \mathrm{x}}=4$
Hence, eq. (i) is exact and solution is given by
$\begin{aligned}
& \int(3 x+4 y+1) d x+\int(5 y+1) d y=C \\
\Rightarrow & \frac{3 x^{2}}{2}+4 x y+x+\frac{5 y^{2}}{2}+y-C=0 \\
\Rightarrow & 3 x^{2}+8 x y+2 x+5 y^{2}+2 y-2 C=0 \\
\Rightarrow & 3 x^{2}+2.4 x y+2 x+5 y^{2}+2 y+C^{\prime}=0 \\
& \text { where, } C^{\prime}=-2 C
\end{aligned}$
On comparing eq. (ii) with standard form of conic section
$a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+C=0$
We get,
$\mathrm{a}=3, \mathrm{~h}=4, \mathrm{~b}=5$
Here, $h^{2}-a b=16-15=1>0$
Hence, the solution of differential equation represents family of hyperbolas.

Topic: Mathematics, Differential Equations, VITEEE, VITEEE 2015
The differential equation of all circles whose centres are at the origin is $\quad$

Options:
A $\frac{d y}{d x}=\frac{y}{x}$
B $\frac{d y}{d x}=\frac{x}{y}$
C $\frac{d y}{d x}=-\frac{x}{y}$
D None of the above

The correct answer is:
$\frac{d y}{d x}=-\frac{x}{y}$

Solution:
$\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2} \quad$ [equation of circle]
Differentiating both sides w.r.t. $\mathrm{x}$.
$2 x+2 y \frac{d y}{d x}=0$
$\frac{d y}{d x}=-\frac{x}{y}$

Topic: Mathematics, Differential Equations, NDA, NDA 2013 (Phase 2)
The differential equation of minimum order by eliminating the arbitrat constants $A$ and $C$ in the equatic $\mathrm{y}=\mathrm{A}[\sin (\mathrm{x}+\mathrm{C})+\cos (\mathrm{x}+\mathrm{C})]$ is

Options:
A $\mathrm{y}^{n}+(\sin \mathrm{x}+\cos \mathrm{x}) \mathrm{y}^{\prime}=1$
B $y^{\prime \prime}=(\sin x+\cos x) y^{\prime}$
C $y^{n}=\left(y^{\prime}\right)^{2}+\sin x \cos x$
D $\mathrm{y}^{\prime \prime}+\mathrm{y}=0$

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

Solution:
$\mathrm{y}=\mathrm{A}[\sin (\mathrm{x}+\mathrm{c})+\cos (\mathrm{x}+\mathrm{c})]$
$\mathrm{y}^{\prime}=\mathrm{A}[\cos (\mathrm{x}+\mathrm{c})-\sin (\mathrm{x}+\mathrm{c})]$
$\mathrm{y}^{\prime \prime}=-\mathrm{A}[\sin (\mathrm{x}+\mathrm{c})+\cos (\mathrm{x}+\mathrm{c})]$
$=-\mathrm{y}$
$\therefore \mathrm{y}^{\prime \prime}+\mathrm{y}=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 2)
The differential equation of the curve $\mathrm{y}=\sin \mathrm{x}$ is

Options:
A $\frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}+x=0$
B $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{y}=0$
C $\frac{d^{2} y}{d x^{2}}-y=0$
D $\frac{d^{2} y}{d x^{2}}+x=0$

The correct answer is:
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{y}=0$

Solution:
Given curve is $\mathrm{y}=\sin \mathrm{x}$
Differentiate both the sides w.r.t 'x'.
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \mathrm{x} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=-\sin \mathrm{x}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{y}$
$\therefore \quad \mathrm{y}+\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2013 (Phase 1)