datavorous/entrance-exam-dataset huggingface dataset:
Let α,β,γ be the real roots of the equation, x3+ax2+bx+c=0, (a,b,c∈R and a,b≠0). If the system of equations (in, u,v,w) given by αu+βv+γw=0, βu+γv+αw=0, γu+αv+βw=0 has non-trivial solution, then the value of a2b is

Options:
A 5
B 3
C 1
D 0

The correct answer is:
3

Solution:
Equation x3+ax2+bx+c=0 has roots α , β , γ. Therefore,

α+β+γ=-a

αβ+βγ+γα=b

Since the given system of equations has non-trivial solutions,

we have

αβγβγαγαβ=0

or α3+β3+γ3-3αβγ=0

or α+β+γ α2+β2+γ2-αβ-βγ-γα=0

or α+β+γ α+β+γ2-3αβ+βγ+γα=0

⇒ -aa2-3b=0

or a2/b=3

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Let $\alpha, \beta$ be the roots of the equation $a x^{2}+b x+c=0$. A root of the equation $a^{3} x^{2}+a b c x+c^{3}=0$ is

Options:
A $\alpha+\beta$
B $\alpha^{2}+\beta$
C $\alpha^{2}-\beta$
D $\alpha^{2} \beta$

The correct answer is:
$\alpha^{2} \beta$

Solution:
We have, $a^{3} x^{2}+a b c x+c^{3}=0$
$\begin{aligned}
&\Rightarrow x^{2}+\frac{b}{a} \cdot \frac{c}{a} x+\left(\frac{c}{a}\right)^{3}=0 \\
&\Rightarrow x^{2}-(\alpha+\beta) \alpha \beta x+\alpha^{3} \beta^{3}=0 \\
&\Rightarrow x=\alpha^{2} \beta, \alpha \beta^{2} \Rightarrow \alpha^{2} \beta \text { is one of the root. }
\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main
Let $\alpha, \beta$ be the roots of the equation $x^2+2 \sqrt{2} x-1=0$. The quadratic equation, whose roots are $\alpha^4+\beta^4$ and $\frac{1}{10}\left(\alpha^6+\beta^6\right)$, is :

Options:
A $x^2-190 x+9466=0$
B $x^2-180 x+9506=0$
C $x^2-195 x+9506=0$
D $x^2-195 x+9466=0$

The correct answer is:
$x^2-195 x+9506=0$

Solution:
$\begin{aligned} & x^2+2 \sqrt{2} x-1=0 \\ & \alpha+\beta=-2 \sqrt{2} \\ & \alpha \beta=-1 \\ & \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2 \\ & =\left((\alpha+\beta)^2-2 \alpha \beta\right)^2-2(\alpha \beta)^2 \\ & =(8+2)^2-2(-1)^2 \\ & =100-2=98 \\ & \alpha^6+\beta^6=\left(\alpha^3+\beta^3\right)^2-2 \alpha^3 \beta^3 \\ & =\left((\alpha+\beta)\left((\alpha+\beta)^2-3 \alpha \beta\right)^2-2(\alpha \beta)^3\right.\end{aligned}$
$\begin{aligned} & =(-2 \sqrt{2}(8+3))^2+2 \\ & =(8)(121)+2=970 \\ & \frac{1}{10}\left(\alpha^6+\beta^6\right)=97 \\ & x^2-(98+97) x+(98)(97)=0 \\ & \Rightarrow x^2-195 x+9506=0\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let α, β be the roots of the equation x2-2x+2=0 Then α14+β14 is equal to

Options:
A -64
B -642
C -128
D -1282

The correct answer is:
-128

Solution:
Given,

x2-2x+2=0

⇒x=2±6i2

⇒x=21±3i2

⇒x=212±3i2

⇒x=2cosπ3±isinπ3

So, α=2cosπ3+isinπ3=2eiπ3

And β=2cosπ3-isinπ3=2e-iπ3

Now α14+β14=2eiπ314+2e-iπ314

⇒α14+β14=214ei14π3+e-i14π3

⇒α14+β14=214cos14π3+isin14π3+cos14π3-isin14π3

⇒α14+β14=272cos14π3

⇒α14+β14=28cos4π+2π3

⇒α14+β14=28cos2π3

⇒α14+β14=-28×12=-128

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let $\alpha, \beta ; \alpha>\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let $\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$. Then $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to

Options:
A $10 \sqrt{3} \mathrm{P}_9$
B $11 \sqrt{3} P_9$
C $10 \sqrt{2} \mathrm{P}_9$
D $11 \sqrt{2} \mathrm{P}_9$

The correct answer is:
$10 \sqrt{3} \mathrm{P}_9$

Solution:
$\begin{aligned}
& \mathrm{x}^2-\sqrt{2 \mathrm{x}}-\sqrt{3}=0\left\langle_\beta^\alpha\right. \\
& \alpha^{\mathrm{n}+2}-\sqrt{2} \alpha^{\mathrm{n}+1}-\sqrt{3} \alpha^{\mathrm{n}}=0 \\
& \text { and } \beta^{\mathrm{n}+2}-\sqrt{2} \beta^{\mathrm{n}+1}-\sqrt{3} \beta^{\mathrm{n}}=0
\end{aligned}$
Subtracting
$\begin{aligned}
& \left(\alpha^{\mathrm{n}+2}-\beta^{\mathrm{n}+2}\right)-\sqrt{2}\left(\alpha^{\mathrm{n}+1}-\beta^{\mathrm{n}+1}\right)-\sqrt{3}\left(\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}\right)=0 \\
& \Rightarrow \mathrm{P}_{\mathrm{n}+2}-\sqrt{2} \mathrm{P}_{\mathrm{n}+1}-\sqrt{3} \mathrm{P}_{\mathrm{n}}=0
\end{aligned}$
$\begin{aligned} & \text { Put } \mathrm{n}=10 \\ & \mathrm{P}_{12}-\sqrt{2} \mathrm{P}_{11}-\sqrt{3} \mathrm{P}_{10}=0 \\ & \mathrm{n}=9 \\ & \mathrm{P}_{11}-\sqrt{2} \mathrm{P}_{10}-\sqrt{3} \mathrm{P}_9=0 \\ & 11\left(\sqrt{3} \cdot \mathrm{P}_{10}+\sqrt{2} \mathrm{P}_{11}-\mathrm{P}_{11}\right)-10\left(\sqrt{2} \mathrm{P}_{10}-\mathrm{P}_{11}\right) \\ & =0-10\left(-\sqrt{3} \mathrm{P}_9\right)=10 \sqrt{3} \mathrm{P}_9\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let α,β be the roots of the equation x2-2x+6=0 and 1α2+1,1β2+1 be the roots of the equation x2+ax+b=0. Then the roots of the equation x2-a+b-2x+a+b+2=0 are :

Options:
A non-real complex numbers
B real and both negative
C real and both positive
D real and exactly one of them is positive

The correct answer is:
real and both negative

Solution:
Given α,β be the roots of the equation x2-2x+6=0

So sum of roots will be α+β=2 and product of roots will be αβ=6

And also given 1α2+1 and 1β2+1 are roots of x2+ax+b=0

So sum of roots will be -a=1α2+1+1β2+1

⇒a=-1α2-1β2-2 .....1

And similarly product of roots will be,

b=1α2+1β2+1+1α2β2 ....2

Now adding equation 1 & 2 we get,

a+b=1αβ2-1=16-1=-56 {as αβ=6}

Now putting the value of a+b in x2-a+b-2x+a+b+2=0

⇒x2--56-2x+2-56=0

⇒6x2+17x+7=0

⇒x=-73,x=-12 are the roots, both roots are real and negative.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let α,β be the roots of the equation x2-4λx+5=0 and α,γ be the roots of the equation x2-32+23x+7+3λ3=0. If β+γ=32, then α+2β+γ2 is equal to

Options:

The correct answer is:
98

Solution:
Given α & β are roots of x2-4λx+5=0

So, α+β=4λ and αβ=5 .......(i)

And α & γ are roots of x2-32+23x+7+3λ3=0

So, α+y=32+23 and αγ=7+3λ3 .........(ii)

Given, if β+γ=32

So, from equation (i) and (ii) we get, α=2λ+3 and

β=2λ-3 , Now by product of roots we get, 4λ2-3=5⇒λ=2

∴ α+2β+λ2= α+β+β+λ2=4λ+322

=42+322=722=98

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let α,β be the roots of the equation x2-x+2=0 with Im(α)>Im(β). Then α6+α4+β4-5α2 is equal to

Options:

The correct answer is:
13

Solution:
Given equation x2-x+2 has roots α & β,

So, α+β=1, αβ=2

⇒α4+β4=α2+β22-2α2β2

⇒α4+β4=α+β2-2αβ2-2α2β2

⇒α4+β4=1-42-2×4

⇒α4+β4=1

It is given that, α is a root of x2-x+2.

⇒α2-α+2=0

⇒α2=α-2

⇒α4=α2+4-4α

⇒α4=α-2+4-4α

⇒α4=2-3α

Now, α6-5α2=α2α4-5

⇒α6-5α2=α-22-3α-5

⇒α6-5α2=α-2-3α-3

⇒α6-5α2=-3α2-α-2

⇒α6-5α2=-3α-2-α-2

⇒α6-5α2=12

⇒α6+α4+β4-5α2=13

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let α, β be the roots of the quadratic equation x2+6x+3=0. Then α23+β23+α14+β14α15+β15+α10+β10 is equal to

Options:
A 81
B 9
C 72
D 729

The correct answer is:
81

Solution:
To find the value of α23+β23+α14+β14α15+β15+α10+β10,

Let an=αn+βn

Hence,

α23+β23+α14+β14α15+β15+α10+β10=a23+a14a15+a10

Now, x2+6x+3=0 has roots α & β

So, x=-6±-62

⇒x=6-1±i2

⇒x=3-1±i2

Hence, α=3 ei3π4 and β=3ei5π4

Now, solving α23+β23+α14+β14α15+β15+α10+β10

=323ei23×3π4+ei23×5π4+314ei14×3π4+ei14×5π4315ei15×3π4+ei15×5π4+310ei10×3π4+ei×10×5π4

=3439ei23×3π4+ei23×5π4+ei14×3π4+ei14×5π435ei15×3π4+ei15×5π4+ei10×3π4+ei×10×5π4

=9×391+i-1+i2+0351+i-1+i2+0as ei21π2+ei35π2=0+isin21π2+0+isin35π2=i-i=0

=9×392i2352i2

=9×34

=81

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let α,βα>β be the roots of the quadratic equation x2-x-4=0. If Pn=αn-βn,n∈ℕ, then P15P16-P14P16-P152+P14P15P13P14 is equal to _____.

Options:

The correct answer is:
16

Solution:
Given,

Pn=αn-βn and α & β are roots of x2-x-4=0

Now putting n-1 in Pn=αn-βn we get,

Pn-1=αn-1-βn-1

Now subtracting Pn-Pn-1 we get,

Pn-Pn-1=αn-βn-αn-1-βn-1

⇒Pn-Pn-1=αn-2α2-α-βn-2β2-β

Now using the equation α2-α-4=0 & β2-β-4=0 we get,

Pn-Pn-1=4αn-2-βn-2

Pn-Pn-1=4Pn-2

Now putting the value in given expression

P15P16-P14P16-P152+P14P15P13P14

=P16P15-P14-P15P15-P14P13P14

=P15-P14P16-P15P13P14=4P134P14P13P14=16

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Let Ω be the sample space and A⊆Ω be an event. Given below are two statements:

(S1): If P(A)=0, then A=ϕ

(S2): If P(A)=, then A=Ω

Then

Options:
A only (S1) is true
B only (S2) is true
C both (S1) and (S2) are true
D both (S1) and (S2) are false

The correct answer is:
both (S1) and (S2) are true

Solution:
Given:

Ω=sample space

A=be an event

So,

If P(A)=0⇒A=ϕ

If P(A)=1⇒A=Ω

Hence, both statements are true.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let ℂ be the set of all complex numbers. Let S1=z∈ℂ:z-2≤1 and S2=z∈ℂ:z1+i+z¯1-i≥4. Then, the maximum value of z-522 for z∈S1∩S2 is equal to :

Options:
A 3+224
B 5+222
C 3+222
D 5+224

The correct answer is:
5+224

Solution:
Here, z-2≤1 Put z=x+iy

x-22+y2≤1

Also, z1+i+z¯1-i≥4

Gives x-y≥2

Let point on circle be A2+cosθ,sinθ

θ∈-3π4,π4

Let P be 52,0

So,

z-522=AP2=2+cosθ-522+sin2θ

=cos2θ-cosθ+14+sin2θ

=54-cosθ

For AP2 to be maximum θ=-3π4

AP2=54+12=52+442=5+224

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let ℝ be the set of all real numbers.

Statement I : The function f:-π2,π2→ℝ defined by fx=secx+tanx is a one - one function.

Statement II : The function f:[0,∞)→ℝ defined by fx=x2 is a one-one function.

Which of the above statements is(are) true?

Options:
A Statement I is true, but Statement II is false
B Statement II is true, but Statement I is false
C Both Statement I and Statement II are true
D Both Statement I and Statement II are false

The correct answer is:
Both Statement I and Statement II are true

Solution:
Given,

ℝ be the set of all real numbers.

Now solving statement I : The function f:-π2,π2→ℝ defined by fx=secx+tanx

Now differentiating the function with respect to x we get,

f'x=secxsecx+tanx

Now we know that secx is always positive in given domain and secx+tanx will also be positive as tanx is negative in -π2,0 but secx≥tanx in -π2,0

So, f'x≥0 hence it is a one - one function.

Now solving statement II : The function f:[0,∞)→ℝ defined by

fx=x2

Differentiating the function we get,

f'x=2x

⇒f'x≥0 ∀ x∈[0,∞)

Hence it is also a one-one function.

So both are true.

Topic: Mathematics, Functions, JEE Main
Let ℝ be the set of all real numbers.

Statement I : The function f:-π2,π2→ℝ defined by fx=secx+tanx is a one - one function.

Statement II : The function f:[0,∞)→ℝ defined by fx=x2 is a one-one function.

Which of the above statements is(are) true?

Options:
A Statement I is true, but Statement II is false
B Statement II is true, but Statement I is false
C Both Statement I and Statement II are true
D Both Statement I and Statement II are false

The correct answer is:
Both Statement I and Statement II are true

Solution:
Given,

ℝ be the set of all real numbers.

Now solving statement I : The function f:-π2,π2→ℝ defined by fx=secx+tanx

Now differentiating the function with respect to x we get,

f'x=secxsecx+tanx

Now we know that secx is always positive in given domain and secx+tanx will also be positive as tanx is negative in -π2,0 but secx≥tanx in -π2,0

So, f'x≥0 hence it is a one - one function.

Now solving statement II : The function f:[0,∞)→ℝ defined by

fx=x2

Differentiating the function we get,

f'x=2x

⇒f'x≥0 ∀ x∈[0,∞)

Hence it is also a one-one function.

So both are true.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 2)
Let α, β, γ be the three roots of the equation x3+bx+c=0 if βγ=1=-α then b3+2c3-3α3-6β3-8γ3 is equal to

Options:
A 1558
B 21
C 1698
D 19

The correct answer is:
19

Solution:
Given that roots of x3+bx+c=0 are α, β, γ

Also given that

βγ=1=-α

⇒α=-1...(i)

Now let us apply the relation between the roots of the cubic equation with the coefficients.

⇒α+β+γ=0...(ii)

⇒αβγ=-c...(iii)

⇒-11=-c

∴c=1...(iv)

On substituting the value of α=-1 in eq ii we get,

β+γ=1...(v)

∵αβ+βγ+γα=b

⇒αβ+γ+βγ=b

∴b=0...(vi)

Hence, equation will be,

x3+1=0

Whose roots will be, -1,-ω, -ω2

We know that 1+ω+ω2=0

⇒α+β+γ=0

⇒-1+β+γ=0

∴β=-ω, γ=-ω2...(vii)

⇒β3=-ω3=-1 and ⇒γ3=-ω6=-1

Hence the value of

b3+2c3-3α3-6β3-8γ3 is

=0+2+3+6+8

=19

Hence this is the correct option.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be the three unit vectors such that $\vec{\alpha} \cdot \vec{\beta}=\vec{\alpha} \cdot \vec{\gamma}=0$ and the angle between $\vec{\beta}$ and $\vec{\gamma}$
is $30^{\circ} .$ Then $\vec{\alpha}$ is

Options:
A $2(\vec{\beta} \times \vec{\gamma})$
B $-2(\vec{\beta} \times \vec{y})$
C $\pm 2(\vec{\beta} \times \vec{\gamma})$
D $(\vec{\beta} \times \vec{y})$

The correct answer is:
$\pm 2(\vec{\beta} \times \vec{\gamma})$

Solution:
$\vec{\alpha}=\lambda(\vec{\beta} \times \vec{\gamma})=\lambda\left(|\vec{\beta}||\vec{\gamma}|\left|\sin 30^{\circ}\right|\right.$
$\Rightarrow \quad|\vec{\alpha}|=|\lambda|\left(|\beta||\vec{\gamma}| \cdot \frac{1}{2}\right)$
$\begin{array}{lc}\Rightarrow & 1=|\lambda| \cdot 1 \cdot 1 \cdot \frac{1}{2} \\ \Rightarrow & |\lambda|=2 \\ \rightarrow & \lambda-\pm 2\end{array}$
$\therefore \quad \vec{\alpha}=\pm 2(\vec{\beta} \times \vec{\gamma})$

Topic: Mathematics, Vector Algebra, JEE Main
Let σ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region EI , EII and EIII

Options:
A E→I=2σ∈0n^,E→II=0,E→III=2σ∈0n^
B E→I=0,E→II=σ∈0n^,E→III=0
C E→I=σ2∈0n^,E→II=0,E→III=σ2∈0n^
D E→I=-σ∈0n^,E→II=0,E→III=σ∈0n^

The correct answer is:
E→I=-σ∈0n^,E→II=0,E→III=σ∈0n^

Solution:
Electric field due to charged sheet for charge density +σ is E→=+σ2∈0 and for -σ is E→=-σ2∈0.

Assuming RHS to be n^

E→I=σ2∈0-n^+σ2∈0-n^=-σ∈0n^ (leftwards)

E→II=σ2∈0-n^+σ2∈0n^=0

E→III=σ2∈0n^+σ2∈0n^=σ∈0n^ (rightwards)

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let Î±,Î²,Î³ be three real numbers satisfying Î± Î² Î³ 2-1 1-1-1-2-1 2 1=0 0 0. If the point AÎ±,Î²,Î³ lies on the plane 2x+y+3z=2, then 3Î±+2Î²-6Î³ is equal to

Options:
A 0
B -13
C 1
D -3

The correct answer is:
-13

Solution:
Î± Î² Î³2-11-1-1-2-121=0 0 0
2Î±-Î²-Î³-Î±-Î²+2Î³Î±-2Î²+Î³=0 0 0
â2Î±-Î²-Î³=0
-Î±-Î²+2Î³=0
Î±-2Î²+Î³=0
Using cramerâs rule
â=2-1-1-1-121-21=2-1+4+1-1-2-12+1
=6-3-3=0
So, the system of equations has infinite solutions
Put Î³=k
2Î±-Î²=kÎ±+Î²=2kâÎ³=k,Î²=k
âÎ±,Î²,Î³â¡k,k,k
âµÎ±,Î²,Î³ satisfies 2x+y+3z=2
â6k=2âk=13
Î±,Î²,Î³â¡13,13,13
â3Î±+2Î²-6Î³=-13

Topic: Mathematics, Matrices, JEE Main
Let α,β be two roots of the equation x2+201/4x+51/2=0. Then α8+β8 is equal to

Options:
A 10
B 100
C 50
D 160

The correct answer is:
50

Solution:
Given,

x2+201/4x+51/2=0

⇒x2+5=-x2014

⇒x2+52=20x2

⇒x4+5+2×x2×5=20x2

⇒x4=-5⇒x8=25

So, α8+β8=25+25=50

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let α, β be two roots of the quadratic equation x2+ax-b=0, b≠0. If the straight line xcosθ+ysinθ=c touches the curve xαn+yβn=2 at the point α,β, then ab2+2b=

Options:
A 12c2
B 4c2
C 2c2
D 1c2

The correct answer is:
4c2

Solution:
Given α and β are roots of x2+ax-b=0

α+β=-a ...(1)

αβ=-b ...(2)

Now, differentiating the curve xαn+yβn=2

nαxαn-1+nβyβn-1dydx=0

at x=α and y=β

dydx=-βα ...3

Now, xcosθ+ysinθ=c

y=-xcotθ+csinθ

As, -cotθ=-βα

y=-βαx+csinθ

Put =α, β

⇒β=c2sinθ and α=c2cosθ ...4

Now, ab2+2b=

By Equations (1) & (2)

⇒-α+β-αβ2-2αβ

⇒α2+β2+2αβ-2αβ(αβ)2

⇒α2+β2α2β2

By Equation (4)

⇒c24cos2θ+c24sin2θc2×c24cos2θ4sin2θ

⇒c24cos2θ+sin2θcos2θsin2θc416cos2θsin2θ⇒4c2

ab2+2b=4c2

Topic: Mathematics, Application of Derivatives, JEE Main
Let binding energy per nucleon of nucleus is denoted by $\mathrm{E}_{\mathrm{bn}}$ and radius of nucleus is denoted as $r$. If mass number of nuclei $\mathrm{A}, \mathrm{B}$ and 64 and 125 respectively then

Options:
A \(r_{A} < r_{B}, E_{b n A} < E_{b n B}\)
B \(r_{A} < r_{B}, E_{b n A} < E_{b n A}\)
C \(r_{A}=\frac{4}{5} r_{B}, E_{b n A} < E_{b n B}\)
D \(r_{A} < r_{B}, E_{b n A} > E_{b n B}\)

The correct answer is:
\(r_{A} < r_{B}, E_{b n A} > E_{b n B}\)

Solution:
$\mathrm{r}=\mathrm{r}_{0}(\mathrm{~A})^{1 / 3}$

r increases with increasing A mass number So, $r_{A}
$\mathrm{E}_{\mathrm{bn}}$ decreases with increasing $\mathrm{A}$ for $\mathrm{A}>56$ ${ }^{56} \mathrm{Fe}$ has highest $\mathrm{E}_{\mathrm{bn}}$ value. So, $E_{b n}$ for $A=64$ is larger as compared to $\mathrm{E}_{\mathrm{bn}}$ for nucleus with $\mathrm{A}=125$

$\mathrm{E}_{\mathrm{bnA}}>\mathrm{E}_{\mathrm{bnB}}$

Topic: Physics, Nuclear Physics, JEE Main
Let $C_{0}$ be a circle of radius 1 .for $n \geq 1$, let $C_{n}$ be a circle whose area equals the area of a square inscribed in $\mathrm{C}_{\mathrm{n}-1} .$ Then $\sum_{\mathrm{i}=0}^{\infty}$ Area $\left(\mathrm{C}_{\mathrm{i}}\right)$ equals:

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

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

Solution:
Let $a, a_{1}, a_{2--}$ be the length of siden of square inscribed in circle $c_{0}, c_{1}, c_{2}=$ and $\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}, \ldots_{----}$be radius of circle $\mathrm{c}_{1}, \mathrm{c}_{2}$, then, $a_{0}^{2}=2=\pi r_{1}^{2}$
$$
\begin{array}{l}
r_{1}=\sqrt{\frac{2}{\pi}} \\
2 a_{1}^{2}=\frac{8}{\pi} \quad a_{1}^{2}=\frac{4}{\pi}=\pi r_{2}^{2} \quad r_{2}=\frac{2}{\pi} \\
\sum_{i=0}^{\infty} \text { Area }(C i)=\pi . i \\
=\pi+2+\frac{4}{\pi}+\frac{8}{\pi^{2}}=\frac{\pi}{1-\frac{2}{\pi}}=\frac{\pi^{2}}{\pi-2}
\end{array}
$$

Topic: Mathematics, Circle, JEE Main
Let C1 and C2 be the centres of the circles x2+y2-2x-2y-2=0 and x2+y2-6x-6y+14=0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1 QC2 is :

Options:
A 6
B 4
C 8
D 9

The correct answer is:
4

Solution:
Equation of given circles are

x-12+y-12=4 and x-32+y-32=4.

Hence, C11, 1 and r1=2; C23, 3 and r2=2

⇒PC1=PC2=2

Now, by distance formula,

C1C2=3-12+3-12=22+22=8

⇒PC12+PC22=C1C22

⇒∠C1PC2=π2 (by converse of pythagoras theorem in △PC1C2)

Hence, area of quadrilateral PC1 QC2=2×area of △PC1C2=2×area of △QC1C2

=2×12×2×2=4

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Let C1 and C2 be two circles of radius r1 and r2 respectively r1>r2 touching both the axes. If the two circles are orthogonal, then r1r2 is equal to

Options:
A 2
B 2+3
C 3+2
D 3

The correct answer is:
2+3

Solution:
Let the equation of a circle touching both the axes be
x-r2+y-r2=r2
x2+y2-2rx-2ry+r2=0
If the two given circles are orthogonal, then
2-r1-r2+2-r1-r2=r12+r22
âr12-4r1r2+r22=0
âr1r2=2+3

Topic: Mathematics, Circle, JEE Main
Let C1 and C2 be two circles of radius r1 and r2 units respectively touching both the axes. If the two circles are orthogonal, then r1r2+r2r1 is equal to

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

The correct answer is:
4

Solution:
Let the equation of the circle be
x-r2+y-r2=r2 and
x2+y2-2rx-2ry+r2=0
If they are orthogonal
2-r1-r2+2-r1-r2=r12+r22
âr12-4r1r2+r22=0
âr1r2+r2r1=2-3+2+3=4

Topic: Mathematics, Circle, JEE Main
Let C1 be the curve obtained by the solution of differential equation 2xydydx=y2-x2, x>0. Let the curve C2 be the solution of 2xyx2-y2=dydx. If both the curves pass through 1,1, then the area (in sq. units) enclosed by the curves C1 and C2 is equal to :

Options:
A π-1
B π2-1
C π+1
D π4+1

The correct answer is:
π2-1

Solution:
dydx=y2-x22xy, x∈0,∞

put y=vx

xdvdx+v=v2-12v

2vv2+1dv=-dxx

Integrate,
lnv2+1=-lnx+C

lny2x2+1=-lnx+C

put x=1,y=1,C=ln2

lny2x2+1=-lnx+ln2

⇒x2+y2-2x=0 (Curve C1)

Similarly,

dydx=2xyx2-y2

Put y=v x

x2+y2-2y=0

Required area =2∫012x-x2-xdx=π2-1 sq. units

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let C1 be the graph of xy=1 and the reflection of C1 in the line y=2x is C2. If the equation of C2 is expressed as 12x2+bxy+cy2+d=0, then the value of (b+c+d) is equal to

Options:

The correct answer is:
6

Solution:
Any point on the hyperbola xy=1 is t,1t
Now, the image of t,1t by the line mirror y=2x is
x-t2=y-1t-1=-22t-1t5
⇒5x=4t-3t and 5y=4t+3t
Now eliminating t from above two equations we get,
12x2-12y2-7xy+25=0
∴b+c+d=6

Topic: Mathematics, Hyperbola, JEE Main
Let C1, C2, C3... are the usual binomial coefficients where Cr= nCr. Let S=C1+2C2+3C3+...+nCn, then S is equal to

Options:
A n2n
B 2n-1
C n2n-1
D 2n+1

The correct answer is:
n2n-1

Solution:
Let S=C1+2C2+3C3+…+nCn= Σ r = 1 n r· nCr
= Σ r = 1 n r·nr n-1Cr-1 ∵ nCr=nr n-1Cr-1
=nΣr=1n n-1Cr-1
=n n-1C0+ n-1C1+ n-1C2+…+ n-1Cn-1
=n2n-1

Topic: Mathematics, Binomial Theorem, JEE Main
Let C1: x2+y2=5 and C2: x2+y2=9. The tangent at (1, 2) to C1 meets the circle C2 at A and B. The tangents drawn at A and B to circle C2 meet at T. Then the coordinates of T are

Options:
A 4,−5
B 35,65
C 4,5
D 95,185

The correct answer is:
95,185

Solution:
Equation of tangent at 1,2 to C 1 is
x+2y−5=0 …(1)
Let point T is h,k
∴ Equation of chord of contact w.r.t. C 2 is
xh+yk−9=0 …(2)
⇒ h1=k2=95
⇒ h=95,k=185

Topic: Mathematics, Circle, JEE Main
Let $C$ be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line $x+y=2$ intersects the circle $\mathrm{C}$ at the points $\mathrm{P}$ and $\mathrm{Q}$. Let $\mathrm{MN}$ be a chord of $\mathrm{C}$ of length 2 unit and slope -1. Then, a distance (in units) between the chord PQ and the chord $\mathrm{MN}$ is

Options:
A $\begin{aligned}& 3-\sqrt{2} \end{aligned}$
B $\begin{aligned} &\sqrt{2}+1 \end{aligned}$
C $\begin{aligned} &\sqrt{2}-1 \end{aligned}$
D $\begin{aligned}& 2-\sqrt{3} \end{aligned}$

The correct answer is:
$\begin{aligned}& 3-\sqrt{2} \end{aligned}$

Solution:
$\begin{aligned}
& C: x^2+y^2=10 \\
& A N=\frac{M N}{2}=1 \\
& \therefore \text { In } \triangle O A N \rightarrow(O N)^2=(O A)^2+(A N)^2 \\
& 10=(O A)^2+1 \rightarrow O A=3
\end{aligned}$
Perpendicular distance of center from
$\mathrm{PQ}=\frac{|0+0-2|}{\sqrt{2}}=\sqrt{2}$
Perpendicular distance between $\mathrm{MN}$ and
$\begin{aligned}
& \mathrm{PQ}=\mathrm{OA}+\sqrt{2} \text { or }|\mathrm{OA}-\sqrt{2}| \\
& =3+\sqrt{2} \text { or } 3-\sqrt{2}
\end{aligned}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Let C be a curve given by yx=1+ 4x-3,x>34. If P is a point on C, such that the tangent at P has slope 23, then a point through which the normal at P passes, is :

Options:
A 1, 7
B 3, -4
C 4, -3
D 2, 3

The correct answer is:
1, 7

Solution:
Given,

yx=1+ 4x-3

So, dydx=124x-3 ×4=23
⇒ 4x-3=9
⇒ x=3
So, y=4
Equation of normal at P3, 4 is y-4=-32x-3
i.e., 2y-8=-3x+9
⇒ 3x+2y-17=0

Hence only option A is satisfying the equation.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2016 (10 Apr Online)
Let c→ be a vector perpendicular to the vectors a→=i^+j^-k^ and b→=i^+2j^+k^. If c→·(i^+j^+3k^)=8, then the value of c→·(a→×b→) is equal to

Options:

The correct answer is:
28

Solution:
c→=λ(a→×b→)

a→×b→=i^j^k^11-1121

(a→×b→)=3i^-2j^+k^

c→·(i^+j^+3k^)=λ(3i^-2j^+k^)·(i^+j^+3k^)

⇒λ(4)=8⇒λ=2

c→=2(a→×b→)

Hence, c→·(a→×b→)=28

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let $C$ be the capacitance of a capacitor discharging through a resistor R. Suppose $t_1$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_2$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_1 / t_2$ will be

Options:
A
1

B
$\frac{1}{2}$

C
$\frac{1}{4}$

D
2

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

Solution:
$\mathrm{U}=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{1}{2 \mathrm{C}}\left(\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}\right)^2=\frac{\mathrm{q}_0^2}{2 \mathrm{C}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{T}} \quad($ where $\tau=\mathrm{CR})$
$\mathrm{U}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{\tau}}$
$\frac{1}{2} \mathrm{U}_{\mathrm{i}}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t}_1 / \mathrm{\tau}}$
$\frac{1}{2}=\mathrm{e}^{-2 \mathrm{t}_1 / \tau} \Rightarrow \mathrm{t}_1=\frac{\mathrm{T}}{2} \ln 2$
Now $\quad \mathrm{q}=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}$
$\frac{1}{4} \mathrm{q}_0=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / 2 \mathrm{~T}}$
$\mathrm{t}_2=\operatorname{T} \ln 4=2 \operatorname{T} \ln 2$
$\therefore \quad \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{1}{4}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2010