datavorous/entrance-exam-dataset huggingface dataset:
If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is :

Options:
A x2+y2x+y=R2xy
B x2+y23=4R2x2y2
C x2+y22=4R2x2y2
D x2+y22=4Rx2y2

The correct answer is:
x2+y23=4R2x2y2

Solution:
Since, circle passing through origin intersect the coordinate axes at A& B, hence AB must be diameter and AB=2R.

Now, let foot of the perpendicular from origin upon AB be Ph, k.

Slope of line OP=k-0h-0=kh

Since, line AB⊥OP⇒ slope of AB=-hk

Thus, equation of line AB is y-k=-hkx-h

For co-ordinates of A, put y=0⇒0-k=-hkx-h⇒x=h2+k2h⇒A h2+k2h, 0.

For co-ordinates of B, put x=0⇒y-k=-hk0-h⇒y=h2+k2k⇒B 0,h2+k2k.

Now, given AB=2R

Applying distance formula,

⇒h2+k2h-02+0-h2+k2k2=2R

⇒h2+k22 1h2+1k2=4R2⇒h2+k23=4R2h2k2

Hence, locus is x2+y23=4R2x2y2

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If a circle of unit radius is divided into two parts by an arc of another circle subtending an angle $60^{\circ}$ on the circumference of the first circle, then the radius of the arc is:

Options:
A
$\sqrt{3}$

B
$\frac{1}{2}$

C
1

D
None of these

The correct answer is:
None of these

Solution:
Given information is incomplete in the question.

Topic: Mathematics, Circle, JEE Main, JEE Main 2013 (25 Apr Online)
If a circle passes through points $(4,0)$ and $(0,2)$ and its centre lies on $\mathrm{Y}$-axis. If the radius of the circle is $r$, then the value of $r^2-r+1$ is

Options:
A $25$
B $21$
C $20$
D $10$

The correct answer is:
$21$

Solution:
Let $(0, y)$ be the centre of the circle.
$\begin{aligned}
& \therefore \quad \sqrt{(0-4)^2+(y-0)^2}=\sqrt{(0-0)^2+(y-2)^2} \\
& \therefore \quad 16+y^2=(y-2)^2 \\
& \therefore \quad 16+y^2=y^2-4 y+4 \\
& \begin{aligned}
\therefore & y=-3 \\
\therefore & \text { centre of the circle is }(0,-3) . \\
\therefore & \text { Radius of the circle }=\mathrm{r}=\sqrt{(0-0)^2+(-3-2)^2} \\
& =5 \text { units } \\
\therefore & \mathrm{r}^2-\mathrm{r}+1=25-5+1=21
\end{aligned}
\end{aligned}$

Topic: Mathematics, Circle, JEE Main
If a circle passes through the point 1,1 and cuts the circle x2+y2=1 orthogonally, then the locus of its centre is

Options:
A x2+y2-3x-3y+1=0
B 2x+2y-1=0
C x2+y2-2x-2y+1=0
D 2x+2y-3=0

The correct answer is:
2x+2y-3=0

Solution:
Let the centre of the circle be α,β.
Since it cuts the circle x2+y2=1 orthogonally,
2-α×0+2-β×0=c1-1
⇒c1=1
∴ The equation of the circle is
x2+y2-2αx-2βy+1=0
It passes through 1,1. So 1+1-2α-2β+1=0
Hence, the locus of α,β is 2x+2y-3=0

Topic: Mathematics, Circle, JEE Main
If a circle passes through the point $(a, b)$ and cuts the circle $x^2+y^2=4$ orthogonally, then the locus of its centre is

Options:
A
$2 a x+2 b y+\left(a^2+b^2+4\right)=0$

B
$2 a x+2 b y-\left(a^2+b^2+4\right)=0$

C
$2 a x-2 b y+\left(a^2+b^2+4\right)=0$

D
$2 a x-2 b y-\left(a^2+b^2+4\right)=0$

The correct answer is:
$2 a x+2 b y-\left(a^2+b^2+4\right)=0$

Solution:
Let the circle be $x^2+y^2+2 g x+2 f y+c=0 \Rightarrow c=4$ and it passes through $(a, b)$ $\Rightarrow \mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+4=0$
Hence locus of the centre is $2 a x+2 b y-\left(a^2+b^2+4\right)=0$

Topic: Mathematics, Circle, JEE Main, JEE Main 2004
If a circle passes through the point $(a, b)$ and cuts the circle $x^2+y^2=p^2$ orthogonally, then the equation of the locus of its centre is

Options:
A
$x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0$

B
$2 a x+2 b y-\left(a^2-b^2+p^2\right)=0$

C
$x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0$

D
$2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$

The correct answer is:
$2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$

Solution:
Let the centre be $(\alpha, \beta)$
$\because$ It cut the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{p}^2$ orthogonally
$2(-\alpha) \times 0+2(-\beta) \times 0=c_1-p^2$
$\mathrm{c}_1=\mathrm{p}^2$
Let equation of circle is $x^2+y^2-2 \alpha x-2 \beta y+p^2=0$
It pass through $(a, b) \Rightarrow a^2+b^2-2 \alpha a-2 \beta b+p^2=0$
Locus $\therefore 2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$

Topic: Mathematics, Circle, JEE Main
If a circle passes through the point $(a, b)$ and cuts the circle $x^2+y^2=p^2$ orthogonally, then the equation of the locus of its centre is

Options:
A
$x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0$

B
$2 a x+2 b y-\left(a^2-b^2+p^2\right)=0$

C
$x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0$

D
$2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$

The correct answer is:
$2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$

Solution:
Let the centre be $(\alpha, \beta)$
$\because$ It cut the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{p}^2$ orthogonally
$2(-\alpha) \times 0+2(-\beta) \times 0=c_1-p^2$
$\mathrm{c}_1=\mathrm{p}^2$
Let equation of circle is $x^2+y^2-2 \alpha x-2 \beta y+p^2=0$
It pass through $(a, b) \Rightarrow a^2+b^2-2 \alpha a-2 \beta b+p^2=0$
Locus $\therefore 2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$

Topic: Mathematics, Circle, JEE Main, JEE Main 2005
If a circle passing through the point -1, 0 touches y-axis at 0, 2, then the x-intercept of the circle is

Options:
A 52
B 5
C 32
D 3

The correct answer is:
3

Solution:
Let a, 2 be the centre of the circle
Equation of the circle is
x-a2+y-22=a2
Put -1,0 in the equation of circle -1-a2+-22=a2
1+2a+a2+4=a2
2a=-5
a=-52
Equation of circle is
x+522+y-22=254
x intercept =2g2-c
=2254-4
=294
=3

Topic: Mathematics, Circle, JEE Main, JEE Main 2015 (11 Apr Online)
If a circular iron sheet of radius $30 \mathrm{~cm}$ is heated such that its area increases at the uniform rate of $6 \pi \mathrm{cm}^2 / \mathrm{hr}$, then the rate (in $\mathrm{mm} / \mathrm{hr}$ ) at which the radius of the circular sheet increases is

Options:
A
$1.0$

B
$0.1$

C
$1.1$

D
$2.0$

The correct answer is:
$0.1$

Solution:
Let $A=\pi r^2$.
$$
\begin{aligned}
\frac{d A}{d t} & =2 \pi r \cdot \frac{d r}{d t} \\
6 \pi & =2 \pi(30) \cdot \frac{d r}{d t} \\
\Rightarrow \quad \frac{3}{30}= & \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{10}=0.1
\end{aligned}
$$
Thus, the rate at which the radius of the circular sheet increases is $0.1$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (07 May Online)
If a complex number \( \mathrm{z} \) lie on a circle of radius \( \frac{1}{2} \) units, then the complex number \( \omega=-1+4 z \) will always lie on a circle of radius \( k \) units, where \( k \) is equal to

Options:

The correct answer is:
2

Solution:
Let us assume that z lies on a circle with centre z0 (fixed point) and radius 1 2 units.
⇒ z-z0=12
Now, ω=-1+4z⇒ω+1=4z
⇒ ω+1-4z0=4z-4z0
Now, taking modulus on both sides, we get,
ω+1-4z0=4z-z0⇒ω+1-4z0=2
Locus of ω represents the circle having centre -1+4z0 and radius 2 units.

Topic: Mathematics, Complex Number, JEE Main
If a complex number z lie on a circle of radius 1 2 units, then the complex number ω = - 1 + 4 z will always lie on a circle of radius k units, where k is equal to

Options:

The correct answer is:
2

Solution:
Let us assume that z lies on a circle with centre z0 (fixed point) and radius 1 2 units.
⇒ z - z 0 = 1 2
Now, ω = - 1 + 4 z ⇒ ω + 1 = 4 z
⇒ ω + 1 - 4 z 0 = 4 z - 4 z 0
Now, taking modulus on both sides, we get,
ω + 1 - 4 z 0 = 4 z - z 0 ⇒ ω + 1 - 4 z 0 = 2
Locus of ω represents the circle having centre - 1 + 4 z 0 and radius 2 units.

Topic: Mathematics, Complex Number, JEE Main
If a complex number $z$ statisfies the equation $x+\sqrt{2}|z+1|+i=0$, then $|z|$ is equal to :

Options:
A
2

B
$\sqrt{3}$

C
$\sqrt{5}$

D
1

The correct answer is:
$\sqrt{5}$

Solution:
Given equation is
$$
\begin{aligned}
& z+\sqrt{2}|z+1|+i=0 \\
& \text { put } z=x+i y \text { in the given equation. } \\
& (x+i y)+\sqrt{2}|x+i y+1|+i=0 \\
& \Rightarrow x+i y+\sqrt{2}\left[\sqrt{(x+1)^2+y^2}\right]+i=0
\end{aligned}
$$
Now, equating real and imaginary part, we get
$$
\begin{aligned}
& x+\sqrt{2} \sqrt{(x+1)^2+y^2}=0 \text { and } \\
& y+1=0 \Rightarrow y=-1 \\
& \Rightarrow x+\sqrt{2} \sqrt{(x+1)^2+(-1)^2}=0 \\
& \Rightarrow \sqrt{2} \sqrt{(x+1)^2+1}=-x \\
& \Rightarrow 2\left[(x+1)^2+1\right]=x^2 \\
& \Rightarrow x^2+4 x+4=0 \\
& \Rightarrow x=-2
\end{aligned}
$$
Thus, $z=-2+i(-1) \Rightarrow|z|=\sqrt{5}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2013 (22 Apr Online)
If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ___ molal. (Rounded-off to the nearest integer) Kb=0.52K kg mol-1.

Options:

The correct answer is:
3

Solution:
α=0.75,n=2

i=1-α+nα=1-0.75+2×0.75=1.75

ΔTb=ikbm

or 2.5=1.75×0.52×m

or, m=2.51.75×0.52=2.74

∴ nearest integer answer will be 3

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If a conductor has a potential $\mathrm{V} \neq 0$ and there are no charge anywhere else outside, then

Options:
A
there must be charges on the surface or inside itself

B
there cannot be any charge in the body of the conductor

C
there must be charges only on the surface

D
there must be charges inside the surface

The correct answers are:
there must be charges on the surface or inside itself
,
there cannot be any charge in the body of the conductor

Solution:
The main concept used are :
1. The charge resides on the outer surface of a closed charged conductor
2. Net charge inside the conductor is zero.
The excess charge can reside only on the surface of conductor and inside net positive and negative charge is zero, so any charge can reside inside the hollow shell or body. So, inside the solid material of conducting body there is no charge, it comes to outer surface.

Topic: Physics, Electrostatic Potential and Capacitance
If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is _____%.

Options:

The correct answer is:
44

Solution:
Before heating, resistance of wire is R0=ρl0A0

After heating, resistance is R=ρlA

As volume = constant

Therefore,

A0l0=1.2l0A

⇒A=A01.2

So,

R=ρlA=ρ×1.2l0A01.2=ρl0A0×1.44=1.44 R0

Percentage increase in R =R-R0R0×100=44%

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (24 Jan Shift 2)
If $a \cos \theta+b \sin \theta=m$ and $a \sin \theta-b \cos \theta=n$, then $a^2+b^2=$

Options:
A $m+n$
B $m^2-n^2$
C $m^2+n^2$
D None of these

The correct answer is:
$m^2+n^2$

Solution:
Given that $a \cos \theta+b \sin \theta=m$
and $a \sin \theta-b \cos \theta=n$.
Squaring and adding, we get
$\begin{aligned}
& (a \cos \theta+b \sin \theta)^2+(a \sin \theta-b \cos \theta)^2=m^2+n^2 \\
& \Rightarrow a^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+b^2\left(\cos ^2 \theta+\sin ^2 \theta\right)
\end{aligned}$
$+2 a b(\cos \theta \sin \theta-\sin \theta \cos \theta)=m^2+n^2$
Hence, $a^2+b^2=m^2+n^2$.
Trick : Here we can guess that the value of $a^2+b^2$ is independent of $\theta$, so put any suitable value of $\theta$ i.e. $\frac{\pi}{2}$, so that $b=m$ and $a=n$. Hence $a^2+b^2=m^2+n^2$. (Also check for other value of $\theta$ ).

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
IfA=cosθ isinθisinθcosθ, θ=π24 and A5=abcd, where i=-1, then which one of the following is not true?

Options:
A 0≤a2+b2≤1
B a2-d2=0
C a2-c2=1
D a2-b2=12

The correct answer is:
a2-b2=12

Solution:
A2=cosθ     isinθisinθ      cosθcosθ     isinθisinθ     cosθ

=cos2θ−sin2θ     2isinθcosθ2isinθcosθ         cos2θ−sin2θ

=cos2θ      isin2θisin2θ       cos2θ

A3=cos2θ     isin2θisin2θ      cos2θcosθ     isinθisinθ     cosθ

=cos3θ     isin3θisin3θ      cos3θ

Similarly, A5=cos5θ     isin5θisin5θ      cos5θ=a    bc    d

a=cos5θ, b=isin5θ, c=isin5θ

d=cos5θ

a=d, b=c

(1) a2+b2=cos25θ−sin25θ=cos10θ∈0,1 as θ=π24

(2) a2−d2=0

(3) a2−b2=cos25θ+sin25θ=1

(4) a2−c2=a2−b2=1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (04 Sep Shift 1)
If $A=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$, the cofactor of $a_{32}$ are

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

The correct answer is:
0,0

Solution:
We have,
$A=\left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right]$
Minor of $a_{32}=\left|\begin{array}{cc}\cos \theta & 0 \\ \sin \theta & 0\end{array}\right|=0-0=0$ and cofactor of $a_{32}=-$ Minor of $a_{32}=-0=0$

Topic: Mathematics, Determinants, JEE Main
If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$, then $(\operatorname{Adj} A)^{-1}=$

Options:
A $A+I$
B $A-I$
C A
D $\operatorname{Adj}\left(A^{-1}\right)$

The correct answer is:
A

Solution:
We have,
$$
\begin{aligned}
& A=\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \therefore|A|=\cos ^2 \alpha+\sin ^2 \alpha=1
\end{aligned}
$$
Now, $|\operatorname{adj} A|=|A|^{n-1}=|A|^{3-1}=|A|^2=(1)^2=1$
Also, $\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A=|A|^{3-2} \cdot A=|A| A=A$
Now, $(\operatorname{adj} A)^{-1}=\frac{\operatorname{adj}(\operatorname{adj} A)}{|\operatorname{adj} A|} \quad\left[\because A^{-1}=\frac{\operatorname{adj} A}{|A|}\right]$
$$
=A
$$
Hence, option (c) correct.

Topic: Mathematics, Matrices, JEE Main
If $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ and $A$ adj $A=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$,then $k$ is equal to

Options:
A $0$
B $1$
C $\sin \alpha \cos \alpha$
D $\cos 2 \alpha$

The correct answer is:
$1$

Solution:
\(\begin{aligned}
& A=\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] \\
& \therefore|A|=\left|\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right|=\cos ^2 \theta+\sin ^2 \theta=1 \neq 0
\end{aligned}\)
Thus, \(A^{-1}\) exists.
Now,
\(\begin{aligned}
& \Rightarrow A^{-1}=\operatorname{adj} A \\
& \Rightarrow A A^{-1}=\operatorname{Aadj} A \\
& \Rightarrow A A^{-1}=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\left[\because A A^{-1}=I\right] \\
& \Rightarrow k=1
\end{aligned}\)

Topic: Mathematics, Matrices, JEE Main
If $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$ and $B=\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$ then the correct relation is

Options:
A $A^2=B^2$
B $A+B=B-A$
C $A B=B A$
D None of these

The correct answer is:
$A B=B A$

Solution:
$A B=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
Clearly,
$=\left[\begin{array}{cc}\cos (\alpha+\beta) & -\sin (\alpha+\beta) \\ \sin (\alpha+\beta) & \cos (\alpha+\beta)\end{array}\right]=B A \quad$ (verify by using trigonometric properties).

Topic: Mathematics, Matrices, JEE Main
If $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$, then which of the following statements is not correct ?

Options:
A $A$ is orthogonal matrix
B $A^{\prime}$ is orthogonal matrix
C Determinant $A=1$
D $A$ is not invertible

The correct answer is:
$A$ is not invertible

Solution:
$|A|=1 \neq 0$, therefore $A$ is invertible.
Thus (4) is not correct.

Topic: Mathematics, Matrices, JEE Main
If a current $I$ given by $I_0 \sin \left(\omega t-\frac{\pi}{2}\right)$ flows in an ac circuit across which an ac potential of $E=E_0 \sin \omega t$ has been applied, then the power consumption $P$ in the circuit will be

Options:
A $P=\frac{E_0 I_0}{\sqrt{2}}$
B $P=\sqrt{2} E_0 I_0$
C $P=\frac{E_0 I_0}{2}$
D $P=0$

The correct answer is:
$P=0$

Solution:
\(\begin{aligned}
& \mathrm{I}=\mathrm{I}_0\left(\sin \omega t-\frac{\pi}{2}\right) \\
& E=E_0 \sin \omega t
\end{aligned}\)
Here \(E\) and \(I\) are in out of phase \(\left[\theta=\frac{\pi}{2}\right]\)
\(\therefore\) Power factor \(\cos \theta=\cos 90 =0\)
\(\therefore\) Power consumption become zero.

Topic: Physics, Alternating Current, JEE Main
If a current is passed in a spring, it

Options:
A Gets compressed
B Get expanded
C Oscillates
D Remains unchanged

The correct answer is:
Gets compressed

Solution:
In adjoining loops of spring, the current being in the same direction, there will be attraction. Due to which the spring gets compressed.

Topic: Physics, Magnetic Effects of Current, NEET
If a current is passed through a spring then the spring will

Options:
A
expand

B
compress

C
insulator

D
none of these

The correct answer is:
compress

Solution:
It will compress due to the force of attraction between two adjacent coils carrying current in the same direction

Topic: Physics, Current Electricity, JEE Main, JEE Main 2002
If a current is passed through a spring then the spring will

Options:
A expand
B compress
C remains same
D none of these

The correct answer is:
compress

Solution:
The spring will compress because the force of attraction between two adjacent turns carrying currents in the same direction. Will compress it.

Topic: Physics, Magnetic Effects of Current, NEET
If a current of 3.0 amperes flowing in the primary coil is reduced to zero in 0.001 second, then the induced e.m.f. in the secondary coil is 15000 volts. The mutual inductance between the two coils is

Options:
A $0.5$ henry
B $5$ henry
C $1.5$ henry
D $10$ henry

The correct answer is:
$5$ henry

Solution:
$e=M \frac{d i}{d t} \Rightarrow$ $M=\frac{15000}{3} \times 0.001=5 \mathrm{H}$

Topic: Physics, Electromagnetic Induction, JEE Main
If a curve passes through the origin and the slope of the tangent to it at any point x, y is x2-4x+y+8x-2, then this curve also passes through the point:

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

The correct answer is:
5,5

Solution:
Given

y0=0

and dydx=x-22+y+4x-2

⇒dydx-yx-2=x-2+4x-2

⇒ I.F.=e-∫1x-2dx=1x-2

Solution of L.D.E.

⇒ y1x-2=∫1x-2x-2+4x-2·dx

⇒yx-2=x-4x-2+C

Now, at x=0,y=0⇒C=-2

y=xx-2-4-2x-2

⇒y=x2-4x

This curve passes through 5, 5

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (25 Feb Shift 1)
If a curve passes through the point 1, -2 and has slope of the tangent at any point x, y on it as x2-2yx, then the curve also passes through the point

Options:
A 3, 0
B (-1, 2)
C (-2,1)
D 3, 0

The correct answer is:
3, 0

Solution:
dydx=x2-2yx

⇒dydx+2xy=x. It is in linear form.

I.F.=e∫2xdx=e2lnx=x2

∴y.x2=∫x.x2 dx=x44+c

∵At x=1, y=-2,

∴-2×12=144+c

⇒c=-2-14=-94

∴curve is y.x2=x44-94

∴It passes through 3, 0.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If a curve passes through the point $\left(2, \frac{7}{2}\right)$ and has slope $\left(1-\frac{1}{x^2}\right)$ at any point $(x, y)$ on it, then the ordinate of the point on the curve whose abscissa is $-2$ is :

Options:
A
$-\frac{3}{2}$

B
$\frac{3}{2}$

C
$\frac{5}{2}$

D
$-\frac{5}{2}$

The correct answer is:
$-\frac{3}{2}$

Solution:
$$
\begin{aligned}
& \text { Slope }=\frac{d y}{d x}=1-\frac{1}{x^2} \\
& \Rightarrow \int d y=\int\left(1-\frac{1}{x^2}\right) d x \\
& \Rightarrow y=x+\frac{1}{x}+\mathrm{C}, \text { which is the equation of }
\end{aligned}
$$
the curve
since curve passes through the point
$$
\begin{aligned}
& \left(2, \frac{7}{2}\right) \\
\therefore \quad & \frac{7}{2}=2+\frac{1}{2}+C \Rightarrow C=1 \\
\therefore \quad & y=x+\frac{1}{x}+1
\end{aligned}
$$
when $x=-2$, then $y=-2+\frac{1}{-2}+1=\frac{-3}{2}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (23 Apr Online)
If a curve $y=a \sqrt{x}+b x$ passes through the point $(1,2)$ and the area bounded by the curve, line $x=4$ and $x$-axis is $8 \mathrm{sq}$. unit, then

Options:
A $a=3, b=-1$
B $a=3, b=1$
C $a=-3, b=1$
D $a=-3, b=-1$

The correct answer is:
$a=3, b=-1$

Solution:
Given curve $y=a \sqrt{x}+b x$. This curve passes through (1, 2), $\therefore 2=a+b...(i)$ and area bounded by this curve and line $x=4$ and $\quad x$-axis is 8 sq. unit, then $\int_0^4(a \sqrt{x}+b x) d x=8$
$\begin{aligned}& \Rightarrow \quad \frac{2 a}{3}\left[x^{3 / 2}\right]_0^4+\frac{b}{2}\left[x^2\right]_0^4=8, \frac{2 a}{3} \cdot 8+8 b=8 \\& \Rightarrow \quad 2 a+3 b=3 \\&\end{aligned}$
From equation (i) and (ii), we get $a=3, b=-1$

Topic: Mathematics, Area Under Curves, JEE Main
If a curve y=fx passes through the point (1,-1) and satisfies the differential equation, y 1+xydx=x dy, then f-12 is equal to

Options:
A 25
B 45
C -25
D -45

The correct answer is:
45

Solution:
Given differential equation,
ydx+xy2dx=xdy
⇒ xdy-ydxy2=xdx
⇒ -dxy=dx22
On integrating we get,
⇒-xy=x22+C
∵ It passes through (1,-1).
∴ 1=12+C ⇒ C=12
∴ x2+1+2xy=0 ⇒ y=-2xx2+1
∴ f-12=45

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2016 (03 Apr)
If a curve y=fx passes through the point 1, 2 and satisfies xdy dx+y=bx4, then for what value of b,∫12fxdx=625?

Options:
A 315
B 10
C 5
D 625

The correct answer is:
10

Solution:
dydx+yx=bx3

I.F.=e∫1 xdx=x

So, solution of D.E. is given by

y·x=∫b·x3·xdx+c

y=cx+bx45

Passes through 1, 2

2=c+b5 …1

∫12fxdx=625

c ln x+bx52512=625

c ln 2+31 b25=625 …2

By equation 1 and 2

c=0 and b=10

Topic: Mathematics, Differential Equations, JEE Main
If a curve y=fx passes through the point 1, 2 and satisfies xdy dx+y=bx4, then for what value of b,∫12fxdx=625?

Options:
A 315
B 10
C 5
D 625

The correct answer is:
10

Solution:
dydx+yx=bx3

I.F.=e∫1 xdx=x

So, solution of D.E. is given by

y·x=∫b·x3·xdx+c

y=cx+bx45

Passes through 1, 2

2=c+b5 …1

∫12fxdx=625

c ln x+bx52512=625

c ln 2+31 b25=625 …2

By equation 1 and 2

c=0 and b=10

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If a curve y=fx, passing through the point 1, 2, is the solution of the differential equation 2x2dy=2xy+y2dx, then f12 is equal to

Options:
A 11+loge2
B 11-loge2
C 1+loge2
D -11+loge2

The correct answer is:
11+loge2

Solution:
dydx=yx+y22x2

⇒y-2dydx-1y·1x=12x2

Put -1y=t ⇒ 1y2dydx=dtdx

⇒ dtdx+1xt=12x2

This is a linear differential equation,

with Integrating Factor : e∫1xdx=elnx=x

So, solution of the linear differential equation is tx=∫12x2·xdx+C

⇒-xy=12lnx+C

The curve passes through 1,2

⇒-12=12ln1+C⇒C=-12

Hence, the particular solution to the differential equation is -xy=12lnx-12

⇒xy=1-lnx2⇒y=2x1-lnx

⇒f12=2×121-ln12=11+ln2=11+loge2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (02 Sep Shift 2)
If a curve \( y=f(x) \) passing through \( (1,2) \) satisfies the differential equation \( y(1+x y) d x-x d y=0 \), then which of the following option is correct?

Options:
A \( f(x)=\frac{2 x}{2-x^{2}} \)
B \( f(x)=\frac{x+1}{x^{2}+1} \)
C \( f(x)=\frac{x-1}{4-x^{2}} \)
D \( f(x)=\frac{4 x}{1-2 x^{2}} \)

The correct answer is:
\( f(x)=\frac{2 x}{2-x^{2}} \)

Solution:
The given differential equation is,

y(1+xy)dx-xdy=0

⇒ ydx-xdy+xy2dx=0

⇒ydx-xdyy2+xdx=0

⇒dxy+xdx=0
On integrating both sides, we get
∫dxy+∫xdx=0
⇒xy+x22=C
If the above curve passes through the point (1, 2), then,
12+12=C⇒C=1
The given curve is xy+x22=1
∴ y=2x2-x2
⇒fx=2x2-x2

Topic: Mathematics, Differential Equations, JEE Main
If a denotes the number of permutations of $x+2$ things taken all at a time, $b$ the number of permutations of $x$ things taken 11 at a time and $c$ the number of permutations of $x-11$ things taken all at a time such that $a=182 b c$, then the value of $x$ is

Options:
A $15$
B $12$
C $10$
D $18$

The correct answer is:
$12$

Solution:
We have $a={ }^{x+2} P_{x+2}=(x+2)!, b={ }^x P_{11}=\frac{x!}{(x-11)!}$
and $c={ }^{x-11} P_{x-11}=(x-11)!$
Now $a=182 b c \Rightarrow(x+2)!=182 \cdot \frac{x!}{(x-11)!}(x-11)!$
$\Rightarrow(x+2)!=182 x!\Rightarrow(x+2)(x+1)=182 \Rightarrow x=12$.

Topic: Mathematics, Permutation Combination, JEE Main
If A denotes the sum of all the coefficients in the expansion of 1-3x+10x2n and B denotes the sum of all the coefficients in the expansion of 1+x2n, then :

Options:
A A=B3
B 3 A=B
C B=A3
D A=3 B

The correct answer is:
A=B3

Solution:
We have,

A=Sum of the coefficients in the expansion of

⇒1-3x+10x2n=(1-3+10)n=(8)n=(2)3n

(putting x=1)

Now, B=Sum of the coefficients in the expansion of

⇒1+x2n=(1+1)n=2n

Clearly, A=B3.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (27 Jan Shift 1)