datavorous/entrance-exam-dataset huggingface dataset:
Let P1, P2…, P15 be 15 points on a circle. The number of distinct triangles formed by points Pi, Pj, Pk such that i+j+k≠15, is :

Options:
A 455
B 419
C 12
D 443

The correct answer is:
443

Solution:
Given:P1, P2…, P15 be 15 points on circle.

∴Total number of triangles=C315

When ​​​​​​​i+j+k=15 then possible cases are

i=1, j+k=14⇒(2, 12),(3, 11),(4, 10),(5, 9),(6, 8)=5 ways

i=2, j+k=13⇒(3,10),……….,(6,7)=4 ways

i=3, j+k=12⇒(4 ,8),(5, 7)=2 ways

i=4, j+k=11⇒(5, 6)=1 way

Hence, there are total 12 ways for i+j+k=15.

∴ The number of possible triangles using the vertices Pi, Pj, Pk such that i+j+k≠15 is equal to C315-12=455-12=443.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let P1: x-2y+3z=5 and P2: 2x-3y+z+4=0 be two planes. The equation of the plane perpendicular to the line of intersection of P1=0 and P2=0 and passing through 1,1,1 is

Options:
A 11x-5y+7z-13=0
B 7x+5y+z=13
C x+2y+z-4=0
D x-2y+4z+3=0

The correct answer is:
7x+5y+z=13

Solution:
The normal vector of plane P1=0 is n1→=i^-2j^+3k^
The normal vector of plane P2=0 is n2→=2i^-3j^+k^
A vector parallel to the line of intersection of P1=0 and P2=0 is

i^j^k^1-232-31
=7i^+5j^+k^
It is also a normal vector for the required plane
Hence, the equation of the required plane is
7x-1+5y-1+z-1=0
7x+5y+z-13=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P1:x+y+2z=3 and P2:x-2y+z=4 be two planes. Let A2,4,5 and B4,3,8 be two points in space. The equation of plane P3 through the line of intersection of P1 and P2 such that the length of the projection upon it of the line segment AB is the least, is

Options:
A 2x-y+3z=7
B 3y+z+1=0
C x+3y+z+2=0
D 3x-3y+4z-11=0

The correct answer is:
2x-y+3z=7

Solution:
Equation of plane P3 is P1+λP2=0
x1+λ+y1-2λ+z2+λ=3+4λ
AB→=<2,-1,3>
Length of projection of AB→ on plane is least
⇒AB→ must be perpendicular to the plane
⇒1+λ2=1-2λ-1=2+λ3⇒λ=1
So, the equation of the plane is 2x-y+3z=7

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P1: x+y+2z-4=0 and P2: 2x-y+3z+5=0 be two planes. Let A1,3,4 and B3,2,7 be two points in space. The equation of a third plane P3 through the line of intersection of P1 and P2 and parallel to AB is

Options:
A x-4y-2z+3=0
B x-4y-2z+9=0
C 2x-3y+4z+9=0
D 3y+z-13=0

The correct answer is:
3y+z-13=0

Solution:
Equation of P3 is P1+λP2=0
x1+2λ+y1-λ+z2+3λ-4+5λ=0
AB→=<2,-1,3>
a normal vector to plane P3 is <1+2λ,1-λ,2+3λ>
If AB is parallel to P3=0
⇒21+2λ-11-λ+32+3λ=0
⇒2+4λ-1+λ+6+9λ=0
⇒14λ+7=0
⇒λ=-12
Equation of required plane is 0+y×32+z12-132=0

i.e. 3y+z-13=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P1=x+y+z+1=0, P2=x-y+2z+1=0 and P3=3x+y+4z+7=0 be three planes. The distance of the line of intersection of planes P1=0 and P2=0 from the plane P3=0 is

Options:
A 226 units
B 126 units
C 426 units
D 726 units

The correct answer is:
426 units

Solution:
A vector parallel to the line of intersection is
i^j^k^1111-12=i^3-j^1+k^-2=3i^-j^-2k^
for a point on the line of intersection
Put z=0
x+y+1=0x-y+1=0⇒x=-1,y=0
So, equation line to the intersection is
x+13=y-0-1=z-0-2
∵ line of intersection is parallel to the plane P3=0
So, distance =-3+0+0+726=426

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P237, 67, Q, R and S be four points on the ellipse 9x2+4y2=36. Let PQ and RS be mutually perpendicular and pass through the origin. If 1PQ2+1RS2=pq, where p and q are coprime, then p+q is equal to

Options:
A 147
B 143
C 137
D 157

The correct answer is:
157

Solution:
Given,

P237, 67, Q, R and S be four points on the ellipse x24+y29=1

Now, OP=r1=2372+672=487 where O is origin,

Let P be r1cosθ, r1sinθ

P lies on ellipse, so we get,

r12cos2θ4+r12sin2θ9=1

⇒ cos2θ4+sin2θ9=748 ... i

Let R be -r2sinθ, r2cosθ as PQ & RS are perpendicular and pass through origin,

So, r22sin2θ4+r22cos2θ9=1

⇒sin2θ4+cos2θ9=1r22 ...ii

Now adding equation i & ii we get,

1r22=14+19-748=31144

Now solving,

1PQ2+1RS2=141OP2+1OR2

⇒1PQ2+1RS2=141r12+1r22

⇒1PQ2+1RS2=14748+31144=13144=pm

∴ p+m=157

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let P= 2α 5-3α2 and Q=2l-m5n are two matrices, where l, m, n, α∈R, then the value of determinant PQ is equal to

Options:
A 0
B -1
C 2
D not possible

The correct answer is:
0

Solution:
PQ= 4lα-2mα 10nα 10l-5m 25n-6lα2 3mα2 -15nα2
PQ=4lα-2mα 10nα10l-5m 25n-6lα2 3mα2-15nα2=2lm5n2α-2α 2α5-5 5-3α2 3α2-3α2
Here, C1 and C3 are identical
⇒PQ=0

Topic: Mathematics, Determinants, JEE Main
Let p→=2i^+3j^+k^ and q→=i^+2j^+k^ be two vectors. If a vector r→=αi^+βj^+γk^ is perpendicular to each of the vectors (p→+q→) and (p→-q→), and |r→|=3, then |α|+|β|+|γ| is equal to

Options:

The correct answer is:
3

Solution:
Given,

p→=2i^+3j^+k^

q→=i^+2j^+k^

So, p→+q→=3i^+5j^+2k^ & p→-q→=i^+j^

Now, p→+q→×p→-q→=i^j^k^352110

=i^0-2-j^0-2+k^3-5

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

⇒r→=±3((p→+q→)×(p→-q→))|(p→+q→)×(p→-q→)|=±3(-2i^+2j^-2k^)22+22+22

r→=±-i^+j^-k^

According to question

r→=αi^+βj^+γk^

So |α|=1,|β|=1,|γ|=1

⇒|α|+|β|+|γ|=3

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let pâ=2i^+j^-2k^,qâ=i^+j^. If râ is a vector such that pâ.râ=râ, râ-pâ=22 and the angle between pâÃqâ and râ is Ï6, then the value of pâÃqâÃrâ is equal to

Options:
A 32
B 12
C 332
D 3

The correct answer is:
32

Solution:
p→×q→×r→=p→×q→r→sin⁡30o
=|p→×q→||r→|12 … 1
p→×q→=i^j^k^21-2110
=2i^-2j^+k^
⇒p→×q→=3
Given, r→-p→2=8
⇒r→2+p→2-2r→.p→=8
⇒r→2+9-2r→=8
⇒r→=1
from 1,

p→×q→×r→=32

Topic: Mathematics, Vector Algebra, JEE Main
Let P(3,2,3),Q(4,6,2) and R(7,3,2) be the vertices of ∆PQR. Then, the angle ∠QPR is

Options:
A π6
B cos-1718
C cos-1118
D π3

The correct answer is:
π3

Solution:
Given,

⇒QP→=-i^-4j^+k^ and RP→=-4i^-j^+k^

cos∠QPR=QP→·RP→QP→RP→

⇒cos∠QPR=4+4+11+16+116+1+1

⇒cos∠QPR=918=12

⇒∠QPR=π3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let P3,3 be a point on the hyperbola, x2a2-y2b2=1. If the normal to it at P intersects the x-axis at 9,0 and e is its eccentricity, then the ordered pair a2,e2 is equal to:

Options:
A 92,3
B 32,2
C 92,2
D 9,3

The correct answer is:
92,3

Solution:
Given hyperbola is x2a2−y2b2=1.

P3,3 lies on hyperbola then 1a2−1b2=19 ...1

Normal at 3, 3 is a2x3+b2y3=a2+b2

Normal passes through 9, 0

3a2=a2+b2 ⇒2a2=b2

Then using (1), 1a2−12a2=19

2a2=9⇒a2=92 and b2=9

e2=1+b2a2=1+2=3.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Let $\mathrm{P}(3 \sec \theta, 2 \tan \theta)$ and $\mathrm{Q}(3 \sec \phi, 2 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$, be two distinct points on the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$. Then the ordinate of the point of intersection of the normals at $\mathrm{P}$ and $\mathrm{Q}$ is:

Options:
A
$\frac{11}{3}$

B
$-\frac{11}{3}$

C
$\frac{13}{2}$

D
$-\frac{13}{2}$

The correct answer is:
$-\frac{13}{2}$

Solution:
Let the coordinate at point of intersection of normals at $\mathrm{P}$ and $\mathrm{Q}$ be $(h, k)$ Since, equation of normals to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ At point $\left(x_1, y_1\right)$ is $\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2$ therefore equation of normal to the hyperbola $\frac{x^2}{3_1^2}-\frac{y^2}{2^2}=1$ at point $P(3 \sec \theta, 2 \tan \theta)$ is
$$
\begin{aligned}
&\frac{3^2 x}{3 \sec \theta}+\frac{2^2 y}{2 \tan \theta}=3^2+2^2 \\
&\Rightarrow 3 x \cos \theta+2 y \cot \theta=3^2+2^2
\end{aligned}
$$
Similarly, Equation of normal to the hyperbola $\frac{x^2}{3^2}-\frac{y^2}{2^2}$ at point Q $(3 \sec \phi, 2 \tan \phi)$ is
$$
\begin{aligned}
&\frac{3^2 x}{3 \sec \phi}+\frac{2^2 y}{2 \tan \phi}=3^2+2^2 \\
&\Rightarrow 3 x \cos \phi+2 y \cot \phi=3^2+2^2
\end{aligned}
$$
Given $\theta+\phi=\frac{\pi}{2} \Rightarrow \phi=\frac{\pi}{2}-\theta$ and these passes through $(h, k)$
$\therefore$ From eq. (2)
$$
\begin{aligned}
&3 x \cos \left(\frac{\pi}{2}-\theta\right)+2 y \cot \left(\frac{\pi}{2}-\theta\right)=3^2+2^2 \\
&\Rightarrow 3 h \sin \theta+2 k \tan \theta=3^2+2^2
\end{aligned}
$$
and $3 h \cos \theta+2 k \cot \theta=3^2+2^2$
Comparing equation (3) \& (4), we get
$3 h \cos \theta+2 k \cot \theta=3 h \sin \theta+2 k \tan \theta$
$3 h \cos \theta-3 h \sin \theta=2 k \tan \theta-2 k \cot \theta$
$3 h(\cos \theta-\sin \theta)=2 k(\tan \theta-\cot \theta)$
$3 h(\cos \theta-\sin \theta)$
$=2 k \frac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$
or, $3 \mathrm{~h}=\frac{-2 k(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$
Now, putting the value of equation (5) in eq. (3)
$$
\begin{aligned}
&\frac{-2 k(\sin \theta+\cos \theta) \sin \theta}{\sin \theta \cos \theta}+2 k \tan \theta=3^2+2^2 \\
&\Rightarrow 2 k \tan \theta-2 k+2 k \tan \theta=13 \\
&-2 k=13 \Rightarrow k=\frac{-13}{2}
\end{aligned}
$$
Hence, ordinate of point of intersection
of normals at $P$ and $Q$ is $\frac{-13}{2}$

Topic: Mathematics, Hyperbola, JEE Main
Let $\mathrm{P}(3 \sec \theta, 2 \tan \theta)$ and $\mathrm{Q}(3 \sec \phi, 2 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$, be two distinct points on the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$. Then the ordinate of the point of intersection of the normals at $\mathrm{P}$ and $\mathrm{Q}$ is:

Options:
A
$\frac{11}{3}$

B
$-\frac{11}{3}$

C
$\frac{13}{2}$

D
$-\frac{13}{2}$

The correct answer is:
$-\frac{13}{2}$

Solution:
Let the coordinate at point of intersection of normals at $\mathrm{P}$ and $\mathrm{Q}$ be $(h, k)$ Since, equation of normals to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ At point $\left(x_1, y_1\right)$ is $\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2$ therefore equation of normal to the hyperbola $\frac{x^2}{3_1^2}-\frac{y^2}{2^2}=1$ at point $P(3 \sec \theta, 2 \tan \theta)$ is
$$
\begin{aligned}
&\frac{3^2 x}{3 \sec \theta}+\frac{2^2 y}{2 \tan \theta}=3^2+2^2 \\
&\Rightarrow 3 x \cos \theta+2 y \cot \theta=3^2+2^2
\end{aligned}
$$
Similarly, Equation of normal to the hyperbola $\frac{x^2}{3^2}-\frac{y^2}{2^2}$ at point Q $(3 \sec \phi, 2 \tan \phi)$ is
$$
\begin{aligned}
&\frac{3^2 x}{3 \sec \phi}+\frac{2^2 y}{2 \tan \phi}=3^2+2^2 \\
&\Rightarrow 3 x \cos \phi+2 y \cot \phi=3^2+2^2
\end{aligned}
$$
Given $\theta+\phi=\frac{\pi}{2} \Rightarrow \phi=\frac{\pi}{2}-\theta$ and these passes through $(h, k)$
$\therefore$ From eq. (2)
$$
\begin{aligned}
&3 x \cos \left(\frac{\pi}{2}-\theta\right)+2 y \cot \left(\frac{\pi}{2}-\theta\right)=3^2+2^2 \\
&\Rightarrow 3 h \sin \theta+2 k \tan \theta=3^2+2^2
\end{aligned}
$$
and $3 h \cos \theta+2 k \cot \theta=3^2+2^2$
Comparing equation (3) \& (4), we get
$3 h \cos \theta+2 k \cot \theta=3 h \sin \theta+2 k \tan \theta$
$3 h \cos \theta-3 h \sin \theta=2 k \tan \theta-2 k \cot \theta$
$3 h(\cos \theta-\sin \theta)=2 k(\tan \theta-\cot \theta)$
$3 h(\cos \theta-\sin \theta)$
$=2 k \frac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$
or, $3 \mathrm{~h}=\frac{-2 k(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$
Now, putting the value of equation (5) in eq. (3)
$$
\begin{aligned}
&\frac{-2 k(\sin \theta+\cos \theta) \sin \theta}{\sin \theta \cos \theta}+2 k \tan \theta=3^2+2^2 \\
&\Rightarrow 2 k \tan \theta-2 k+2 k \tan \theta=13 \\
&-2 k=13 \Rightarrow k=\frac{-13}{2}
\end{aligned}
$$
Hence, ordinate of point of intersection
of normals at $P$ and $Q$ is $\frac{-13}{2}$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2014 (11 Apr Online)
Let $P\left(\frac{\pi}{4}\right), Q\left(\frac{5 \pi}{4}\right), R\left(\frac{3 \pi}{4}\right), T\left(\frac{7 \pi}{4}\right)$ be the points on the hyperbola $x^2-4 y^2-4=0$ in the parametric form. Then the area of the quadrilateral $P Q R T$ is (in square units)

Options:
A $4 \sqrt{2}$
B $16 \sqrt{2}$
C $32 \sqrt{2}$
D $8 \sqrt{2}$

The correct answer is:
$8 \sqrt{2}$

Solution:
Equation of hyperbola $x^2-4 y^2-4=0$
$$
\frac{x^2}{4}-\frac{y^2}{1}=1
$$
Then, $a=2, b=1$
Parametric point on hyperbola $=(a \sec \theta, b \tan \theta)$
Then, $\mathrm{P}\left(\frac{\pi}{4}\right)=\left(2 \sec \left(\frac{\pi}{4}\right), 1 \tan \left(\frac{\pi}{4}\right)\right)$
$$
=(2 \sqrt{2}, 1)
$$
Similarly, Q $\left(\frac{5 \pi}{4}\right)=\left(2 \sec \left(\frac{5 \pi}{4}\right), 1 \tan \left(\frac{5 \pi}{4}\right)\right)$
$$
\begin{aligned}
& =(-2 \sqrt{2}, 1) \\
& \mathrm{R}\left(\frac{3 \pi}{4}\right)=(-2 \sqrt{2},-1) \\
& \mathrm{T}\left(\frac{7 \pi}{4}\right)=(2 \sqrt{2},-1)
\end{aligned}
$$
Area of quadrilateral PQRT(sq. unit)
$$
\begin{aligned}
& =\frac{1}{2}\left\{\left(x_1-x_3\right)\left(y_2-y_4\right)-\left(x_2-x_4\right)\left(y_1-y_3\right)\right\} \\
& =\frac{1}{2}\{(2 \sqrt{2}+2 \sqrt{2})(1+1)-(-2 \sqrt{2}-2 \sqrt{2})(1+1)\} \\
& =\frac{1}{2}\{4 \sqrt{2} \times 2+4 \sqrt{2} \times 2\} \\
& =\frac{1}{2} \times 16 \sqrt{2}=8 \sqrt{2}
\end{aligned}
$$
So, option (d) is correct.

Topic: Mathematics, Hyperbola, JEE Main
Let P=1,0 and Q=-1,0 are two fixed points and R is a variable point on one side of the line PQ such that ∠RPQ-∠RQP=45o, then the locus of R is

Options:
A y2-x2+2xy-1=0
B x2-y2+2xy+1=0
C y2+x2-2xy=1
D y2-x2-2xy+1=0

The correct answer is:
y2-x2-2xy+1=0

Solution:
Let, ∠RPQ=θ & ∠RQP=ϕ, where θ-ϕ=45∘
Now, tanθ-ϕ=1
⇒tan⁡θ-tan⁡ϕ1+tan⁡θtan⁡ϕ=1, where tan⁡θ=y1-x & tan⁡ϕ=yx+1
⇒y1-x-yx+1=1+y1-xyx+1
⇒2yx=1-x2+y2
⇒y2-x2-2xy+1=0

Topic: Mathematics, Straight Lines, JEE Main
Let P1,2,3 be a point in space and Q be a point on the line x-12=y-35=z-13 such that PQ is parallel to 5x-4y+3z=1. If the length of PQ is equal to k units, then the value of k2 is equal to

Options:

The correct answer is:
3825

Solution:
Given, P1,2,3
Any general point on the line is Q2λ+1,5λ+3,3λ+1
PQ→=2λi^+5λ+1j^+3λ-2k^
Since, PQ→ is parallel to the given plane, therefore
10λ-20λ-4+9λ-6=0
⇒-λ-10=0⇒λ=-10
Hence, the length of PQ=400+2401+1024=3825

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P≡a,0, Q≡-1,0 and R≡2,0 are three given points. If the locus of the point S satisfying the relation SQ2+SR2=2SP2 is 2x+3=0, then the sum of all possible values of a is

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

The correct answer is:
-3

Solution:
We have, P≡a,0,Q≡-1,0,R≡2,0 .

Let, S≡x,y
Now, given that SQ2+SR2=2SP2. Hence,
x+12+y2+x-22+y2=2x-a2+y2
or 2x2+2y2-2x+5=2x2+2y2-4ax+2a2
or 4a-2x=2a2-5 or 2x+3=0
or x=-32=2a2-54a-2

⇒a=1,-4

Topic: Mathematics, Straight Lines, JEE Main
Let Pa1, b1 and Qa2, b2 be two distinct points on a circle with center C(2,3). Let O be the origin and OC be perpendicular to both CP and CQ. If the area of the triangle OCP is 352, then a12+a22+b12+b22 is equal to __________

Options:

The correct answer is:
24

Solution:
Given,

Pa1, b1 and Qa2, b2 be two distinct points on a circle with center C(2,3),

And O be the origin and OC be perpendicular to both CP and CQ,

So, PCQ will be a straight line

Now on plotting the diagram we get,

So, OC=22+32=5

Now let CP=CQ=I
Now by using area of triangle OCP=12×OC×I we get,

352=12×5×I⇒I=7

And OP=OQ=OC2+I2=12

So, a12+b12+a22+b22=OP2+OQ2=12+12=24

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let Pa,b be a point on the parabola y2=8x such that the tangent at P passes through the centre of the circle x2+y2-10x-14y+65=0. Let A be the product of all possible values of a and B be the product of all possible values of b. Then the value of A+B is equal to

Options:
A 0
B 25
C 40
D 65

The correct answer is:
65

Solution:
Given Pa,b is point on y2=8x, such that tangent at P pass through centre of x2+y2-10x-14y+65=0 i.e. 5,7

Now tangent at P2t2,4t will be ty=x+2t2

and it pass through 5,7

So, 7t=5+2t2

⇒t=1,t=52

So, point P2t2,4t⇒2,4 when t=1

and 252,10 when t=52

Now product of all possible value of a will be A=2×252=25

And product of all possible value of b will be B=4×10=40

∴ A+B=65

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let $P=\left[a_{i j}\right]$ be a $3 \times 3$ matrix and let $Q=\left[b_{i j}\right]$, where $b_{i j}=2^{i+j} a_{i j}$ for $1 \leq i, j \leq 3$. If the determinant of $P$ is 2 , then the determinant of the matrix $Q$ is

Options:
A $2^{10}$
B $2^{11}$
C $2^{12}$
D $2^{13}$

The correct answer is:
$2^{13}$

Solution:
$|Q|=\left|\begin{array}{lll}2^{2} a_{11} & 2^{3} a_{12} & 2^{4} a_{13} \\ 2^{3} a_{21} & 2^{4} a_{22} & 2^{5} a_{23} \\ 2^{4} a_{31} & 2^{5} a_{32} & 2^{6} a_{33}\end{array}\right|$

$=2^{2} \cdot 2^{3} \cdot 2^{4}\left|\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ 2 a_{21} & 2 a_{22} & 2 a_{23} \\ 2^{2} a_{31} & 2^{2} a_{32} & 2^{2} a_{33}\end{array}\right|$

$=2^{9} \cdot 2 \cdot 2^{2}\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|$

$=2^{12} \times|P|=2^{12} \times 2=2^{13}$

Topic: Mathematics, Determinants, JEE Main
Let P and Q are two points in the xy plane on the curve y=x11â2x7+7x3+11x+6 such that OPââi^=5, OQââi^=-5, then the magnitude of OPâ+OQâ is

Options:
A 10
B 12
C 14
D 8

The correct answer is:
12

Solution:
Let P and Q be x1,y1 and x2,y2
OP→⋅i^=5⇒x1=5
OQ→⋅i^=-5⇒x2=-5
Let y=fx=x11-2x7+7x3+11x+6
∴ y1=fx1=f5
y2=fx2=f-5
OP→+OQ→=x1+x2i^+y1+y2j^
=f5+f-5j^=12j^
OP→+OQ→=12

Topic: Mathematics, Vector Algebra, JEE Main
Let P and Q be 2 circles externally touching each other at point X. Line segment AB is a direct common tangent to circles P and Q at points A and B respectively. Another common tangent to P and Q at X intersects line AB at a point Y. If BY=10 units and the radius of P is 9 units, then the value of the reciprocal of the radius of the circle Q is equal to

Options:

The correct answer is:
0.09

Solution:
AY=BY (∵Y lies on the radical axis)
From the figure,
AB2=r1+r22-r1-r22=4r1r2
AB=2r1r2
Now, BY=10 (given)

⇒AB=20=29r2

⇒r2=1009⇒1r2=0.09

Topic: Mathematics, Circle, JEE Main
Let $P$ and $Q$ be $3 \times 3$ matrices with $P \neq Q$. If $P^3=Q^3$ and $P^2 Q=Q^2 P$, then determinant of $\left(P^2+Q^2\right)$ is equal to

Options:
A
$-2$

B
$1$

C
$0$

D
$-1$

The correct answer is:
$0$

Solution:
$P^3=Q^3$
$P^3-P^2 Q=Q^3-Q^2 P$
$P^2(P-Q)=Q^2(Q-P)$
$P^2(P-Q)+Q^2(P-Q)=O$
$\left(P^2+Q^2\right)(P-Q)=O \quad \Rightarrow\left|P^2+Q^2\right|=0$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2012 (Offline)
Let $p$ and $q$ be any two logical statements and $\mathrm{r}: \mathrm{p} \rightarrow(\sim p \vee q)$. If $r$ has a truth value $F$, then the truth values of $p$ and $q$ are respectively:

Options:
A
$\mathrm{F}, \mathrm{F}$

B
$\mathrm{T}, \mathrm{T}$

C
T, F

D
F, T

The correct answer is:
T, F

Solution:
$p \rightarrow(\sim p \vee q)$ has truth value $\mathrm{F}$.
It means $p \rightarrow(\sim p \vee q)$ is false.
It means $p$ is true and $\sim p \vee q$ is false.
$\Rightarrow p$ is true and both $\sim p$ and $q$ are false.
$\Rightarrow p$ is true and $q$ is false.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2013 (25 Apr Online)
Let $p$ and $q$ be real numbers such that $p \neq 0, p^3 \neq q$ and $p^3 \neq-q$. If $\alpha$ and $\beta$ are non-zero complex numbers satisfying $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$, then $a$ quadratic equation having $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ as its roots is

Options:
A
$\left(p^3+q\right) x^2-\left(p^3+2 q\right) x$
$$
+\left(p^3+q\right)=0
$$

B
$\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$
$$
+\left(p^3+q\right)=0
$$

C
$\left(p^3-q\right) x^2-\left(5 p^3-2 q\right) x$
$$
+\left(p^3-q\right)=0
$$

D
$\left(p^3-q\right) x^2-\left(5 p^3+2 q\right) x$
$$
+\left(p^3-q\right)=0
$$

The correct answer is:
$\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$
$$
+\left(p^3+q\right)=0
$$

Solution:
Sum of roots $=\frac{\alpha^2+\beta^2}{\alpha \beta}$ and product $=1$
Given, $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$
$$
\begin{aligned}
& \Rightarrow(\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)=q \\
& \therefore \quad \alpha^2+\beta^2-\alpha \beta=\frac{-q}{p}
\end{aligned}
$$
and $\quad(\alpha+\beta)^2=p^2$
$$
\Rightarrow \quad \alpha^2+\beta^2+2 \alpha \beta=p^2
$$
From Eqs. (i) and (ii), we get
$$
\alpha^2+\beta^2=\frac{p^3-2 q}{3 p}
$$
and $\alpha \beta=\frac{p^3+q}{3 p}$
$\therefore$ Required equation is
$$
\begin{gathered}
x^2-\frac{\left(p^3-2 q\right) x}{\left(p^3+q\right)}+1=0 \\
\Rightarrow\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0
\end{gathered}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
Let p and q be the length of two chords of a circle which subtend angles 36o and 60o respectively at the centre of the circle. Then, the angle (in radian) subtended by the chord of length p+q at the centre of the circle is

(use Ï=3.1)

Options:

The correct answer is:
1.86

Solution:
If l is the length of the chord, then l=2rsinθ2 .
p=2rsin 18o, q=2rsin⁡30o
⇒p+q=2rsin18o+sin⁡30o
=2r5-14+12=2r5+14
=2rcos36o=2rsin54o=2rsin108o2
⇒p+q subtends an angle 108o at the centre
108o=1.86 radian

Topic: Mathematics, Circle, JEE Main
Let P and Q be two distinct points on a circle which has center at C2,3 and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set P,Q is equal to

Options:
A 4,0,0,6
B 2+22,3-5,2-22,3+5
C 2+22,3+5,2-22,3-5
D -1,5,5,1

The correct answer is:
-1,5,5,1

Solution:
Given the circle has centre 2,3 and it passes through the point 0,0.

So, the radius of the circle is 22+32=13

The equation of the line which passes through the point 0,0 & 2,3 is y-0=3-02-0x-0⇒y=32x

So, the slope of OC=32

From the diagram, the line PQ is perpendicular to the line OC.

So, the slope of PQ is -23

So, tanθ=-23

Using symmetric from of line

The coordinate of P=2+13cosθ,3+13sinθ

=2+13·-313,3+13213

=-1,5

The coordinate of Q=2-13cosθ,3-13sinθ

=2-13·-313,3-13213

=5,1

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let P and Q be two points on the curves x2+y2=2 and x28+y24=1 respectively. Then the minimum value of the length PQ is

Options:
A 1
B 2-2
C 22
D 2

The correct answer is:
2-2

Solution:
The shortest distance PQ occurs along the common normal;
i.e., when P is 0,2 & Q is 0,2.
Hence PQminimum=2-2

Topic: Mathematics, Application of Derivatives, JEE Main
Let p and q be two positive numbers such that p+q=2 and p4+q4=272. Then p and q are roots of the equation:

Options:
A x2-2x+2=0
B x2-2x+8=0
C x2-2x+136=0
D x2-2x+16=0

The correct answer is:
x2-2x+16=0

Solution:
Consider p2+q22-2p2q2=272

p+q2-2pq2-2p2q2=272

16-16pq+2p2q2=272

pq2-8pq-128=0

pq2-8pq-128=0

pq=8±242=16,-8

∴ pq=16

∴ Required equation

: x2-2x+16=0

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Let $p$ and $q$ be two Statements. Amongst the following, the Statement that is equivalent to $p \rightarrow q$ is

Options:
A
$p \wedge \sim q$

B
$\sim p \vee q$

C
$\sim p \wedge q$

D
$p \vee \sim q$

The correct answer is:
$\sim p \vee q$

Solution:
Let $p$ and $q$ be two statements. $p \rightarrow q$ is equivalent to $\sim p \vee q$.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2012 (19 May Online)
Let p and q be two statements. Then ~(p∧(p→~q) is equivalent to

Options:
A p∨(p∧(~q))
B p∨((~p)∧q)
C ~p∨q
D p∨(p∧q)

The correct answer is:
~p∨q

Solution:
Given,

~(p∧(p→~q)

≡~p∨~(~p∨~q) as A→B≡~A∨B

≡~p∨(p∧q) as ~A∨B≡~A∧~B

≡(~p∨p)∧(~p∨q)

≡t∧(~p∨q)

≡~p∨q

≡(~p)∨q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let $p$ and $q$ denote the following statements $p$ : The sun is shining
$q$ : I shall play tennis in the afternoon
The negation of the statement "If the sun is shining then I shall play tennis in the afternoon", is

Options:
A
$q \Rightarrow \sim p$

B
$q \wedge \sim p$

C
$p \wedge \sim q$

D
$\sim q \Rightarrow \sim p$

The correct answer is:
$p \wedge \sim q$

Solution:
Let $p$ : The sun is shining. $q$ : I shall play tennis in the afternoon.
Negation of $p \rightarrow q$ is $\sim(p \rightarrow q)=p \wedge \sim q$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2012 (26 May Online)
Let P be a non-singular matrix such that I+P+P2+â¦..+Pn= O (where O denotes the null matrix), then P-1 is

Options:
A Pn
B - Pn
C - 1+P+P2+ ….+ Pn
D None of these

The correct answer is:
Pn

Solution:
Given, I+P+P2+ â¦â¦+Pn= O
P-1+P-1P+P-1PP+ â¦â¦.+P-1PPn-1=O
â P-1+ I+IP+â¦â¦..+ IPn-1= O
â P-1= Pn

Topic: Mathematics, Matrices, JEE Main