datavorous/entrance-exam-dataset huggingface dataset:
If $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are the roots of the equation $x^{2}+p x+q=0,$ where $\alpha, \beta, p$ and $q$ are real, then the roots of the equation $\left(p^{2}-4 q\right)\left(p^{2} x^{2}+4 p x\right)-16 q=0$ are

Options:
A $\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)$
B $\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)$ and $\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)$
C $\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)$
D $(\sqrt{\alpha}+\sqrt{\beta})$ and $(\sqrt{\alpha}-\sqrt{\beta})$

The correct answer is:
$\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)$

Solution:
Since, $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are the roots of the equation $x^{2}+p x+q=0$
$\therefore \quad$ Sum of roots $=-p$
$\Rightarrow \quad(\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-P$
$\Rightarrow$
$\mathbf{2}\alpha=-p \Rightarrow \alpha=-\frac{\underline{p}}{2}$
and produet of roots $=\mathrm{q}$
$(\alpha+\sqrt{\beta})(\alpha-\sqrt{\beta})=q$
$\Rightarrow \alpha^{2}-\beta=q$
$\Rightarrow \quad \beta=\alpha^{2}-q=\left(-\frac{p}{2}\right)^{2}-q=\frac{p^{2}}{4}-q$
$\Rightarrow \quad p^{2}-4 q=4 \beta$
$\therefore$ The given equation $\left(p^{2}-4 q\right)\left(p^{2} x^{2}+4 p x\right)-16 q=0$
$4 \beta\left(p^{2} x^{2}+4 p x\right)-16\left(\alpha^{2}-\beta\right)=0$
$\left[\alpha^{2}-\beta=q\right]$
$\beta\left(4 a^{2} x^{2}-8 x x\right)-4\left(a^{2}-\beta\right)=0$
$\alpha^{2} \beta x^{2}-2 x f\left(x+\beta=\alpha^{2}\right.$
$\Rightarrow \quad(\alpha x \sqrt{\beta}-\sqrt{\beta})^{2}=\alpha^{2}$
$\Rightarrow \quad \operatorname{cox} \sqrt{\beta}-\sqrt{\beta}=\pm \alpha$
$\therefore$
$x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}$
$\Rightarrow x=\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)$

Topic: Mathematics, Quadratic Equation, JEE Main
If α and β are the roots of the equation x2+αx+β=0 such that α≠β, then the number of integral values of x satisfying x-β-α<1 is

Options:
A 0
B 1
C 2
D more than 2

The correct answer is:
2

Solution:
α,β are roots of x2+αx+β=0
⇒α+β=-α and αβ=β⇒α=1,β=-2
Now, x+2-1<1
⇒-1<x+2-1<1
⇒0<x+2<2
⇒-2<x+2<2, x+2≠0
⇒-4<x<0, x≠-2
⇒x∈-4,-2∪-2,0
So, there are two integral values of x, i.e. -3,-1

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+1=0$, then $\alpha^{2009}+\beta^{2009}=$

Options:
A
$-1$

B
1

C
2

D
$-2$

The correct answer is:
1

Solution:
$x^2-x+1=0 \quad \Rightarrow x=\frac{1 \pm \sqrt{1-4}}{2}$
$x=\frac{1 \pm \sqrt{3} i}{2}$
$\alpha=\frac{1}{2}+i \frac{\sqrt{3}}{2}, \quad \beta=\frac{1}{2}-\frac{i \sqrt{3}}{2}$
$\alpha=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}, \quad \beta=\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}$
$\alpha^{2009}+\beta^{2009}=2 \cos 2009\left(\frac{\pi}{3}\right)$
$=2 \cos \left[668 \pi+\pi+\frac{2 \pi}{3}\right]=2 \cos \left(\pi+\frac{2 \pi}{3}\right)$
$=-2 \cos \frac{2 \pi}{3}=-2\left(-\frac{1}{2}\right)=1$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2010
If Î± and Î² are the roots of the equation x2+x+c=0 such that Î±+Î², Î±2+Î²2 and Î±3+Î²3 are in arithmetic progression, then c is equal to

Options:
A 1
B 43
C 45
D 47

The correct answer is:
47

Solution:
Given, Î±+Î²=-1,Î±Î²=c
Also, 2a2+Î²2=Î±+Î²+Î±3+Î²3
â2Î±+Î²2-2Î±Î²=Î±+Î²+Î±+Î²3-3Î±Î²Î±+Î²
â2-4c=-2+3câc=47

Topic: Mathematics, Quadratic Equation, JEE Main
If α, β and γ are the roots of the equation x3-3x2+4x+4=0, then the value of α2+1111β2+1111γ2+1 is equal to

Options:
A 32
B 16
C 56
D 64

The correct answer is:
56

Solution:
α+β+γ=3
αβ+βγ+γα=4
αβγ=-4
α2+1111β2+1111γ2+1=α2+1β2+1γ2+1-1-1γ2+1-β2
=α2+1β2+γ2+β2γ2-γ2-β2
=α2β2+α2γ2+α2β2γ2+β2+γ2+β2γ2-γ2-β2
=α2β2+β2γ2+γ2α2+αβγ2
=αβ+βγ+γα2-2αβγα+β+γ+αβγ2
=16+83+16
=56

Topic: Mathematics, Determinants, JEE Main
If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}-8 x+8=0$, then $\Sigma \alpha^{2}$ and $\Sigma \frac{1}{\alpha \beta}$ are respectively is equal to

Options:
A 16 and 0
B $-16$ and 0
C 16 and 8
D 0 and $-16$

The correct answer is:
16 and 0

Solution:
Since, $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}-8 x+8=0$, then
$\begin{gathered}
\alpha+\beta+\gamma=0, \alpha \beta+\beta \gamma+\gamma \alpha=-8 \\
\alpha \beta \gamma=-8 ...(i)
\end{gathered}$
Therefore, $(\alpha+\beta+\gamma)^{2}=0$
$\Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2}+2(\alpha \beta+\beta \gamma+\gamma \alpha)=0$
$\begin{aligned} \Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2} &=-2(-8)=16 \\ \Sigma \alpha^{2} &=16 \end{aligned}$
and
$\begin{gathered}
\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=\frac{0}{-8}=0 \\
\Sigma \frac{1}{\alpha \beta}=0
\end{gathered}$

Topic: Mathematics, Quadratic Equation, JEE Main
If α,β and γ are the roots of the equation x3-13x2+15x+189=0 and one root exceeds the other by 2, then the value of α+β+γ is equal to

Options:
A 23
B 17
C 13
D 19

The correct answer is:
19

Solution:
Let, the roots be α,β,α+2.
S1=α+β+α+2=2α+β+2=13⇒2α+β=11⇒β=11-2α
S2=αβ+βα+2+α+2α=15
⇒βα+α+2+αα+2=15
⇒11-2α2α+2+αα+2=15
⇒22α+22-4α2-4α+α2+2α=15
⇒3α2-20α-7=0⇒α-73α+1=0
⇒α=7 or -13.
α=7,β=11-2α=11-14=-3,γ=α+2=9
α=-13,β=11-2α=11+23=353,γ=α+2=53.
Since, αβγ=-189, hence we will take the first case.
α+β+γ=7+-3+9=19

Topic: Mathematics, Quadratic Equation, JEE Main
If Î±, Î² and Î³ are the roots of the equation x3-px2+qx-r=0, then the value of Î±2Î²+Î±2Î³+Î²2Î±+Î²2Î³+Î³2Î±+Î³2Î² is equal to

Options:
A pq+3r
B pq+r
C pq-3r
D q2r

The correct answer is:
pq-3r

Solution:
x3-px2+qx-r=0
α+β+γ=p
αβ+βγ+γα=q
αβγ=r
⇒α2β+α2γ+β2α+β2γ+γ2α+γ2β
=α+β+γαβ+βγ+γα-3αβγ
=pq-3r

Topic: Mathematics, Quadratic Equation, JEE Main
If Î±, Î² and Î³ are the roots of the equation x3-px2+qx-r=0, then the value of Î±Î²Î³+Î²Î³Î±+Î³Î±Î² is equal to

Options:
A pq+3r
B pq+r
C pq-3r
D q2-2prr

The correct answer is:
q2-2prr

Solution:
Since α,β,γ are the roots of x3-px2+qx-r=0
α+β+γ=p, αβ+βγ+γα=q, αβγ=r
αβγ+βγα+γαβ=α2β2+β2γ2+γ2α2αβγ
=αβ+βγ+γα2-2αβγα+β+γαβγ
=q2-2rpr

Topic: Mathematics, Quadratic Equation, JEE Main
If Î±, Î² and Î³ are the roots of the equation x3+x+2=0, then the equation whose roots are Î±-Î²Î±-Î³, Î²-Î³Î²-Î± and Î³-Î±Î³-Î² is

Options:
A x3-6x2+216=0
B x3-3x2+112=0
C x3+6x2-216=0
D x3+3x2-112=0

The correct answer is:
x3+3x2-112=0

Solution:
Putting y=Î±-Î²Î±-Î³, we get,
y=Î±2-Î±Î²+Î³+Î²Î³
Now, Î±+Î²+Î³=0 and Î±Î²Î³=-2
ây=Î±2-Î±-Î±+-2Î±
=2Î±2-2Î±=2Î±3-1Î±
Also, Î± is a root of the given equation.
âÎ±3=-Î±-2
ây=2-Î±-2-1Î±=2-Î±-3Î±=-2-6Î±
ây+2=-6Î±âÎ±=-6y+2
Î± is a root of the given equation

â-6y+23+-6y+2+2=0
-216-6y+22+2y+23=0
ây+23-3y+22-108=0
ây3+6y2+12y+8-3y2-12-12y-108=0
ây3+3y2-112=0
Hence, the required equation is x3+3x2-112=0

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha$ and $\beta$ are the roots of the equation $(x-a)(x-b)=c, c \neq 0$; then the roots of the equation $(x-\alpha)(x-\beta)+c=0$ are

Options:
A a, c
B $b, c$
C $a, b$
D $a+c, b+c$

The correct answer is:
$a, b$

Solution:
Option 3 a,b
Concept:
Quadratic equations whose roots are $\alpha$ and $\beta$ is given by
$(x-\alpha)(x-\beta)$
Calculation:
Given that,
$(x-a)(x-b)-c$ ...(i)
According to the question, $\alpha$ and $\beta$ are the roots of the equation.
Therefore,
$\begin{aligned} & (x-a)(x-b)-c=(x-\alpha)(x-\beta) \\ & \Rightarrow(x-a)(x-b)=(x-\alpha)(x-\beta)+c\end{aligned}$
which represents, root of equation $(x-\alpha)(x-\beta)+c$ are $a$ and $b$.

Topic: Mathematics, Quadratic Equation, JEE Main
If Î± and Î² are the solutions of cotâ¡x=-3 in 0,2Ï and Î± and Î³ are the solutions of cosecâ¡x=-2 in 0,2Ï, then the value of Î±-Î²Î²+Î³ is equal to

Options:
A 12
B 2
C 13
D 3

The correct answer is:
12

Solution:
cot⁡x=-3
⇒x=5π6,11π6 and

cosec⁡x=-2
⇒x=7π6,11π6
So, α=11π6, β=5π6, γ=7π6
⇒α-β=π and β+γ=2π
⇒α-ββ+γ=π2π=12

Topic: Mathematics, Trigonometric Equations, JEE Main
If α and β are the solutions of sinx=-12 in 0,2π and α and γ are the solutions of cos⁡x=-32 in 0,2π, then the value of α+ββ-γ is equal to

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
sinâ¡x=-12⇒x=7Ï6,11Ï6
cosâ¡x=-32⇒x=5Ï6,7Ï6
⇒Î±=7Ï6,Î²=11Ï6,Î³=5Ï6
⇒Î±+Î²=7Ï6+11Ï6=3Ï,Î²-Î³=11Ï6-5Ï6=Ï
⇒Î±+Î²Î²-Î³=3ÏÏ=3

Topic: Mathematics, Trigonometric Equations, JEE Main
If α,β and γ are three consecutive terms of a non-constant G.P. Such that the equations αx2+2βx+γ=0 and x2+x-1=0 have a common root, then αβ+γ is equal to:

Options:
A βγ
B Î±Î²
C αγ
D 0

The correct answer is:
βγ

Solution:
Given α, β,γ are in G.P.⇒β2=αγ
For the equation αx2+2βx+γ=0
D=4β2-4αγ=0
Hence roots of the equation are equal and equals to –βα=-γβ
Since given equation have common roots, hence -γβ must be a root of x2+x-1=0
⇒ γ2β2-γβ-1=0
⇒ γ2-γ β-β2=0
⇒ γ2=ββ+γ
⇒ γ.β2α=ββ+γ
⇒ αβ+γ=βγ

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (12 Apr Shift 2)
If α and β, be the coefficients of x4 and x2, respectively in the expansion of x+x2-16+x-x2-16, then

Options:
A α+β=60
B α+β=-30
C α-β=60
D α-β=-132

The correct answer is:
α-β=-132

Solution:
x+x2-16+x-x2-16=2C0x6+C2x4x2-1+C4x2x2-12+C6x2-13

=2x6+15x6-x4+15x2x4-2x2+1+-1+3x2-3x4+x6

=232x6-48x4+18x2-1

Now given that α & β are the coefficients of x4 & x2 respectively, hence

α=-96 and β=36

∴α-β=-132

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (08 Jan Shift 2)
If α and β be the roots of the equation x2-2x+2=0, then the least value of n for which αβn=1 is

Options:
A 5
B 4
C 2
D 3

The correct answer is:
4

Solution:
x2-2x+2=0

⇒ (x-1)2+1=0

⇒x=1±i , i=-1

Now let the roots α=i+i & β=1-i

αβn=1+i1-in=i1-i1-in

⇒αβn=(i)n

Using given information,

⇒(i)n=1

⇒ The minimum value of n∈N=4 because i4=1.

Topic: Mathematics, Complex Number, JEE Main
If α and β be the roots of the equation x2-2x+2=0, then the least value of n for which αβn=1 is

Options:
A 5
B 4
C 2
D 3

The correct answer is:
4

Solution:
x2-2x+2=0

⇒ (x-1)2+1=0

⇒x=1±i , i=-1

Now let the roots α=i+i & β=1-i

αβn=1+i1-in=i1-i1-in

⇒αβn=(i)n

Using given information,

⇒(i)n=1

⇒ The minimum value of n∈N=4 because i4=1.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If Î» Î¿ and Î» be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

Options:
A 2 h c m λ ο - λ λ λ ο
B 2 h c m Î» Î¿ - λ
C 2 h m Î» Î¿ - λ
D 2 h m 1 λ ο - 1 λ

The correct answer is:
2 h c m λ ο - λ λ λ ο

Solution:
Einstein photo Electric Equation
hν - hν 0 = 1 2 mV max 2
h C λ - h C λ 0 = 1 2 mV max 2
2 hC m 1 λ - 1 λ 0 = V max 2
V max = 2 hC m λ 0 - λ λ λ 0

Topic: Chemistry, Structure of Atom, JEE Main
If α and β be two roots of the equation x2-64x+256=0. Then the value of α3β518+β3α518 is :

Options:
A 2
B 3
C 1
D 4

The correct answer is:
2

Solution:
Given equation is x2-64x+256=0 whose roots are α, β.

α+β=64, αβ=256
Now α3β518+β3α518
=α38+58+β38+58(αβ)5/8=α+β(αβ)5/8
=64(256)5/8=6432=2

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (06 Sep Shift 1)
If any tangent to the ellipse 25x2+9y2=225 meets the coordinate axes at A and B such that OA=OB then, the length AB is equal to (where, O is the origin)

Options:
A 17 units
B 34 units
C 217 units
D 234 units

The correct answer is:
217 units

Solution:
Given, equation of the ellipse is x29+y225=1
Let, a point on it be P3cos⁡θ,5sin⁡θ
and equation of the tangent at P is x3cos⁡θ+y5sin⁡θ=1
which meets the x-axis at A3cosθ,0 and y -axis at B0,5sinθ
Let, OA=OB=λ
cos⁡θ=3λ, sin⁡θ=5λ
Now, cos2θ+sin2θ=1
9λ2+25λ2=1⇒λ2=34
∴λ=34
Now, AB=2λ=217 units

Topic: Mathematics, Ellipse, JEE Main
If any two adjacent rows or columns of a determinant are intercharged in position, the value of the determinant:

Options:
A Becomes zero
B Remains the same
C Changes its sign
D Is doubled

The correct answer is:
Changes its sign

Solution:
If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant changes its sign.

Topic: Mathematics, Matrices, NDA, NDA 2014 (Phase 1)
If $\arg (\mathrm{z})<0$, then $\arg (-\mathrm{z})-\arg (\mathrm{z})$ equals

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

The correct answer is:
$\pi$

Solution:
$\begin{aligned} & \arg (-\mathrm{z})-\arg (\mathrm{z}) \\ & =\arg (-1 \times \mathrm{z})-\arg (\mathrm{z}) \\ & =\arg (-1)+\arg (\mathrm{z})-\arg (\mathrm{z}) \\ & =\pi\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\alpha, \beta$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^2+4 y^2=4$ then $\sqrt{3} \cos \frac{\alpha+\beta}{2}=$

Options:
A $2 \cos \frac{\alpha-\beta}{2}$
B $2 \sin \frac{\alpha-\beta}{2}$
C $2 \sec \frac{\alpha+\beta}{2}$
D $2 \sin \frac{\alpha+\beta}{2}$

The correct answer is:
$2 \cos \frac{\alpha-\beta}{2}$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Ellipse, JEE Main
If $\alpha, \beta$ are the roots of the equation $a x^2+b x+c=0$ then the equation whose roots are $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$,is

Options:
A $a c x^2+(a+c) b x+(a+c)^2=0$
B $a b x^2+(a+c) b x+(a+c)^2=0$
C $a c x^2+(a+b) c x+(a+c)^2=0$
D None of these

The correct answer is:
$a c x^2+(a+c) b x+(a+c)^2=0$

Solution:
If roots are $\alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha}$, then sum of roots are $=\left(\alpha+\frac{1}{\beta}\right)+\left(\beta+\frac{1}{\alpha}\right)=(\alpha+\beta)+\frac{\alpha+\beta}{\alpha \beta}=-\frac{b}{a c}(a+c)$
and product $=\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$
$\begin{aligned} & =\alpha \beta+1+1+\frac{1}{\alpha \beta}=2+\frac{c}{a}+\frac{a}{c} \\ & =\frac{2 a c+c^2+a^2}{a c}=\frac{(a+c)^2}{a c}\end{aligned}$
Hence required equation is given by
$x^2+\frac{b}{a c}(a+c) x+\frac{(a+c)^2}{a c}=0$
$\Rightarrow a c x^2+(a+c) b x+(a+c)^2=0$
Trick: Let $a=1, b=-3, c=2$, then $\alpha=1, \beta=2$
$\therefore \quad \alpha+\frac{1}{\beta}=\frac{3}{2}$ and $\beta+\frac{1}{\alpha}=3$
Therefore, required equation must be
$(x-3)(2 x-3)=0$ i.e.$2 x^2-9 x+9=0$
Here (1) gives this equation on putting
$a=1, b=-3, c=2$

Topic: Mathematics, Quadratic Equation, JEE Main
If α, β are the roots of the equation x2 –3x+4=0, then the equation whose roots are α-2α+2, β-2β+2 is

Options:
A 7x2 –1=0
B 7x2+1=0
C 7x2+2=0
D 7x2 –2=0

The correct answer is:
7x2+1=0

Solution:
Replace x by 21+x1-x
41+x1-x2-321+x1-x+4=0
⇒7x2+1=0

Topic: Mathematics, Quadratic Equation, JEE Main
If α,β are the roots of the equation x2-5+3log35-5log53x+33log3513-5log5323-1=0 then the equation, whose roots are α+1β and β+1α,

Options:
A 3x2-20x-12=0
B 3x2-10x-4=0
C 3x2-10x+2=0
D 3x2-20x+16=0

The correct answer is:
3x2-10x-4=0

Solution:
Given α,β are the roots of the equation x2-5+3log35-5log53x+33log3513-5log5323-1=0

Now 3log35-5log53=3log35-3log35log53

=3log35-3log35=0

Also 3log3513-5log5323=5log5323-5log5323

=0

So, given equation becomes x2-5x-3=0

i.e. α+β=5; αβ=-3

Now if the roots are α+1β and β+1α

i.e., αβ+1β&αβ+1α

i.e., -2β & -2α

Then let -2α=t⇒α=-2t

As α2-5α-3=0

⇒-2t2-5-2t-3=0

⇒4t2+10t-3=0

⇒3t2-10t-4=0

i.e., 3x2-10x-4=0 is the required equation.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (27 Jul Shift 2)
If $\alpha, \beta$ are the roots of the equation $\ell \mathrm{x}^{2}-\mathrm{mx}+\mathrm{m}=0$, $\ell \neq \mathrm{m}, \ell \neq 0$, then which one of the following statements is
correct?

Options:
A $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$
B $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{\mathrm{m}}{\ell}}=0$
C $\sqrt{\frac{\alpha+\beta}{\alpha \beta}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$
D The arithmetic mean of $\alpha$ and $\beta$ is the same as their geometric mean

The correct answer is:
$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$

Solution:
As given : $\alpha$ and $\beta$ are the roots of the quadratio equation $\ell \mathrm{x}^{2}-\mathrm{mx}+\mathrm{m}=0$
So, sum of roots,
$\alpha+\beta=\frac{\mathrm{m}}{\ell}$ and product of roots, $\alpha \beta=\frac{\mathrm{m}}{\ell}$
$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=\frac{\alpha+\beta}{\sqrt{\alpha \beta}}=\frac{\mathrm{m} / \ell}{\sqrt{\mathrm{m} / \ell}}$
$\Rightarrow \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=\sqrt{\frac{\mathrm{m}}{\ell}}=0 \Rightarrow \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6 x^2+11 x-6=0$ and if $a=\alpha^2+\beta^2+\gamma^2$, $b=\alpha \beta+\beta \gamma+\gamma \alpha$ and $c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$, then the correct inequality among the following is

Options:
A $a < b < c$
B $b < a < c$
C $b < c < a$
D $c < a < b$

The correct answer is:
$b < a < c$

Solution:
Given equation $x^3-6 x^2+11 x-6=0$
has the roots $\alpha, \beta \gamma$.
Given, $\quad a=\alpha^2+\beta^2+\gamma^2$ ...(i)
$b=\alpha \beta+\beta \gamma+\gamma \alpha$ ...(ii)
$c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$ ...(iii)
In cubic equation the sum of the roots
$\alpha+\beta+\gamma=-\left(\frac{-6}{1}\right)=6$
$\alpha \beta+\beta \gamma+\gamma \alpha=\left(\frac{11}{1}\right)=11$
product of the roots
$\alpha \cdot \beta \cdot \gamma=-\left(\frac{-6}{1}\right)=6$
From Eq. (ii), $b=11$
From Eq. (i), $a=\alpha^2+\beta^2+\gamma^2$
$\Rightarrow \quad a=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)$
$\Rightarrow \quad a=(6)^2-2(11) \Rightarrow 36-22$
$\Rightarrow \quad a=14$
From Eq. (iii)
$c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$
$=\left(\alpha \beta+\beta^2+\alpha \gamma+\beta \gamma\right)(\gamma+\alpha)$
$=\alpha \beta \gamma+\beta^2 \gamma+\alpha \gamma^2+\beta \gamma^2+\alpha^2 \beta$ $+\alpha \beta^2+\alpha^2 \gamma+\alpha \beta \gamma$
$\Rightarrow c=[(\alpha+\beta+\gamma)-\gamma][(\alpha+\beta+\gamma)-\alpha]$ $[(\alpha+\beta+\gamma)-\beta]$
$=(6-\gamma)(6-\alpha)(6-\beta)$
$=(36-6 \gamma-6 \alpha+\alpha \gamma)(6-\beta)$
$=216-36 \gamma-36 \alpha+6 \alpha \gamma-36 \beta+6 \gamma \beta$ $+6 \alpha \beta-\alpha \beta \gamma$
$=216-\alpha \beta \gamma+6(\alpha \beta+\beta \gamma+\gamma \alpha)-36(\alpha+\beta+\gamma)$
$=216-6+6(11)-36(6)$
$=210+66-216=60$
Hence, $\quad c=60$
$\Rightarrow \quad b < a < c$Alternate method
Given equation is $x^3-6 x^2+11 x-6=0$
$\Rightarrow \quad(x-1)(x-2)(x-3)=0$
The roots of this equation are 1,2 , and 3 .
Let $\alpha=1, \beta=2, \gamma=3$
Now, $\quad a=\alpha^2+\beta^2+\gamma^2$
$=1+4+9=14$
$b=\alpha \beta+\beta \gamma+\gamma \alpha$
$=2+6+3=11$
$c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$
$=3 \cdot 5 \cdot 4=60$
From the values of $a, b$ and $c$ it is clear that $b < a < c$.

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+a x^2+b x+c=0$, then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

Options:
A $\frac{a}{c}$
B $\frac{c}{a}$
C $-\frac{b}{c}$
D $\frac{b}{a}$

The correct answer is:
$-\frac{b}{c}$

Solution:
Let $\alpha, \beta, \gamma$ are roots of equation
$$
\begin{aligned}
& x^3+a x^2+b x+c=0 \\
& \alpha+\beta+\gamma=-a \\
& \therefore \quad \alpha \beta+\beta \gamma+\gamma \alpha=b \\
& \text { Now, } \alpha^{-1}+\beta^{-1}+\gamma^{-1}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} \\
&=\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}=\frac{b}{-c}=-\frac{b}{c}
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}+p x+q=0$, then the value of the determinant $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta\end{array}\right|$ is

Options:
A $q$
B 0
C $p$
D $p^{2}-2 q$

The correct answer is:
0

Solution:
We have, $\alpha, \rho$ and $\gamma$ are roots of equation
$$
\begin{gathered}
x^{3}+p x+q=0 \\
\alpha+\beta+\gamma=0 \\
\text { Here, } \alpha \beta+\beta \gamma+\gamma \alpha=p \\
\alpha \beta \gamma=-q \\
\text { Applying } C_{1} \rightarrow C_{1}+C_{2}+C_{3} \text {, we get } \\
\left|\begin{array}{lll}
\alpha+\beta+\lambda & \beta & \gamma \\
\alpha+\beta+\lambda & \gamma & \alpha \\
\alpha+\beta+\lambda & \alpha & \beta
\end{array}\right|=(\alpha+\beta+\lambda)\left|\begin{array}{ccc}
1 & \beta & \gamma \\
1 & \gamma & \alpha \\
1 & \alpha & \beta
\end{array}\right|
\end{gathered}
$$
Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$, we get
$$
\begin{aligned}
&(\alpha+\beta+\lambda)\left|\begin{array}{ccc}
1 & \beta & \gamma \\
0 & \gamma-\beta & \alpha-\gamma \\
0 & \alpha-\beta & \beta-\gamma
\end{array}\right| \\
&=(\alpha+\beta+\lambda)[(\gamma-\beta)(\beta-\gamma)-(\alpha-\gamma)(\alpha-\beta)] \\
&=0 \times[(\gamma-\beta)(\beta-\gamma)-(\alpha-\gamma)(\alpha-\beta)]=0
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha, \beta$ are the roots of the equations $x^{2}-2 x-1=$ $0,$ then what is the value of $\alpha^{2} \beta^{-2}+\alpha^{-2} \beta^{2}$

Options:
A -2
B 0
C 30
D 34

The correct answer is:
34

Solution:
Since, \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x-1=0\), then

Sum of roots, \(\alpha+\beta=2\) and

product of the roots \(\alpha \beta=-1\)

Since, \((\alpha+\beta)=\alpha^{2}+\beta^{2}+2 \alpha \beta\)

\(\Rightarrow 4=\alpha^{2}+\beta^{2}-2\)

\(\Rightarrow \alpha^{2}+\beta^{2}=6\)

Now, \(\alpha^{2} \beta^{-2}+\alpha^{-2} \beta^{2}=\frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}}=\frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}\)

\(\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}=6^{2}\)

\(\Rightarrow \alpha^{4}+\beta^{4}+2 \alpha^{2} \beta^{2}=36\)

\(\Rightarrow \alpha^{4}+\beta^{4}+2=36\)

\(\Rightarrow \alpha^{4}+\beta^{4}=34 \ldots . .(\mathrm{i})\)

\(\Rightarrow \frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}=\frac{34}{(-1)^{2}}=34\)

[Putting value of \(\alpha^{4}+\beta^{4}=34\) from Equation (i) \(]\)

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha, \beta$ are the roots of $x^2-2 x+4=0$, then $\alpha^5+\beta^5$ is equal to

Options:
A $16$
B $32$
C $64$
D None of these

The correct answer is:
$32$

Solution:
Here $\alpha^2+\beta^2=-4, \alpha^3+\beta^3=-16$
$\therefore\left(\alpha^2+\beta^2\right)\left(\alpha^3+\beta^3\right)=\alpha^5+\beta^5+\alpha^2 \beta^2(\alpha+\beta)$
$\Rightarrow(-4)(-16)=\alpha^5+\beta^5+(16)(2) \Rightarrow \alpha^5+\beta^5=32$

Topic: Mathematics, Quadratic Equation, JEE Main
If area bounded by the curves $y^2=4 a x$ and $y=m x$ is $\frac{a^2}{3}$, then the value of $m$ is

Options:
A $2$
B $-2$
C $\frac{1}{2}$
D None of these

The correct answer is:
$2$

Solution:
The two curves $y^2=4 a x$ and $y=m x$ intersect at $\left(\frac{4 a}{m^2}, \frac{4 a}{m}\right)$ and the area enclosed by the two curves is given by $\int_0^{4 a / m^2}(\sqrt{4 a x}-m x) d x$
$\begin{aligned} & \therefore \int_0^{4 a / m^2}(\sqrt{4 a x}-m x) d x=\frac{a^2}{3} \\ & \Rightarrow \frac{8}{3} \frac{a^2}{m^3}=\frac{a^2}{3} \Rightarrow m^3=8 \Rightarrow m=2\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main