datavorous/entrance-exam-dataset huggingface dataset:
Let \( A(2,3,5), B(-1,3,2) \& C(\lambda, 5, \mu) \) be the vertices of \( \Delta A B C \). If the median through \( A \) is equally inclined to the coordinate axes, then

Options:
A \( 5 \lambda-8 \mu=0 \)
B \( 10 \lambda-7 \mu=0 \)
C \( 7 \lambda-10 \mu=0 \)
D \( 8 \lambda-5 \mu=0 \)

The correct answer is:
\( 10 \lambda-7 \mu=0 \)

Solution:
A≡(2,3,5), B≡(-1,3,2), C≡(λ,5,μ)
Centroid G ≡ 2 - 1 + λ 3 , 3 + 3 + 5 3 , 5 + 2 + μ 3
≡1+λ3,113,7+μ3
D.R. of line AG are < 2 - 1 + λ 3 , 3 - 1 1 3 , 5 - 7 + μ 3 >
≡ < 5 - λ 3 , - 2 3 , 8 - μ 3 >
Line is equally inclined to axes.
⇒ 5 - λ 3 = - 2 3 & 8 - μ 3 = - 2 3
⇒λ=7 & μ=10
∴10λ-7μ=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let $\mathrm{AB}$ be the latus rectum of the parabola $y^{2}=4 a x$ in the $x y$-plane. Let $T$ be the region bounded by the finite arc $\mathrm{AB}$ of the parabola and the line segment $\mathrm{AB}$. A rectangle PQRS of maximum possible area is inscribed in $\mathrm{T}$ with $\mathrm{P}, \mathrm{Q}$ on line $\mathrm{AB}$, and $\mathrm{R}, \mathrm{S}$ on arc $\mathrm{AB}$. Then area(PQRS)/area(T) equals

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

The correct answer is:
$\frac{1}{\sqrt{3}}$

Solution:
$\begin{array}{l}
T=\int_{0}^{a} 2 \sqrt{a} \sqrt{x} \cdot d x=\frac{8 a^{2}}{3} \\
t_{1}=t_{2}
\end{array}$
Area of PQRS
$=2 \mathrm{a}\left|\mathrm{t}_{1}-\mathrm{t}_{2}\right| \times\left|\mathrm{a}-\mathrm{at}_{1}^{2}\right|$
But $\mathrm{t}_{1}=-\mathrm{t}_{2}$
$=4 \mathrm{a}^{2} \mathrm{t}_{1}\left(1-\mathrm{t}_{1}^{2}\right)$
Differentiation with respect $\mathrm{t}_{1}$
We will get $\mathrm{t}_{1}^{2}=\frac{1}{3}$
Now put $\mathrm{t}_{1}=\frac{1}{\sqrt{3}}$ get Area of $\mathrm{PQRS}=\frac{8 \mathrm{a}^{2}}{3 \sqrt{3}}$
ratio Becomes $\frac{1}{\sqrt{3}}$

Topic: Mathematics, Parabola, JEE Main
Let $\mathrm{ABC}$ be a triangle with vertices at points $\mathrm{A}$ $(2,3,5)$, B $(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$ in three dimensional space. If the median through $\mathrm{A}$ is equally inclined with the axes, then $(\lambda, \mu)$ is equal to:

Options:
A
$(10,7)$

B
$(7,5)$

C
$(7,10)$

D
$(5,7)$

The correct answer is:
$(7,10)$

Solution:
Since $\mathrm{AD}$ is the median
$$
\therefore \quad \mathrm{D}=\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right)
$$
Now, dR's of AD is
$$
a=\left(\frac{\lambda-1}{2}-2\right)=\frac{\lambda-5}{2}
$$
$$
b=4-3=1, c=\frac{\mu+2}{2}-5=\frac{\mu-8}{2}
$$
Also, $a, b, c$ are $\mathrm{dR}$ 's
$\therefore a=k l, b=k m, c=k n$ where $l=m=n$ and $l^2+m^2+n^2=1$
$$
\Rightarrow l=m=n=\frac{1}{\sqrt{3}}
$$
Now, $a=1, b=1$ and $c=1$
$\Rightarrow \lambda=7$ and $\mu=10$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (25 Apr Online)
Let $\mathrm{ABC}$ be an acute scalene triangle, and $\mathrm{O}$ and $\mathrm{H}$ be its circumcentre and orthocenter respectively. Further let $\mathrm{N}$ be the midpoint of OH. The value of the vector sum $\overrightarrow{\mathrm{NA}}+\overrightarrow{\mathrm{NB}}+\overrightarrow{\mathrm{NC}}$ is

Options:
A $\overrightarrow{0}$ (zero vector)
B $\overrightarrow{\mathrm{HO}}$
C $\frac{1}{2} \overrightarrow{\mathrm{HO}}$
D $\frac{1}{2} \overrightarrow{\mathrm{OH}}$

The correct answer is:
$\frac{1}{2} \overrightarrow{\mathrm{HO}}$

Solution:
Circumcenter (origin $\mathrm{O}$ )
$(\overrightarrow{\mathrm{O}})=\left(\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{3}\right) ; \mathrm{H}=\left(\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{2}\right)$
$\overrightarrow{\mathrm{N}}=\frac{1}{4}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})$

Topic: Mathematics, Straight Lines, JEE Main
Let $\mathrm{ABCD}$ be a square of a side length 1 , Let $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ be points in the interiors of the sides $A D, B C, A B, C D$, respectively, such that $P Q$ and RS intersect at right angles. If $P Q=\frac{3 \sqrt{3}}{4}$ then RS equals

Options:
A $\frac{2}{\sqrt{3}}$
B $\frac{3 \sqrt{3}}{4}$
C $\frac{\sqrt{2}+1}{2}$
D $4-2 \sqrt{2}$

The correct answer is:
$\frac{3 \sqrt{3}}{4}$

Solution:
$$
\mathrm{PQ} \perp \mathrm{RS} \Rightarrow
$$
$$
c-a=b-d
$$
$$
\begin{array}{l}
P Q=\frac{3 \sqrt{3}}{4} \\
P Q^{2}=\frac{27}{16} \\
1+(a-c)^{2}=\frac{27}{16} \\
R S=\sqrt{(b-d)^{2}+1}
\end{array}
$$
By equation $(1),(2)$ and (3)
$$
R S=\frac{3 \sqrt{3}}{4}
$$

Topic: Mathematics, Straight Lines, JEE Main
Let Ajay will not appear in JEE exam with probability p=27, while both Ajay and Vijay will appear in the exam with probability q=15. Then the probability, that Ajay will appear in the exam and Vijay will not appear is:

Options:
A 935
B 1835
C 2435
D 335

The correct answer is:
1835

Solution:
The probability that Ajay will not appear in exam is, PA'=p=27.

⇒PA=57

The probability that Ajay and Vijay will appear in exam is, PA∩V=q=15.

⇒PA×PV=15

⇒PV=15×75

⇒PV=725

The probability that Ajay will appear and Vijay will not appear in the exam is,

⇒PA∩V'=57×1825

⇒PA∩V'=1835

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Let \( \alpha \) and \( \beta \) are roots of equation \( x^{2}-2 x-7=0 \) and \( a_{n}=\alpha^{n}-\beta^{n} \), then value of \( \frac{a_{2021}-2\left(a_{2020}+a_{2019}\right)}{a_{2019}} \) is

Options:

The correct answer is:
5

Solution:
For equation x2-2x-7=0

α+β=2 and αβ=-7

Now, a2021-2a2020+a2019a2019

=α2021-β2019-2α2020-β2020+α2019-β2019α2019-β2019

=α2019α2-2α-β2019β2-2β-2α2019-β2019α2019-β2019

As α and β are roots of x2-2x-7=0

So, α2-2α=7 and β2-2β=7

=7α2019-7β2019-2α2019-β2019α2019-β2019

=5α2019-β2019α2019-β2019

=5

Topic: Mathematics, Quadratic Equation, JEE Main
Let \( \alpha \) be a root of the equation \( x^{2}+x+1=0 \) and the matrix \( A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \alpha^{2} & \alpha^{4}\end{array}\right] \), then the matrix \( A^{31} \) is equal to

Options:
A \( A^{3} \)
B \( I_{3} \)
C \( A^{2} \)
D \( A \)

The correct answer is:
\( A^{3} \)

Solution:
Here, α=ω,ω2 where ω is complex cube root of unity.

A2=131111ωω21ω2ω1111ωω21ω2ω=100001010

⇒A4=100001010100001010=100010001

⇒A31=A28×A3=I×A3=A3

Topic: Mathematics, Matrices, JEE Main
Let \( \alpha \) be the angle in radians between \( \frac{x^{2}}{36}+\frac{y^{2}}{4}=1 \) and the circle \( x^{2}+y^{2}=12 \) at their points of intersection. If tan \( ^{-1} \frac{k}{2 \sqrt{3}} \)
, then find the value of \( \frac{k^{2}}{4} \).

Options:

The correct answer is:
4

Solution:
Given ellipse equation x236+y24=1 ...i

and circle equation x2+y2=12 ...ii

After solving equation i & ii the points of intersection we get,

y=±3 & x=± 3
consider the point P3,3
Equation of the tangent at P to the circle is
3x+3y=12
Equation of the tangent at P to the ellipse is x12+34y=1
if α is angle between these tangents, then
tanα=23=423

tan-1423=α

compare with α=tan-1k23 we get k=4,

so, k24=424=4

Topic: Mathematics, Application of Derivatives, JEE Main
Let an angle of a triangle is 60°. If the variance of the angles of the triangle is 4614°, then the other two angles are

Options:
A 23° and 97°
B 22° and 98°
C 21° and 99°
D 20° and 100°

The correct answer is:
21° and 99°

Solution:
Let three angles of triangle be x, y & z.

As sum of three angles of triangle=180°.

⇒x+y+z=180°

⇒60°+y+z=180° (Given one angle is 60∘)

⇒y+z=120° …1

Now, it is given that variance of the angles of triangle is equal to 4614°.

∴x2+y2+z23=4614°

⇒60∘2+y2+z23=4614°

⇒y2+z2=10242°

⇒y+z2-2yz=10242°

⇒14400°-2yz=10242° (Using 1)

⇒yz=2079° …2

Now, y+z2=y-z2+4yz

⇒120°=y-z2+4×2079° (Using 1 & 2)

⇒y-z2=14400°-8316°=6084°

⇒y-z=78° …3

Solving equations 1 & 3, we get,

y=99° & z=21°

Thus, other two angles are 99° & 21°.

Topic: Mathematics, Statistics, JEE Main
Let an ant starts from the origin O and travels 2 units on negative x-axis, 3 units on positive y-axis and travels 3 units on negative z-axis to reach at point A. If aâ=i^-3j^+2k^ and bâ be such that the resultant of aâ and bâ is 3i^-4j^+k^, then OAâÃaâÃbâ2 is equal to

Options:

The correct answer is:
1550

Solution:
OAâ=-2i^+3j^-3k^,aâ=i^-3j^+2k^
aâ+bâ=3i^-4j^+k^ (given)
âbâ=2i^-j^-k^
OAâÃaâÃbâ=OAâ⋅bâaâ-OAâ⋅aâbâ
=-4aâ+17bâ
=-4i^-3j^+2k^+172i^-j^-k^
=30i^-5j^-25k^
OAâÃaâÃbâ2=900+25+625=1550

Topic: Mathematics, Vector Algebra, JEE Main
Let an ellipse E:x2a2+y2 b2=1,a2>b2, passes through 32,1 and has eccentricity 13. If a circle, centered at focus F(α,0),α>0, of E and radius 23, intersects E at two points P and Q, then PQ2 is equal to :

Options:
A 83
B 43
C 163
D 3

The correct answer is:
163

Solution:
The ellipse x2a2+y2b2=1 passes through the point 32,1

So, 32a2+1b2=1 ...1 and 1-b2a2=13 ...2

On solving equations 1 and 2, we get

⇒a2=3 & b2=2

⇒x23+y22=1 ...3

We know that, focii of the ellipse x2a2+y2b2=1 is ±ae, 0

Here, a=3 & e=13

∴ Its focus is 1,0 ∵α>0

Now, equation of circle is

(x-1)2+y2=43 ...4

Solving 3 and 4 we get

y=±23,x=1

⇒PQ2=1-12+23+232

=432=163

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let an ellipse with centre 1, 0 and latus rectum of length 12 have its major axis along x-axis. If its minor axis subtends an angle 60∘ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to _______.

Options:

The correct answer is:
9

Solution:
Given,

An ellipse with centre 1, 0 and latus rectum of length 12 have its major axis along x-axis

And its minor axis subtends an angle 60∘ at the foci,

So, on plotting the diagram we get,

Now from diagram we get,

2b2a=12⇒b2=a4 ...1

And tan30∘=bae

⇒132=bae2

⇒13=b2a2-b2

⇒a2-b2=3b2

⇒b2=a24 .......2

Now from equation 1 & 2 we get,

⇒a=1, b2=14⇒b=12

Hence, 2a+2b2=9

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let Î± and Î² are the roots of equation ax2+bx+c=0 aâ 0. If 1, Î±+Î², Î±Î² are in arithmetic progression and Î±, 2, Î² are in harmonic progression, then the value of Î±2+Î²2-2Î±2Î²22Î±2+Î²2 is equal to

Options:
A 0
B 0.5
C 1
D 1.5

The correct answer is:
1.5

Solution:
1, Î±+Î²,Î±Î² are in A.P. â1,-ba,ca are in A.P.
â1+ca=-2baâa+c+2b=0 â¦ 1
1Î±,12,1Î² are in A.P. â1Î±+1Î²=1âÎ±+Î²=Î±Î²
â-ba=caâb+c=0 â¦ 2
From 1 & 2 we get,
a=-b=c
âÎ±,Î² are roots of equation x2-x+1=0
Now, Î±2+Î²2-2Î±2Î²22Î±2+Î²2=12-(Î±Î²)2(Î±+Î²)2-2Î±Î²
=12-1212-21=12+1=1.5

Topic: Mathematics, Quadratic Equation, JEE Main
Let Î±, Î² and Î³ are the roots of the equation 2x3+9x2-27x-54=0. If Î±,Î²,Î³ are in geometric progression, then the value of Î±+Î²+Î³=

Options:
A 192
B 212
C 13
D 11

The correct answer is:
212

Solution:
Let, α=βr and γ=βr
α⋅β⋅γ=βr⋅β⋅βr=542=27
⇒β3=27⇒β=3
α+β+γ=βr+β+βr=-92
⇒3r+3+3r=-92
⇒1r+1+r=-32
⇒2+2r+2r2=-3r
⇒2r2+5r+2=0
⇒2r+1r+2=0⇒r=-12,-2
α=βr=3-2=-32
β=3
γ=βr=3-2=-6
α+β+γ=32+3+6=212

Topic: Mathematics, Sequences and Series, JEE Main
Let ⊕ and ⊗ are two mathematical operators. If p⊕ p⊗q is not a tautology, then ⊕ and ⊗ can be

Options:
A ∨ and ⇒ respectively
B â and â§ respectively
C ⇒ and â¨ respectively
D None of these

The correct answer is:
⇒ and ∧ respectively

Solution:
A pâ¨pâq is false only if p is false and pâq is false which is never possible simultaneously. Hence, pâ¨pâq is a tautology
C pâpâ¨q is false only if p is true and pâ¨q is false which is never possible simultaneously. Hence, pâpâ¨q is a tautology
B pâpâ§q is false only if p is true & pâ§q is false which is possible when p is true & q is false. Hence, pâpâ§q is not a tautology

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let ⊕ and ⊗ are two mathematical operators. If p⊕q⊗r is equivalent to p∧q⇒r, then ⊕ and ⊗

Options:
A can be â¨ and â§ respectively
B can be â§ and â¨ respectively
C can both be â
D can be ⇒ and â respectively

The correct answer is:
can both be ⇒

Solution:
p∧q⇒r
A p∨(q∧r) A is incorrect option
B p∧q∨r B is incorrect option
C p⇒q⇒r C is correct option
(D) p⇒q⇔r D is incorrect option

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let Î± and Î² are two positive roots of x2-2ax+ab=0 where 0<b<a, then the value of Sn=1+2ba+3ba2+.....+nban-1, ânâN
cannot exceed

Options:
A αβ
B α+βα-β
C βα
D α+βα-β4

The correct answer is:
α+βα-β4

Solution:
Sn<S∞
⇒Sn<1+2ba+3ba2+.... upto ∞

⇒Sn<11-ba+1×ba1-ba2=11-ba2=aa-b2
⇒Sn<4a24a2-4ab2
Now, α+β=2a and αβ=ab
⇒Sn<α+β2α+β2-4αβ2=α+β2α-β22
⇒Sn<α+βα-β4

Topic: Mathematics, Sequences and Series, JEE Main
Let α and β be real numbers. Consider a 3×3 matrix A such that A2=3A+αI. If A4=21A+βI, then

Options:
A α=1
B α=4
C β=8
D β=-8

The correct answer is:
β=-8

Solution:
Given, A2=3A+αI

∴A3=A2.A

⇒A3=3A2+αA

⇒A3=33A+αI+αA

⇒A3=9A+αA+3αI

Now, A4=A3.A

A4=9+αA2+3αA

=9+α3A+αI+3αA

=A27+6α+α9+α.......1

Given, A4=21A+βI.........2

On comparing equation 1 and equation 2,

⇒ 27+6α=21⇒α=-1

⇒β=α9+α=-8

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let $\alpha$ and $\beta$ be the distinct roots of $a x^2+b x+c=0$, then $\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2}$ is equal to

Options:
A
$\frac{a^2}{2}(\alpha-\beta)^2$

B
0

C
$-\frac{a^2}{2}(\alpha-\beta)^2$

D
$\frac{1}{2}(\alpha-\beta)^2$

The correct answer is:
$\frac{a^2}{2}(\alpha-\beta)^2$

Solution:
$$
\begin{aligned}
& \text { Given limit }=\lim _{x \rightarrow \alpha} \frac{1-\cos a(x-\alpha)(x-\beta)}{(x-\alpha)^2}=\lim _{x \rightarrow \alpha} \frac{2 \sin ^2\left(a \frac{(x-\alpha)(x-\beta)}{2}\right)}{(x-\alpha)^2} \\
& =\lim _{x \rightarrow \alpha} \frac{2}{(x-\alpha)^2} \times \frac{\sin ^2\left(a \frac{(x-\alpha)(x-\beta)}{2}\right)}{a^2(x-\alpha)^2(x-\beta)^2} \times \frac{a^2(x-\alpha)^2(x-\beta)^2}{4} \\
& =\frac{a^2(\alpha-\beta)^2}{2} .
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Main, JEE Main 2005
Let α and β be the roots of equation px2+qx+r = 0, p≠0. If p, q, r are in A.P. and 1 α + 1 β = 4 , then the value of α - β is

Options:
A 3 4 9
B 2 1 3 9
C 6 1 9
D 2 1 7 9

The correct answer is:
2 1 3 9

Solution:
1 α + 1 β = 4
α + β α β = 4
∴ -qprp=4
∴ -qr=4
p, q, r are in A.P.

Let p=a-d ; q=a, r=a+d.
∴ -aa+d=4 ⇒ -a=4a+4d ⇒ d=-5a4
p=a+5a4=9a4; q=a, r=a-5a4=-a4
∴ 94x2+44x-14=0
∴ 9x2+4x-1=0
α - β 2 = α + β 2 - 4 α β
= - 4 9 2 - 4 - 1 9
= 1 6 8 1 + 4 9 = 5 2 8 1
∴ α - β = 2 1 3 9 .

Hence, the correct answer is 2 1 3 9 .

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2014 (06 Apr)
Let α and β be the roots of equation x2-6x-2=0. If an=αn-βn, ∀ n≥1, then the value of a10-2a82a9 is equal to

Options:
A -3
B 6
C -6
D 3

The correct answer is:
3

Solution:
Given, α, β are roots of x2-6x-2=0.

⇒ α2-6α-2=0 and β2-6β-2=0

⇒ α2-2=6α andβ2-2=6β ...1

a10-2a82a9=α10-β10-2α8-β82α9-β9

=α10-2α8-β10-2β82α9-β9

= α 8 α 2 - 2 - β 8 β 2 - 2 2 α 9 - β 9

= α 8 6 α - β 8 6 β 2 α 9 - β 9 [ ∵ From (1)]

=6α9-6β92α9-β9=3

Topic: Mathematics, Quadratic Equation, JEE Main
Let α and β be the roots of equation x2-6x-2=0. If an=αn-βn, ∀ n≥1, then the value of a10-2a82a9 is equal to

Options:
A -3
B 6
C -6
D 3

The correct answer is:
3

Solution:
Given, α, β are roots of x2-6x-2=0.

⇒ α2-6α-2=0 and β2-6β-2=0

⇒ α2-2=6α andβ2-2=6β ...1

a10-2a82a9=α10-β10-2α8-β82α9-β9

=α10-2α8-β10-2β82α9-β9

= α 8 α 2 - 2 - β 8 β 2 - 2 2 α 9 - β 9

= α 8 6 α - β 8 6 β 2 α 9 - β 9 [ ∵ From (1)]

=6α9-6β92α9-β9=3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2015 (04 Apr)
Let Î±, Î² and Î³ be the roots of fx=0, where fx=x3+x2-5x-1. Then the value of Î±+Î²+Î³ is equal to (where â denotes the greatest integer function)

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

The correct answer is:
-3

Solution:
fx=x3+x2-5x-1
f â² x =3 x 2 +2xâ5= xâ1 3x+5
x=-53â point of local maxima,
x=1â point of local minima,
Increasing function in -â,-53âª1,â
Decreasing function in -53,1
Also, f0=-1,f1=-4,f2=1,f-1=5
f-2=-8+4+10-1=5
f-3=-27+9+15-1=-4
âÎ±â-3,-2; Î²â-1,0; Î³â1,2
âÎ±=-3, Î²=-1, Î³=1
âÎ±+Î²+Î³=-3

Topic: Mathematics, Application of Derivatives, JEE Main
Let α and β be the roots of the equation, 5x2+6x-2=0. If Sn=αn+βn, n=1, 2, 3, ...., then

Options:
A 6S6+5S5=2S4
B 5S6+6S5+2S4=0
C 5S6+6S5=2S4
D 6S6+5S5+2S4=0

The correct answer is:
5S6+6S5=2S4

Solution:
If α, β are roots of quadratic equation 5x2+6x-2=0

⇒ 5α2+6α-2=0 ⇒5α2+6α=2

⇒ 5β2+6β-2=0 ⇒5β2+6β=2

5S6+6S5=5α6+β6+6α5+β5

=α45α2+6α+β45β2+6β

=2α4+β4=2S4

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Let α and β be the roots of the equation x2+2i-1=0. Then, the value of α8+β8 is equal to

Options:
A 50
B 250
C 1250
D 1550

The correct answer is:
50

Solution:
Since α and β are the roots of the equation x2+2i-1=0

so α2=-2i+1=β2

⇒α4=-2i+12=4i2+1-4i=-3-4i

⇒α8=-3-4i2=-7+24i

Similarly, β8=-7+24i

∴α8+β8=-14+48i=50

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Let α and β be the roots of the equation x2+2x+2=0, then α15+β15 is equal to

Options:
A -512
B 128
C 512
D -256

The correct answer is:
-256

Solution:
Let the two roots of x2+2x+2=0 be α and β
⇒α,β=-2±4-82=-1±i
⇒α,β=2-12±i12
=2cos3π4±isin3π4
⇒α15+β15=2cos3π4+isin3π415+2cos 3π4-isin3π415
=2152cos45π4+isin45π4+cos45π4-isin45π4
=2.2152.cos45π4=2172.cos5π4=-2172.12
=-256

Topic: Mathematics, Complex Number, JEE Main
Let Î± and Î² be the roots of the equation x2+ax+1=0, aâ 0 . Then the equation whose roots are -Î±+1Î² and -1Î±+Î² is

Options:
A x2=0
B x2+2ax+4=0
C x2-2ax+4=0
D x2-ax+1=0

The correct answer is:
x2-2ax+4=0

Solution:
α+β=-a and αβ=1
Let S and P be the sum and product of the roots of the required equation. Then,
S=-α-1β-1α-β=-(α+β)-1α+1β
=-(α+β)-α+βαβ=-(-a)--a1=2a
P=-α+1β-1α+β
=1+αβ+1αβ+1=1+1+1+1=4
So, the required equation is
x2-Sx+P=0
i.e. x2-2ax+4=0

Topic: Mathematics, Quadratic Equation, JEE Main
Let α and β be the roots of the equation x2-x-1=0 . If pk=αk+βk,k≥1, then which one of the following statements is not true?

Options:
A p3=p5-p4
B p5=11
C p1+p2+p3+p4+p5=26
D p5=p2⋅p3

The correct answer is:
p5=p2⋅p3

Solution:
α5=5α+3
β5=5β+3

P5=5α+β+6
=51+6
P5=11 and

P2=α2+β2=α+1+β+1=3
and P3=α3+β3=2α+1+2β+1=21+2=4
P2×P3=12 and P5=11⇒P5≠P2×P3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Let α and β be the roots of the equation x2+x+1=0. Then for y≠0 in R, y+1αβαy+β1β1y+α is equal to

Options:
A y3
B y(y2–1)
C y3–1
D y(y2–3)

The correct answer is:
y3

Solution:
Roots of the equation x2+x+1=0 are α and β and we know that the sum of roots of a quadratic equation ax2+bx+c=0 is -ba and ca respectively.

⇒α+β=-1 and αβ=1.

Now, let ∆=y+1αβαy+β1β1y+α

Applying R1→R1+R2+R3, we get

∆=y+1+α+βy+1+α+βy+1+α+βαy+β1β1y+α

=y+1+α+β111αy+β1β1y+α

=y+1+-11y+βy+α-1-1αy+α-β+1α-βy+β

=yy2+α+βy+αβ-yα-α2+β+α-βy-β2

=yy2+α+βy+αβ-yα+β+1-α2+β2+α+β

On putting the values of α+β & αβ, we get

=yy2+-1y+1-y-1+1-α+β2-2αβ+-1

=yy2-y+1--12-2-1

=yy2-y+1

=y3 (on simplifying)

Topic: Mathematics, Determinants, JEE Main
Let α and β be the roots of the equation x2+x+1=0. Then for y≠0 in R, y+1αβαy+β1β1y+α is equal to

Options:
A y3
B y(y2–1)
C y3–1
D y(y2–3)

The correct answer is:
y3

Solution:
Roots of the equation x2+x+1=0 are α and β and we know that the sum of roots of a quadratic equation ax2+bx+c=0 is -ba and ca respectively.

⇒α+β=-1 and αβ=1.

Now, let ∆=y+1αβαy+β1β1y+α

Applying R1→R1+R2+R3, we get

∆=y+1+α+βy+1+α+βy+1+α+βαy+β1β1y+α

=y+1+α+β111αy+β1β1y+α

=y+1+-11y+βy+α-1-1αy+α-β+1α-βy+β

=yy2+α+βy+αβ-yα-α2+β+α-βy-β2

=yy2+α+βy+αβ-yα+β+1-α2+β2+α+β

On putting the values of α+β & αβ, we get

=yy2+-1y+1-y-1+1-α+β2-2αβ+-1

=yy2-y+1--12-2-1

=yy2-y+1

=y3 (on simplifying)

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Let α, β and γ be the roots of the equation x3+6x2-px-42=0. If α, β and γ are in arithmetic progression then α+β+γ=

Options:
A 10
B 11
C 12
D 13

The correct answer is:
12

Solution:
Let, α=β-d and γ=β+d
β-d, β, β+d are the roots
β-d+β+β+d=-6⇒β=-2
β-dββ+d=42⇒-2-d-2-2+d=42
⇒2+d22-d=-42⇒4-d2=-21
⇒d2=25⇒d=±5
⇒ The roots are -7,-2,3
⇒α+β+γ=12

Topic: Mathematics, Sequences and Series, JEE Main
Let α, β and γ be the roots of the equation x3+x+1=0, then αβα+β+βγβ+γ+γαγ+αα2+β2+γ2 is equal to

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

The correct answer is:
-32

Solution:
α+β+γ=0,∑αβ=1,αβγ=-1
∑αβα+β=∑αβ-γ=-3αβγ=3
α2+β2+γ2=α+β+γ2-2αβ+βγ+γα=-2

Hence, ∑αβα+βα2+β2+γ2=-32

Topic: Mathematics, Quadratic Equation, JEE Main
Let $\alpha$ and $\beta$ be the roots of the quadratic equation $a x^2+b x+c=0$. Observe the lists given below
The correct match of List-I from List-II is
(i) (ii) (iii) (iv)

Options:
A $\begin{array}{llll}\mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{F}\end{array}$
B $\begin{array}{llll}\mathrm{E} & \mathrm{B} & \mathrm{A} & \mathrm{D}\end{array}$
C $\begin{array}{llll}\mathrm{E} & \mathrm{D} & \mathrm{B} & \mathrm{F}\end{array}$
D E $\quad$ B $\quad$ D $\quad$ A

The correct answer is:
E $\quad$ B $\quad$ D $\quad$ A

Solution:
Using the condition that the roots of $a x^2+b x+c=0$ may be in the ratio $m: n$ is $m n b^2=a c(m+n)^2$.
(i) If the roots are $\alpha=\beta$, then
$$
\begin{aligned}
\alpha \cdot \alpha b^2 & =a c(\alpha+\alpha)^2 \\
\Rightarrow \quad b^2 & =4 a c
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad b^2=4 a c \\
& \text { (ii) If the roots are } \alpha=2 \beta \text {, then } \\
& \\
& \Rightarrow \quad \beta \cdot 2 \beta b^2=a c(\beta+2 \beta)^2 \\
& 2 b^2=9 a c
\end{aligned}
$$
(iii) If the roots are $\alpha=3 \beta$, then
$$
\Rightarrow \quad \begin{aligned}
\beta \cdot 3 \beta b^2 & =a c(\beta+3 \beta)^2 \\
\Rightarrow \quad 3 b^2 & =16 a c
\end{aligned}
$$
(iv) If the roots are $\alpha=\beta^2$, then
$$
\begin{aligned}
& \left(a^2 c\right)^{\frac{1}{2+1}}+\left(a c^2\right)^{\frac{1}{2+1}}=-b \\
\Rightarrow \quad & \left(a^2 c\right)^{\frac{1}{3}}+\left(a c^2\right)^{\frac{1}{3}}=-b
\end{aligned}
$$
Hence, option (d) is correct.

Topic: Mathematics, Quadratic Equation, JEE Main
Let α and β be the roots of x2-3x+p=0 and γ and δ be the roots of x2-6x+q=0. If α,β,γ,δ from a geometric progression. Then ratio (2 q+p):(2 q-p) is

Options:
A 3:1
B 9:7
C 5:3
D 33:31

The correct answer is:
9:7

Solution:
Let α=a, β=ar, γ=ar2, δ=ar3

α+β=3  ⇒a+ar=3 ...........1

γ+δ=6 ⇒ar2+ar3=6 .........2

By 1 and 2 ⇒ ar21+ra1+r=63

⇒ r2=2

∴ 2q+p2q−p=2γδ+αβ2γδ−αβ=2a2r5+a2r2a2r5−a2r=2r4+12r4−1=8+18−1=97

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Let α and β be the roots of x2-6x-2=0. If an=αn-βn for n⩾1, then the value of a10-2a83a9 is:

Options:
A 1
B 3
C 2
D 4

The correct answer is:
2

Solution:
α2-6α-2=0

α10-6α9-2α8=0β10-6β9-2β8=0α10-β10-6α9-β9-2α8-β8=0

⇒a10-6a9-2a8=0

⇒a10-2a83a9=2

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Let α and β be the roots of x2+x+1=0, then the equation whose roots are α2020 and β2020 is

Options:
A x2+x+1=0
B x2-x-1=0
C x2+x-1=0
D x2-x+1=0

The correct answer is:
x2+x+1=0

Solution:
x2+x+1=0
⇒x=ω,ω2
⇒α=ω and ⇒β=ω2
⇒α2020=ω2020=ω3673.ω=ω
⇒β2020=ω22020=ω32×673×ω2=ω2
⇒ The required equation is x2+x+1=0

Topic: Mathematics, Complex Number, JEE Main
Let α and β be two numbers where α<β . The geometric mean of these numbers exceeds the smaller number α by 12 and the arithmetic mean of the same numbers is smaller by 24 than the larger number β, then the value of |β-α| is

Options:

The correct answer is:
48

Solution:
Given, Î± and Î² be two number where Î±<Î² .
âµ The geometric mean of these numbers exceeds the smaller number Î± by 12 .
i.e. Î±Î²=Î±+12 .....(i)
And the arithmetic mean of the same numbers is smaller by 24 than the number Î² .
i.e. Î±+Î²2=Î²-24 ......(ii)
By solving Equations. (i) and (ii), we get
Î±=6, Î²=54
â´ Î²-Î±=54-6=48

Topic: Mathematics, Sequences and Series, JEE Main
Let α and β be two real roots of the equation k+1tan2x-2⋅λtan⁡x=1-k, where k≠-1 and λ are real numbers. If tan2α+β=50, then a value of λ is

Options:
A 102
B 10
C 5
D 52

The correct answer is:
10

Solution:
k+1tan2⁡x-2λtan⁡x+k-1=0
tan⁡α+tan⁡β=2λk+1
tanαtan⁡β=k-1k+1
tan⁡α+β=2λk+11-k-1k+1=2λ2=λ2
tan2⁡α+β=λ22=50
λ=±10

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (07 Jan Shift 1)