datavorous/entrance-exam-dataset huggingface dataset:
The function $f: N \rightarrow N, N$ being the set of of natural numbers. defined by $f(x)=2 x+3$ is

Options:
A injective and surjective
B injective but not surjective
C not injective but surjective
D neither injective nor surjective

The correct answer is:
injective but not surjective

Solution:
Given $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$
$\therefore \mathrm{f}(\mathrm{x})=2 \mathrm{x}+3 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2>0$
So, $\mathrm{f}(\mathrm{x})$ is increasing, $\forall \mathrm{x} \in \mathrm{N}$.
Hence, $\mathrm{f}(\mathrm{x})$ is injective. Let $\mathrm{f}(\mathrm{x})=\mathrm{y}$
$\Rightarrow \mathrm{y}=2 \mathrm{x}+3 \Rightarrow \mathrm{x}=\frac{\mathrm{y}-3}{2}$
This is injective
$\therefore \mathrm{x}=\frac{1}{2}$
and $\mathrm{y} \in \mathrm{N}$ but $\mathrm{x} \notin \mathrm{N}$
Hence, $\mathrm{f}(\mathrm{x})$ is not surjective.

Topic: Mathematics, Functions, NDA, NDA 2014 (Phase 2)
The function $f: R \rightarrow R$ defined by $f(x)=\left(x^{2}+1\right)^{35}$ for all $x$ $\in R$ is

Options:
A one-one but not onto
B onto but not one-one
C neither one-one nor onto
D both one-one and onto

The correct answer is:
neither one-one nor onto

Solution:
$f(-1)=f(1)=2^{35}$
Here, two real numbers 1 and $-1$ have the same image. So, the function is not one-one and let $y=\left(x^{2}+1\right)^{35}$
$\Rightarrow x=\sqrt{(y)^{1 / 35}-1}$
Thus, every real number has no pre image. So, the function is not onto. Hence, the function is neither one-one nor onto.

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 2)
The function $\mathrm{f}(\mathrm{x})=\frac{1-\sin \mathrm{x}+\cos \mathrm{x}}{1+\sin \mathrm{x}+\cos \mathrm{x}}$ is not defined at $\mathrm{x}=\pi$.
The value of $f(\pi)$ so that $f(x)$ is continuous at $x=\pi$ is

Options:
A $-\frac{1}{2}$
B $\frac{1}{2}$
C $-1$
D 1

The correct answer is:
$-1$

Solution:
$\begin{aligned} & \lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x} \\ & \text { Using L'hospital's rule } \\ & \Rightarrow \lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x} \\ & \Rightarrow \frac{-\cos \pi-\sin \pi}{\cos \pi-\sin \pi} \\ & \Rightarrow \frac{-(-1)-0}{-1-0} \\ & \Rightarrow-1 \end{aligned}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 2)
The function $f(x)=k \sin x+\frac{1}{3} \sin 3 x$ has maximum value at
$x=\frac{\pi}{3}$, what is the value of $k ?$

Options:
A 3
B $\frac{1}{3}$
C 2
D $\frac{1}{2}$

The correct answer is:
2

Solution:
Let $f(x)=k \sin x+\frac{1}{3} \sin 3 x$
Differentiate both side w.r.t. ' $x$ '
$\Rightarrow \quad f^{\prime}(x)=k \cos x+\frac{3}{3} \cos 3 x$
$\begin{array}{ll}\text { Put } & f^{\prime}(x)=0 \\ \Rightarrow & k \cos x+\cos 3 x=0\end{array}$
Since the $f(x)$ has maximum value at $x=\frac{\pi}{3}$
$\therefore \quad k \cos \frac{\pi}{3}+\cos 3\left(\frac{\pi}{3}\right)=0$
$\Rightarrow \quad \frac{k}{2}+(-1)=0 \Rightarrow k=2$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 1)
The function $f(x)=\frac{x}{x^{2}+1}$ from $R$ to $R$ is

Options:
A one $-$ one as well as onto
B onto but not one-one
C neither one-one nor onto
D one-one but not onto

The correct answer is:
one-one but not onto

Solution:
Given $f: R \rightarrow R$ defined as $f(x)=\frac{x}{x^{2}+1}$
Which is a injective function
-ie- $f(x)=\frac{x}{x^{2}+1}$ is one-one but not onto.

Topic: Mathematics, Functions, NDA, NDA 2010 (Phase 1)
The function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}$ increases for all $\quad$

Options:
A $\mathrm{x}>-1$ only
B $\mathrm{x} < -1$ only
C $\mathrm{x}>1$ only
D $x < 1$ only

The correct answer is:
$\mathrm{x}>1$ only

Solution:
Given $f(x)=x^{2}-2 x$ On differentiating w.r.t $^{\circ} x^{\prime}$, we get $f^{\prime}(x)=2 x-2$ $f(x)$ is increasing, if $f^{\prime}(x)>0$ $\Rightarrow 2 x-2>0$
$\Rightarrow x>1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
The function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-4 \mathrm{x}, \mathrm{x} \in[0,4]$ attains minimum
value at

Options:
A $x=0$
B $x=1$
C $\quad x=2$
D $\mathrm{x}=4$

The correct answer is:
$\quad x=2$

Solution:
Given function attains minimum at $\mathrm{x}=2$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)
The function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-3 \mathrm{x}^{2}+6$ is an increasing function for:

Options:
A $\quad 0 < x < 2$
B $x < 2$
C $x>2$ or $x < 0$
D all $x$

The correct answer is:
$x>2$ or $x < 0$

Solution:
For increasing function, $f^{\prime}(x)=3 x^{2}-6 x>0$
$\Rightarrow x(x-2)>0$
Either both factors $\mathrm{x}$ and $(\mathrm{x}-2)$ are simultaneously $+\mathrm{ve}$ simultaneously-ve. Case - I: $\mathrm{x}>0, \mathrm{x}-2>0$
$\Rightarrow \quad \mathrm{x}>0,, \mathrm{x}>2$
i.e. $x>2$
Case - II: $x < 0, \quad x-2 < 0$
$\Rightarrow \quad x < 0, \quad x < 2$
ie. $x < 0$ From both cases $\mathrm{x}>2$ or $\mathrm{x} < 0$.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 2)
The function $f(x)=x \operatorname{cosec} x$ is

Options:
A continuous for all values of $x$
B discontinuous everywhere
C continuous for all $x$ except at $x=n \pi$, where $n$ is an integer
D continuous for all $x$ except at $x=n \pi / 2$, where $n$ is an integer

The correct answer is:
discontinuous everywhere

Solution:
The function $f(x)=x$ cosec $x$ is discontinuous everywhere.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2010 (Phase 1)
The function $f(\mathrm{x})=|\mathrm{x}|-\mathrm{x}^{3}$ is

Options:
A odd
B even
C both even and odd
D neither even nor odd

The correct answer is:
neither even nor odd

Solution:
$f(x)=|x|-x^{3}$
$f(x)=\left\{\begin{array}{l}x-x^{3}, x \geq 0 \\ -x-x^{3}, x < 0\end{array} \quad f(-x)=\left\{\begin{array}{l}x+x^{3}, x \geq 0 \\ x+x^{3}, x < 0\end{array}\right.\right.$
Neither even nor odd.

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 2)
The function $f: \mathrm{X} \rightarrow \mathrm{Y}$ defined by $f(\mathrm{x})=\cos \mathrm{x}$, where $\mathrm{x} \in \mathrm{X}$,
is one-one and onto if $\mathrm{X}$ and $\mathrm{Y}$ are respectively equal to

Options:
A $[0, \pi]$ and $[-1,1]$
B $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $[-1,1]$
C $[0, \pi]$ and $(-1,1)$
D $[0, \pi]$ and $[0,1]$

The correct answer is:
$[0, \pi]$ and $[-1,1]$

Solution:
Observe the cosx graph in the figure.
It is clear that, function is one $-$ one and onto when x and y are $[0, \pi]$ and $[-1,1]$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 1)
The function $\mathrm{y}=f(x)=m x+c$ has $\quad$

Options:
A maximum point but no minimum point
B minimumpoint but no maximum point
C both maximum and minimum points
D neither maximum point nor minimum point

The correct answer is:
neither maximum point nor minimum point

Solution:
Let $f(x)=m x+c$
then, $f^{\prime}(x)=m$
So, $f^{\prime}(x) \neq 0$ for any real value of $x$. Hence, $f(x)$ has neither maximum point nor minimum point.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
The function $y=\tan ^{-1} x-x \quad$

Options:
A is always decreasing
B is always increasing
C first increases and then decreases
D first decreases and then increases

The correct answer is:
is always decreasing

Solution:
Let $y=\tan ^{-1} x-x$
On differentiating w.r.t $x$, we get $\frac{d y}{d x}=\frac{1}{1+x^{2}}-1=\frac{1-1-x^{2}}{1+x^{2}}=\frac{-x^{2}}{1+x^{2}}$
$\Rightarrow \frac{d y}{d x} < 0, \forall x \in R$
Hence, function is always decreasing.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
The fundamental cause of the sickle-cell disease is a change in the structure of:

Options:
A blood
B capillaries
C hemoglobin
D red cells

The correct answer is:
hemoglobin

Solution:
Sickle cell anemia is an autosome linked recessive trait that can be transmitted from parents to the offspring when both the partners are carriers for the gene (or heterozygous). Sickle cell anemia is a disease that affects the hemoglobin molecule. An atypical hemoglobin molecule called hemoglobin S will be formed, which distorts the structure of red blood cells into a sickle or crescent shape.

Topic: Biology, Principles of Inheritance and Variation, NEET
The fundamental frequency in an open organ pipe is equal to the third harmonic of closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

Options:
A 12.5 cm
B 8 cm
C 13.2 cm
D 16 cm

The correct answer is:
13.2 cm

Solution:
For closed organ pipe, third harmonic 3v4lcop
For open organ pipe, fundamental frequency v2loop.

According to question
foop=3fcop
⇒v2loop=3v4lcop
loop=23×lcop=23×20≈13.2 cm

Topic: Physics, Waves and Sound, NEET, NEET 2018
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is

Options:
A 80 cm
B 100 cm
C 120 cm
D 140 cm

The correct answer is:
120 cm

Solution:
For closed organ pipe,
\(n_c=\frac{V}{4 l}\)
For open organ pipe,
\(n_0=\frac{V}{2 l^{\prime}}\)
For the second overtone of an open organ pipe,
\(\begin{aligned}
& n^{\prime}=3 n_0=\frac{3 V}{2 l^{\prime \prime}} \\
& n_c=n^{\prime} \\
& \frac{V}{4 l}=\frac{3 V}{2 l^{\prime}} \\
& l^{\prime}=6 l=6 \times 20 \\
& l^{\prime}=120 \mathrm{~cm}
\end{aligned}\)

Topic: Physics, Waves and Sound, NEET, NEET 2015 (Phase 1)
The fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an open pipe. The ratio of their lengths is

Options:
A $1: 2$
B $1: 4$
C $1: 8$
D $1: 16$

The correct answer is:
$1: 4$

Solution:
Let the length of closed pipe is $L_{1}$ and that of pipe is $L_{2}$. Fundamental frequency of closed pipe, $v_{1}=\frac{v}{4 L_{1}}$
and frequency of second harmonic of open pipe.
$$
v_{2}=2 \times \frac{v}{2 L_{2}}
$$
Given,
$$
v_{1}=v_{2}
$$
$i, \theta$
$$
\frac{v}{4 L_{1}}=\frac{2 v}{2 L_{2}} \Rightarrow \frac{L_{1}}{L_{2}}=\frac{1}{4}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2013
The fundamental frequency of a sonometer wire carrying a block of mass ' $M$ ' and density ' $\rho$ ' is ' $n$ ' Hz. When the block is completely immersed in a liquid of density ' $\sigma$ ' then the new frequency will be

Options:
A $\mathrm{n}\left[\frac{\rho-\sigma}{\rho}\right]^{\frac{1}{2}}$
B $\mathrm{n}\left[\frac{\rho-\sigma}{\sigma}\right]^{\frac{1}{2}}$
C $\mathrm{n}\left[\frac{\rho}{\rho-\sigma}\right]^{\frac{1}{2}}$
D $n\left[\frac{\sigma}{\rho-\sigma}\right]^{\frac{1}{2}}$

The correct answer is:
$\mathrm{n}\left[\frac{\rho-\sigma}{\rho}\right]^{\frac{1}{2}}$

Solution:
$$
\begin{aligned}
& \mathrm{n} \propto \sqrt{\mathrm{T}} \\
& \mathrm{T}=\mathrm{mg}=\rho \mathrm{Vg} \\
\therefore \quad \mathrm{T} & \propto \sqrt{\rho \mathrm{Vg}}
\end{aligned}
$$
After immersion in the liquid,
$$
\begin{array}{ll}
\therefore & \frac{\mathrm{n}_2}{\mathrm{n}} \propto \frac{\sqrt{\mathrm{V}(\rho-\sigma) \mathrm{g}}}{\sqrt{\mathrm{V} \rho \mathrm{g}}} \\
\therefore & \mathrm{n}_2=\mathrm{n}\left[\frac{\rho-\sigma}{\rho}\right]^{\frac{1}{2}}
\end{array}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (13 May Shift 1)
The fundamental frequency of a sonometer wire is $50 \mathrm{~Hz}$ for some length and
tension. If the length is increased by $25 \%$ by keeping tension same, then frequency
change of second harmonic is

Options:
A decreased by $10 \%$
B decreased by $20 \%$
C decreased by $5 \%$
D decreased by $15 \%$

The correct answer is:
decreased by $20 \%$

Solution:
(D)
$\begin{aligned} & \mathrm{n}_{2} \ell_{2}=\mathrm{n}_{1} \ell_{1} \\ \therefore & \mathrm{n}_{2}=\frac{\ell_{1}}{\ell_{2}} \cdot \mathrm{n}_{1}=\frac{\ell_{1}}{1.25 \mathrm{R}_{1}} \mathrm{n}_{1}=0.8 \mathrm{n}_{1} \\ \therefore & \mathrm{n}_{1}-\mathrm{n}_{2}=\mathrm{n}_{1}(1-08)=0.2 \mathrm{n}_{1} \\ & \frac{\mathrm{n}_{1}-\mathrm{n}_{2}}{\mathrm{n}_{1}}=0.2 \end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (19 Oct Shift 1)
The fundamental frequency of a sonometer wire is n. If the length and diameter of the wire are double keeping the tension same, then the new fundamental frequency is

Options:
A n22
B 2n
C n4
D n2

The correct answer is:
n4

Solution:
Frequency of vibration of sonometer wire is given as
n=12lTm=12lTÏr2d
ânâ1dnâ1rand n â1l
If the length and diameter of the wire are doubled.
The new frequency will be n4 .

Topic: Physics, Waves and Sound, NEET
The fundamental frequency of a sonometer wire is n. if the length and diameter of the wire are doubled keeping the tension same, then the new fundamental frequency is

Options:
A 2n2
B n22
C 2n
D n4

The correct answer is:
n4

Solution:
The fundamental frequency of a sonometer wire is

f=12l4Tπρd2

⇒f∝1ld

So if the length and diameter of the wire are doubled, then its frequency becomes one-fourth

Topic: Physics, Waves and Sound, NEET
The fundamental frequency of a string stretched with a weight of 4 kg is 256 Hz. The weight required to produce its octave is

Options:
A 16 kg-wt
B 12 kg-wt
C 24 kg-wt
D 8 kg-wt

The correct answer is:
16 kg-wt

Solution:
As nâT
â´ To produce octave of the note (of double the frequency), T has to be made 4 times, ie, weight required =4Ã4 kg=16kg

Topic: Physics, Waves and Sound, NEET
The fundamental frequency of a wire stretched by $2 \mathrm{kgwt}$ is $100 \mathrm{~Hz}$. The weight required to produce its octave is

Options:
A $12 \mathrm{kgwt}$
B 8 kgwt
C 4 kgwt
D $16 \mathrm{kgwt}$

The correct answer is:
8 kgwt

Solution:
Octave means the new frequency is $2 n$
The velocity of sound is proportional to the square-root of the tension in the wire.
The frequency is directly proportional to the velocity of sound in the wire:
$\begin{aligned} & \therefore n \propto \sqrt{T} \\ & \Rightarrow \frac{n_2}{n_1}=\sqrt{\frac{T_2}{T_1}} \\ & \therefore T_2=T_1\left[\frac{n_2^2}{n_1^2}\right]=4 T_1=8 \mathrm{kgwt}\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (10 Aug Shift 1)
The fundamental frequency of air column in pipe 'A' closed at one end is in unison with second overtone of an air column in pipe 'B' open at both ends. The ratio of length of air column in pipe 'A' to that of air column in pipe ' $\mathrm{B}$ ' is

Options:
A $1: 6$
B $3: 8$
C $2: 3$
D $3: 4$

The correct answer is:
$1: 6$

Solution:
Fundamental frequency of ta closed pipè, $\mathrm{n}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1}$
Frequency of the second overtone, $\mathrm{n}_2=\frac{3 \mathrm{v}}{2 \mathrm{~L}_2}$
Given $\mathrm{n}_1=\mathrm{n}_2$
$\Rightarrow \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{3 \mathrm{v}}{2 \mathrm{~L}_2}$
$\therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{1}{6}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (10 May Shift 2)
The fundamental frequency of an air column in a pipe closed at one end is 100 Hz. If the same pipe is open at both the ends, the frequencies produced in Hz are

Options:
A 100, 200, 300, 400, .....
B 100, 300, 500, 700, .....
C 200, 300, 400, 500 ,....
D 200, 400, 600, 800, .....

The correct answer is:
200, 400, 600, 800, .....

Solution:
For a closed pipe fundamental frequency n1=V4L=100 Hz
For an open pipe fundamental frequency n'1=V2L=2n1=200 Hz
In an open pipe all multiples of the fundamental are produced.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2017
The fundamental frequency of an air column in a pipe closed at one end is 100 Hz. If the same pipe is open at both the ends, the frequencies produced in Hz are

Options:
A 100, 200, 300, 400, .....
B 100, 300, 500, 700, .....
C 200, 300, 400, 500 ,....
D 200, 400, 600, 800, .....

The correct answer is:
200, 400, 600, 800, .....

Solution:
For a closed pipe fundamental frequency n1=V4L=100 Hz
For an open pipe fundamental frequency n'1=V2L=2n1=200 Hz
In an open pipe all multiples of the fundamental are produced.

Topic: Physics, Waves and Sound, NEET
The fundamental frequency of an air column in pipe ' $\mathrm{A}$ ' closed at one end coincides with second overtone of pipe ' $\mathrm{B}$ ' open at both ends. The ratio of length of pipe ' $\mathrm{A}$ ' to that of pipe ' $\mathrm{B}$ ' is

Options:
A $3: 8$
B $3: 4$
C $1: 6$
D $2: 3$

The correct answer is:
$1: 6$

Solution:
Fundamental frequency of pipe closed at one end
$$
\mathrm{n}=\frac{\mathrm{v}}{4 \ell}
$$
Second overtone of pipe open at both the ends
$$
\begin{aligned}
& \mathrm{n}^{\prime}=\left(\frac{3 \mathrm{v}}{2 \ell^{\prime}}\right) \\
& \mathrm{n}^{\prime}=\mathrm{n} \\
& \therefore \frac{3 \mathrm{v}}{2 \ell^{\prime}}=\frac{\mathrm{v}}{4 \ell} \\
& \therefore \frac{\ell}{\ell^{\prime}}=\frac{1}{6}
\end{aligned}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (23 Sep Shift 2)
The fundamental frequency of open pipe is 'n'. If it is closed from one end then frequency of the $2^{\text {nd }}$ harmonic of closed pipe is higher by $200 \mathrm{~Hz}$ than ${ }^{\prime} \mathrm{n}^{\prime}$. The value of 'n' is

Options:
A $800 \mathrm{~Hz}$
B $200 \mathrm{~Hz}$
C $100 \mathrm{~Hz}$
D $400 \mathrm{~Hz}$

The correct answer is:
$400 \mathrm{~Hz}$

Solution:
$\mathrm{n}=\frac{\mathrm{v}}{2 \ell}$
$2^{\mathrm{nd}}$ harmonic $=1^{\mathrm{st}}$ overtone $=(2 \mathrm{p}+1) \mathrm{n}_{0}=3 \mathrm{n}_{0}$
$\frac{3 \mathrm{v}}{4 \ell}-\frac{\mathrm{v}}{2 \ell}=200$
$\frac{v}{2 \ell}\left(\frac{3}{2}-1\right)=200$
$\frac{v}{2 \ell}=400$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (20 Oct Shift 2)
The fundamental frequency of sonometer wire increases by 9 Hz, if its tension is increased by 69% , keeping the length constant. The frequency of the wire is

Options:
A 42 Hz
B 24 Hz
C 30 Hz
D 36 Hz

The correct answer is:
30 Hz

Solution:
We know that, frequency of vibration of a stretched string is given by
v=12lTm
where, T = tension in string
m = mass /length
and l = length of string.
New frequency,
v'=v+9=12lT+69/100Tm
⇒vv'=T169100T⇒vv+9=100169=1013
⇒13v=10v+90
v=30Hz

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2019 (Shift 1)
The fundamental radio antenna is a metal rod which has a length equal to

Options:
A $\lambda$ in free space at the frequency of operation
B $\frac{\lambda}{2}$ in free space at the frequency of operation
C $\frac{\lambda}{4}$ in free space at the frequency of operation
D $\frac{3 \lambda}{4}$ in free space at the frequency of operation

The correct answer is:
$\frac{\lambda}{4}$ in free space at the frequency of operation

Solution:
The common antenna is a straight conductor of length $1=\frac{\lambda}{4}$ held vertically with its lower end touching the ground.

Topic: Physics, Communication System, VITEEE, VITEEE 2006
The general solution of $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ax}+\mathrm{h}}{\mathrm{by}+\mathrm{k}}$ represents a circle
only when $\quad$

Options:
A $a=b=0$
B $a=-b \neq 0$
C $a=b \neq 0, h=k$
D $a=b \neq 0$

The correct answer is:
$a=-b \neq 0$

Solution:
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ax}+\mathrm{h}}{\mathrm{by}+\mathrm{k}}$
$\Rightarrow(\mathrm{by}+\mathrm{k}) \mathrm{dy}=(\mathrm{ax}+\mathrm{h}) \mathrm{d} \mathrm{x}$
Intergrating, we get $\int(\mathrm{by}+\mathrm{k}) \mathrm{dy}=\int(\mathrm{ax}+\mathrm{h}) \mathrm{dx}$
$\Rightarrow \frac{\mathrm{by}^{2}}{2}+\mathrm{ky}=\frac{\mathrm{ax}^{2}}{2}+\mathrm{hx}$
$\Rightarrow b y^{2}+2 k y=a x^{2}+2 h x$
$\Rightarrow a x^{2}-b y^{2}+2 h x-2 k y=0$
This represents circle only when $\mathrm{a}=-\mathrm{b} \neq 0$.

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 2)
The general solution of the differential equation \elln $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\mathrm{x}=0$ is? $\quad$

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

The correct answer is:
$\mathrm{y}=-\mathrm{e}^{-\mathrm{x}}+\mathrm{c}$

Solution:
Let $\ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\mathrm{x}=0 \Rightarrow \ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=-\mathrm{x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}$
Integrate both the side, $y=-e^{-x}+c$

Topic: Mathematics, Differential Equations, NDA, NDA 2013 (Phase 1)
The general solution of the differential equation
$\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$ is?

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

The correct answer is:
$x y=c$

Solution:
Given differential equation is
$$
\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0
$$
$\Rightarrow x d y+y d x=0$
$\Rightarrow x d y=-y d x$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=-\frac{\mathrm{dx}}{\mathrm{x}}$
On integrating both side we get $\ell \mathrm{n} \mathrm{y}=-\ln \mathrm{x}+\mathrm{enc}$
$\Rightarrow\left(\mathrm{y}=\frac{\mathrm{c}}{\mathrm{x}}\right)$

Topic: Mathematics, Differential Equations, NDA, NDA 2013 (Phase 1)
The genotypes of a Husband and Wife are IA IB and IAi .
Among the blood types of their children, how many different genotypes and phenotypes are possible?

Options:
A 3 genotypes; 3 phenotypes
B 3 genotypes; 4 phenotypes
C 4 genotypes; 3 phenotypes
D 4 genotypes; 4 phenotypes

The correct answer is:
4 genotypes; 3 phenotypes

Solution:
Husband × Wife I A I B I A i

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Topic: Biology, Principles of Inheritance and Variation, NEET
The geometric mean and harmonic mean of two non negative observations are 10 and 8 respectively. Then what is the arithmetic mean of the observations equal to?

Options:
A 4
B 9
C 12,5
D 2

The correct answer is:
12,5

Solution:
Let 'a' and 'b' be two non-negative numbers. $\mathrm{GM}=\sqrt{\mathrm{ab}}=10$
$\Rightarrow \quad a b=100$
and $\mathrm{H.M.}=\frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}}=8$
$\Rightarrow \frac{200}{a+b}=8$
$\Rightarrow \quad a+b=25$
Consider $(a-b)^{2}=(a+b)^{2}-4 a b=625-400=225$
$\Rightarrow a-b=15$
and $a+b=25$
$\Rightarrow 2 a=40 \Rightarrow a=20$ and $b=5$
A.M. $=\frac{20+5}{2}=12.5$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 1)
The geometric mean of the observations $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots \ldots \mathrm{x}_{\mathrm{n}}$ is
$\mathrm{G}_{1}$, The geometric mean of the observations $\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}, \ldots . \mathrm{y}_{\mathrm{n}}$
is $\mathrm{G}_{2}$. The geometric mean of observations $\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}, \frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}, \frac{\mathrm{x}_{3}}{\mathrm{y}_{3}}, \ldots \cdot \frac{\mathrm{x}_{\mathrm{n}}}{\mathrm{y}_{\mathrm{n}}}$ is

Options:
A $\mathrm{G}_{1} \mathrm{G}_{2}$
B $\ln \left(\mathrm{G}_{1} \mathrm{G}_{2}\right)$
C $\frac{\mathrm{G}_{1}}{\mathrm{G}_{2}}$
D $\ln \left(\frac{G_{1}}{G_{2}}\right)$

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

Solution:
$\quad \mathrm{G}_{1}=\left[\mathrm{x}_{1} \times \mathrm{x}_{2} \times \mathrm{x}_{3} \times \ldots \ldots \ldots \times \mathrm{x}_{\mathrm{n}}\right]^{y_{1}}$
$\mathrm{G}_{2}=\left[\mathrm{y}_{1} \times \mathrm{y}_{2} \times \mathrm{y}_{3} \times \ldots \ldots . \times \mathrm{y}_{\mathrm{s}}\right]^{1 / \mathrm{m}}$
$\Rightarrow \frac{\mathrm{G}_{1}}{\mathrm{G}_{2}}=\left[\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}} \times \frac{\mathrm{x}_{2}}{\mathrm{y}_{2}} \times \frac{\mathrm{x}_{3}}{\mathrm{y}_{3}} \times \ldots \ldots \ldots \times \frac{\mathrm{x}_{\mathrm{n}}}{\mathrm{y}_{\mathrm{n}}}\right]^{1 / \mathrm{n}}$
$\frac{\mathrm{G}_{1}}{\mathrm{G}_{2}}$ is the G.M of $\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}, \frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}, \frac{\mathrm{x}_{3}}{\mathrm{y}_{3}}, \ldots \ldots \ldots \ldots, \frac{\mathrm{x}_{\mathrm{a}}}{\mathrm{y}_{\mathrm{a}}}$

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 2)