datavorous/entrance-exam-dataset huggingface dataset:
Let the coefficients of x-1 and x-3 in the expansion of 2x15-1x1515, x>0, be mand n respectively. If r is a positive integer such mn2=Cr.152r, then the value of r is equal to ______.

Options:

The correct answer is:
5

Solution:
The rth term of 2x15-1x1515

⇒Tr+1=Cr152x1515-r-1x15r

For term having x-1

15-r5-r5=-1⇒r=10 ⇒m=C1015·25·-110

For term having x-3

15-2r5=-3 ⇒r=15 ⇒n=C1515·20·-115

∴mn2=C1015·25=C515·25

On comparing with m·n2=Cr15·2r, we get r=5

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let the complex number z=x+iy be such that 2z-3i2z+i is purely imaginary. If x+y2=0, then y4+y2-y is equal to

Options:
A 23
B 32
C 34
D 43

The correct answer is:
34

Solution:
Given,

The complex number z=x+iy be such that 2z-3i2z+i is purely imaginary,

Now putting the value of z=x+iy in 2z-3i2z+i we get,

2x+iy-3i2x+iy+i=2x+i2y-32x+i2y+12x-i2y+12x-i2y+1

Now taking real part as,

4x2+2y-32y+14x2+2y+12=0

⇒4x2+4y2-4y-3=0

⇒x2+y2-y-34=0

Now using,

x+y2=0⇒x=-y2 we get,

x2+y2-y-34=0

⇒y4+y2-y-34=0

⇒y4+y2-y=34

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let the complex numbers Z1, Z2 and Z3 are the vertices A, B and C respectively of an isosceles right-angled triangle ABC with right angle at C, then the value of Z1-Z22Z1-Z3Z3-Z2 is equal to

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

The correct answer is:
2

Solution:
Z1-Z2Z3-Z2=2eiπ/4 … (1)
Z3-Z1Z2-Z1=12eiπ/4 … 2
On dividing 1 by 2, we get,

Z1-Z2Z2-Z1Z3-Z2Z3-Z1=2
⇒Z1-Z22Z1-Z3Z3-Z2=2

Topic: Mathematics, Complex Number, JEE Main
Let the co-ordinates of one vertex of ∆ABC be A(0,2,α) and the other two vertices lie on the line x+α5=y-12=z+43. For α∈ℤ, if the area of ∆ABC is 21 sq. units and the line segment BC has length 221 units, then α2 is equal to _______.

Options:

The correct answer is:
9

Solution:
Given,

The co-ordinates of one vertex of ∆ABC be A(0,2,α) and the other two vertices lie on the line x+α5=y-12=z+43

Now plotting the diagram we get,

Now any point on the plane x+α5=y-12=z+43=k is given by 5k-α, 2k+1, 3k-4

Now direction ratio of line AD is given by, 5k-α-0, 2k+1-2, 3k-4-α

≡5k-α, 2k-1, 3k-4-α

Now AD is perpendicular to BC,

So, 55k-α+22k-1+33k-4-α=0

⇒19k-4α-7=0 .....1

Also given area, 21=12×221×AD

⇒AD=21

⇒5k-α+2k-1+3k-4-α=21

⇒19k2-8kα+α2-14k+4α=2 ......2

Now on solving equation 1 & 2 we get,

α=3⇒α2=9.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let the coordinates of two points P and Q be 1,2 and 7,5 respectively. The line PQ is rotated through 315° in clockwise direction about the point of trisection of PQ which is nearer to Q. The equation of the line in the new position is

Options:
A 2x-y-6=0
B x-y-1=0
C 3x-y-11=0
D 3x-y-9=0

The correct answer is:
3x-y-11=0

Solution:
Point of trisection nearer to Q is 5,4.
After rotation, the slope of the line is
tan⁡θ1=tan⁡45°+tan⁡θ21-tan⁡45°tan⁡θ2=1+121-12=3 (where, tan⁡θ2 is the slope of AB )
Equation of the required line is y-4=3x-5⇒3x-y-11=0

Topic: Mathematics, Straight Lines, JEE Main
Let the curve be represented by $x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}), y=2(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t})$. Then normal at any point ' $t$ ' of the curve is at a distance of
units from the origin.

Options:
A $1$
B $0$
C $2$
D $4$

The correct answer is:
$2$

Solution:
$\begin{array}{ll} & x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}) \\ \therefore \quad & \frac{\mathrm{d} x}{\mathrm{dt}}=2(-\sin \mathrm{t}+\sin \mathrm{t}+\mathrm{t} \cos \mathrm{t})=2 \mathrm{t} \cos \mathrm{t} \\ & y=2(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}) \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{dt}}=2(\cos \mathrm{t}-\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t})=2 \mathrm{t} \sin \mathrm{t} \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{2 \mathrm{t} \sin \mathrm{t}}{2 \mathrm{t} \cos \mathrm{t}}=\mathrm{tan} \mathrm{t} \\ & \text { Slope of normal }=-\frac{1}{\frac{\mathrm{d} y}{\mathrm{~d} x}}=-\frac{1}{\tan \mathrm{t}}=-\frac{\cos \mathrm{t}}{\sin \mathrm{t}}\end{array}$
$\therefore \quad$ Equation of the normal is
$$
\begin{aligned}
& y-2(\sin t-t \cos t)=-\frac{\cos t}{\sin t}[x-2(\cos t+t \sin t)] \\
& \Rightarrow y \sin \mathrm{t}-2 \sin ^2 \mathrm{t}+2 \mathrm{t} \sin \mathrm{t} \cos \mathrm{t} \\
& =-x \cos t+2 \cos ^2 t+2 t \sin t \cos t \\
& \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2\left(\sin ^2 \mathrm{t}+\cos ^2 \mathrm{t}\right) \\
& \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2 \\
& \therefore \quad \text { Distance from origin }=\left|\frac{-2}{\sqrt{\cos ^2 \mathrm{t}+\sin ^2 \mathrm{t}}}\right|=2 \text { units } \\
&
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
Let the curve y=fx satisfies the equation dydx=1-1x2 and passes through the point 2,72, then the value of f1 is

Options:
A 3
B 2
C 72
D 1

The correct answer is:
3

Solution:
dydx=1-1x2
So, y=x+1x+C
Since, it passes through 2,72
72=2+12+C ⇒C=1
y=x+1x+1=fx
f1=3

Topic: Mathematics, Differential Equations, JEE Main
Let the curve y=yx be the solution of the differential equation, dydx=2x+1. If the numerical value of area bounded by the curve y=yx and x-axis is 483, then the value of y1 is equal to ________.

Options:

The correct answer is:
2

Solution:
dydx=2x+1

⇒∫dy=∫2x+1dx

⇒yx=x2+2x+C

Area =483

⇒2∫-1-1+1-C-x+12-C+1dx=483

⇒2-x+133-Cx+x-1-1+1-C=483

⇒-1-C3+3C-3C1-C-3C-3+31-C+3=28

⇒C=-1

⇒fx=x2+2x-1, f1=2

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let the determinant of a square matrix A of order m be m-n, where m and n satisfy 4m+n=22 and 17m+4n=93. If detn adjadjmA=3a5b6c, then a+b+c is equal to

Options:
A 84
B 96
C 101
D 109

The correct answer is:
96

Solution:
We have been given that

A=m-n, where 4m+n=22 .......i

and 17m+4n=93 .......ii

Solving i and ii

m=5, n=2

Order =5

⇒A=3

We know that adjA=Aw-1adjadjA=Aw-12 where w×w is the order of the matrix.

∴ det n adj adjmA

=2adj adj 5A

=255A16

=25 580 A16=25·316·580

=311 580 65

So, a+b+c=96

Hence this is the required option.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let the domain of the function fx=log4log5log318x-x2-77 be a,b. Then the value of the integral ∫absin3xsin3x+sin3a+b-x is equal to _____.

Options:

The correct answer is:
1

Solution:
For domain log5log318x-x2-77>0

log318x-x2-77>1

⇒18x-x2-77>3

⇒x2-18x+80<0

⇒x∈8,10

⇒a=8 and b=10

I=∫absin3xsin3x+sin3a+b-xdx

Apply ∫abfxdx=∫abfa+b-xdx

I=∫absin3a+b-xsin3x+sin3a+b-xdx

⇒2I=b-a⇒I=b-a2∵ a=8 & b=10

⇒I=10-82=1

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let the eccentricity of an ellipse x2a2+y2b2=1,a>b, be 14. If this ellipse passes through the point -425,3, then a2+b2 is equal to

Options:
A 29
B 31
C 32
D 34

The correct answer is:
31

Solution:
Given,

x2a2+y2b2=1 a>b

Now using eccentricity formula, e2=1-b2a2

We get, 116=1-b2a2

⇒b2a2=1-116=1516⇒b2=1516a2

Now again x2a2+y2b2=1 is passing through -425,3 on satisfying the point we get,

⇒16×25a2+9b2=1

⇒325a2+9b2=1

Now putting the value b2=1516a2 in above equation we get,

325a2+91516a2=1

805a2=1

16=a2

So, b2=15 and a2+b2=15+16=31

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Let the eccentricity of an ellipse x2a2+y2b2=1 is reciprocal to that of the hyperbola 2x2-2y2=1. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.

Options:

The correct answer is:
2

Solution:
Given,

The eccentricity of an ellipse x2a2+y2b2=1 is reciprocal to that of the hyperbola 2x2-2y2=1,

Now eccentricity of rectangular hyperbola 2x2-2y2=1 is eH=2

So, ee=12

Now, focus of hyperbola =±1, 0

Now given that both curve intersect orthogonally, so ellipse and hyperbola are confocal

So, for ellipse aee=1⇒a=2

Now length of latusrectum L.R.=2b2a=2a1-ee2

=22.12=2

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+9 y^{2}=9,$ then the ratio $a^{2}: b^{2}$ equals

Options:
A 8: 1
B 1: 8
C 9: 1
D 1: 9

The correct answer is:
8: 1

Solution:
Given equation of ellipses is $x^{2}+9 y^{2}=9$
$\Rightarrow \quad \frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
Here, $a=3, b=1$
$c=\sqrt{(3)^{2}-(1)^{2}}=\sqrt{8}$
$\therefore$ Eccentricity of ellipse, $e=\frac{c}{a}$ $e=\frac{\sqrt{8}}{3}$
$\therefore$ Eccentricity of hyperbola $=\frac{3}{\sqrt{8}}$
$\Rightarrow$
$1+\frac{b^{2}}{a^{2}}=\frac{9}{8}$
$\Rightarrow$
$\frac{b^{2}}{a^{2}}=\frac{1}{8}$
$a^{2}: b^{2}=8: 1$

Topic: Mathematics, Hyperbola, JEE Main
Let the eccentricity of the hyperbola H:x2a2-y2 b2=1 be 52 and length of its latus rectum be 62. If y=2x+c is a tangent to the hyperbola H, then the value of c2 is equal to

Options:
A 18
B 20
C 24
D 32

The correct answer is:
20

Solution:
Given equation of the hyperbola is x2a2-y2b2=1

If e is eccentricity of the hyperbola, then e2=1+b2a2

1+b2a2=52⇒b2=3a22

Length of the latus-rectum is given by 2b2a

i.e. 2b2a=62

2a×3a22=62

⇒a=22

and b2=32×8=12

⇒b=23

If y=mx+c is a tangent to the hyperbola, then c2=a2m2-b2

∴ c2=4×8-12

Hence c2=20

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Let the eccentricity of the hyperbola with the principal axes along the coordinate axes and passing through 3,0 and 32,2 is e, then the value of e2+1e2-1 is equal to

Options:

The correct answer is:
5.5

Solution:
One vertex of the hyperbola is 3,0.

So, assuming the equation of hyperbola as x29-y2b2=1
and passing through 32,2, we get,
⇒929-4b2=1⇒1=4b2⇒b2=4
⇒e2=1+b2a2=1+49=139
⇒e2+1e2-1=139+1139-1=224=112=5.5

Topic: Mathematics, Hyperbola, JEE Main
Let the eccentricity of the hyperbola x2a2-y2b2=1 be 54. If the equation of the normal at the point 85,125 on the hyperbola is 85x+βy=λ, then λ-β is equal to _____.

Options:

The correct answer is:
85

Solution:
Using the formula of eccentricity we get,

e2=1+b2a2=2516⇒b2a2=916 ...1

A85,125 satisfies x2a2-y2 b2=1

⇒645a2-14425 b2=1 ...2

Solving 1 & 2 b=65a=85

Normal at A is 5a2x8+5 b2y12=a2+b2

Comparing it 85x+βy=λ

Gives λ=100,β=15

So, λ-β=85

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Let the ellipse E:x2+9y2=9 intersect the positive x- and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is mn, where m and n are coprime, then m-n is equal to

Options:
A 16
B 15
C 17
D 18

The correct answer is:
17

Solution:
Given,

Ellipse E:x2+9y2=9 ........i

Now point A3,0 & B0,1 which is the intersection of given ellipse with positive axis,

So, equation of line passing through A & B is given by,

L:x3+y1=1⇒x=3-3y .......ii

Now equation of Circle with diametric point -3,0 & 3,0 is given by,

C:x2+y2=9 ........iii

Let Q be foot of perpendicular from P upon major axis,

So, from ii & iii we get,

3-3y2+y2=9

⇒y=95, 0

Hence, PQ=95

Now Area of triangle will be,

=12×OA×PQ=12×3×95=2710

Hence, m-n=17

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let the end error on the LHS and RHS be equal to 1 cm. For the balance point at O, find out the % error in the value of X? (If the end error is 1 cm from both sides then it means the correct reading will become 10 cm + 1 cm from LHS and 90 cm + 1 cm from the RHS)

Options:
A 4.2 %
B 8.1%
C 9.2%
D None

The correct answer is:
8.1%

Solution:
x 9 0 = lâ¡ 1 lâ¡ 2 = 1 1 9 1
x = 1 1 9 1 9 0
x ′ 9 0 = 1 0 9 0
⇒ x ′ = 1 0
⇒ Δ x = 1 1 9 1 9 0 - 1 0
= 9 9 0 - 9 1 0 9 1
= 8 0 9 1
Δ x % = Δ x x × 1 0 0 = 8 0 9 1 9 9 0 9 1 × 1 0 0
= 8 0 0 9 9 %
= 8.1 %

Topic: Physics, Experimental Physics, JEE Main
Let the equation of a line is x-21=y-32=z-43. An insect starts flying from P1,3,2 in a straight line meeting the given line at a point Ra,b,c and then goes to the point Q6,7,5 in a straight line such that PR is perpendicular to RQ. Then, the least value of 7a+b+c is equal to

Options:
A 105
B 45
C 10
D 7

The correct answer is:
45

Solution:
Any general point on the line is Î»+2,2Î»+3,Î»+4
PRâ=<Î»+1,2Î»,3Î»+2>
RQâ=<4-Î»,4-2Î»,1-3Î»>
From the diagram, PRââRQâ=0

âÎ»+14âÎ»+2Î»4â2Î»+3Î»+21â3Î»=0
âÎ»2-3Î»-4+4Î»2-8Î»+9Î»2+3Î»-2=0
â14Î»2-8Î»-6=0
âÎ»=1,-37
Coordinates of R are 3,5,7 or 117,157,197
Least value of 7a+b+c=11+15+19=45

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let the equation of a line through 3,6,-2 and parallel to the planes x-y+2z=5 and 3x+y+2z=6 is L=0. If point α,β,2 satisfy L=0, then α+2β is equal to

Options:
A 10
B 13
C 15
D 19

The correct answer is:
19

Solution:
A normal vector to the 1st plane is n1→=<1,-1,2>
A normal vector to the 2nd plane is n2→=<3,1,2>
A vector parallel to the required line =i^j^k^1-12312
=i^-4-j^-4+k^4
=-4i^+4j^+4k^
Equation of the required line is x-3-1=y-61=z+21
∵α,β,2 satisfy the required line
⇒α-3-1=β-61=4
⇒α=-1, β=10⇒α+2β=19

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let the equation of plane passing through the line of intersection of the planes x+2y+az=2 and x−y+z=3 be 5x−11y+bz=6a−1. For c∈ℤ, if the distance of this plane from the point a,−c,c is 2a, then a+bc is equal to

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

The correct answer is:
-4

Solution:
Let the equation of plane passing through the intersection of two given planes:

x+2y+az-2+λ(x-y+z-3)=0

⇒x(λ+1)+y(2-λ)+z(a+λ)-2-3λ=0

This is same as 5x-11y+bz=6a-1

λ+15=2-λ-11=a+λb=2+3λ6a-1

⇒-11λ-11=10-5λ

⇒6λ=-21⇒λ=-72

2-λ-11=2+3λ6a-1⇒2+72-11=2-2126a-1⇒-12=-172(6a-1)⇒6a-1=17⇒a=3

2-λ-11=a+λb⇒-12=3-72b

⇒-b2=-12⇒b=1

∴ point (a,-c,c)≡(3,-c,c)

Distance=2a=23

Plane:5 x-11 y+z=17

The distance of the plane from (3,-c,c) is

15+11c+c-17147=23

⇒|12c-2|=14

⇒c=-1,43

c=-1 c∈ℤ

∴a+bc=3+1-1=-4

Hence this is the required option.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let the equation of the pair of lines, y=px and y=qx, can be written as y-pxy-qx=0. Then the equation of the pair of the angle bisectors of the lines x2-4xy-5y2=0 is:

Options:
A x2-3xy+y2=0
B x2+4xy-y2=0
C x2+3xy-y2=0
D x2-3xy-y2=0

The correct answer is:
x2+3xy-y2=0

Solution:
We have, equation of a pair of lines x2-4xy-5y2=0

Now, equation of bisector of a pair of lines x2-y21--5=xy-2

⇒x2-y26=xy-2

⇒x2-y2=-3xy

⇒x2+3xy-y2=0.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Let the equation of the plane P containing the line x+10=8-y2=z be ax+by+3z=2(a+b) and the distance of the plane P from the point (1,27,7) be c. Then a2+b2+c2 is equal to

Options:

The correct answer is:
355

Solution:
Given,

The line x+101=y-8-2=z1 which passes through the point (-10,8,0)

Also given the equation of plane ax+by+3z=2(a+b) and the point (-10,8,0) will lie on it as plane contains the line,

So, a-10+b×8+3×0=2(a+b)

⇒b=2a ......1

Now product of direction ratio of line and direction ratio of plane will be zero as they both are perpendicular,

So, a×1+b×-2+3×1=0

⇒a-2b+3=0 ........2

Now on solving equation 1 & 2 we get, a=1and b=2

Hence, the equation of plane will be,

x+2y+3z-6=0

Now distance from (1,27,7) from plane x+2y+3z-6=0 will be,

c=1+54+21-614=7014=514

Hence, a2+b2+c2=1+4+350=355

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let the equation of the plane, that passes through the point (1,4,-3) and contains the line of intersection of the planes 3 x-2 y+4 z-7=0 and x+5 y-2 z+9=0, be αx+βy+γz+3=0, then α+β+γ is equal to :

Options:
A -15
B 15
C -23
D 23

The correct answer is:
-23

Solution:
Given plane equations are

3x-2y+4z-7 = 0 & x+5y-2z+9=0

Family of planes ⇒P1+λP2=0

⇒ (3x-2y+4z-7)+λ(x+5y-2z+9)=0
Passing through the point (1,4,-3)
⇒ (3-8-12-7)+λ(1+20+6+9)=0

⇒ λ=2436=23

Required equation of plane is 3+23x+y-2+103+z4-43-7+6=0

⇒ -11x-4y-8z+3=0

Given equation is αx+βy+γz+3=0

By comparing above equations

So, α+β+γ=-23

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Let the equation x2+y2+px+(1-p)y+5=0 represent circles of varying radius r∈(0,5]. Then the number of elements in the set S={q:q=p2 and q is an integer} is ___________

Options:

The correct answer is:
61

Solution:
r=p24+1-p24-5

=2p2-2p-192

Since r∈(0,5]

So, 0<2p2-2p-192≤5

0<2p2-2p-19≤100

⇒2p2-2p-19>0 ...i

For, 2p2-2p-19>0

p=2±4-4×2×-194

=2±1564

⇒p-1+392p-1-392>0....ii

p2=10±392

Also, 2p2-2p-119≤0 ...iii

For, 2p2-2p-119=0

p=2±4-4×2×-1194

=1±2392

⇒p-1+2392p-1-2392≤0...iv

p2=60±2392

So, from ii & iv

⇒p∈1-2392, 1-392∪1+392, 1+2392

⇒p2∈10-392,60-2392∪10+392,60+2392

So, number of integral values of p2 is 61.

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Let the equations of side BC and the internal angle bisector of angle B of ÎABC are 2x-5y+a=0 and y+x=0 respectively. If A=2,3, then the value of a is equal to

Options:
A 4
B 2
C –2
D –4

The correct answer is:
–4

Solution:
Reflection of A2,3 about line y+x=0 is A'-3,-2 which lies on side BC
⇒ 2-3-5-2+a=0⇒a=-4

Topic: Mathematics, Straight Lines, JEE Main
Let the equations of the sides PQ, QR, RS and SP of a quadrilateral PQRS are x+2y-3=0, x-1=0, x-3y-4=0 and 5x+y+12=0 respectively. If Î¸ is the angle between the diagonals PR and QS, then the value of tanâ¡Î¸ is equal to

Options:
A 2
B -2
C 1
D Not defined

The correct answer is:
Not defined

Solution:
Solving x+2y-3=0 and 5x+2y+12=0, we get,

P=-3,3
Solving x+2y-3=0 and x-1=0, we get,

Q=1,1
Solving x-1=0 and x-3y-4=0, we get,

R=1,-1
Solving x-3y-4=0 and 5x+y+12=0, we get,

S=-2,-2
m1=slope of PR=-1, m2=slope of QS=1
⇒m1m2=-1
⇒ the angle between diagonals PR & QS is 90o

Hence, tan90° is not defined

Topic: Mathematics, Straight Lines, JEE Main
Let the equations of two adjacent sides of a parallelogram ABCD be 2x-3y=-23 and 5x+4y=23. If the equation of its one diagonal AC is 3x+7y=23 and the distance of A from the other diagonal is d, then 50d2 is equal to ______________

Options:

The correct answer is:
529

Solution:
Given,

The equations of two adjacent sides of a parallelogram ABCD be 2x-3y=-23 and 5x+4y=23,

So, AB≡2x-3y=-23 and BC≡5x+4y=23

Also given, AC≡3x+7y=23

Solving the above lines we get, A(–4, 5), B(–1, 7), C(3, 2)

We know that,

Diagonal of parallelogram have same midpoint,

So AC and BD have same mid-point and let point D be x,y,

So midpoint formula we get,

x-12=-4+32⇒x=0 and y+72=2+52⇒y=0

Hence, point D is (0, 0)

Now Equation of BD will be 7x+y=0

Now finding the distance of A-4,5 from 7x+y=0 we get,

d=7(-4)+572+12=2350

Hence, 50d2=232=529

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let the equations of two ellipses be
$$
E_1: \frac{x^2}{3}+\frac{y^2}{2}=1 \text { and } E_2: \frac{x^2}{16}+\frac{y^2}{b^2}=1 \text {, }
$$
If the product of their eccentricities is $\frac{1}{2}$, then the length of the minor axis of ellipse $E_2$ is :

Options:
A
8

B
9

C
4

D
2

The correct answer is:
4

Solution:
Given equations of ellipses
$$
\begin{aligned}
& \mathrm{E}_1: \frac{x^2}{3}+\frac{y^2}{2}=1 \\
& \Rightarrow e_1=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}} \\
& \text { and } \mathrm{E}_2: \frac{x^2}{16}+\frac{y^2}{b^2}=1 \\
& \Rightarrow e_2=\sqrt{\frac{1-b^2}{16}}=\sqrt{\frac{16-b^2}{4}} \\
& \text { Also, given } e_1 \times e_2=\frac{1}{2} \\
& \Rightarrow \frac{1}{\sqrt{3}} \times \sqrt{\frac{16-b^2}{4}}=\frac{1}{2} \Rightarrow 16-b^2=12 \\
& \Rightarrow b^2=4 \\
& \therefore \text { Length of minor axis of } \\
& \mathrm{E}_2=2 b=2 \times 2=4
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2013 (22 Apr Online)
Let the first term a and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Options:
A 241
B 231
C 210
D 220

The correct answer is:
231

Solution:
Given that,

a2+ar2+ar22=33033, a,r∈N

⇒a21+r2+r4=112×273

On comparing both sides of the equation we get,

a=11 and 1+r2+r4=273

⇒r2+r4=272

⇒r21+r2=16×17

⇒r=4

Now a+ar+ar2=11+11×4+11×16

=11+44+176=231.

Therefore, the required value is 231

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let the first three terms $2, p$ and $q$, with $q \neq 2$, of a G.P. be respectively the $7^{\text {th }}, 8^{\text {th }}$ and $13^{\text {th }}$ terms of an A.P. If the $5^{\text {th }}$ term of the G.P. is the $n^{\text {th }}$ term of the A.P., then $n$ is equal to:

Options:
A 163
B 151
C 177
D 169

The correct answer is:
163

Solution:
$p^2=2 q$
$2=a+6 d$ ...(i)
$\mathrm{p}=\mathrm{a}+7 \mathrm{~d}$ ...(ii)
$q=a+12 d$ ...(iii)
$\mathrm{p}-2=\mathrm{d}$ ((ii) - (i))
$q-p=5 d$ ((iii) - (ii))
$q-p=5 d$
$\begin{aligned} & q-p=5(p-2) \\ & q=6 p-10 \\ & p^2=2(6 p-10) \\ & p^2-12 p+20=0 \\ & p=10,2 \\ & p=10 ; q=50 \\ & d=8 \\ & a=-46 \\ & 2,10,50,250,1250 \\ & a^4=a+(n-1) d \\ & 1250=-46+(n-1) 8 \\ & n=163\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let the foci and length of the latus rectum of an ellipse x2a2+y2b2=1, a>b be ±5, 0 and 50, respectively. Then, the square of the eccentricity of the hyperbola x2b2−y2a2b2=1 equals

Options:

The correct answer is:
51

Solution:
Given,

Ellipse x2a2+y2b2=1

And its focii ≡±5, 0 and its latusrectum 2b2a=50

So, ae=5 and b2=52a2

Now, using the formula eccentricity we get,

b2=a21−e2=52a2

⇒a1−e2=522 as a≠0

⇒5e1−e2=522

⇒2−2e2=e

⇒2e2+e−2=0

⇒2e2+2e−e−2=0

⇒2ee+2−11+2=0

⇒e+22e−1=0

∴e≠−2; e=12

⇒a=52 and b=5

Now, finding eccentricity of hyperbola,

x2b2−y2a2b2=1

We know that formula of eccentricity of hyperbola is given by, B2=A2eH2-1 where A2=b2 & B2=a2b2

⇒a2b2=b2eH2−1

⇒a2=eH2−1

⇒50=eH2−1

⇒eH2=51

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let the foci of a hyperbola $H$ coincide with the foci of the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$ and the eccentricity of the hyperbola $H$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $H$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3 \alpha^2+2 \beta^2$ is equal to

Options:
A 237
B 242
C 205
D 225

The correct answer is:
225

Solution:
$\begin{aligned} & \mathrm{e}_1=\sqrt{1-\frac{75}{100}}=\frac{5}{10}=\frac{1}{2} \\ & \mathrm{e}_2=2 \\ & \mathrm{~F}_1(6,1), \mathrm{F}_2(-4,1) \\ & 2 \mathrm{ae}_2=10 \Rightarrow \mathrm{a}=\frac{5}{2} \Rightarrow 2 \mathrm{a}=5 \\ & \Rightarrow \alpha=5 \\ & 4=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{b}^2=3 \mathrm{a}^2 \\ & \mathrm{~b}=\sqrt{3} \times \frac{5}{2} \\ & \beta=5 \sqrt{3} \\ & 3 \alpha^2+2 \beta^2=3 \times 25+2 \times 25 \times 3 \\ & =225\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
Let the foci of a hyperbola $H$ coincide with the foci of the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$ and the eccentricity of the hyperbola $H$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $H$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3 \alpha^2+2 \beta^2$ is equal to

Options:
A 237
B 242
C 205
D 225

The correct answer is:
225

Solution:
$\begin{aligned} & \mathrm{e}_1=\sqrt{1-\frac{75}{100}}=\frac{5}{10}=\frac{1}{2} \\ & \mathrm{e}_2=2 \\ & \mathrm{~F}_1(6,1), \mathrm{F}_2(-4,1) \\ & 2 \mathrm{ae}_2=10 \Rightarrow \mathrm{a}=\frac{5}{2} \Rightarrow 2 \mathrm{a}=5 \\ & \Rightarrow \alpha=5 \\ & 4=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{b}^2=3 \mathrm{a}^2 \\ & \mathrm{~b}=\sqrt{3} \times \frac{5}{2} \\ & \beta=5 \sqrt{3} \\ & 3 \alpha^2+2 \beta^2=3 \times 25+2 \times 25 \times 3 \\ & =225\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (09 Apr Shift 2)