datavorous/entrance-exam-dataset huggingface dataset:
Let the mirror image of the point (1,3,a) with respect to the plane r→·(2i^-j^+k^)-b=0 be (-3,5,2). Then the value of a+b is equal to ___ .

Options:

The correct answer is:
1

Solution:
plane =2x-y+z=b

Mid-point R≡-1,4,a+22→ lies on plane

∴-2-4+a+22=b

⇒a+2=2b+12⇒a=2b+10

Direction ratios of PQ→=4,-2, a-2

PQ→ is parallel to the normal vector of plane.

∴24=-1-2=1a-2

⇒a-2=2⇒a=4,b=-3

∴|a+b|=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm ), about its axis be I . The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is:

Options:
A 16 cm
B 14Â cm
C 12Â cm
D 18Â cm

The correct answer is:
16 cm

Solution:
Moment of inertia of a hollow cylinder whose inner radius id r and outer radius is R is given as

I=m(R2+r2)2

m202+1022=mk2
K=400+1002; K=250
K=510 cm

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Let the nth term of a series be given by tn=n2-n-2n2+3n,n≥3. The product t3t4t5.....t50 is equal to

Options:
A 152⋅7⋅13⋅53
B 15⋅72⋅12⋅53
C 152⋅7⋅12⋅51
D 15⋅72⋅13⋅53

The correct answer is:
15⋅72⋅13⋅53

Solution:
tn=n-2n+1nn+3
t3t4t5...tn=1â43â62â54â73â65â8...n-3nn-1n+2n-2n+1nn+3
=1â4â2â5n-1n+2nn+3
Put n=50 we get,
1â4â2â549â52â50â53
=172â5â13â53

Topic: Mathematics, Sequences and Series, JEE Main
Let the normal at a point P on the curve y2-3x2+y+10=0 intersect the y -axis at 0,32. If m is the slope of the tangent at P to the curve, then m is equal to ___________.

Options:

The correct answer is:
4

Solution:
P≡x1, y1
2yy'-6x+y'=0

⇒y'=6x11+2y1
⇒32-y1-x1=-1+2y16x1
⇒9-6y1=1+2y1⇒ y1=1
∴ x1=±2
∴ Slope of tangent =±123
=±4
∴ n=4

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let the normal at the point P on the parabola y2=6x pass through the point 5,-8. If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is

Options:
A -94
B 94
C -52
D -3

The correct answer is:
-94

Solution:
Given y2=6x comparing with y2=4a→a=32

Now normal at P is given by P32t2,3t

y+tx=3t+32t3

Now this normal is passing through 5,-8

So -8+5t=3t+32t3

⇒-16+10t=6t+3t3

⇒3t3-4t+16=0

Now solving above equation we get t=-2

So tangent at P is given by ty=x+at2

putting a & t we get -2y=x+32×4

-2y=x+6

Now tangent is intersecting at directrix which is x=-32

So -2y=-32+6 ⇒y=34-3=-94

So point of intersection is -32,-94

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let the normals at all the points on a given curve pass through a fixed point a,b. If the curve passes through 3,-3 and 4,-22, given that a-22 b=3, then a2+b2+ab is equal to _____.

Options:

The correct answer is:
9

Solution:
All normals of circle passes through centre

Radius =CA=CB

CA2=CB2

a-32+b+32

=a-42+b+222

a+3-22b=3

a-22b+3b=3 ...1

given that a-22b=3 ...2

from 1 and 2 ⇒a=3,b=0

a2+b2+ab=9

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let the normals at points A4a,-4a and B9a,-6a on the parabola y2=4ax meet at the point P. The equation of the normal from P on y2=4ax (other than PA and PB) is

Options:
A 5x+y-135a=0
B 5x-y+115a=0
C 5x+y+115=0
D 5x-y-115a=0

The correct answer is:
5x+y-135a=0

Solution:
Let, A=at12,2at1, B=at22,2at2
and the foot of the required normal is at32,2at3, then
2at1=-4a, 2at2=-6a and t1+t2+t3=0
ât1=-2, t2=-3 and -2-3+t3=0
ât3=5
Hence, the equation of the required normal is y=-t3x+2at3+at32
ây=-5x+10a+125a
ây+5x-135a=0

Topic: Mathematics, Parabola, JEE Main
Let the normals at the four points $\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)$ and $\left(x_4, y_4\right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be concurrent at some point (called as conormal point). Then $\left(x_1+x_2+x_3+x_4\right)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)$ is equal to

Options:
A 4
B 3
C -4
D 2

The correct answer is:
4

Solution:
The correct option is (1) 4
Let the point of concurrency be ($h, k$)
The equation of normal at $\left(x^{\prime}, y^{\prime}\right)$ is $\frac{x-x^{\prime}}{\frac{x^{\prime}}{a^2}}=\frac{y-y^{\prime}}{\frac{y^{\prime}}{b^2}}$
It passes through ( $h, k)$
Thus, $\left(a^2-b^2\right) x^{\prime} y^{\prime}+b^2 k x^{\prime}-a^2 h y^{\prime}=0 \cdots$
Equation (1) is satisfied by $\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)$ and $\left(x_4, y_4\right)$
Thus the curve passing through these points is $\left(a^2-b^2\right) x y+b^2 k x-a^2 h y=0$ From equation (1),
From equation (1),
$y^{\prime}=\frac{b^2 k x^{\prime}}{a^2 h-\left(a^2-b^2\right) x^{\prime}}$
But, $\frac{x^{\prime 2}}{a^2}+\frac{y^{\prime 2}}{b^2}=1 \cdots$ (3)
Substituting equation (2) in equation (3), Thus eliminating $y^{\prime}$ we get, $-\left(a^2-b^2\right) x^{\prime 4}+2 h a^2\left(a^2-b^2\right) x^{\prime 3}+\ldots-2 a^4 h\left(a^2-b^2\right) x^{\prime}+a^6 h^2=0$
Then $x_1+x_2+x_3+x_4=\frac{2 h a^2}{a^2-b^2}$
$x_1 x_2 x_3+x_2 x_3 x_4+x_3 x_4 x_1+x_1 x_2 x_4=-2 a^4 h$ and, $x_1 x_2 x_3 x_4=\frac{a^6 h^2}{-\left(a^2-b^2\right)}$
So, $\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)=\frac{2\left(a^2-b^2\right)}{a^2 h}$
Hence, $\left(x_1+x_2+x_3+x_4\right)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)=4$

Topic: Mathematics, Ellipse, JEE Main
Let the number (22)2022 + (2022)22 leave the remainder α when divided by 3 and β when divided by 7. Then (α2 + β2 ) is equal to

Options:
A 20
B 13
C 5
D 10

The correct answer is:
5

Solution:
Given that α be the remainder when 222022+202222 is divided by 3 and β be the remainder when the same is divided by 7.

⇒222022+202222=21+12022+202222.

Here 202222 is divisible by 3 as 2022 is divisible by 3.

So on expanding 21+12022, we get

⇒21+12022=C02022212022+C12022212021+......+C2022202212022

=332021×72022+C12022×32020×72021+......+1

=3k1+1

In this case the remainder is 1

Hence, α=1

Now, ⇒222022+202222=21+12022+2023-122

Take 2023-122

⇒2023-122=C022202322-C122202321+......+C2222-122

=7C02272128922-C12272028921+......+1

=7k2+1

⇒21+12022+2023-122=7k1+1+7k2+1

=7μ+2

⇒β=2.

Hence, α2+β2=12+22=5.

Therefore, the required answer is 5.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let the number of elements in sets A and B be five and two respectively. Then the number of subsets of A×B each having at least 3 and at most 6 elements is

Options:
A 752
B 782
C 792
D 772

The correct answer is:
792

Solution:
Given,

The number of elements in sets A and B be five and two respectively,

So, n(A)=5 and n(B)=2

Now Cartesian product will be, nA×B=10

Number of subsets having three elements =10C3

Number of subsets having four elements=10C4

Number of subsets having five elements =10C5

Number of subsets having six elements =10C6

So, number of subsets having atleast 3 and at most 6 elements will be =10C3+10C4+10C5+10C6

=120+210+252+210

=792

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let the number of elements of the sets $A$ and be $p$ and $q$, respectively. Then, the number relations from the set $A$ to the set $B$ is

Options:
A $2^{p+q}$
B $2^{\rho q}$
C $p+q$
D $p q$

The correct answer is:
$2^{\rho q}$

Solution:
Given number of elements of $A$ and $B$ are $p$ and q, respectively.
$\therefore$ The number of relations from the set $A$ to the $\operatorname{set} B$ is $2^{p q}$

Topic: Mathematics, Sets and Relations, JEE Main
Let the numbers 2, b, c be in an A.P. and A=1112bc4b2c2 . If det(A) ∈[2,16], then c lies in the interval:

Options:
A 2,3
B 4,6
C 3,2+234
D 2+234,4

The correct answer is:
4,6

Solution:
1112bc4b2c2→C2→C2-C1, C3→C3-C11002b-2c-24b2-4c2-4=b-2c-2c-b
Let common difference of AP=d
∴ b=2+d, c=2+2d
A=d×2d×d=2d3∈2, 16
⇒d3∈1, 8⇒d∈1, 2⇒2d∈2, 4
⇒2+2d∈4, 6
∴ c∈4, 6

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Let the orthocentre and centroid of a triangle be A-3, 5 and B3, 3 respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is:

Options:
A 352
B 10
C 210
D 352

The correct answer is:
352

Solution:
Centroid B divides the line joining Orthocentre A and circumcentre C in the ratio 2:1
AB= 3+32+5-32= 62+22= 40=210
AC=32×210=310
Radius =AC2=3104=352

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2018 (08 Apr)
Let the plane ax+by+cz+d=0 bisect the line joining the points 4,-3,1 and 2,3,-5 at the right angles. If a, b, c, d are integers, then the minimum value of a2+b2+c2+d2 is

Options:

The correct answer is:
28

Solution:
Mid-point of 4,-3,1 and 2,3,-5 is 3,0,-2 which lies on the plane.

Direction ratios of normal vector to the plane are 1,-3,3

Equation of Plane is 1(x-3)-3(y-0)+3(z+2)=0

x-3y+3z+3=0

a2+b2+c2+d2min=28

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Let the plane containing the line of intersection of the planes P1: x+λ+4y+z=1 and P2: 2x+y+z=2 pass through the points 0,1,0 and 1,0,1. Then the distance of the point 2λ,λ,-λ from the plane P2 is

Options:
A 56
B 46
C 26
D 36

The correct answer is:
36

Solution:
Given,

Equation of planes P1: x+λ+4y+z=1 and P2: 2x+y+z=2

Now we know that equation of plane passing through the planes P1 & P2 is given by P1+kP2=0

Now given it is passing through the points 0,1,0 and 1,0,1,

So, x+λ+4y+z-1+k12x+y+z-2=0 when passing through 0,1,0 we get,

0+λ+41+0-1+k12×0+1+0-2=0

⇒λ+4-1-k=0\

⇒λ+3-k=0 ........1

And when passing through 1,0,1 we get, k=-1

So, from equation 1 we get, λ=-4

So, point 2λ,λ,-λ will be -8,-4,4

And distance of point-8,-4,4 from the plane P2 will be,

D=-2×8-1×4+1×4-222+12+12=186=36

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let the plane P : 4x-y+z=10 be rotated by an angle π2 about its line of intersection with the plane x+y-z=4. If α is the distance of the point (2, 3, -4) from the new position of the plane P, then 35α is equal to

Options:
A 85
B 105
C 126
D 90

The correct answer is:
126

Solution:
We know that if a plane passes through the line of intersection of planes P1 and P2 then the family of planes is given by P1+λP2 where λ∈R.

Equation of plane after rotation :

4x-y+z-10+λx+y-z-4=0

⇒4+λx+yλ-1+z1-λ-4λ-10=0

Now angle between the new plane and 4x-y+z-10 is 90°.

⇒n→1·n→2=0

⇒4+λ4+λ-1-1+1-λ1=0

⇒16+4λ-λ+1+1-λ=0

⇒2λ=-18

⇒λ=-9

∴ Equation of plane : –5x–10y+10z+26=0

Distance of plane from (2, 3, –4)

=-10-30-40+2625+100+100=5415=α

35α=35·5415=7×543=7×18=126

Hence this is the required option.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let the plane P:8x+α1y+α2z+12=0 be parallel to the line L:x+22=y-33=z+45. If the intercept of P on the y-axis is 1 , then the distance between P and L is

Options:
A 27
B 614
C 72
D 14

The correct answer is:
14

Solution:
Given,

The plane P:8x+α1y+α2z+12=0 be parallel to the line L:x+22=y-33=z+45 and the intercept of P on the y-axis is 1,

Now normal of the plane will be perpendicular to the given line, so by perpendicular condition we get,

⇒8×2+3α1+5α2=0 …1

For Y intercept of plane put x=z=0 in equation of plane

⇒y=-12α1=1⇒α1=-12

Put α1 in equation 1 we get α2=4

Hence, equation of plane P: 2 x-3 y+z+3=0

Now distance of point A-2,3,-4 from the plane P is =2-2-33-4+322+-32+12=14.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let the plane P pass through the intersection of the planes 2x+3y-z=2 and x+2y+3z=6, and be perpendicular to the plane 2x+y-z+1=0. If d is the distance of P from the point -7,1,1, then d2 is equal to :

Options:
A 25083
B 1553
C 2583
D 25082

The correct answer is:
25083

Solution:
A plane P pass through the intersection of planes P1 and P2 can be written as,

P≡P1+λP2=0

2+λx+3+2λy+3λ-1z-2+6λ=0

Given, P⊥P3(2x+y-z+1=0)

∴ n→.n→3=0

⇒2λ+2+2λ+3-3λ-1=0

⇒4+2λ+2λ+3-3λ+1=0

⇒λ=-8

∴P≡-6x-13y-25z+46=0

6x+13y+25z-46=0

Since, the distance d from (x0,y0,z0) to the plane Ax+By+Cz+D=0 is,

d=Ax0+By0+Cz0+DA2+B2+C2

Now the distance of plane P from point -7,1,1,

d=-42+13+25-4636+169+625=50830

∴d2=50×50830=25083

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let the plane passing through the point -1,0,-2 and perpendicular to each of the planes 2x+y-z=2 and x-y-z=3 be ax+by+cz+8=0. Then the value of a+b+c is equal to:

Options:
A 3
B 8
C 5
D 4

The correct answer is:
4

Solution:
The plane is normal to x-y-z=3 & 2x+y-z=2

So, the normal of required plane is 2i^+j^-k^×i^-j^-k^

=i^j^k^21-11-1-1

=-2i^+j^-3k^

The required plane is passing through the point -1,0,-2.

So, the equation of plane is

-2x+1+1y-0-3z+2=0

⇒-2x+y-3z-8=0

⇒2x-y+3z+8=0

So, a=2,b=-1 & c=3

Hence, a+b+c=4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let the plane x+3y-2z+6=0 meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle ABC is α, β, 67, then 98α+β2 is equal to __________.

Options:

The correct answer is:
288

Solution:
Given,

The plane x+3y-2z+6=0 meet the co-ordinate axes at the points A, B, C,

So, A-6,0,0, B0,-2,0 & C0,0,3

And the orthocenter of the triangle ABC is α, β, 67,

We know that, circumcentre of triangle is point of intersection of perpendicular bisectors,

So, plotting the diagram we get,

Now from diagram we can see that O to midpoint of BC ⊥ BC

So, by perpendicular vector formula we get,

x-0×0+y+1×2+z-32×3=0

⇒4y+6z-5=0 .......1

Similarly, for side AC we get,

x+3×-6+y-0×0+z-32-3=0

⇒4x+2z+9=0 .........2

Similarly, for side AB we get,

⇒x+3×-6+y+1×2+z-0×0=0

⇒3x-y+8=0 .........3

Now from equation 1, 2 & 3 we get,

x=-94-z2, y=54-32z, z=z

Now using the relation between centroid, circumcentre and orthocentre, we get,

⇒G=α-92-z3, β+52-3z3, 67+2z3

⇒-2,-23,1=α-92-z3, β+52-3z3, 67+2z3

Now on comparing both side we get,

⇒z=1514, α=-37, β=-97

⇒98α+β2=98×14449=288

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to___________

Options:

The correct answer is:
25

Solution:
$\begin{aligned}
& \mathrm{P}(-3 \lambda-2,4 \lambda+2,2 \lambda+5) \\
& \mathrm{Q}(-\mu-2,2 \mu-6,1) \\
& \mathrm{DRS} \text { of } \mathrm{PQ}=(3 \lambda-\mu, 2 \mu-4 \lambda-8,-2 \lambda-4) \\
& \mathrm{DRS} \text { of } \mathrm{PQ}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-1 & 2 & 0 \\
-3 & 4 & 2
\end{array}\right| \\
& =(4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})
\end{aligned}$
OR
$\begin{aligned}
& (2,1,1) \\
& \frac{3 \lambda-\mu}{2}=\frac{2 \mu-4 \lambda-8}{1}=\frac{-2 \lambda-4}{1} \\
& \Rightarrow \mu=\lambda+2 \& 7 \lambda=\mu-8 \\
& \lambda=-1 \quad \mu=1 \\
& Q:(-3,-4,1) \\
& L_{P Q}=\frac{x+3}{2}=\frac{y+4}{1}=\frac{z-1}{1} \\
& (-1, \alpha, \beta) \Rightarrow 1=\frac{\alpha+4}{1}=\frac{\beta-1}{1} \\
& \Rightarrow \alpha=-3, \beta=2 \\
& (\alpha-\beta)^2=25
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let the point A lies on 3x-4y+1=0, the point B lies on 4x+3y-7=0 and the point C is -2,5. If ABCD is a rhombus, then the locus of D is

Options:
A 25x+22+y-52=3x-4y+12
B 3x-4y+12+4x+3y-72=1
C 3x-4y+12-4x+3y-72=1
D 4x+3y-72-3x-4y+12=1

The correct answer is:
25x+22+y-52=3x-4y+12

Solution:
C-2,5 lies on 4x+3y-7=0
⇒ BC is perpendicular to 3x-4y+1=0
⇒ DA is perpendicular to 3x-4y+1=0
⇒ Because DA=DC
so, the distance of D from 3x-4y+1=0 is equal to DC
⇒h+22+k-52=3h-4k+15
⇒25x+22+y-52=3x-4y+12

Topic: Mathematics, Straight Lines, JEE Main
Let the point at which the circle passing through 0,0 and 1,0 touches the circle x2+y2=9 is Ph,k, then k is equal to

Options:
A 5
B 22
C 6
D 7

The correct answer is:
22

Solution:
From the figure,

for the small circle, the radius is 32 units and the centre is 12,λ
Therefore,
122+λ2=32⇒λ=±2

Since, OP is the diameter of the circle,
⇒P=1,±22⇒k=22

Topic: Mathematics, Circle, JEE Main
Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at distance of 9 units from the point $\mathrm{P}$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Options:
A 165
B 160
C 155
D 150

The correct answer is:
155

Solution:
$P Q$ line
$\begin{aligned}
& \frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4} \\
& \operatorname{pt}(4 t+1,-2 t-2,4 t+3) \\
& \text { distance }{ }^2=16 t^2+4 t^2+16 t^2=81 \\
& t= \pm \frac{3}{2} \\
& \operatorname{pt}(7,-5,9) \\
& \alpha^2+\beta^2+\gamma^2=155
\end{aligned}$
option (1)

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let the point Pα,β be at a unit distance from each of the two lines L1:3x-4y+12=0, and L2:8x+6y+11=0. If P lies below L1 and above L2, then 100α+β is equal to

Options:
A -14
B 42
C -22
D 14

The correct answer is:
14

Solution:
Given, the point Pα,β be at a unit distance from each of the two lines L1:3x-4y+12=0, and L2:8x+6y+11=0, so on plotting the diagram we get,

âââââââ

Now, L1:3x-4y+12=0 and L2:8x+6y+11=0

Since L1 & L2 are perpendicular so it will form square of unit length, so equation of angle bisector of L1 and L2 of angle containing origin and will pass through α,β, so

Equation of angle bisector will be, 3x-4y+1232+42=8x+6y+1182+62

⇒23x-4y+12=8x+6y+11

⇒2x+14y-13=0

⇒2α+14β-13=0 â¯i

Also given perpendicular distance is one unit so, 3α-4β+125=1

⇒3α-4β+7=0 â¯ii

Solving equation i & ii

2α+14β-13=0

3α-4β+7=0

⇒Pα,β,α=-2325,β=5350

So, 100α+β=14

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Let the point p,p+1 lie inside the region E=x,y:3-x≤y≤9-x2,0≤x≤3. If the set of all values of p is the interval a,b, then b2+b-a2 is equal to ________.

Options:

The correct answer is:
3

Solution:
Given,

The point p,p+1 lie inside the region E=x,y:3-x≤y≤9-x2,0≤x≤3,

And the set of all values of p is the interval a,b,

Now by observation p,p+1 lies on y-x=1,

Also point p,p+1 lies on x+y=3

So, solving y-x=1 and x+y=3 we get, P1,2

Again given point p,p+1 lies on x2+y2=9

So, solving y-x=1 and x2+y2=9, we get

x2+(1+x)2=9

⇒2x2+2x-8=0

⇒x2+x-4=0.........i

⇒x=-1±1+4·42=17-12

Hence, p∈1,17-12

Now on comparing with p∈a,b we get,

⇒a=1, b2+b=4 [using i]

Hence, the value of b2+b-a2=4-1=3

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let the points A:0,a, B:-2,0 and C:1,1 form an obtuse angled triangle (obtuse angled at angle A), then the complete set of values of a is

Options:
A -2,1
B -2,1-23
C -1,2
D -1,2-23

The correct answer is:
-1,2-23

Solution:
0,a is a point that lies inside a circle whose end points of diameter are -2,0 and 1,1, but does not lie on this diameter.
⇒0,a lies inside circle x+2x-1+yy-1=0
⇒0+20-1+aa-1<0
⇒a2-a-2<0⇒a-2a+1<0
⇒a∈-1,2
But a≠23 because at a=23
points 0,23,-2,0 & 1,1 are collinear
⇒a∈-1,2-23

Topic: Mathematics, Circle, JEE Main
Let the points A,B,C and D are represented by complex numbers Z1,Z2,Z3 and Z4 respectively. If A,B and C are not collinear and 2Z1+Z2+Z3-4Z4=0, then the value of Area of ÎDBCArea of ÎABC is equal to _____. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
1

Solution:
Here, Z4=Z1+Z2+Z322
Let, midpoint of the line joining B & C is EZ5, where Z5=Z2+Z32, then
Z4 is midpoint of Z1 & Z5
âArea of ÎBDCArea of ÎBAC=12=0.5

Topic: Mathematics, Complex Number, JEE Main
Let the points A,B,C and D are represented by complex numbers Z1,Z2,Z3 and Z4 respectively. If A,B and C are not collinear and 2Z1+Z2+Z3-4Z4=0, then the value of Area of ÎDBCArea of ÎABC is equal to

Options:

The correct answer is:
0.5

Solution:
Here, Z4=Z1+Z2+Z322
Let, midpoint of the line joining B & C is EZ5, where Z5=Z2+Z32, then
Z4 is midpoint of Z1 & Z5
âArea of ÎBDCArea of ÎBAC=12=0.5

Topic: Mathematics, Complex Number, JEE Main
Let the points of intersections of the lines x-y+1=0, x-2y+3=0 and 2x-5y+11=0 are the mid points of the sides of a triangle ABC. Then the area of the triangle ABC is

Options:

The correct answer is:
6

Solution:
If P, Q, R are mid-points of sides of ∆ABC then 4×Area(ΔPQR)=Area(∆ABC)

Area of ΔABC=4×12a1 b1c1a2 b2c2a3 b3c32C1C2C3

Where C1,C2,C3 are cofactors of c1,c2,c3

ΔABC=4×121-111-232-5112(-1)(3)(-1)

=2×9÷3

=6

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let the points on the plane P be equidistant from the points -4,2,1 and 2,-2,3. Then the acute angle between the plane P and the plane 2x+y+3z=1 is

Options:
A π6
B π4
C π3
D 5π12

The correct answer is:
π3

Solution:
Plotting the diagram of the given information we get,

Normal vector of plane will be =AB→=OB→-OA→

=6i^-4j^+2k^ =23i^-2j^+k^

Now equation of plane is given by,

P≡3x+1-2y+1z-2=0

P≡3x-2y+z+1=0

Assuming other plane to be P'≡2x+y+3z-1=0

Now the angle between them is given by, cosθ=n^1·n^2n1n2

θ=cos-16-2+314×14

θ=cos-1714==cos-112=π3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Let the points P and Q lie on the lines y=x and y=2x respectively. The equation of the locus of the mid-point of the line segment joining P and Q, if PQ=4, is

Options:
A 25x2+36xy+13y2=4
B 25x2-36xy+13y2=4
C 25x2-36xy-13y2=4
D 25x2+36xy-13y2=4

The correct answer is:
25x2-36xy+13y2=4

Solution:
Let, Pa,a and Qb,2b and the midpoint of PQ is h,k
2h=a+b
2k=a+2b
a=22h-k and b=2k-h
Now, a-b2+a-2b2=16
⇒6h-4k2+8h-6k2=16
⇒9h2+4k2-12hk+16h2+9k2-24hk=4
⇒25x2+13y2-36xy=4

Topic: Mathematics, Straight Lines, JEE Main
Let the population of ants surviving at a time t be governed by the differential equation dptdt-pt=-100. If p0=50, then p-ln2 is equal to

Options:
A 100
B 75
C 90
D 40

The correct answer is:
75

Solution:
dptdt-pt=-100
I.F. = e-∫dt=e-t
e-tpt=100e-t+C
Now P0=50⇒C=-50
P-ln2=100–50eln12=100–25=75.

Topic: Mathematics, Differential Equations, JEE Main
Let the population of rabbits surviving at a time t be governed by the differential equation dptdt=12{pt-400}. If p(0)=100, then p(t) equals

Options:
A 600 - 500 et2
B 400-300 e-t2
C 400 - 300 et/2
D 300-200 e-t2

The correct answer is:
400 - 300 et/2

Solution:
dptdt=12{pt-400}
p0=100
2dptpt-400=dt
Integrating,
⇒∫0p(t)dp(t)pt-400=∫0t12dt⇒lnpt-4000pt=t2
⇒lnpt-400300=t2
⇒pt-400=300et2
⇒400-pt=300et2 (∴pt<400)
⇒pt=400-300et2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2014 (06 Apr)
Let the position vectors of points 'A' and 'B' be i^+j^+k^ and 2i^+j^+3k^, respectively. A point 'P' divides the line segment AB internally in the ratio λ:1λ>0. If O is the origin and OB→·OP→-3OA→×OP→2=6 then λ is equal to

Options:

The correct answer is:
0.8

Solution:
Position vector of P is OP→=a→+λb→λ+1 ∵ OB→·OP→-3|OA→×OP→|2=6

⇒ b→·a→+λb→λ+1-3a→×a→+λb→λ+12=6

⇒a→·b→+λ|b→|2λ+1-3λ2(λ+1)2|a→×b→|2=6

⇒ 6+λ.14λ+1-3λ2(λ+1)2·6=6

⇒ 18λ2(λ+1)2+6=6+8λλ+1

⇒ 18λλ+12-8λλ+1=0 λλ+1≠0

⇒ 10λ=8⇒ λ=0.8

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let the position vectors of the points A,B,C and D be 5i^+5j^+2λk^,i^+2j^+3k^,-2i^+λj^+4k^ and -i^+5j^+6k^. Let the set S={λ∈ℝ: the points A,B,C and D are coplanar}. The ∑λ∈S(λ+2)2 is equal to

Options:
A 25
B 372
C 14
D 41

The correct answer is:
41

Solution:
Given that:

A(5,5,2λ), B1,2,3, C-2,λ,4 and

D-1,5,6.

Now,

AB→≡(-4,-3,3-2λ)

AC→≡(-7,λ-5,4-2λ)

AD→≡(-6,0,6-2λ)

Given that A,B,C and D are coplanar.

∴[AB→ AC→ AD→]=0

⇒-4-33-2λ-7λ-54-2λ-606-2λ=0

Expand the determinant along the first row.

⇒-4λ-56-2λ-0--3-76-2λ--64-2λ+3-2λ0--6λ-5

On solving the above equation we get,

∴λ=3,λ=2

We have two values of λ.

⇒∑λ∈s(λ+2)2=3+22+2+22

=41

Hence, ∑λ∈s(λ+2)2=41

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (06 Apr Shift 1)