datavorous/entrance-exam-dataset huggingface dataset:
If two organisms can't interbreed to produce fertile offsprings, you can't place them under a

Options:
A genus
B species
C order
D phylum

The correct answer is:
species

Solution:
The classification of all living organisms has been done by dividing them into different levels based on their features and characteristics. There is a hierarchy in this classification and each step represents a rank or category known as a taxonomic category. There are around seven main taxonomic ranks or categories - species, genus, family, order, class, phylum/division, and kingdom. Groups of individuals having similar fundamental qualities are called species. Species contain a group of interacting and interbreeding organisms. When the group of related species has more common characters among them they are referred to as genus. Families that exhibit few similar characters are placed under the same Order. All the fishes, amphibians, reptiles, birds along with mammals constitute the taxonomic level phylum.

Topic: Biology, The Living World, NEET
If two parabolas y2=4ax-k and x2=4ay-k have only one common point P, then the coordinates of P are

Options:
A 2k,2k
B k,k
C a,2k
D k,2a

The correct answer is:
2k,2k

Solution:
Parabolas y2=4ax-k and x2=4ay-k touch each other at the line y=x (âµ both parabolas are inverse of each other)
ây=x is the common tangent at P
â we get point P by solving y2=4ax-k and y=x
â point of contact P=2a,2a
â2a2=4a2a-kâk=aâP=2k,2k

Topic: Mathematics, Parabola, JEE Main
If two parabolas y2=4ax–k and x2=4ay–k have only one common point P, then the equation of normal to y2=4ax–k at P is

Options:
A y+x=4a
B y+x=2a
C y+x=4
D y+x=2

The correct answer is:
y+x=4a

Solution:
Parabolas y2=4ax–k and x2=4ay–k touche each other at line y=x
(∵ both parabola are inverse of each other)
⇒ y=x is a common tangent at P
⇒ we get point P by solving y2=4ax–k and y=x
⇒ point of contact P is 2a,2a
⇒ equation of normal at P2a,2a is x+y=4a

Topic: Mathematics, Parabola, JEE Main
If two parallel chords of a circle, having diameter 4 units, lie on the opposite sides of the center and subtend angles cos-1⁡17 and sec-1⁡7 at the center respectively, then the distance between these chords is:

Options:
A 87
B 167
C 47
D 87

The correct answer is:
87

Solution:
Given condition,

∴ cos⁡2θ=17 ⇒2cos2θ-1=17
⇒2cos2⁡θ=87

⇒cosθ=27

Then, CP1=rcosθ=47
Since, cos−117=sec−17, hence both the chords are at equal distance from center of the circle.
Hence, the distance between the chords =2 × 4 7 = 8 7 .

Topic: Mathematics, Circle, JEE Main, JEE Main 2017 (08 Apr Online)
If two points A and B lie on the curve y=x2 such that OA→⋅i^=1 and OB→⋅j^=4, where O is the origin and A and B lie in the 1st and 2nd quadrant respectively, then OA→⋅OB→ is equal to

Options:
A 0
B 2
C 4
D 5

The correct answer is:
2

Solution:
Let the points A,B be α,α2,β,β2 respectively where α>0, β<0
OA→=αi^+α2j^
OB→=βi^+β2j^
OA→⋅i^=α=1⇒OA→=i^+j^
OB→⋅j^=β2=4⇒β=±2
∵β<0⇒β=-2⇒OB→=-2i^+4j^
OA→⋅OB→=i^+j^-2i^+4j^=-2+4=2

Topic: Mathematics, Vector Algebra, JEE Main
If two points are taken on the minor axis of the ellipse x225+y29=1 at the same distance from the centre as the foci, then the sum of the squares of the perpendicular distances from these points on any tangent to the ellipse is

Options:
A 25
B 18
C 50
D 80

The correct answer is:
50

Solution:
Given ellipse is x225+y29=1 â¦ 1
e=1-925=45
S0,545=S(0,4)
S'0,-545=S'(0,-4)
Let, the tangent to the ellipse be y=mx+25m2+9 where, m is a parameter.

Now, the sum of the squares of â¥ distances on this tangent from the points S and S' is
=m0-4+25m2+9m2+12+m0+4+25m2+9m2+12
=2(m2+1)42+(25m2+9)=50

Topic: Mathematics, Ellipse, JEE Main
If two points P and Q are on the curve y=2x+2, such that OP â â i ^ =â1 and OQ â â i ^ =2, where i^ is a unit vector along the x-axis, then OQâ-4OPâ is equal to

Options:

The correct answer is:
10

Solution:
Let Px1,y1 and Qx2,y2 be two points on y=2x+2
∵ OP → ⋅ i ^ =−1⇒ x 1 =−1 and y1=2
OQ → ⋅ i ^ =2⇒ x 2 =2 and y2=16
⇒OP→=-i^+2j^ and OQ→=2i^+16j^
⇒OQ→-4OP→=6i^+8j^
⇒OQ→-4OP→=10

Topic: Mathematics, Vector Algebra, JEE Main
If two points P and Q lie on the hyperbola x2a2-y2b2=1 (a<b), whose centre C be such that CP is perpendicular to CQ, then the value of 1CP2+1CQ2 is

Options:
A b2-a22ab
B 1a2+1b2
C 2abb2-a2
D 1a2-1b2

The correct answer is:
1a2-1b2

Solution:
Let CP=r1 be inclined to transverse axis at an angle θ such that Pr1cos⁡θ, r1sin⁡θ lies on the hyperbola. Therefore,
r12 cos2⁡θa2-sin2⁡θb2=1
Replacing θ by 90o+θ, we get,
r22 sin2⁡θa2-cos2⁡θb2=1

(where, CQ=r2)
⇒ 1r12+1r22=cos2⁡θa2 -sin2⁡θb2+sin2⁡θa2-cos2⁡θb2
⇒ 1r12+1r22=cos2⁡θ1a2-1b2+sin2⁡θ1a2-1b2
⇒ 1r12+1r22=1a2-1b2
⇒ 1CP2+1CQ2=1a2-1b2

Topic: Mathematics, Hyperbola, JEE Main
If two sets $A$ and $B$ have 99 elements in common, then the number of elements common to the sets $A \times B$ and $B \times A$ is

Options:
A $2^{99}$
B $99^{2}$
C 100
D 18

The correct answer is:
$99^{2}$

Solution:
Since, $n(A \cap B)=99$
$n((A \times B) \cap(B \times A))$
$$
\begin{aligned}
&=n((A \cap B) \times(B \cap A)) \\
&=n(A \cap B) \cdot n(B \cap A) \\
&=n(A \cap B) \cdot n(A \cap B)=99 \cdot 99=(99)^{2}
\end{aligned}
$$

Topic: Mathematics, Sets and Relations, JEE Main
If two similar springs each of spring constant K1 are joined in series, the new spring constant and time period would be changed by a factor:

Options:
A 12,22
B 14,2
C 14,22
D 12,2

The correct answer is:
12,2

Solution:
1keq=1k1+1k2

1keq=1k+1k⇒keq=k2

k'=k2

T=2πMk

T'=2πMk'

⇒T'=2πMk×2

T'=2T

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (26 Feb Shift 1)
If two soap bubbles of different radii are connected by a tube,

Options:
A
air flows from the bigger bubble to the smaller bubble till the sizes are interchanged.

B
air flows from bigger bubble to the smaller bubble till the sizes are interchanged

C
air flows from the smaller bubble to the bigger.

D
there is no flow of air.

The correct answer is:
air flows from the smaller bubble to the bigger.

Solution:
The pressure inside the smaller bubble will be more $\left(P_i=P_0+\frac{4 T}{r}\right)$
Therefore, if the bubbles are connected by a tube, the air will flow from smaller bubble to the bigger.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2004
If two straight lines whose direction cosines are given by the relations l+m-n=0,3l2+m2+cnl=0 are parallel, then the positive value of c is

Options:
A 6
B 4
C 3
D 2

The correct answer is:
6

Solution:
Given, l+m-n=0 & 3l2+m2+cln=0

Now taking n=l+m and putting in 3l2+m2+cln=0 we get,

3l2+m2+cll+m=0

⇒3l2+m2+cl2+clm=0

⇒3+cl2+clm+m2=0

⇒3+clm2+clm+1=0 ..equation (i)

∵ lies are parallel.

So, roots of equation (i) must be equal

⇒ D=0

⇒c2-43+c=0

⇒c2-4c-12=0

⇒c-6c+2=0

c=6 or c=-2

So, positive value of c=6

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (27 Jun Shift 1)
If two tangents drawn from a point α,β lying on the ellipse 25x2+4y2=1 to the parabola y2=4x are such that the slope of one tangent is four times the other, then the value of 10α+52+16β2+502 equals ______

Options:

The correct answer is:
2929

Solution:
Let α=15cosθ, β=12sinθ

Equation of tangent to y2=4x with slope m will be y=mx+1m

Since the tangent passes through α,β, so

12sinθ=m5cosθ+1m

⇒m2cosθ5-m12sinθ+1=0

As it is a quadratic equation in m so it will have two roots m1 and m2

Given m1=4m2

Now, m1+m2=12sinθcosθ5

m1m2=5cosθ

After eliminating m1 and m2, we get,

cosθ=-5±292⇒cos2θ=54±10294

i.e. α=-5±2910⇒10α+5=±29

and β2=14sin2θ⇒16β2=-50±1029

Hence, 10α+52+16β2+502=2929

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (24 Jun Shift 1)
If two tangents drawn from a point $P$ to the parabola $y^2=4 x$ are at right angles, then the locus of $P$ is

Options:
A
$2 x+1=0$

B
$x=-1$

C
$2 x-1=0$

D
$x=1$

The correct answer is:
$x=-1$

Solution:
The locus of perpendicular tangents is directrix
$$
\text { i.e, } x=-a ; x=-1
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2010
If two tangents drawn from a point P to the parabola y2=16(x-3) are at right angles, then the locus of point P is:

Options:
A x+4=0
B x+2=0
C x+3=0
D x+1=0

The correct answer is:
x+1=0

Solution:
If two tangents drawn from a point P to the parabola y2=16(x-3) are at right angles.

Given the equation of parabola y2=16(x-3)

Then, the locus point is P

Locus is directrix of parabola is x+a=0
x-3+4=0

x+1=0

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (27 Aug Shift 2)
If two tangents drawn from the point P(h,k) to the parabola y2=8x are such that the slope of one of the tangent is 3 times the slope of the other, then the locus of point P is

Options:
A 3y2=16x
B 3y2=8x
C y2=32x
D 3y2=32x

The correct answer is:
3y2=32x

Solution:
Equation of tangent in slope form is y=mx+am
Let the tangent passes through Ph,k
⇒k=mh+am⇒m2h-km+a=0
⇒m1+m2=kh & m1m2=ah
Where, a=2 & m2=3m1
⇒4m1=kh & 3m12=2h
⇒3k4h2=2h⇒3y2=32x

Topic: Mathematics, Parabola, JEE Main
If two vectors $2 \hat{i}+3 \hat{j}-\hat{k}$ and $-4 \hat{i}-6 \hat{j}-2 \hat{k}$ are parallel to each other then value of $\lambda$ be

Options:
A 0
B 2
C 3
D 4

The correct answer is:
2

Solution:
Let $\vec{A}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{B}=-4 \hat{i}-6 \hat{j}+$,
$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$ i.e. $\frac{2}{-4}=\frac{3}{-6}=\frac{-1}{\lambda} \Rightarrow \lambda=2$.
$\vec{A}$ and $\vec{B}$ are parallel to each other.

Topic: Physics, Mathematics in Physics, JEE Main
If two vectors $\mathbf{A}$ and $\mathbf{B}$ are mutually perpendicular, then the component of $\mathbf{A}-\mathbf{B}$ along the direction of $\mathbf{A}+\mathbf{B}$ is

Options:
A $\sqrt{|\mathbf{A}|^2+|\mathbf{B}|^2}$
B $\sqrt{|A|^2-|B|^2}$
C $\frac{|A|^2-|B|^2}{\sqrt{|A|^2+|B|^2}}$
D $\frac{|A|^2+|B|^2}{\sqrt{|A|^2-|B|^2}}$

The correct answer is:
$\frac{|A|^2-|B|^2}{\sqrt{|A|^2+|B|^2}}$

Solution:
The two vectors $\mathbf{A}$ and $\mathbf{B}$ are mutually perpendicular to each other
A. $\mathbf{B}=0$
$\Rightarrow \quad A B \cos \theta=0$
$\Rightarrow \quad \cos \theta=0 \Rightarrow \theta=90^{\circ}$
$|\mathbf{A}-\mathbf{B}|$
$\begin{aligned} & =\sqrt{A^2+B^2-2 A B \cos 90^{\circ}} \\ & =\sqrt{A^2+B^2}\end{aligned}$
Unit vector along $\mathbf{A}-\mathbf{B}$
$=\frac{\mathbf{A}-\mathbf{B}}{|\mathbf{A}-\mathbf{B}|}=\frac{\mathbf{A}-\mathbf{B}}{\sqrt{A^2+B^2}}$
component of $(\mathbf{A}+\mathbf{B})$ in direction of $(\mathbf{A}-\mathbf{B})$ will be given as,
$=\frac{(\mathbf{A}+\mathbf{B}) \cdot(\mathbf{A}-\mathbf{B})}{|\mathbf{A}-\mathbf{B}|}$
$=\frac{\mathbf{A} \cdot \mathbf{A}-\mathbf{A} \cdot \mathbf{B}+\mathbf{B} \cdot \mathbf{A}-\mathbf{B} \cdot \mathbf{B}}{\sqrt{A^2+B^2}}$
$=\frac{A^2-B^2}{\sqrt{A^2+B^2}}=\frac{|\mathbf{A}|^2-|\mathbf{B}|^2}{\sqrt{|\mathbf{A}|^2+|\mathbf{B}|^2}}$

Topic: Physics, Mathematics in Physics, JEE Main
If two vectors A→ and B→ having equal magnitude R are inclined at an angle θ, then

Options:
A A→−B→=2Rsinθ2
B A→+B→=2Rsinθ2
C A→+B→=2Rcosθ2
D A→−B→=2Rcosθ2

The correct answer is:
A→+B→=2Rcosθ2

Solution:
The magnitude of resultant vector A→+B→ is given by,

R'=A2+B2+2ABcosθ

Here A=B=R

Then R'=R2+R2+2R2cosθ

=R21+cosθ

=2R2cos2θ2

=2Rcosθ2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (31 Jan Shift 2)
If two vectors P→=i^+2mj^+mk^ and Q→=4i^-2j^+mk^ are perpendicular to each other. Then, the value of m will be

Options:
A -1
B 2
C 3
D 1

The correct answer is:
2

Solution:
For two vectors to be perpendicular, their dot product is equal to zero.

So, P→·Q→=0

⇒i^+2mj^+mk^·4i^-2j^+mk^=0⇒4-4m+m2=0⇒m-22=0⇒m=2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (24 Jan Shift 2)
If two vertical poles $20 \mathrm{~m}$ and $80 \mathrm{~m}$ high stand apart on a horizontal plane, then the height (in $\mathrm{m}$ ) of the point of intersection of the lines joining the top of each pole to the foot of other is

Options:
A
16

B
18

C
50

D
15

The correct answer is:
16

Solution:
We put one pole at origin. $B C=80 \mathrm{~m}, O A=20 \mathrm{~m}$ Line $O C$ and $A B$ intersect at $M$. To find: Length of $M N$.
Eqn of $O C: y=\left(\frac{80-0}{a-0}\right) x$ $\Rightarrow y=\frac{80}{a} x$
Eqn of $A B: y=\left(\frac{20-0}{0-a}\right)(x-a)$ $\Rightarrow y=\frac{-20}{a}(x-a)$
At $M:(1)=(2)$
$$
\begin{aligned}
& \Rightarrow \quad \frac{80}{a} x=\frac{-20}{a}(x-a) \\
& \Rightarrow \quad \frac{80}{a} x=\frac{-20}{a} x+20 \Rightarrow x=\frac{a}{5} \\
& \therefore \quad y=\frac{80}{a} \times \frac{a}{5}=16
\end{aligned}
$$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2012 (07 May Online)
If two vertices of a triangle are $(5,-1)$ and $(-2,3)$ and its orthocentre is at $(0,0)$, then the third vertex is

Options:
A
$(4,-7)$

B
$(-4,-7)$

C
$(-4,7)$

D
$(4,7)$

The correct answer is:
$(-4,-7)$

Solution:
Let the third vertex of $\triangle A B C$ be $(a, b)$.
Orthocentre $=H(0,0)$
Let $A(5,-1)$ and $B(-2,3)$ be other two vertices of $\triangle A B C$.
Now, (Slope of $A H) \times($ Slope of $B C)=-1$
$$
\begin{aligned}
& \Rightarrow\left(\frac{-1-0}{5-0}\right)\left(\frac{b-3}{a+2}\right)=-1 \\
& \Rightarrow b-3=5(a+2)
\end{aligned}
$$
Similarly,
$($ Slope of $B H) \times($ Slope of $A C)=-1$
$$
\begin{aligned}
& \Rightarrow \quad-\left(\frac{3}{2}\right) \times\left(\frac{b+1}{a-5}\right)=-1 \\
& \Rightarrow \quad 3 b+3=2 a-10 \\
& \Rightarrow 3 b-2 a+13=0
\end{aligned}
$$
On solving equation (1) and (2) we get $a=-4, b=-7$
Hence, third vertex is $(-4,-7)$.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (12 May Online)
If two vertices of an equilateral triangle are $\mathrm{A}(-a, 0)$ and $B(a, 0), a>0$, and the third vertex C lies above $x$-axis then the equation of the circumcircle of $\triangle \mathrm{ABC}$ is :

Options:
A
$3 x^2+3 y^2-2 \sqrt{3} a y=3 a^2$

B
$3 x^2+3 y^2-2 a y=3 a^2$

C
$x^2+y^2-2 a y=a^2$

D
$x^2+y^2-\sqrt{3} a y=a^2$

The correct answer is:
$3 x^2+3 y^2-2 \sqrt{3} a y=3 a^2$

Solution:
Let $\mathrm{C}=(x, y)$
Now, $\mathrm{CA}^2=\mathrm{CB}^2=\mathrm{AB}^2$
$$
\begin{aligned}
& \Rightarrow(x+a)^2+y^2=(x-a)^2+y^2=(2 a)^2 \\
& \Rightarrow x^2+2 a x+a^2+y^2=4 a^2 \\
& \text { and } x^2-2 a x+a^2+y^2=4 a^2
\end{aligned}
$$
From (i) and (ii), $x=0$ and $y=\pm \sqrt{3} a$
Since point $\mathrm{C}(x, y)$ lies above the $x$-axis and $a>0$, hence $y=\sqrt{3} a$
$$
\therefore \quad \mathrm{C}=(0, \sqrt{3} a)
$$
Let the equation of circumcircle be
$$
x^2+y^2+2 g x+2 f y+\mathrm{C}=0
$$
Since points A( $-a, 0), \mathrm{B}(a, 0)$ and $\mathrm{C}(0, \sqrt{3} a)$ lie on the circle, therefore
$$
\begin{aligned}
& a^2-2 g a+\mathrm{C}=0 \\
& a^2+2 g a+\mathrm{C}=0
\end{aligned}
$$
and $3 a^2+2 \sqrt{3} a f+\mathrm{C}=0$
From (iii), (iv), and (v)
$$
g=0, c=-a^2, f=-\frac{a}{\sqrt{3}}
$$
Hence equation of the circumcircle is
$$
\begin{aligned}
& x^2+y^2-\frac{2 a}{\sqrt{3}} y-a^2=0 \\
& \Rightarrow x^2+y^2-\frac{2 \sqrt{3} a y}{3}-a^2=0 \\
& \Rightarrow 3 x^2+3 y^2-2 \sqrt{3} a y=3 a^2
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2013 (22 Apr Online)
If two waves having amplitudes $2 A$ and $A$ and same frequency and velocity, propagate in the same direction in the same phase, the resulting amplitude will be

Options:
A $3 A$
B $\sqrt{5} A$
C $\sqrt{2} A$
D $A$

The correct answer is:
$3 A$

Solution:
In the same phase $\varphi=0$ so resultant amplitude $=a_1+a_2=2 A+A=3 A$

Topic: Physics, Waves and Sound, JEE Main
If two waves represented by $y_1=4 \sin \omega t$ and $y_2=3 \sin \left(\omega t+\frac{\pi}{3}\right)$ interfere at a point, the amplitude of the resulting wave will be about

Options:
A $7$
B $6$
C $5$
D $3.5$

The correct answer is:
$6$

Solution:
$\phi=\pi / 3, a_1=4, a_2=3$
So, $A=\sqrt{a_1^2+a_2^2+2 a_1 \cdot a_2 \cos \phi} \Rightarrow A \approx 6$

Topic: Physics, Wave Optics, JEE Main
If $u=\sqrt{a^2 \cos ^2 \theta+b^2 \sin ^2 \theta}+\sqrt{a^2 \sin ^2 \theta+b^2 \cos ^2 \theta}$, then the difference between the maximum and minimum values of $u^2$ is given by

Options:
A
$2\left(a^2+b^2\right)$

B
$2 \sqrt{a^2+b^2}$

C
$(a+b)^2$

D
$(a-b)^2$

The correct answer is:
$(a-b)^2$

Solution:
$u=\sqrt{a^2 \cos ^2 \theta+b^2 \sin ^2 \theta}+\sqrt{a^2 \sin ^2 \theta+b^2 \cos ^2 \theta}$
$=\sqrt{\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2} \cos 2 \theta}+\sqrt{\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2} \cos 2 \theta}$
$\Rightarrow u^2=a^2+b^2+2 \sqrt{\left(\frac{a^2+b^2}{2}\right)^2-\left(\frac{a^2-b^2}{2}\right)^2 \cos ^2 2 \theta}$
min value of $u^2=a^2+b^2+2 a b$
max value of $u^2=2\left(a^2+b^2\right)$
$\Rightarrow u_{\max }^2-u_{\min }^2=(a-b)^2$.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2004
If $\mathrm{u}=\mathrm{a}-\mathrm{b}$ and $\mathrm{v}=\mathrm{a}+\mathrm{b}$ and $|\mathrm{a}|=|\mathrm{b}|=2$, then
$|\mathbf{u} \times \mathbf{v}|$ is equal to

Options:
A $2 \sqrt{16-(a \cdot b)^{2}}$
B $\sqrt{16-(\mathrm{a} \cdot \mathbf{b})^{2}}$
C $2 \sqrt{4-(\mathbf{a} \cdot \mathbf{b})^{2}}$
D $2 \sqrt{4+(\mathbf{a} \cdot \mathbf{b})^{2}}$

The correct answer is:
$2 \sqrt{16-(a \cdot b)^{2}}$

Solution:
$|\mathbf{u} \times \mathbf{v}|=|(\mathbf{a}-\mathbf{b}) \times(\mathbf{a}+\mathbf{b})|$
$$
=2|\mathbf{a} \times \mathbf{b}| \quad(\because \mathbf{a} \times \mathbf{a}=\mathbf{b} \times \mathbf{b}=0)
$$
and $|\mathbf{a} \times \mathbf{b}|^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=(a b \sin \theta)^{2}+(a b \cos \theta)^{2}$
$$
=a^{2} b^{2}
$$
$$
\begin{array}{l}
\Rightarrow \quad|\mathbf{a} \times \mathbf{b}|=\sqrt{a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
\text { So, } \quad|\mathbf{u} \times \mathbf{v}|=2|\mathbf{a} \times \mathbf{b}|
\end{array}
$$
$$
\begin{array}{l}
=2 \sqrt{a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
=2 \sqrt{2^{2} 2^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}} \\
\left.=2 \sqrt{16-(\mathbf{a} \cdot \mathbf{b})^{2}} \because|\mathbf{a}|=|\mathbf{b}|=2\right)
\end{array}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them, then $2 \hat{u} \times 3 \hat{v}$ is a unit vector for

Options:
A
exactly two values of $\theta$

B
more than two values of $\theta$

C
no value of $\theta$

D
exactly one value of $\theta$

The correct answer is:
exactly one value of $\theta$

Solution:
$|2 \hat{u} \times 3 \hat{v}|=1$
$\begin{aligned}
& 6|\hat{u}\|\hat{v} \| \sin \theta|=1 \\
& \sin \theta=\frac{1}{6}
\end{aligned}$
Hence there is exactly one value of $\theta$ for which $2 \hat{u} \times 3 \hat{v}$ is a unit vector.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2007
If $u=e^{x^2-y^2}$, then

Options:
A $x u_x=y u_y$
B $y u_x=x u u_y$
C $y u_x+x u_y=0$
D $x^2 u_y+y^2 u_x=0$

The correct answer is:
$y u_x+x u_y=0$

Solution:
Given that, $u=e^{x^2-y^2}$
On differentiating partially
$u_x=e^{x^2-y^2}(2 x)$
On differentiating partially w.r.t. $y$.
$\begin{aligned} u_y & =e^{x^2-y^2}(-2 y) \\ y u_x & =e^{x^2-y^2} 2 x y \\ x u_y & =e^{x^2-y^x}(-2 x y)\end{aligned}$
On adding Eqs. (i) and (ii)
$y u_x+x u_y=0$

Topic: Mathematics, Differentiation, JEE Main
If $u+i v=\frac{3 i}{x+i y+2}$, then $y=$

Options:
A $\frac{9 u}{u^2+v^2}$
B $\frac{3 u}{u^2+v^2}$
C $\frac{6 u}{u^2+v^2}$
D $\frac{12 u}{u^2+v^2}$

The correct answer is:
$\frac{3 u}{u^2+v^2}$

Solution:
$$
\begin{aligned}
& \text {We have, } u+i v=\frac{3 i}{x+i y+2} \\
& \Rightarrow x+i y+2=\frac{3 i}{u+i v} \\
& \Rightarrow(x+2)+i y=\frac{3 i}{u+i v}+\frac{u-i v}{u-i v}=\frac{3 u i-3 v\left(i^2\right)}{u^2-(i v)^2} \\
& \Rightarrow(x+2)+i y=\frac{3 u i+3 v}{u^2+v^2} \\
& \Rightarrow(x+2)+i y=\frac{3 v}{u^2+v^2}+\frac{3 u}{u^2+v^2} i
\end{aligned}
$$
by comparing Imaginary parts
$$
y=\frac{3 u}{u^2+v^2}
$$

Topic: Mathematics, Complex Number, JEE Main
If \( U \) is the universal set with \( 100 \) elements; \( A \) and \( B \) are two sets such that \( n(A)=50, n(B)=60 \),
\( n(A \cap B)=20 \) then \( n\left(A^{\prime} \cap B^{\prime}\right)= \)

Options:
A \( 20 \)
B \( 10 \)
C \( 40 \)
D \( 90 \)

The correct answer is:
\( 10 \)

Solution:
$(B)$
$n(U)=100$
$n(A)=50$
$n(B)=60$
$n(A \cap B)=20$
$n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=50+60-20$
$=110-20$
$=90$
$n\left(A^{\prime} \cap B^{\prime}\right)=n(A \cup B)^{\prime}$
$=n(U)-n(A \cup B)$
$=100-90$
$=10$

Topic: Mathematics, Sets and Relations, JEE Main
If $\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}+3 \hat{k}$ and $\vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j}$ are vectors in 3-dimensional space, then the maximum possible value of $|\vec{u} \times \vec{v} \cdot \vec{w}|$ is

Options:
A
$\sqrt{3}$

B
5

C
$\sqrt{14}$

D
7

The correct answer is:
5

Solution:
Let $\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}-3 \hat{k}$ and
$$
\vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j}
$$
Now, $\vec{u} \times \vec{v}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 4 \\ 1 & 0 & -3\end{array}\right|$
$$
=\hat{i}(-3+\hat{j}-4(+\hat{k})-1 \quad(\quad)
$$
$$
=-3 \hat{i}+4 \hat{j}-\hat{k}
$$
Now,
$$
\begin{aligned}
& (\vec{u} \times \vec{v}) \cdot \vec{w}=(-3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(\cos \theta \hat{i}+\sin \theta \hat{j}) \\
& =-3 \cos \theta+4 \sin \theta
\end{aligned}
$$
Now, maximum possible value of
$$
|-3 \cos \theta+4 \sin \theta|=\sqrt{(-3)^2+(4)^2}=\sqrt{25}=5
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (12 May Online)
If $\vec{u}, \vec{v}$ and $\vec{w}$ are three non-coplanar vectors, then $(\vec{u}+\vec{v}-\vec{w}) .(\vec{u}-\vec{v}) \times(\vec{v}-\vec{w})$ equals

Options:
A
$3 \vec{u} . \vec{v} \times \vec{w}$

B
0

C
$\overrightarrow{\mathrm{u}} \cdot \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}}$

D
$\overrightarrow{\mathrm{u}} \cdot \overrightarrow{\mathrm{w}} \times \overrightarrow{\mathrm{v}}$

The correct answer is:
$\overrightarrow{\mathrm{u}} \cdot \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}}$

Solution:
$(\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{w}}) \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{w}}-\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}})$
$(\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{w}}) \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{w}}+\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}})=\frac{\overrightarrow{\mathrm{u}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})}{0}$
$-\frac{\overrightarrow{\mathrm{u}} .(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{w}})}{0}+\overrightarrow{\mathrm{u}} .(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}})+\frac{\overrightarrow{\mathrm{v}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})}{0}-\overrightarrow{\mathrm{v}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{w}})$
$+\frac{\overrightarrow{\mathrm{v}} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}})}{0}-\overrightarrow{\mathrm{w}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})+\frac{\overrightarrow{\mathrm{w}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{w}})}{0}-\frac{\overrightarrow{\mathrm{w}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{w}})}{0}=\overrightarrow{\mathrm{u}} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}})-\overrightarrow{\mathrm{v}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{w}})-\overrightarrow{\mathrm{w}} \cdot(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})$
$=[\overrightarrow{\mathrm{u}} \overrightarrow{\mathrm{v}} \overrightarrow{\mathrm{w}}]+[\overrightarrow{\mathrm{v}} \overrightarrow{\mathrm{w}} \overrightarrow{\mathrm{u}}]-[\overrightarrow{\mathrm{w}} \overrightarrow{\mathrm{u}} \overrightarrow{\mathrm{v}}]=\overrightarrow{\mathrm{u}} .(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{w}})$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003