datavorous/entrance-exam-dataset huggingface dataset:
Let P be the point of intersection of the common tangents to the parabola y2=12x and the hyperbola 8x2-y2=8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:

Options:
A 5:4
B 2:1
C 13:11
D 14:13

The correct answer is:
5:4

Solution:
Let m is slope of common tangent and an equation of tangent to hyperbola x21-y28=1 is

y=mx±1⋅m2-8 ...1

equation of tangent to parabola y2=12x is

y=mx+3m ...2

∵ 1 & 2 is same

∴ ±m2-8=3m

⇒m4-8m2-9=0⇒m2=-1 (Reject)

⇒m4-8m2-9=0⇒m2=9⇒m=±3

∵ equation of common tangents is y=3x+1 & y=-3x-1

∴ point of intersection of common tangents is R-13,0 & Foci of hyperbolas are 3,0 & -3,0

⇒-3λ+3λ+1=-13⇒λ=54

Hence, the required ratio is 5 : 4.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Let P be the point of intersection of the line x+33=y+21=1-z2 and the plane x+y+z=2. If the distance of the point P from the plane 3x-4y+12z=32 is q, then q and 2q are the roots of the equation

Options:
A x2-18x-72=0
B x2-18x+72=0
C x2+18x+72=0
D x2+18x-72=0

The correct answer is:
x2-18x+72=0

Solution:
Given,

P be the point of intersection of the line x+33=y+21=1-z2 and the plane x+y+z=2

Now let, x+33=y+21=1-z2=λ

Say, point A3λ-3, λ-2, 1-2λ

Since point lie on the plane also,

So, 3λ-3+λ-2+1-2λ=2

⇒2λ=6

⇒λ=3

Hence, the point will be P6, 1, -5

Now finding the distance of the point P6, 1, -5 from the plane 3x-4y+12z=32 we get,

q=18-4-60-329+16+144=7813=6

Equation with roots q and 2q will be,

x2-3qx+2q2=0

⇒x2-18x+72=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let $\mathrm{P}$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then, the shortest distance of P from the line $4 x=2 y=z$ is

Options:
A $\frac{5 \sqrt{14}}{7}$
B $\frac{3 \sqrt{14}}{7}$
C $\frac{ \sqrt{14}}{7}$
D $\frac{6 \sqrt{14}}{7}$

The correct answer is:
$\frac{3 \sqrt{14}}{7}$

Solution:
$\begin{aligned}
& \mathrm{L}_1 \equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda \\
& \mathrm{P}(\lambda+2,5 \lambda+4, \lambda+2) \\
& \mathrm{L}_2 \equiv \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{2} \\
& \mathrm{P}(2 \mu+3,3 \mu+2,2 \mu+3) \\
& \lambda+2=2 \mu+3 \quad 3 \mu+2=5 \lambda+4 \\
& \lambda=2 \mu+1 \quad 3 \mu=5 \lambda+2 \\
& 3 \mu=5(2 \mu+1)+2 \\
& 3 \mu=10 \mu+7 \quad \\
& \mu=-1 \quad \lambda=-1
\end{aligned}$
Both satisfies (P)
$\begin{aligned}
& \mathrm{P}(1,-1,1) \\
& \mathrm{L}_3 \equiv \frac{\mathrm{x}}{1 / 4}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{1} \\
& \mathrm{~L}_3=\frac{\mathrm{x}}{\mathrm{l}}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{4}=\mathrm{k}
\end{aligned}$
Coordinates of $\mathrm{Q}(\mathrm{k}, 2 \mathrm{k}, 4 \mathrm{k})$
$\mathrm{DR}$ 's $\mathrm{PQ}=\langle\mathrm{k}-1,2 \mathrm{k}+1,4 \mathrm{k}-1>$
$\mathrm{PQ} \perp$ to $\mathrm{L}_3$
$\begin{aligned}
& (\mathrm{k}-1)+2(2 \mathrm{k}+1)+4(4 \mathrm{k}-1)=0 \\
& \mathrm{k}-1+4 \mathrm{k}+2+16 \mathrm{k}-4=0 \\
& \mathrm{k}=\frac{1}{7} \\
& \mathrm{Q}\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right) \\
& \mathrm{PQ}=\sqrt{\left(1-\frac{1}{7}\right)^2+\left(-1-\frac{2}{7}\right)^2+\left(1-\frac{4}{7}\right)^2} \\
& =\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7} \\
& P Q=\frac{3 \sqrt{14}}{7}
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Let P be the point on the parabola, y2=8x which is at a minimum distance from the center C of the circle x2+y+62=1. Then the equation of the circle, passing through C and having its center at P is

Options:
A x2+y2-x4+2y-24=0
B x2+y2-4x+9y+18=0
C x2+y2-4x+8y+12=0
D x2+y2-x+4y-12=0

The correct answer is:
x2+y2-4x+8y+12=0

Solution:
y2=8x is the equation of the given parabola. If P is a point at a minimum distance from '0,-6', then it should be normal to the parabola at P.

Normal to parabola y2=8x is

y=mx-2·2·m-2·m3
It passes through 0,-6
⇒ m3+2m-3=0⇒m=1
Pam2,-2am=P(2,-4)
Equation of circle with centre P and passes through 'C' is

x-22+y+42=8
⇒x2+y2-4x+8y+12=0

Topic: Mathematics, Parabola, JEE Main
Let P be the point on the parabola, y2=8x which is at a minimum distance from the center C of the circle x2+y+62=1. Then the equation of the circle, passing through C and having its center at P is

Options:
A x2+y2-x4+2y-24=0
B x2+y2-4x+9y+18=0
C x2+y2-4x+8y+12=0
D x2+y2-x+4y-12=0

The correct answer is:
x2+y2-4x+8y+12=0

Solution:
y2=8x is the equation of the given parabola. If P is a point at a minimum distance from '0,-6', then it should be normal to the parabola at P.

Normal to parabola y2=8x is

y=mx-2·2·m-2·m3
It passes through 0,-6
⇒ m3+2m-3=0⇒m=1
Pam2,-2am=P(2,-4)
Equation of circle with centre P and passes through 'C' is

x-22+y+42=8
⇒x2+y2-4x+8y+12=0

Topic: Mathematics, Parabola, JEE Main, JEE Main 2016 (03 Apr)
Let $p$ be the proposition : Mathematics is a interesting and let $q$ be the propositions that Mathematics is difficult,then the symbol $p \wedge q$ means

Options:
A Mathematics is interesting implies that Mathematics is difficult
B Mathematics is interesting implies and is implied by Mathematics is difficult
C Mathematics is interesting and Mathematics is difficult
D Mathematics is interesting or Mathematics is difficult

The correct answer is:
Mathematics is interesting and Mathematics is difficult

Solution:
$p \wedge q$ means Mathematics is interesting and Mathematics is difficult. As "$\wedge$" stands for logical AND.

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let P be the relation defined on the set of all real numbers such that P=a,b:sec2a-tan2b=1. Then, P is

Options:
A reflexive and symmetric but not transitive
B symmetric and transitive but not reflexive
C reflexive and transitive but not symmetric
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
P=a,b:sec2a-tan2b=1

P=a,a:sec2a-tan2a=1

its reflexive relation.

Now P=a,b:sec2a-tan2b=1
sec2a-tan2b=1
⇒sec2a=1+tan2b
⇒sec2a=sec2b
⇒sec a=sec b
sec2a-tan2b=1
1+tan2a-sec2b+1=1
sec2b-tan2a=1
Hence, it is symmetric.
If sec a=sec b and sec b=sec c then sec a=sec c⇒ transitive
So, it is an equivalence relation.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2014 (09 Apr Online)
Let P be the relation defined on the set of all real numbers such that P = a, b : sec 2 a- tan 2 b = 1 . Then P is

Options:
A reflexive and symmetric but not transitive
B symmetric and transitive but not reflexive
C reflexive and transitive but not symmetric
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
P = a, b : sec 2 a- tan 2 b = 1
sec 2 a - tan 2 b = 1
â sec 2 a = 1 + tan 2 b
â sec 2 a = sec 2 b
â sec a = sec b
sec 2 a - tan 2 b = 1
1 + tan 2 a - sec 2 b + 1 = 1
sec 2 b - tan 2 a = 1
Hence, it is Symmetric.
If sec a = sec b and sec b = sec c then sec a = sec c â transitive
So it is an equivalence relation.

Topic: Mathematics, Sets and Relations, JEE Main
Let \( p \) be the statement 'Ravi races' and let \( q \) be the statement 'Ravi wins'. Then, the verbal translation of \( \sim(p \vee(\sim q)) \) is?

Options:
A Ravi does not race and Ravi does not win.
B It is not true that Ravi races and that Ravi does not win.
C Ravi races and Ravi wins.
D It is not true that Ravi races or that Ravi does not win.

The correct answer is:
It is not true that Ravi races or that Ravi does not win.

Solution:
Given, p: Ravi races, q: Ravi wins.

We know that the symbol '∨' is used for the logical connective word "or".

∴The statement of the given proposition ~p∨~q is "It is not true that Ravi races or that Ravi does not win".

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let $p$ be the statement " $x$ is an irrational number", $q$ be the statement " $y$ is a transcendental number", and $r$ be the statement " $\mathrm{x}$ is a rational number iff $\mathrm{y}$ is a transcendental number".
Statement $-1$ : $r$ is equivalent to either $q$ or $p$
Statement $-2: r$ is equivalent to $\sim(p \leftrightarrow \sim q)$.

Options:
A
Statement $-1$ is false, Statement $-2$ is true

B
Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$

C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.

D
Statement $-1$ is true, Statement $-2$ is false.

The correct answer is:
Statement $-1$ is true, Statement $-2$ is false.

Solution:
Given statement $r=\sim p \leftrightarrow q$
Statement $-1: r_1=(p \wedge \sim q) \vee(\sim p \wedge q)$
Statement $-2: r_2=\sim(p \leftrightarrow \sim q)=(p \wedge q) \vee(\sim q \wedge \sim p)$
From the truth table of $r, r_1$ and $r_2$, $r=r_1$.
Hence Statement $-1$ is true and Statement $-2$ is false.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2008
Let Ph,k be a point on an argand plane equidistant from the roots of the equation z+14=16z4, then the value of h is equal to

Options:
A 0
B 23
C 13
D 12

The correct answer is:
13

Solution:
z+1z4=16

⇒z+1z=2 or -2 or 2i or -2i
Hence, the roots are 1, -13, -15-25i and -15+25i

i.e. the roots are equidistant from the centre of the circle as shown in the figure.

13,0 is the centre of the circle.

Hence, h=13

Topic: Mathematics, Complex Number, JEE Main
Let Ph, k be a point on the curve y=x2+7x+2, nearest to the line, y=3x-3. Then the equation of the normal to the curve at P is

Options:
A x+3y+26=0
B x+3y-62=0
C x-3y-11=0
D x-3y+22=0

The correct answer is:
x+3y+26=0

Solution:
Tangent at Ph,k will be parallel to given line

dydxh,k= 2h+7 = 3

⇒ h=–2
Point P lies on curve k=(-2)2-7×2+2=-8

For normal at P(-2,-8)

Slope of normal=-13(Negative inverse of slope of tangent)
Equation of normal : x+3y+26=0

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Let pâ=i^+j^+k^ and râ be a variable vector such that rââi^, rââj^ and rââk^ are even natural numbers. If rââpââ¤20, then the number of values of râ is

Options:
A 20
B 60
C 75
D 120

The correct answer is:
120

Solution:
Let, râ=xi^+yj^+zk^
rââi^=xrââj^=yrââk^=z Integers (even natural numbers)
rââpâ=x+y+zâ¤20
Let, x=2k1, y=2k2, z=2k3 âk1,k2,k3âN
âk1+k2+k3â¤10
Now, adding a dummy variable Ï, such that
k1+k2+k3+Ï=10
k1=Î±+1
k2=Î²+1
k3=Î³+1 Î±,Î²,Î³âw
Î±+Î²+Î³+Ï=7ânumber of ways=7+4-1C4-1=10C3
Hence, the number of values of râ=120

Topic: Mathematics, Vector Algebra, JEE Main
Let P is a point on the line y+2x=2 and Q and R are two points on the line 3y+6x=3. If the triangle PQR is an equilateral triangle, then its area (in sq. units) is equal to

Options:
A 115
B 153
C 135
D 1215

The correct answer is:
153

Solution:
The distance between the given parallel lines h is 2-15=15
â´ Length of the side of the triangle is =2h3
Hence, the area of triangle =34Ã4h23=h23=153 sq. units

Topic: Mathematics, Straight Lines, JEE Main
Let P=limx→0+1+tan2x 12x, then logP is equal to

Options:
A 12
B 14
C 2
D 1

The correct answer is:
12

Solution:
Let P=1+tan2x12x (1∞ form)

If l=limx→afxgx, where limx→afx→1 and limx→agx→∞, then

l=elimx→afx-1·gx

⇒ P=elimx→0+1+tan2x-1·12x

⇒ P=elimx→0+12tanxx2

Using limx→0tanxx=0, we get

P=e12

⇒ logP=loge12=12.

Topic: Mathematics, Limits, JEE Main, JEE Main 2016 (03 Apr)
Let p: Maths is interesting and q: Maths is easy, then p⇒~p∨q is equivalent to

Options:
A If Maths is easy then it is interesting
B Either Maths is interesting or it easy
C If Maths is interseting then it is easy
D Maths is neither interesting nor easy

The correct answer is:
If Maths is interseting then it is easy

Solution:
p⇒~pâ¨q
â¡~pâ¨~pâ¨q
â¡~pâ¨~pâ¨q
â¡~pâ¨q
â¡p⇒q

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let Pn be a square matrix of order 3 such that Pn=aij, where aij=3i+j42n for 1≤i≤3, 1≤j≤3. Then the value of limn→∞Tr4P1+42P2.....4nPn is
(where TrA denotes trace of matrix A i.e sum of principal diagonal elements of A)

Options:
A 7
B 8
C 253
D 9

The correct answer is:
8

Solution:
∵Pn=3i+j42n=142n3i+j
4nPn=14n3i+j
Tr4P1+42P2...4nPn=Tr4P1+Tr42P2....+Tr4nPn
=244+2442.........244n
limn→∞Tr4P1+42P2....+4nPn
limn→∞244+2442.......244n
=61-14=8

Topic: Mathematics, Matrices, JEE Main
Let p, q and r are three statements in which p is true. Consider two compound statements
S1: p⇒q⇒r≡p⇒q⇒r
S2: p⟺q⟺r≡p⟺q⟺r
State in order, whether S1,S2 are true (T) or false (F)

Options:
A TT
B TF
C FT
D FF

The correct answer is:
FT

Solution:
p	q	r	p⇒q⇒r	p⇒q⇒r	p⟺q⟺r	p⟺q⟺r

T	T	T	T	T	T	T

T	F	T	T	T	F	F

T	T	F	F	F	F	F

T	F	F	T	T	T	T

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let p,q and r be real numbers p≠q,r≠0, such that the roots of the equation 1x+p+1x+q=1r are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to

Options:
A p2+q2
B p2+q22
C 2p2+q2
D p2+q2+r2

The correct answer is:
p2+q2

Solution:
Given quadratic equation is 1x+p +1x+q=1r.

Let α and β be the roots of given equation.

⇒2x+p+qr=x+px+q

⇒x2+p+q-2rx+pq-pr-qr=0

Now, sum of roots α+β=-ba=-p+q-2r

⇒-p+q-2r=0 (∵Given that roots are equal in magnitude and opposite in sign)

⇒p+q=2r ...1

Product of roots αβ=ca=pq-pr-qr

Now, α2+β2=α+β2-2αβ

=0-2pq-pr-qr=-2pq+2rp+q

=-2pq+p+q2=p2+q2 (∵ from 1)

=p2+q2

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2018 (16 Apr Online)
Let p, q and r be three statements. Consider two compound statements
S1: pâqâr â¡ pâqâr
S2: pâqâr â¡ pâqâr
State in order, whether S1,S2 are true or false.

(where, T represents true and F represents false)

Options:
A T T
B T F
C FT
D FF

The correct answer is:
FT

Solution:
p	q	r	p⇒q⇒r	p⇒q⇒r	p⇔q⇔r	p⇔q⇔r

T	T	T	T	T	T	T

T	F	T	T	T	F	F

F	T	T	T	T	F	F

F	F	T	T	T	T	T

T	T	F	F	F	F	F

T	F	F	T	T	T	T

F	T	F	F	T	T	T

F	F	F	F	T	F	F

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let p, q and r be three statements, then ~pâqâr is equivalent to

Options:
A ~pâ¨râ§qâ¨r
B pârâ§qâr
C ~p∧r∨q∨r
D pâqâr

The correct answer is:
p→r∧q→r

Solution:
~p→q→r≡p∨q→r
≡~p∨q∨r
≡~p∧~q∨r
≡~p∨r∧~q∨r≡p→r∧q→r

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let p, q and r be three statements, then pâqâr is equivalent to

Options:
A p∨r∧q∨r
B p∨r∧~q∨r
C p∧r∨q∨r
D p∨q→r

The correct answer is:
p∨r∧~q∨r

Solution:
p→q→r≡~p∨q→r
≡~~p∨q∨r
≡p∧~q∨r
≡p∨r∧~q∨r

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let PQ and RS be the tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then the value of 2r is equal to

Options:
A PQ Â· RS
B PQ + RS 2
C 2 PQ · RS PQ + RS
D PQ 2 + RS 2 2

The correct answer is:
PQ · RS

Solution:
From the figure, it is clear that Δ PRQ and Δ RSP are similar.

∴ PR RS = PQ RP
⇒ PR 2 = PQ · RS
⇒ PR = PQ · RS
⇒ 2 r = PQ · RS

Topic: Mathematics, Circle, JEE Main
Let PQ be a diameter of the circle x2+y2=9. If α and β are the lengths of the perpendiculars from P and Q on the straight line, x+y=2 respectively, then the maximum value of αβ is _______

Options:

The correct answer is:
7

Solution:
Let P(3cosθ,3sinθ)∴Q(-3cosθ,-3sinθ)
given line x+y-2=0 ∴α=|3cosθ+3sinθ-2|2
β=|-3cosθ-3sinθ-2|2
∴ αβ=(3cosθ+3sinθ-2)·(3cosθ+3sinθ+2)2=9(1+sin2θ)-42
∴ maximum of αβ=7

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Let PQ be a focal chord of the parabola y2=36x of length 100, making an acute angle with the positive x-axis. Let the ordinate of P be positive and M be the point on the line segment PQ such that PM:MQ=3:1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line PQ?

Options:
A -6,45
B 6,29
C 3,33
D -3,43

The correct answer is:
-3,43

Solution:
We have been given the length of focal chord PQ=100.

Length of focal chord at t=at+1t2=100

Where a=9

t+1t=±103

∴t=3,13,-3,-13

Since ordinate of P is +ve

∴t=3 or 13

(t=3 as slope of PQ is positive)

∴M(21,9)

Required line:y-9=-8060(x-21)

⇒4x+3y=111

-3,43 does not lie on line.

Hence this is the correct option.

Topic: Mathematics, Parabola, JEE Main
Let PQ be a focal chord of the parabola y2=36x of length 100, making an acute angle with the positive x-axis. Let the ordinate of P be positive and M be the point on the line segment PQ such that PM:MQ=3:1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line PQ?

Options:
A -6,45
B 6,29
C 3,33
D -3,43

The correct answer is:
-3,43

Solution:
We have been given the length of focal chord PQ=100.

Length of focal chord at t=at+1t2=100

Where a=9

t+1t=±103

∴t=3,13,-3,-13

Since ordinate of P is +ve

∴t=3 or 13

(t=3 as slope of PQ is positive)

∴M(21,9)

Required line:y-9=-8060(x-21)

⇒4x+3y=111

-3,43 does not lie on line.

Hence this is the correct option.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let PQ be the common chord of the circles S1:x2+y2+2x+3y+1=0 and S2:x2+y2+4x+3y+2=0, then the perimeter (in units) of the triangle C1PQ is equal to where, C1=-1,-32

Options:
A 92
B 22+3
C 32+2
D 32+22

The correct answer is:
22+3

Solution:
Equation of the common chord is 2x+1=0
C1M=-2+12=12
PM=94-14=2
Length of the common chord =22
Hence, the perimeter of ΔC1PQ=32+32+22
=3+22 units

Topic: Mathematics, Circle, JEE Main
Let PQ be the focal chord of the parabola y2=4x. If the center of the circle having PQ as its diameter lies on the line y=45, then the radius (in units) is equal to

Options:

The correct answer is:
3.6

Solution:
Let, P is t2,2t then,
Q is 1t2,-2t
Midpoint of PQ is t2+1t22,2t-2t2
⇒t-1t=45⇒t2+1t2-2=165
⇒t2+1t2+2=365
⇒t2+1+1t2+1=7.2

If S is the focus, then
PS+SQ=7.2⇒PQ=7.2

Hence, radius =3.6 units

Topic: Mathematics, Parabola, JEE Main
Let PQ be the focal chord of the parabola y2=4x. If the centre of the circle having PQ as its diameter lies on the line y=-45, then the length (in units) of chord PQ is

Options:

The correct answer is:
7.2

Solution:
Let P is t2,2t then Q is 1t2,-2t
mid point of PQ is t2+1t22,2t-2t2
⇒t-1t=-45⇒t2+1t2-2=165
⇒t2+1t2+2=365
⇒t2+1+1t2+1=7.2
PS+SQ=7.2⇒PQ=7.2
{ PS=a+at2,PQ=a+at2 where a=1 }

Topic: Mathematics, Parabola, JEE Main
Let p^×q→×p^+p^⋅q→q→=x2+y2q→+14-4x-6yp^, where p^ and q→ are non-collinear vectors ( p^ is a unit vector) and x, y are scalars, then the value of x2+y2 is equal to

Options:
A 10
B 11
C 12
D 13

The correct answer is:
13

Solution:
p^×q→×p^+p^⋅q→q→=x2+y2q→+14-4x-6yp^
p^2q→-p^⋅q→p^+p^⋅q→q→=x2+y2q→+14-4x-6yp^
Comparing both sides, we get,
14-4x-6y=-p^⋅q→
x2+y2=1+p^⋅q→
x2+y2-4x-6y+14=1
x2+y2-4x-6y+13=0
x-22+y-32=0⇒x=2, y=3

Topic: Mathematics, Vector Algebra, JEE Main
Let p, q∈ Q . If 2-3 is a root of the quadratic equation x2+px+q=0, then

Options:
A p2â4q+12=0
B q2+4p+14=0
C p2–4q–12=0
D q2â4pâ16=0

The correct answer is:
p2–4q–12=0

Solution:
In given equation x2+px+q=0, p, q∈Q.

And, we know that the irrational roots of a quadratic equation exist in conjugate pair, if the coefficients are rational.

Thus, if one root of the equation x2+px+q=0 is 2-3, then, the other root will be 2+3.

We know that the sum and product of the roots of a quadratic equation ax2+bx+c=0 are respectively -ba and ca.

Therefore, the sum of roots 2+3+2-3=-p

⇒p=-4

And, the product of roots 2+32-3=q

⇒q=22-32=1

Thus, we have p2-4q-12=-42-4×1-12

=16-16=0.

Thus, the answer is, p2-4q-12=0.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Let PQR be a triangle. The pointsA, B and C are on the sides QR, RP and PQ respectively such that QAAR=RBBP=PCCQ=12. Then Area∆PQRArea∆ABC is equal to

Options:
A 4
B 1
C 2
D 52

The correct answer is:
1

Solution:
Let the position vector of P is O→, Q is q→ and R is r→.

Now, position vector of A is

2q→+r→3

Position vector of B is 2r→3.

Position vector of C is q→3.

AB→=r→-2q→3

AC→=-r→-q→3

So,

AB→×AC→=13r→-2q→×-r→-q→

⇒AB→×AC→=13-r→×q→+2q→×r→

⇒AB→×AC→=q→×r→

Area of ∆PQR is =12q→×r→

Area of ∆ABC is 12AB→×AC→

Area of ∆ABC=12q→×r→

Therefore,

Area ∆PQRArea∆ABC=1

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let PQR be a triangle with R(-1,4,2). Suppose M(2,1,2) is the mid point of PQ. The distance of the centroid of ∆PQR from the point of intersection of the line x-20=y2=z+3-1 and x-11=y+3-3=z+11 is

Options:
A 69
B 9
C 69
D 99

The correct answer is:
69

Solution:
Given:

x-20=y2=z+3-1 and x-11=y+3-3=z+11

⇒x-20=y2=z+3-1=klet

⇒x=2, y=2k, z=-k+3 ...i

Also, x-11=y+3-3=z+11=λlet

⇒x=λ+1, y=-3λ-3. z=λ-1 ...ii

Using i and ii,

⇒2=λ+1, 2k=-3λ-3

⇒λ=1, 2k=-3-3

⇒λ=1, k=-3

⇒x=2, y=-6, z=0

So, the intersection point of two given lines is 2,-6,0.

Now, we know that centroid G divides MR (median) in the ratio 1 : 2.

So, by using section formula, G≡1×-1+2×23,1×4+2×13,1×2+2×23

So, centroid is G(1, 2, 2).

Let A be the point of intersection of given lines.

⇒AG=2-12+2+62+2-02

⇒AG=1+64+4

⇒AG=69

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let p,q,r be three logical statements. Consider the compound statements
S1:~p∨q∨~p∨r and S2:p→q∨r
Then, which of the following is NOT true?

Options:
A If S2 is True, then S1 is True
B If S2 is False, then S1 is False
C IfS2 is False, then S1 is True
D If S1 is False, then S2 is False

The correct answer is:
IfS2 is False, then S1 is True

Solution:
S1:~p∨q∨~p∨r

≡~p∨q∨r

S2:p→q∨r≡~p∨q∨r

So S1 is equivalent to S2

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Let p, q, r be three statements such that the truth value of p∧q→~q∨r is F . Then the truth values of p, q, r are respectively :

Options:
A T, T, F
B T, T, T
C T, F, T
D F, T, F

The correct answer is:
T, T, F

Solution:
Given that the truth value of p∧q→~q∨r is F .

For the above truth value to be possible, p∧q should be TRUE and ~q∨r should be FALSE.

p∧q is TRUE when p & q both are TRUE.

~q∨r is FALSE when q is TRUE and r is FALSE.

So, the truth values of p, q, r are T, T, F respectively.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Let $\mathrm{p}, \mathrm{q}, \mathrm{r}$ denote arbitrary statements. Then the logically equivalent of the statement $\mathrm{p} \Rightarrow(\mathrm{q} \vee \mathrm{r})$ is:

Options:
A
$(\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}$

B
$(\mathrm{p} \Rightarrow \mathrm{q}) \vee(\mathrm{p} \Rightarrow \mathrm{r})$

C
$(\mathrm{p} \Rightarrow \sim \mathrm{q}) \wedge(\mathrm{p} \Rightarrow \mathrm{r})$

D
$(\mathrm{p} \Rightarrow \mathrm{q}) \wedge(\mathrm{p} \Rightarrow \sim \mathrm{r})$

The correct answer is:
$(\mathrm{p} \Rightarrow \mathrm{q}) \vee(\mathrm{p} \Rightarrow \mathrm{r})$

Solution:
Given statement is
$p \Rightarrow(q \vee r)$ which is equivalent to
$$
(p \Rightarrow q) \vee(p \Rightarrow r)
$$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2014 (12 Apr Online)
Let $p, q, r \in R$ and $r>p>0$. If the quadratic equation $p x^2+q x+r=0$ has two complex roots $\alpha$ and $\beta$, then $|\alpha|+|\beta|$ is

Options:
A
equal to 1

B
less than 2 but not equal to 1

C
greater than 2

D
equal to 2

The correct answer is:
greater than 2

Solution:
Given quadratic equation is
$$
\begin{aligned}
& p x^2+q x+r=0 \\
& D=q^2-4 p r
\end{aligned}
$$
Since $\alpha$ and $\beta$ are two complex root
$$
\begin{array}{ll}
\therefore \beta=\bar{\alpha} \Rightarrow|\beta|=|\bar{\alpha}| \Rightarrow|\beta|=|\alpha| \\
(\because|\bar{\alpha}|=|\alpha|) \\
\text { Consider } \\
|\alpha|+|\beta|=|\alpha|+|\alpha| & (\because|\beta|=|\alpha|) \\
=2|\alpha|>2.1=2 & (\because|\alpha|>1)
\end{array}
$$
Hence, $|\alpha|+|\beta|$ is greater than 2 .

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2012 (19 May Online)
Let pâ,qâ,râ,sâ are non-zero vectors in which no two of them are perpendicular and no three of them are coplanar. If qâÃrââpâÃsâ+râÃpââqâÃsâ=kpâÃqââsâÃrâ, then the value of k2 is equal to

Options:

The correct answer is:
0.5

Solution:
q→×r→⋅p→×s→=q→⋅p→q→⋅s→r→⋅p→r→⋅s→
=q→⋅p→r→⋅s→-q→⋅s→r→⋅p→…i
r→×p→⋅q→×s→=r→⋅q→r→⋅s→p→⋅q→p→⋅s→
=r→⋅q→p→⋅s→-r→⋅s→p→⋅q→…ii
p→×q→⋅s→×r→=p→⋅s→p→⋅r→q→⋅s→q→⋅r→
=p→⋅s→q→⋅r→-q→⋅s→p→⋅r→…iii
From i,ii and iii
r→⋅q→p→⋅s→-q→⋅s→r→⋅p→=kp→⋅s→q→⋅r→-q→⋅s→p→⋅r→
⇒k=1

Topic: Mathematics, Vector Algebra, JEE Main