datavorous/entrance-exam-dataset huggingface dataset:
Let n be a positive integer. Let A=∑k=0n-1k×Ckn12k+34k+78k+1516k+3132k. If 63A=1-1230, then n is equal to ______ .

Options:

The correct answer is:
6

Solution:
A=∑k=0nCkn-12k+-34k+-78k+-1516k+-3132k

A=1-12n+1-34n+1-78n+1-1516n+1-3132n

A=12n+14n+18n+116n+132n

A=12n1-12n51-12n⇒A=1-125n2n-1

2n-1A=1-125n,

Given, 63A=1-1230

Clearly 5n=30

n=6

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let n be an odd natural number such that the variance of 1,2,3,4,…,n is 14. Then n is equal to ________.

Options:

The correct answer is:
13

Solution:
Variance of n natural number is given by,

Variance=∑n2n-∑nn2=nn+12n+16n-nn+12n2

=n+12n+16-n+14=n+1n-112=n2-112

And it is given that Variance=14

⇒n2-112=14

⇒n2-1=168

⇒n=13

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Let N be the foot of perpendicular from the point P(1,-2,3) on the line passing through the points (4,5,8) and (1,-7,5). Then the distance of N from the plane 2x-2y+z+5=0 is

Options:
A 8
B 6
C 9
D 7

The correct answer is:
7

Solution:
Given,

N be the foot of perpendicular from the point P(1,-2,3) on the line passing through the points (4,5,8) and (1,-7,5),

Now finding direction ratio of line L we get,

4-1,5+7,8-5≡3,12,3≡1,4,1

Hence, the equation of Line will be,

L:x-11=y+74=z-51=r

Now let point N≡(r+1,4r-7,r+5)

And given P≡(1,-2,3)

So, direction ratio of PN(r, 4r-5, r+2)

Since, PN⊥L so by perpendicular condition we get,

r+4(4r-5)+(r+2)=0

⇒r=1

Hence, N≡(2,-3,6)

Now finding, distance of N(2,-3,6) from plane 2x-2y+z+5=0 we get,

Distance=4+6+6+54+4+1=7

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let Nβ be the number of β particle emitted by 1 gram of Na24 radioactive nuclei having a half life of 15 h. In 7.5 h, the number Nβ is close to NA=6.023×1023 mole-1

Options:
A 1.75×1022
B 6.2×1021
C 7.5×1021
D 1.25×1022

The correct answer is:
7.5×1021

Solution:
∵NN0=12n
N = no. of un-decayed nuclei
N0 = initial no. of nuclei
Also t=n×T12
⇒NN0=12tT12
⇒N=N0127.515
⇒N=N01212⇒N=N02
∴ Number of nuclei decayed
=N0-N02=N02-12
=0.4141.414N0
=0.29N0
∵N0=124×6.023×1023
⇒ No. of nuclei decayed =0.29×124×6.023×1023
=0.0727×1023
=7.5×1021

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2015 (11 Apr Online)
Let \( N \) be the product of non-real roots of \( x^{4}-4 x^{3}+6 x^{2}-4 x=9999 \), then sum of digits of \( N \) equals

Options:

The correct answer is:
2

Solution:
x4−9x3+6x2−4x+1=10000

⇒x−14=1002

⇒x−12=100 or 100i2 where i=−1

For non-real roots, x−12=100i2

⇒x=+10i, 1−10i

So, N=1+10i1−10i=1+100=101

Sum of digits of N=1+0+1=2

Topic: Mathematics, Quadratic Equation, JEE Main
Let $N$ be the set of all natural numbers, $Z$ be the set of all integers and $\sigma: N \rightarrow Z$ be defined by
$$
\sigma(n)=\left\{\begin{array}{ccc}
\frac{n}{2}, & \text { if } & n \text { is even } \\
-\frac{n-1}{2}, & \text { if } & n \text { is odd }
\end{array}\right. \text {. Then, }
$$

Options:
A $\sigma$ is onto but not one-one
B $\sigma$ is one-one but not onto
C $\sigma$ is neither one-one nor onto
D $\sigma$ is one-one and onto

The correct answer is:
$\sigma$ is one-one and onto

Solution:
Given that,
$$
\sigma(n)=\left\{\begin{array}{cl}
\frac{n}{2} & \text { if } n \text { is even } \\
-\frac{(n-1)}{2} & \text { if } n \text { is odd }
\end{array}\right.
$$
Case-I : If $n$ is even
$$
\sigma(n)=1,2,3,4,5, \ldots
$$
Case-II: If $n$ is odd
$$
\sigma(n)=0,-1,-2,-3,-4, \ldots
$$
Thus, for every value of $n \sigma(n)$ has an unique image.
$\therefore \quad \sigma(n)$ is one-one function.
So, range of $\sigma(n)=Z$
$\therefore \quad$ Range of $\sigma(n)=$ codomain of $\sigma(n)=Z$.
Hence, $\sigma(n)$ is one-one and onto function.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 1)
Let $\mathrm{N}$ be the set of natural numbers and $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$, be a function given by $\mathrm{f}(\mathrm{x})=\mathrm{x}+1, \mathrm{x} \in \mathrm{N}$.
Which one of the following is correct ?

Options:
A $\mathrm{f}$ is one-one and onto
B $\mathrm{f}$ is one-one but not onto
C $\mathrm{f}$ is only onto
D $\mathrm{f}$ is neither one-one nor onto

The correct answer is:
$\mathrm{f}$ is one-one but not onto

Solution:
To show $\mathrm{f}$ is one-one. Let $f(x)=f(y)$ (To show: $x=y)$ $\Rightarrow \quad x+1=y+1$
$\Rightarrow \quad x=y$
Hence, $\mathrm{f}$ is one-one, Now, ' $\mathrm{f}$ ' is not onto because every element of codomain does not have it's pre-image in domain.

Topic: Mathematics, Functions, NDA, NDA 2013 (Phase 1)
Let N be the set of natural numbers and two functions f and g be defined as f,g:N→N such that
fn=n+12, if n is odd n2, if n is even
and gn=n--1n. Then fog is:

Options:
A onto but not one-one
B Both one-one and onto
C One-one but not onto
D Neither one-one nor onto

The correct answer is:
onto but not one-one

Solution:
We have, gn=n+1, if n oddn-1, if n even
fg1=f(2)=1
fg2=f(1)=1
∴fgx is many one.
fg2k=f(2k-1)=k
fg2k+1=f2k+2=k+1
∴f(g(x)) is onto.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that N-2,3 N, N+2 are in geometric progression be k48. Then the value of k is

Options:
A 2
B 4
C 16
D 8

The correct answer is:
4

Solution:
Given,

Two fair dices are rolled,

So total sample space will be ns=36

Now given: N-2, 3 N, N+2 are in G.P.

So, 3N=N-2N+2

⇒3N=N2-4

⇒ N2-3N-4=0

⇒N-4N+1=0

⇒N=4 or N=-1 (rejected)

Now favourable cases for sum 4 ≡1,3,3,1,2,2

So, nA=3

Hence, PA=336=112=448⇒k=4

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let n denote the number of solutions of the equation z2+3z¯=0, where z is a complex number. Then the value of ∑k=0∞1nk is equal to

Options:
A 1
B 43
C 32
D 2

The correct answer is:
43

Solution:
Given z2+3z¯=0

Put z=x+iy, then we know that z¯=x-iy, hence, we have

x+iy2+3x-iy=0

⇒ x2+i2y2+2ixy+3x-3iy=0

We also, know that i2=-1, hence, we get

⇒ x2-y2+2ixy+3x-3iy=0

⇒ x2-y2+3x+i(2xy-3y)=0+i0

On comparing the real and imaginary parts, we get

x2-y2+3x=0 …1

And 2xy-3y=0 …2

⇒ y2x-3=0

⇒x=32, y=0

Put x=32 in equation 1, we get 94-y2+92=0

⇒ y2=274

⇒ y=±332.

⇒ x, y=32, 332, 32, -332.

Now, put y=0, in the equation 1, we get x2+3x=0

⇒x=0, -3.

∴ x, y=0, 0, -3, 0.

∴ No of solutions=n=4

Now, ∑k=0∞1nk=∑k=0∞14k

=11+14+116+164+……

The above progression is a geometric progression, with first term a=1 and common ratio r=14 and the sum of infinite terms of the geometric progression is a1-r

Thus, ∑k=0∞1nk=11-14

=134=43.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations

x+y+z=12x+Ny+2z=23x+3y+Nz=3

has unique solution is k6, then the sum of value of k and all possible values of N is

Options:
A 18
B 19
C 20
D 21

The correct answer is:
20

Solution:
Given equations are

x+y+z=12x+Ny+2z=23x+3y+Nz=3

So,

∆=1112N233N

Applying C2↔C2-C1 and C3↔C3-C1

⇒∆=1002N-2030N-3

⇒∆=N-2N-3

For unique solution ∆≠0, so

N≠2,3

Also, N∈1,2,3,4,5,6 and for unique solution N can take values 1,4,5,6.

⇒P(system has unique solution ) =46=k6

So, k=4

Therefore, required sum is

=4+1+4+5+6=20

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let N denote the set of all natural numbers. Define two binary relations on N as R1=x,y∈N×N:2x+y=10 and R2=x,y∈N×N:x+2y=10. Then

Options:
A both R1 and R2 are transitive relations
B range of R2 is 1,2,3,4
C range of R1 is 2,4,8
D both R1 and R2 are symmetric relations

The correct answer is:
range of R2 is 1,2,3,4

Solution:
R1=1,8, 2,6, 3,4, 4,2
R2=8,1, 6,2, 4,3, 2,4 }
Range of R2=1,2,3,4

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2018 (16 Apr Online)
Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that 2N<N! is mn where m and n are coprime, then 4m-3n is equal to

Options:
A 6
B 12
C 10
D 8

The correct answer is:
8

Solution:
Given that,

2N<N!

N is the sum of numbers of two dice.

⇒2≤N≤12

Let us check the given condition 2N<N!

N=1 (not possible) →0

N=2 (not possible) →1

N=3 (not possible) →2

N=4 (possible)

We need to find the probability for N≥4.

⇒PN≥4=1-PN=2-PN=3

=1-136-236

=1112=mn

Now let us find 4m-3n.

=4×11-3×12=8.

Hence, the required answer is 8.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let $n_P$ and $n_e$ be the number of holes and conduction electrons respectively in a semiconductor. Then

Options:
A $n_P \gt n_e$ in an intrinsic semiconductor
B $n_P=n_e$ in an extrinsic semiconductor
C $n_p=n_e$ in an intrinsic semiconductor
D $n_0 \gt n_P$ in an intrinsic semiconductor

The correct answer is:
$n_p=n_e$ in an intrinsic semiconductor

Solution:
In an intrinsic semicondutor, number of holes carriers are equal to number of electrons as no doping is there in intrinsic material.
In extrinsic materials, due to doping, numbers of holes are different to those of electrons.
So,
$\mathrm{n}_{\mathrm{p}}=\mathrm{n}_{\mathrm{e}}$

Topic: Physics, Semiconductors, JEE Main
Let normals to the parabola y2=4x at variable points Pt12,2t1 and Qt22,2t2 meet at the point Rt2,2t, then the line joining P and Q always passes through a fixed point Î±,Î², then the value of Î±+Î² is equal to

Options:

The correct answer is:
2

Solution:
t=-t1-2t1 and t=-t2-2t2
⇒t1,t2 are the roots of the equation t=-x-2x

⇒x2+tx+2=0 has roots t1 & t2
⇒t1+t2=-t and t1t2=2
Now, the equation of PQ is t1+t2y=2x+2t1t2
⇒-ty=2x+4⇒2x+2+ty=0
⇒PQ always passes through the common point of x+2=0 and y=0
⇒α,β=-2,0⇒αβ=0 & α+β=-2

Topic: Mathematics, Parabola, JEE Main
Let $\mathrm{O}=(0,0) ;$ let $\mathrm{A}$ and $\mathrm{B}$ be points respectively on $\mathrm{x}$-axis and $\mathrm{y}$-axis such that $\angle \mathrm{OBA}=60^{\circ}$. Let $\mathrm{D}$ be a point in the first quadrant such that $\mathrm{OAD}$ is an equilateral triangle. Then the slope of $\mathrm{DB}$ is -

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

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

Solution:

Topic: Mathematics, Straight Lines, JEE Main
Let O=0,0, A=3,0, B=0,-1 and C=3,2, then the minimum value of z+z-3+z+i+z-3-2i occurs at the (where, z is a complex number)

Options:
A point of intersection of AB and CO
B point of intersection of AC and BO
C point of intersection of CB and AO
D Mean of O,A,B,C

The correct answer is:
point of intersection of CB and AO

Solution:
The minimum value of z+z-3 occurs at points on the line segment joining O & A
Similarly, the minimum value of z+i+z-3-2i
occurs at points on the line segment joining B & C.
â The minimum value of z+z-3+z+i+z-3-2i
occurs at point of intersection of OA & BC.

Topic: Mathematics, Complex Number, JEE Main
Let O0,0 and A0,1 be two fixed points. Then, the locus of a point P such that the perimeter of ΔAOP is 4 is

Options:
A 8x2+9y2-9y=18
B 9x2-8y2+8y=16
C 8x2-9y2+9y=18
D 9x2+8y2-8y=16

The correct answer is:
9x2+8y2-8y=16

Solution:
Given, OA+PO+PA=4

⇒PO+PA=3

⇒Locus of P is ellipse with foci at O & A and major

axis 2b=3

Distance between foci =2be=1⇒e=13

⇒Minor axis 2a=2b1-e2=3.223=22

⇒ Equation of Locus of P is :

x28+y-1229=14

By simplifying the above equation, we get

9x2+8y2-8y=16

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Let $\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}$ and $\overrightarrow{O C}=3 \vec{b}$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OC}}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $\mathrm{OABC}$ is equal to :

Options:
A 32
B 40
C 38
D 35

The correct answer is:
35

Solution:
Area of parallelogram having sides
$\begin{aligned}
& \overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OC}}=|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OC}}|=|2 \overrightarrow{\mathrm{a}} \times 3 \overrightarrow{\mathrm{b}}|=15 \\
& 6|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=15 \\
& \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2} \ldots \ldots . .(1)
\end{aligned}$
Area of quadrilateral
$\begin{aligned}
& \mathrm{OABC}=\frac{1}{2}\left|\overrightarrow{\mathrm{d}}_1 \times \overrightarrow{\mathrm{d}}_2\right| \\
& =\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{OB}}|=\frac{1}{2}|(3 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}) \times(6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}})| \\
& =\frac{1}{2}|18 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-10 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=14|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\
& =14 \times \frac{5}{2}=35
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let ÎOAB be an equilateral triangle with side length unity (O being the origin). Also, M and N are the points of trisection of AB, M being closer to A and N being closer to B. Position vectors of A, B, M and N are aâ, bâ, mâ and nâ respectively, then the value of mâânâ is equal to

Options:
A 12
B 23
C 1318
D 49

The correct answer is:
1318

Solution:
m→=2a→+b→3,n→=a→+2b→3
m→⋅n→=192a→+b→⋅a→+2b→
=192a→2+2b→2+5a→⋅b→
=192+2+5×1×1×12=1318

Topic: Mathematics, Vector Algebra, JEE Main
Let OABC be a regular tetrahedron with side length unity, then its volume (in cubic units) is

Options:
A 32
B 62
C 132
D 162

The correct answer is:
162

Solution:
Let the position vector of O,A,B,C are 0→,a→,b→,c→ respectively
OA→=a→,OB→=b→,OC→=c→
OA→=OB→=OC→=1 [Given]
Volume =16OA→OB→OC→=16a→b→c→……i
∴a→b→c→2=a→⋅a→a→⋅b→a→⋅c→b→⋅a→b→⋅b→b→⋅c→c→⋅a→c→⋅b→c→⋅c→

⇒a→b→c→2=112121211212121
⇒a→b→c→2=134-1214+12-14=12
⇒a→b→c→=12
From i,

Volume of the tetrahedron =162 cubic units

Topic: Mathematics, Vector Algebra, JEE Main
Let O be an interior point of triangle ABC, such that 2OA→+3OB→+4OC→=0, then the ratio of the area of ∆ABC to the area of ∆AOC is

Options:
A 3:1
B 3:2
C 2:1
D 4:3

The correct answer is:
3:1

Solution:
Let position vectors of O,A,B,C are Oâ,aâ,bâ,câ respectively.
2OAâ+3OBâ+4OCâ=0
â2aâ+3bâ+4câ=0
Area of âAOC=12OAâÃOCâ=12aâÃcâ
Area of âABC=12BAâÃBCâ=12aâÃbâ+bâÃcâ+câÃaâ
=12aâÃ-4câ-2aâ3+-4câ-2aâ3Ãcâ+câÃaâ
=1243câÃaâ+23câÃaâ+câÃaâ
=32|câÃaâ|
Area of âABCArea of âAOC=31

Topic: Mathematics, Vector Algebra, JEE Main
Let O be the origin, and M and N be the points on the lines x-54=y-41=z-53 and x+812=y+25=z+119 respectively such that MN is the shortest distance between the given lines. Then OM→·ON→ is equal to _________.

Options:

The correct answer is:
9

Solution:
Let,

L1:x-54=y-41=z-53=λ

⇒M(4λ+5,λ+4,3λ+5)

And L2:x+812=y+25=z+119=μ

⇒N(12μ-8,5μ-2,9μ-11)

Now, finding MN→=(4λ-12μ+13,λ-5μ+6,3λ-9μ+16) ..…1

Now finding cross product of direction ratios of line we get,

b→1×b→2=i^j^k^4131259=-6i^+8k^ …...2

Now, solving the equation 1 & 2 as they are parallel vectors, we get,

∴4λ-12μ+13-6=λ-5μ+60=3λ-9μ+168

Taking 4λ-12μ+13-6=λ-5μ+60

⇒λ-5μ+6=0 …3

Taking 4λ-12μ+13-6=3λ-9μ+168

λ-3μ+4=0 …4

Solve 3 and 4 we get

λ=-1, μ=1

So, M(1,3,2) and N(4,3,-2)

Hence, OM→·ON→=4+9-4=9

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let O be the origin and OP and OQ be the tangents to the circle x2+y2-6x+4y+8=0 at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point α,12, then a value of α is

Options:
A 32
B -12
C 52
D 1

The correct answer is:
52

Solution:
The given information can be represented in the form of the diagram below.âââââ

Since angle in a semicircle is a right angle.

Hence, the other circle will be passing through the centre of the first circle.

The two ends of diameter of the circle is O0,0 and C3,-2.

Hence, the required equation of circle is x-0x-3+y-0y--2=0

⇒x2+y2-3x+2y=0

Put α,12 in the above equation⇒α2+14-3α+1=0
⇒α2-3α+54=0
⇒4α2-12α+5=0

⇒4α2-10α-2α+5=0

⇒2α(2α-5)-1(2α-5)=0⇒α=12,52

Therefore this is the correct option.

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let $O$ be the origin and the position vector of A and B be 2i^+2j^+k^ and 2i^+4j^+4k^ respectively. If the internal bisector of ∠AOB meets the line AB at C, then the length of OC is

Options:
A 2331
B 2334
C 3434
D 3231

The correct answer is:
2334

Solution:
We know that, the internal bisector divides base in the same ratio as the two remaining sides.

⇒OAOB=ACBC

⇒22+22+1222+42+42=ACBC

⇒36=ACBC

⇒AC:BC=1:2

⇒C≡1×2+2×23,1×4+2×23,1×4+2×13

⇒C≡2,83,2

⇒OC=22+832+22

⇒OC=8+649=1363

⇒OC=2343

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let O be the origin and the position vector of the point P be -i^-2j^+3k. If the position vectors of the points A, B and C are -2i^+j^-3k, 2i^+4j^-2k and -4i^^+2j^-k respectively, then the projection of the vector OP→ on a vector perpendicular to the vectors AB→ and AC→ is

Options:
A 3
B 83
C 73
D 103

The correct answer is:
3

Solution:
Given,

O be the origin and the position vector of the point P be -i^-2j^+3k^,

So, OP→=-i^-2j^+3k^

Also given the position vectors of the points A, B and C are -2i^+j^-3k^, 2i^+4j^-2k^ and -4i^^+2j^-k^

Now finding,

AB→=4i^+3j^+k^

AC→=-2i^+j^+2k^

And AB→×AC→=i^j^k^431-212

⇒AB→×AC→=5i^-10j^+10k^

Now finding the Projection of OP→ on vector perpendicular to AB→ & AC→ we get,

Projection=OP→·AB→×AC→AB→×AC→

=5-i^-2j^+3k^i^-2j^+2k^51+4+4=3

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let O be the vertex and Q be any point on the parabola, x2=8y. If the point P divides the line segment OQ internally in the ratio 1:3, then the locus of P is

Options:
A x2=2y
B x2=y
C y2=x
D y2=2x

The correct answer is:
x2=2y

Solution:
General point on x2=8y is Q4t,2t2

Let Ph,k divide OQ in ratio 1:3
h,k=14t+301+3,12t2+301+3

h,k=t,2t24
h=t & k=t22

k=h22

⇒ x2=2y is the required locus.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2015 (04 Apr)
Let OA→=a→,OB→=12a→+4b→ and OC→=b→, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then area of the quadrilateral OABCarea of S is equal to _____

Options:
A 6
B 10
C 7
D 8

The correct answer is:
8

Solution:
Area of parallelogram, S=|a→×b→|

Area of quadrilateral = Area ∆OAB+Area ∆OBC

=12{|a→×(12a→+4b→)|+|b→×(12a→+4b→)|}

=12{|4a→×b→|+|12b→×a→|}

=12{4|a→×b→|+12|a→×b→|}

=162|a→×b→|

=8|(a→×b→)|

Ratio =8|(a→×b→)||(a→×b→)|=8

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let one observation equal to the mean is added to n observations. If the variance changes from 78 to 72, then n is equal to

Options:

The correct answer is:
12

Solution:
Adding mean to the observations does not change the value of ∑xi-x-2, hence,

∑i=1nxi-x-2n∑i=1n+1xi-x-2n+1=7872
⇒n+1n=7872⇒n=12

Topic: Mathematics, Statistics, JEE Main
Let orthocentre of ΔABC is 4,6. If A=4,7 and B=-2,4, then coordinates of vertex C is

Options:
A 5,4
B 4,5
C -5,-4
D -4,-5

The correct answer is:
5,4

Solution:
Let feet of altitudes through A,B,C are D,E,F respectively.
Now, C lies on the line CF,CA & CB
Now, the slope of AD is 7-64-4=N.D.
⇒ the slope of BC is 0 ⇒ equation of BC is y=4
Now, the slope of BE is 6-44-(-2)=13
⇒ the slope of AC is -3
⇒ equation of the line AC is y-7=-3x-4
Point of intersection of y=4 and y-7=-3x-4 gives point C

⇒C=5,4

Topic: Mathematics, Straight Lines, JEE Main
Let \( \overrightarrow{\mathrm{c}} \) be a vector perpendicular to the vectors \( \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\widehat{\mathrm{k}} \) and \( \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\widehat{\mathrm{k}} \) If \( \overrightarrow{\mathrm{c}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \widehat{\mathrm{k}})=8 \) then the value of \( \overrightarrow{\mathrm{c}} \cdot(\vec{a} \times \vec{b}) \) is equal to

Options:

The correct answer is:
28

Solution:
c→=λ(a→×b→)

a→×b→=i^j^k^11-1121

(a→×b→)=3i^-2j^+k^

c→·(i^+j^+3k^)=λ(3i^-2j^+k^)·(i^+j^+3k^)

⇒λ(4)=8⇒λ=2

c→=2(a→×b→)

Topic: Mathematics, Vector Algebra, JEE Main
Let P=100310931 and Q=qij be two 3×3 matrices such that Q-P5=I3. Then q21+q31q32 is equal to :

Options:
A 10
B 9
C 15
D 135

The correct answer is:
10

Solution:
Given,

P=100310931

⇒P2=100310931100310931=1006102761

⇒P4=P2×P2=10061027611006102761=100121090121

⇒P5=P4×P=100121090121100310931=1001510135151

Let,

Q=q11q12q13q21q22q23q31q31q33

Now, given Q-P5=I3

⇒q21-15=0, q31-135=0,  q32-15=0

⇒q21=15, q31=135, q32=15

∴q21+q31q32=15+13515=10.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Let P=1004101641 and I be the identity matrix of order 3. If Q=qij is a matrix such that P50-Q=I, then the value of q31+q32q21 is equal to

Options:
A 52
B 103
C 201
D 205

The correct answer is:
103

Solution:
P=1004101641 ⇒P2=10081016+3281
So, P3=100121016+32+48121
Using symmetry, we can write
P50=1002001016.50.5122001
As, P 50 −Q=I⇒Q= P 50 −I=[ 0 0 0 200 0 0 16.50.51 2 200 0 ] ,
q31=16.50.512, q32=200 and q21=200
∴ q31+q32q21=16.50.512.200+1=102+1=103

Topic: Mathematics, Matrices, JEE Main
Let $P=(-1,0), Q=(0,0)$ and $R=(3,3 \sqrt{3})$ be three points. The equation of the bisector of the angle $\mathrm{PQR}$

Options:
A
$\sqrt{3} x+y=0$

B
$x+\frac{\sqrt{3}}{2} y=0$

C
$\frac{\sqrt{3}}{2} x+y=0$

D
$x+\sqrt{3} y=0$

The correct answer is:
$\sqrt{3} x+y=0$

Solution:
Slope of the line $\mathrm{QM}$ is $\tan \frac{2 \pi}{3}=-\sqrt{3}$
Hence equation is line $\mathrm{QM}$ is $y=-\sqrt{3} \mathrm{x}$.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2007
Let P1: 2x+y+z+1=0, P2: 2x-y+z+3=0 and P3: 2x+3y+z+5=0 be three planes, then the distance of the line of intersection of planes P1=0 and P2=0 from the plane P3=0 is

Options:
A 314 units
B 614 units
C 37 units
D 67 units

The correct answer is:
614 units

Solution:
A vector normal to P1=0 is n1â=2i^+j^+k^
A vector normal to P2=0 is n2â=2i^-j^+k^
Hence, the line of intersection is parallel to n1âÃn2â
=i^j^k^2112-11=i^2-j^0+k^-4=2i^-4k^
And for point of intersection Put z=0 in
P1=0 and P2=0, i.e.
2x+y+1=0
2x-y+3=0
4x+4=0
âx=â1 and y=1
So point is -1,1,0
Equation of the line of intersection is
x+11=y-10=z-0-2
The line is parallel to P3=0

Hence, distance =-2+3+514
=614

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P1: 3y+z+1=0 and P2: 2x-y+3z-7=0 and the equation of line AB is x-12=y-3-1=z-43 in 3D space. Shortest distance between the line of intersection of planes P1 and P2 and the line AB is equal to

Options:
A 710 units
B 725 units
C 610 units
D 225 units

The correct answer is:
725 units

Solution:
Plane P1 is parallel to line AB
⇒ Shortest distance = Perpendicular distance of any point on line to plane P1=0
=0+9+4+19+1=1410=725 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P1,7,2 be a point and the equation of the line L is x-12=y-71=z-2-1. If PQ is the distance of the plane 2x+y-z=1 from the point P measured along a line inclined at an angle 60Â° with L, then the length of PQ is

Options:
A 3 units
B 32 units
C 6 units
D 8 units

The correct answer is:
6 units

Solution:
Equation of the line through P and parallel to L is
x-12=y-71=z-2-1
∵ This line is perpendicular to the plane
From the diagram,
PR=PQcos⁡60°⇒PQ=2PR
PR=2+7-2-14=3
⇒PQ=6 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let P1, P2 and P3 are the probabilities of a student passing three independent exams A, B and C respectively. If P1, P2 and P3 are the roots of equation 20x3-27x2+14x-2=0, then the probability that the student passes in exactly one of A, B and C is

Options:
A 320
B 720
C 14
D 15

The correct answer is:
14

Solution:
P1+P2+P3=2720
P1P2+P2P3+P3P1=1420
P1P2P3=220
Probability that the student passes in exactly one of A,B,C is
=P1+P2+P3-2P1P2+P2P3+P3P1+3P1P2P3
=2720-21420+620=520=14

Topic: Mathematics, Probability, JEE Main