datavorous/entrance-exam-dataset huggingface dataset:
If y=yx is the solution curve of the differential equation dydx+ytanx=xsecx, 0≤x≤π3, y0=1, then yπ6 is equal to

Options:
A π12-32loge2e3
B π12+32loge23e
C π12-32loge23e
D π12+32loge2e3

The correct answer is:
π12-32loge2e3

Solution:
Given:

dydx+ytanx=xsecx

This is a linear differential equation.

I.F.=e∫tanxdx=secx

Then solution of differential equation is

ysecx=∫xsec2xdx

⇒ysecx=xtanx-∫tanxdx

⇒ysecx=xtanx-lnsecx+C

Given:

y0=1⇒c=1

∴ysecx=xtanx-lnsecx+1

At x=π6, we get

ysecπ6=π6tanπ6-lnsecπ6+1

⇒y23=π613-ln23+1

⇒y=π12-32ln23+32

⇒y=π12-32ln23-lne

⇒y=π12-32loge2e3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (01 Feb Shift 1)
If y=y(x) is the solution curve of the differential equation x2-4dy-y2-3ydx=0, x>2,y(4)=32 and the slope of the curve is never zero, then the value of y(10) equals :

Options:
A 31+(8)14
B 31+22
C 31-22
D 31-(8)14

The correct answer is:
31+(8)14

Solution:
Given,

x2-4dy-y2-3ydx=0

⇒∫dyy2-3y=∫dxx2-4

⇒13∫y-(y-3)y(y-3)dy=∫dxx2-4

⇒13∫1y-3-1ydy=∫dxx2-22

⇒13log|y-3|-log|y|=14logx-2x+2+C

⇒13logy-3y=14logx-2x+2+C

At x=4, y=32

⇒13log32-332=14log4-24+2+C

⇒0=14log13+C

⇒0=-log314+C

⇒C=14log3

⇒13logy-3y=14logx-2x+2+14log(3)

Putting, x=10

⇒13logy-3y=14log23+14log(3)

⇒13logy-3y=14log2

⇒logy-3y=34log2

⇒logy-3y=log234

⇒-y+3=814·y

⇒3=814+1y

⇒31+814=y

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (27 Jan Shift 2)
If y=y(x) is the solution curve of the differential equation x2 dy+y-1xdx=0; x>0 and y(1)=1, then y12 is equal to :

Options:
A 3+e
B 3-e
C 32-1e
D 3+1e

The correct answer is:
3-e

Solution:
x2dy+ydx=dxx

⇒dydx+yx2=1x3

I.F=e∫1x2dx=e-1x

⇒y·e-1x=∫e-1x·1x3dx+C

Let -1x=t⇒1x2dx=dt

⇒y·e-1x=∫-tet·dt+C

=-tet-et+C

⇒y·e-1x=1xe-1x+e-1x+C

Put x=1

⇒(1)·e-1=e-11+e-1+C

⇒C=-e-1

Equation is y·e-1x=1xe-1x+e-1x-e-1

⇒y=1x+1-e1xe

At x=12⇒y12=2+1-e2e⇒y=3-e

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (01 Sep Shift 2)
If y=yx is the solution of the differential equation 1+e2xdydx+21+y2ex=0 and y0=0, then 6y'0+ylogc32 is equal to:

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

The correct answer is:
-4

Solution:
Given,

1+e2xdydx+21+y2ex=0

⇒dy1+y2+2ex1+e2xdx=0 .........(i)

Now integrating both side we get,

⇒∫dy1+y2+∫2ex1+e2xdx=∫0

⇒tan-1y+2tan-1ex=c

∵y0=0

so, C=π2⇒tan-1 y+2 tan-1ex=π2 ....(ii)

Now from equation (i), we get dydxx=0=-1

From equation (ii) we get, y (In 3)=-13

So, 6y'0+(y In 3)2=6-1+13=-4.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (29 Jun Shift 2)
If y=yx is the solution of the differential equation 5+ex2+y⋅dydx+ex=0 satisfying y0=1 then value of y(loge13) is

Options:
A 1
B -1
C 0
D 2

The correct answer is:
-1

Solution:
Given dy2+y=-exdx5+ex

⇒∫dy2+y=-∫exdx5+ex

⇒loge(2+y)=-loge5+ex+logeC

⇒loge(2+y)=logeC5+ex

⇒y=C5+ex-2

∵y(0)=1

∴ c=18

y=185+ex-2

y(loge13)=185+eloge13-2

⇒y(loge13)=185+13-2

∴ yloge13=-1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If $y=y(x)$ is the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x), y(0)=\frac{3}{4}$, then $y\left(\frac{\pi}{8}\right)$ is equal to:

Options:
A $\mathrm{e}^{\pi / 8}$
B $\mathrm{e}^{\pi / 4}$
C $\mathrm{e}^{-\pi / 4}$
D $\mathrm{e}^{-\pi / 8}$

The correct answer is:
$\mathrm{e}^{-\pi / 4}$

Solution:
$\begin{aligned} & \frac{d y}{d x}+2 y=\sin 2 x, y(0)=\frac{3}{4} \\ & \text { I.F }=e^{\int 2 d x}=e^{2 x} \\ & y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x \\ & y \cdot e^{2 x}=\frac{e^{2 x}(2 \sin 2 x-2 \cos 2 x)}{4+4}+C \\ & x=0, y=\frac{3}{4} \Rightarrow \frac{3}{4} \cdot 1=\frac{1(0-2)}{8}+C \\ & \frac{3}{4}=-\frac{1}{4}+C \\ & 1=C \\ & y=\frac{2 \sin 2 x-2 \cos 2 x}{8}+1 \cdot e^{-2 x} \\ & x=\frac{\pi}{8}, y=\frac{1}{8}\left(2 \sin \frac{\pi}{4}-2 \cos \frac{\pi}{4}\right)+e^{-2\left(\frac{\pi}{8}\right)} \\ & y=0+e^{-\frac{\pi}{4}}\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (05 Apr Shift 1)
If y=yx is the solution of the differential equation, dydx+2ytanx=sinx,yπ3=0, then the maximum value of the function yx over R is equal to :

Options:
A 8
B 12
C -154
D 18

The correct answer is:
18

Solution:
dydx+2ytanx=sinx

I.F.=e∫2tanxdx=e2lnsecx

I.F.=sec2x

y.sec2x=∫sinx.sec2xdx+C

y.sec2x=∫secxtanxdx+C

y.sec2x=secx+C

x=π3;y=0

⇒C=-2

⇒y=secx-2sec2x=cosx-2cos2x

Let cosx=t, -1≤t≤1

⇒y=t-2t2⇒dydt=1-4t=0⇒t=14

Second-order derivative is negative

∴ max=14-18=2-18=18

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (16 Mar Shift 1)
If y=y(x) is the solution of the differential equation dydx+4xx2-1y=x+2x2-152,x>1 such that y(2)=29loge2+3 and y2=αlogeα+β+β-γ,α,β,γ∈ℕ, then αβγ is equal to

Options:

The correct answer is:
6

Solution:
Given differential equation is dydx+4xx2-1y=x+2x2-152,x>1

Now IF=e∫4xx2-1dx=x2-12

The required equation will be ⇒y·x2-12=∫x+2x2-11/2dx

⇒y·x2-12=12∫2xx2-11/2dx+2∫dxx2-11/2

⇒y·x2-12=2lnx2-1+x+x2-1+C

Now using, at x=2, y(2)=29loge2+3 we get,

⇒9·29ln2+3=2ln2+3+3+C

⇒C=-3

Now finding the value of function at x=2 we get,

y×1=2ln1+2+1-3

Now on comparing we get,

⇒β=1, α=2, γ=3

⇒αβγ=1×2×3=6

Hence, this is the required answer.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (13 Apr Shift 2)
If y=y(x) is the solution of the differential equation dydx=(tanx-y)sec2x , x∈-π2, π2 , such that y0=0, then y-π4 is equal to:

Options:
A 1e-2
B 2+1e
C e-2
D 12-e

The correct answer is:
e-2

Solution:
The given differential equation can be written as
dydx+ysec2x=tanx.sec2x
Integrating factor =e∫sec2xdx=etanx
Hence, solution of given differential equation,
y.etanx=∫tanx.sec2x.etanxdx
⇒y.etanx=tanx.etanx-∫sec2x.etanxdx (using integration by parts)

⇒y.etanx=tanx.etanx-etanx+c

Given, y0=0⇒c=1

Solution of given differential is

y.etanx=tanx.etanx-etanx+1

Hence, y-π4=-1e-1-e-1+1e-1

⇒y-π4=e-2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If y=yx is the solution of the differential equation dydx+tanxy=sinx,0≤x≤π3, with y0=0, then yπ4 is equal to

Options:
A 14loge2
B

122loge2

C loge2
D 12loge2

The correct answer is:

122loge2

Solution:
dydx+tanxy=sinx; 0≤x≤π3

I.F.=e∫tanxdx=elnsecx=secx

ysecx=∫tanxdx+C

ysecx=lnsecx+C

x=0, y=0 ⇒ C=0

ysecx=lnsecx

y=cosx·lnsecx

yx=π4=12·ln2

yx=π4=122loge2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (16 Mar Shift 2)
If y=yx is the solution of the differential equation ,eydydx-1=ex such that y0=0, then y1 is equal to

Options:
A 1+loge2
B 2+loge2
C 2e
D loge2

The correct answer is:
1+loge2

Solution:
ey=t
eydydx=dtdx
dtdx-t=ex
IF=e-∫dx

⇒t·e-x=x+1
ey-x=x+1
y=x+ln⁡x+1
at x=1,y=1+ln⁡2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If y=y(x) is the solution of the differential equation, xdydx+2y=x2 satisfying y1=1, then y12 is equal to

Options:
A 764
B 14
C 1316
D 4916

The correct answer is:
4916

Solution:
Given differential equation can be written as
dydx+2yx=x , which is a linear differential equation.
If =e∫2xdx=e2 ln x=x2
Solution is
y.x2=∫x.x2dx
⇒y.x2=x44+c
Since, given curve passes through 1,1
⇒c=34
Hence, yx=x24+34x2.
So, y12=4916.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If y=yx is the solution of the differential equation xdydx+2y=xex,y1=0 then the local maximum value of the function zx=x2y(x)-ex,x∈R is

Options:
A 1-e
B 0
C 12
D 4e-e

The correct answer is:
4e-e

Solution:
Given,

dydx+2yx=ex

I.F.=e∫2xdx=e2lnx=x2

∴ General solution: yx2=∫exx2dx

We know that

∫exfxdx==exfx-f'(x)+f''x-f'''x+⋯+-1nfnx+C

So, yx2=exx2-2x+2+C ...i

Given y1=0⇒01=e11-21+2+C

⇒C=-e

∴y=exx2x2-2x+2-e (from eq i)

Hence, z(x)=x2exx2x2-2x+2-e-ex

=exx2-2x+2-e-ex

zx=exx2-2x+1-e

z'x=ex2x-2+x2-2x+1ex

z'x=exx2-1
To find local maxima, put z'x=0

⇒exx2-1=0 ⇒ x-1x+1=0 ∵ ex>0, ∀x∈R

⇒x=1,-1

It is clear from the sign scheme method, zx has local maximum value at x=-1 and has local minimum value at x=1.

∴ The local maximum value is

z-1=e-1-12-2-1+1-e

=1e1+2+1-e =4e-e

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If y=yx, y∈0,π2 is the solution of the differential equation secydydx-sinx+y-sinx-y=0, with y0=0, then 5y'π2 is equal to _____.

Options:

The correct answer is:
2

Solution:
Given,

secydydx-sinx+y-sinx-y=0

⇒secydydx=2sinxcosy

⇒sec2ydy=2sinxdx

⇒∫sec2ydy=∫2sinxdx

⇒tany=-2cosx+c

Given, y0=0

So, 0=-2cos0+c

⇒c=2

So, the particular solution is

tany=-2cosx+2

at x=π2,tany=-2cosπ2+2

⇒tany=2⇒sec2y=5

Here, differentiate both sides of tany=-2cosx+2, we get,

sec2ydydx=2sinx

At x=π2

5dydx=2sinπ2

⇒5dydx=2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (27 Jul Shift 1)
If $\left|\begin{array}{lll}y+z & x-z & x-y \\ y-z & z-x & y-x \\ z-y & z-x & x+y\end{array}\right|=k x y z$ then the value of $k$ is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
$\left|\begin{array}{ccc}y+z & x-z & x-y \\ y-z & z+x & y-x \\ z-y & z-x & x+y\end{array}\right|=\left|\begin{array}{ccc}y+z & x-z & x-y \\ 2 y & 2 x & 0 \\ 2 z & 0 & 2 x\end{array}\right|$ $R_2 \rightarrow R_2+R_1$ and $R_3 \rightarrow R_3+R_1$
$\begin{aligned} & =4\left|\begin{array}{ccc}y+z & x-z & x-y \\ y & x & 0 \\ z & 0 & x\end{array}\right| \\ & =4\left[(y+z)\left(x^2\right)-(x-z)(x y)+(x-y)(-z x)\right] \\ & =4\left[x^2 y+z x^2-x^2 y+x y z-z x^2+x y z\right]=8 x y z\end{aligned}$
Hence, $k=8$

Topic: Mathematics, Determinants, JEE Main
If \( y=a \log |x|+b x^{2}+x \) has its extremum values at \( x=-1 \) and \( x=2 \) then find value of \( a+b \).

Options:

The correct answer is:
1.50

Solution:
dydx=ax+2bx+1 or dydx=0 at x=-1, 2

a-1+2b-1+1=0 or -a-2b+1=0

a2+4b+1=0 or a+8b+2=0

solving we get ; a=2, b=-12

Topic: Mathematics, Application of Derivatives, JEE Main
If yellow light emitted by sodium lamp in Young's double-slit experiment is replaced by a monochromatic blue light of the same intensity

Options:
A fringe width will decrease
B fringe width will increase
C fringe width will remain unchanged
D fringe width will become less intense

The correct answer is:
fringe width will decrease

Solution:
As β=λDd and λb<λy
∴ Fringe width β will decrease

Topic: Physics, Wave Optics, NEET
If \(y=\operatorname{cosec}^{-1}(x)\) and \(\frac{d y}{d x}=\frac{-1}{|x| \sqrt{x^2-1}}\), then

Options:
A \(y \in\left(-\frac{\pi}{2}, 0\right)\)
B \(y \in\left(-\frac{\pi}{2}, 2 \pi\right)\)
C \(y \in\left(-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right)\)
D \(y \in R\)

The correct answer is:
\(y \in\left(-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right)\)

Solution:
It is given that, \(y=\operatorname{cosec}^{-1} x\)
\(\text {and } \frac{d y}{d x}=\frac{-1}{|x| \sqrt{x^2-1}}\)
\(\because\) Domain of \(\operatorname{cosec}^{-1} x\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\) and the \(\frac{d y}{d x}\) is define for \(x \in\left(-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right)\). Because for derivatives we should exclude the end points of a internal.
Hence, option (c) is correct.

Topic: Mathematics, Differentiation, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 1)
If you spill a chemical toilet cleaning liquid on your hand, your first aid would be:

Options:
A vinegar
B aqueous NaOH
C aqueous NaHCO3
D aqueous NH3

The correct answer is:
aqueous NaHCO3

Solution:
In toilet cleaning liquid the main constituent isHCl, which can cause skin burn so it should be treated with NaHCO3 which can easily consume the acid.

If a few drops of a concentrated acid accidentally spill over the hand of a student, wash the hand immediately with plenty of water and apply a paste of sodium hydrogen carbonate.
Water will dilute the acid and wash it out. Any remaining acid will be neutralized with sodium hydrogen carbonate. This will minimize the effect of acid and damage to the skin.

Other given substances are not good for skin.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2020 (02 Sep Shift 2)
If Young's double slit experiment is performed in water instead of air, then

Options:
A no fringes would be seen
B fringe width would decrease
C fringe width would increase
D fringe width would remain unchanged

The correct answer is:
fringe width would decrease

Solution:
Fringe width, $\beta=\frac{D \lambda}{d}$ or $\beta \propto \lambda$
If the experiment is performed in water, wavelength $\lambda$ decreases, so, fringe width also decreases.

Topic: Physics, Wave Optics, JIPMER, JIPMER 2010
If Young's double slit experiment is performed in water instead of air

Options:
A The fringe width will decrease
B The fringe width will increase
C The fringe width will remain unchanged
D There will be no fringe

The correct answer is:
The fringe width will decrease

Solution:
As we know
Î² = D d Î» .....(i)
and Î» â 1 Î¼ .....(ii)
From eqs. (i) and (ii),
Î² â Î» â 1 Î¼
β 1 = β μ

Topic: Physics, Wave Optics, NEET
If Young's double slit experiment, is performed in water

Options:
A the fringe width will decrease
B the fringe width will increase
C the fringe width will remain unchanged
D there will be no change

The correct answer is:
the fringe width will decrease

Solution:
As we know
$$
\beta=\frac{D \lambda}{d}
$$
and
$$
\lambda \propto \frac{1}{\mu}
$$
From Eqs. (i) and (ii)
$\therefore$
$$
\begin{array}{l}
\beta \propto \lambda \propto \frac{1}{\mu} \\
\beta \propto \frac{1}{\mu}
\end{array}
$$
Hence, (a) is correct.

Topic: Physics, Thermodynamics, MHT CET, MHT CET 2007
If \( y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right) \), then \( \frac{d y}{d x} \) is equal to

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

The correct answer is:
\( \frac{1}{\sqrt{1-x^{2}}} \)

Solution:
Given,

y=sin-15x+121-x213

On putting x=sinθ,5=rcosα and 12=rsinα, we get
y=sin-1rcosαsinθ+rsinαcosθ13
=sin-113sinθ+α13 ;∵ r=25+144=13
=θ+α

=sin-1x+tan-1125 ;∵ tanα=125

Differentiating both sides w.r.t. x, we get

dydx=11-x2

Topic: Mathematics, Differentiation, JEE Main
If \(y=\sin ^{-1}[\sqrt{x-a x}-\sqrt{a-a x}]\) then \(\frac{d y}{d x}=\)

Options:
A \(\frac{1}{\sin \sqrt{a-a x}}\)
B \(\sin \sqrt{\mathrm{x}} \cdot \sin \sqrt{\mathrm{a}}\)
C \(\frac{1}{2 \sqrt{\mathrm{x}} \sqrt{1-\mathrm{x}}}\)
D zero

The correct answer is:
\(\frac{1}{2 \sqrt{\mathrm{x}} \sqrt{1-\mathrm{x}}}\)

Solution:
Put \(x=\sin ^2 \theta\) and \(a=\sin ^2 \phi\). then
\(\begin{aligned}
& y=\sin ^{-1}(\sin \theta \cos \phi-\cos \theta \sin \phi) \\
& =\sin ^{-1} \sqrt{x}-\sin ^{-1} \sqrt{a} \\
& \therefore \quad \frac{d y}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}
\end{aligned}\)

Topic: Mathematics, Differentiation, JEE Main
If \(y=x^{x^2}\), then \(\frac{d y}{d x}\) is equal to

Options:
A \((2 \ln x)\)
B \((2 \ln x+1)\)
C \((\ln \ln x+1) x^{x^2}\)
D None of these

The correct answer is:
None of these

Solution:
Consider \(y=x^{x^2} \Rightarrow \ln y=x^2 \ln x\)
\(\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=2 x \ln x+x^2 \cdot \frac{1}{x}=x(1+2 \ln x) \\
& \frac{d y}{d x}=x^{x^2} \cdot x(1+2 \ln x)=x^{x^2+1}(1+2 \ln x)
\end{aligned}\)

Topic: Mathematics, Differentiation, JEE Main
If z≠0 be a complex number such that z-1z=2, then the maximum value of z is

Options:
A 2
B 1
C 2-1
D 2+1

The correct answer is:
2+1

Solution:
Given z-1z=2

We know that z-1z≤z-1z≤z+1z Let z=r

i.e. r-1r≤2 and

r+1r≥2 which is always true.

Now r-1r≤2

⇒r-1r≥-2 & r-1r≤2

⇒r2-1≤2r

⇒r-12≤2

⇒r-1≤2

∴ zmax=1+2

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (29 Jul Shift 2)
If $Z_1 \neq 0$ and $Z_2$ be two complex numbers such that $\frac{Z_2}{Z_1}$ is a purely imaginary number, then $\left|\frac{2 Z_1+3 Z_2}{2 Z_1-3 Z_2}\right|$ is equal to:

Options:
A
2

B
5

C
3

D
1

The correct answer is:
1

Solution:
Let $z_1=1+i$ and $z_2=1-i$
$\frac{z_2}{z_1}=\frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1+i)(1-i)}=-i$
$\frac{2 z_1+3 z_2}{2 z_1-3 z_2}=\frac{2+3\left(\frac{z_2}{z_1}\right)}{2-3\left(\frac{z_2}{z_1}\right)}=\frac{2-3 i}{2+3 i}$
$\left|\frac{2 z_1+3 z_2}{2 z_1-3 z_2}\right|=\left|\frac{2-3 i}{2+3 i}\right|=\left|\frac{2-3 i}{2+3 i}\right|$
$\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right]$
$=\frac{\sqrt{4+9}}{\sqrt{4+9}}=1$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2013 (09 Apr Online)
If z1=1-2i, z2=1+i and z3=3+4i, then 1z1+2z2z3z2=

Options:
A 72
B 52
C 452
D 152

The correct answer is:
452

Solution:
1z1+2z2z3z2

=11-2i+21+iz3z2

=1+2i1-2i1+2i+21-i1+i1-i32+4212+12

=1+2i5+2+2i252

=5215+2i5+1-i

=126-3i=452

Topic: Mathematics, Complex Number, JEE Main
If z=12-2i, is such that |z+1|=αz+β(1+i),i=-1 and α,β∈R, then α+β is equal to

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

The correct answer is:
3

Solution:
Given: z+1=αz+β1+i, z=12-2i

⇒12-2i+1=α12-2i+β1+i

⇒32-2i=α2-2αi+β+βi

⇒94+4=α2+β+i-2α+β

Now, on comparing real and imaginary part we get,

⇒52=α2+β+i-2α+β

⇒2α=β, 52=α2+β

⇒2α=β, 52=5α2

⇒α=1, β=2

⇒α+β=3

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If z=123-i and the least positive integral value of n such that z101+i109106=zn is k, then the value of 25k is equal to

Options:

The correct answer is:
4

Solution:
z=-12i1+i3=iÏ2
z101=iÏ
z101+i109106=iÏ+i106=i(âÏ2)106=âÏ2
as given that z101+i109106=zn
â´-Ï2=iÏ2n=inÏ2n
Ï2n-2in=-1
this is possible only when n=4r+2 and 2nâ2 is a multiple of 3 i.e.,
24r+2-2 is a multiple of 3
i.e., 8r+2 is a multiple of 3 âr=2
â´n=10 â´25k=4

Topic: Mathematics, Complex Number, JEE Main
If z1=2+3i, z2=3-2i and z3=-1-23i, then which of the following is true? (where, i2=-1)

Options:
A argz2z3=argz2-z1z3-z1
B argz2z3=argz3z1
C 12argz2z3=argz2-z1z3-z1
D 2argz3z2=argz3-z1z2-z1

The correct answer is:
12argz2z3=argz2-z1z3-z1

Solution:
Here, z1=z1=z3=13
Hence, z1,z2,z3 lie on a circle with center 0,0
and r=13 as shown
Now, argz2z3=2argz2-z1z3-z1
â´ argz3z2=2argz3-z1z2-z1

Topic: Mathematics, Complex Number, JEE Main
$\text { If } z_{1}=\sqrt{2}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right] \text { and } z_{2}=\sqrt{3}\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right]$ then $\left|z_{1} z_{2}\right|$ is equal to $\sqrt{m}$. Value of $m$ is

Options:
A 6
B 3
C 2
D 5

The correct answer is:
6

Solution:
$z_{1}=\sqrt{2}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right]=\sqrt{2}\left[\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right]=1+i$

$\left|z_{1}\right|=\sqrt{2}$

and $z_{2}=\sqrt{3}\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right]=\sqrt{3}\left[\frac{1}{2}+i \frac{\sqrt{3}}{2}\right]$

$\left|z_{2}\right|=\sqrt{\frac{3}{4}+\frac{9}{4}}=\sqrt{3}$

$\left|z_{1} z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|=\sqrt{2} \cdot \sqrt{3}=6$

Topic: Mathematics, Complex Number, JEE Main
If $Z_1=2+i$ and $Z_2=3-4 i$ and $\frac{\overline{Z_1}}{Z_1}=a+b i$, then the value of $-7 a+b$ is (where $i=\sqrt{-1}$ and $\mathrm{a}, \mathrm{b} \in \mathbb{R})$

Options:
A 1
B -1
C $\frac{-3}{25}$
D $\frac{-9}{25}$

The correct answer is:
-1

Solution:
$\begin{aligned} & \mathrm{Z}_1=2+\mathrm{i} \quad \mathrm{Z}_2=3-4 \mathrm{i} \\ & \overline{Z_1}=2-\mathrm{i} \quad \overline{Z_2}=3+4 \mathrm{i} \\ & \frac{\overline{Z_1}}{\overline{Z_2}}=\frac{2-\mathrm{i}}{3+4 \mathrm{i}} \\ & =\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)} \\ & =\frac{6-8 i-3 i+4 i^2}{(3)^2-(4 i)^2} \\ & =\frac{6-1 \mathrm{li}-4}{9+16} \quad \ldots\left[\mathrm{i}^2=-1\right] \\ & \mathrm{a}+\mathrm{bi}=\frac{2-1 \mathrm{li}}{25} \\ & \therefore \quad a=\frac{2}{25}, b=\frac{-11}{25} \\ & \end{aligned}$
$\begin{aligned} & \text { Now }-7 \mathrm{a}+\mathrm{b} \\ & =-7\left(\frac{2}{25}\right)-\frac{11}{25} \\ & =\frac{-14-11}{25} \\ & =\frac{-25}{25}=-1\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If z(1+a)=b+ic and a2+b2+c2=1, then 1+iz 1−iz =

(where, a,b,c∈R and i2=-1)

Options:
A a+ib 1+c
B b−ic 1+a
C a+ic 1+b
D None of these

The correct answer is:
a+ib 1+c

Solution:
1+iz 1−iz = 1+i(b+ic)/(1+a) 1−i(b+ic)/(1+a) = 1+a−c+ib 1+a+c−ib = (1+a−c+ib)(1+a+c+ib) (1+a+c) 2 + b 2 = 1+2a+ a 2 − b 2 − c 2 +2ib+2iab) 1+ a 2 + c 2 + b 2 +2ac+2(a+c) = a 2 + b 2 + c 2 +2a+ a 2 − b 2 − c 2 +2ib(1+a) 1+1+2ac+2(a+c) = 2a(a+1)+2ib(1+a) 2(1+a)(1+c) = a+ib 1+c .

Topic: Mathematics, Complex Number, JEE Main
If $|z|=1$ and $z \neq \pm 1$, then all the points representing $\frac{z}{1-z^{2}}$ lie on;

Options:
A a line not passing through the origin
B the line $y=x$
C the $x$-axis
D the $y$-axis

The correct answer is:
the $y$-axis

Solution:
Let $z=e^{i \theta}, \theta \neq \pm n \pi ~ n \in I$
Let $w=\frac{e^{i \theta}}{1-e^{i 2 \theta}}=\frac{1}{e^{-i \theta}-e^{i \theta}}=\frac{1}{-2 \cdot i \sin \theta}$

$\therefore$ Locus is y-axis

Topic: Mathematics, Complex Number, JEE Main
If $z \neq 1$ and $\frac{z^2}{z-1}$ is real, then the point represented by the complex number $z$ lies

Options:
A
either on the real axis or on a circle passing through the origin

B
on a circle with centre at the origin

C
either on the real axis or on a circle not passing through the origin

D
on the imaginary axis

The correct answer is:
either on the real axis or on a circle passing through the origin

Solution:
Let $z=x+$ iy $(\because x \neq 1$ as $z \neq 1)$
$z^2=\left(x^2-y^2\right)+i(2 x y)$
$\frac{z^2}{z-1}$ is real
$\Rightarrow$ its imaginary part $=0$
$\Rightarrow 2 x y(x-1)-y\left(x^2-y^2\right)=0$
$\Rightarrow y\left(x^2+y^2-2 x\right)=0$
$\Rightarrow y=0 ; x^2+y^2-2 x=0$
$\therefore \mathrm{z}$ lies either on real axis or on a circle through origin.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2012 (Offline)
If z1 and z2 are two distinct complex numbers satisfying the relation z12-z22=z-12+z-22-2z-1z-2 and arg⁡z1-arg⁡z2=aπb, then the least possible value of a-b is equal to (where, a & b are integers)

Options:

The correct answer is:
1

Solution:
z12-z22=z-12+z-22-2z-1z-2
z12-z22=z12+z22-2z1z2
z1+z2z1-z2=z1-z2z1-z2
⇒z1+z2=z1-z2
z1¯ ⊥ z2¯
⇒arg⁡z1z2=2nπ±π2
i.e. -π2,π2,3π2,……..
The minimum value of a-b=1 (when a=1 & b=2)

Topic: Mathematics, Complex Number, JEE Main
If $z_1$ and $z_2$ are two non-zero complex numbers such that $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$ then $\arg z_1-\arg z_2$ is equal to

Options:
A
$\frac{\pi}{2}$

B
$-\pi$

C
0

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

The correct answer is:
0

Solution:
$\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right| \Rightarrow z_1$ and $z_2$ are collinear and are to the same side of origin; hence $\arg z_1-\arg z_2=0$.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2005
If $z=\frac{4}{1-i}$, then $\bar{z}$ is (where $\bar{z}$ is complex conjugate of $z$ )

Options:
A $2(1+i)$
B $(1+i)$
C $\frac{2}{1-\mathrm{i}}$
D $\frac{4}{1+\mathrm{i}}$

The correct answer is:
$\frac{4}{1+\mathrm{i}}$

Solution:
Hints: $z=\frac{4}{1-i}$
$$
\overline{\mathrm{z}}=\frac{4}{1+\mathrm{i}}
$$

Topic: Mathematics, Complex Number, JEE Main
If z+1z+1=0, then z2003+1z2003 is equal to

Options:
A 1
B −1
C 0
D None of these

The correct answer is:
−1

Solution:
Given equation can be written as z 2 +z+1=0
â z=Ï, Ï2 roots
Now, z2003=Ï2003=Ï2
z2003+1z2003=Ï2+1Ï2=Ï2+Ï=-1

Topic: Mathematics, Complex Number, JEE Main
If z1, z2 and z3 are 3 distinct complex numbers such that 3z1-z2=5z2-z3=7z3-z1, then the value of 9z1-z2+25z2-z3+49z3-z1 is equal to

Options:
A 0
B 1
C -1
D 15

The correct answer is:
0

Solution:
Let, 3z1-z2=5z2-z3=7z3-z1=k
9z1-z22=25z2-z32=49z3-z12=k2
⇒9z1-z2z-1-z-2=k2⇒9z1-z2=k2z-1-z-2
similarly, 25z2-z3=k2z-2-z-3
and 49z3-z1=k2z-3-z-1
⇒9z1-z2+25z2-z3+49z3-z1=k2z-1-z-2+z-2-z-3+z-3-z-1=0

Topic: Mathematics, Complex Number, JEE Main
If z1, z2 and z3 are the vertices of a triangle in the argand plane such that z1-z2=z1-z3, then argâ¡2z1-z2-z3z3-z2 is

Options:
A π3
B 0
C π2
D π6

The correct answer is:
π2

Solution:
z1-z2z3-z2=z1-z2z3-z2eiθz1-z2z2-z3=z1-z3z2-z3e-iθ
⇒z1-z2z3-z2+z1-z3z3-z2= purely imaginary number
arg2z1-z2-z3z3-z2=±π2

Topic: Mathematics, Complex Number, JEE Main
If z1, z2 and z3 are three complex numbers such that z 1 = z 2 = z 3 = 1 z 1 + 1 z 2 + 1 z 3 = 1 , then z1+z2+z3 is

Options:

The correct answer is:
1

Solution:
z 1 = z 2 = z 3 = 1 (given)
Now z1=1⇒z12=1⇒z1z1¯=1 ⇒1z1=z1¯
Similarly, 1z2=z2¯ & 1z3=z3¯
Now 1 z 1 + 1 z 2 + 1 z 3 = 1
⇒z1+z2+z3¯=1 (∵z1¯+z2¯=z1+z2¯)
and |z1+z2+z3¯|=1  (∵|z1¯|=|z|)
⇒ z 1 + z 2 + z 3 = 1

Topic: Mathematics, Complex Number, JEE Main