datavorous/entrance-exam-dataset huggingface dataset:
If a straight line $y-x=2$ divides the region $x^2+y^2 \leq 4$ into two parts, then the ratio of the area of the smaller part to the area of the greater part is

Options:
A
$3 \pi-8: \pi+8$

B
$\pi-3: 3 \pi+3$

C
$3 \pi-4: \pi+4$

D
$\pi-2: 3 \pi+2$

The correct answer is:
$\pi-2: 3 \pi+2$

Solution:
Let I be the smaller portion and II be the greater portion of the given figure then,
Area of $I=\int_{-2}^0\left[\sqrt{4-x^2}-(x+2] d x\right.$
$$
\begin{aligned}
& =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{-2}^0-\left[\frac{x^2}{2}+2 x\right]_{-2}^0 \\
& =\left[2 \sin ^{-1}(-1)\right]-\left[-\frac{4}{2}+4\right]=2 \times \frac{\pi}{2}-2=\pi-2
\end{aligned}
$$
Now, area of $\mathrm{II}=$ Area of circle $-$ area of $\mathrm{I}$.
$$
\begin{aligned}
& =4 \pi-(\pi-2) \\
& =3 \pi+2
\end{aligned}
$$
Hence, required ratio $=\frac{\text { area of I }}{\text { area of II }}=\frac{\pi-2}{3 \pi+2}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2012 (12 May Online)
If a substance 'A' dissolves in solution of a mixture of ' B' and ' C ' with their respective number of moles as nA,nB and nC, mole fraction of C in the solution is:

Options:
A nCnA×nB×nC
B nCnA+nB+nC
C nCnA-nB-nC
D nBnA+nB

The correct answer is:
nCnA+nB+nC

Solution:
Mole fraction is defined as the amount (number of moles or molecules) of a constituent or component divided by the total amount of all the constituent or component present.

Mole fraction=Number of moles of compnentTotal number of moles of all the components
Suppose we have a mixture with 3 components A, B and C.

Mole fraction of C=nCnA+nB+nC

Hence, the answer is option B.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If a substance with half life 3 days is taken at other place in 12 days. What amount of substance is left now

Options:
A $1 / 4$
B $1 / 8$
C $1 / 16$
D $1 / 32$

The correct answer is:
$1 / 16$

Solution:
$N=N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N}{N_0}=\left(\frac{1}{2}\right)^4=\frac{1}{16}$

Topic: Chemistry, Chemical Kinetics, JEE Main
If $A^T$ denotes the transpose of the matrix $A=\left[\begin{array}{lll}0 & 0 & a \\ 0 & b & c \\ d & e & f\end{array}\right]$, where $a, b, c, d, e$ and $f$ are integers such that $a b d \neq 0$, then the number of such matrices for which $A^{-1}=A^T$ is

Options:
A
2(3!)

B
$3(2 !)$

C
$2^3$

D
$3^2$

The correct answer is:
$2^3$

Solution:
$$
\begin{aligned}
& \text { } A=\left[\begin{array}{lll}
0 & 0 & a \\
0 & b & c \\
d & e & f
\end{array}\right],|A|=-a b d \neq 0 \\
& c_{11}=+(b f-c e), c_{12}=-(-c \mathrm{~d})=c d, \\
& c_{21}=-(-c a)=a e, c_{22}=+(-a d)=-a d, \\
& c_{31}=+(-a b)=-a b, c_{32} \quad c_{23}=-(0)=0 \\
& \operatorname{Adj} A=\left[\begin{array}{ccc}
(b f-c e) & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right]
\end{aligned}
$$
$A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{a b d}\left[\begin{array}{ccc}b f-c e & a e & -a b \\ c d & -a d & 0 \\ -b d & 0 & 0\end{array}\right]$
$$
\begin{aligned}
& A^T=\left[\begin{array}{lll}
0 & 0 & d \\
0 & b & e \\
a & c & f
\end{array}\right] \\
& \text { Now } A^{-1}=A^T \\
& \Rightarrow \frac{1}{-a b d}\left[\begin{array}{ccc}
b f-c e & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cccc}
b f-c e & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right] \\
& \therefore b f-c e=\left[\begin{array}{ccc}
0 & 0 & d \\
0 & b & e \\
a & c & f
\end{array}\right]
\end{aligned}
$$
$a b d^2=a b, a b^2 d=a d, a^2 b d=b d \quad$...(ii)
abde $=a b c d=a b d f=0$
From (ii),
$\left(a b d^2\right) \cdot\left(a b^2 d\right) \cdot\left(a^2 b d\right)=a b . a d . b d$
$\Rightarrow(a b d)^4-(a b d)^2=0$
$\Rightarrow(a b d)^2\left[(a b d)^2-1\right]=0$
$\because a b d \neq 0, \therefore a b d=\pm 1$
From (iii) and (iv),
$$
e=c=f=0
$$
From (i) and (v),
$$
b f=a e=c d=0
$$
From (iv), (v) and (vi), it is clear that $a, b$, $d$ can be any non-zero integer such that $a b d=\pm 1$
But it is only possible, if $a=b=d=\pm 1$ Hence, there are 2 choices for each $a, b$ and $d$. there fore, there are $2 \times 2 \times 2$ choices for $a, b$ and $d$. Hence number of required matrices $=2 \times 2 \times 2=(2)^3$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2012 (19 May Online)
If a tangent drawn at Pα,α3 to the curve y=x3 meets it again at Qβ,β3, then 2β+α is equal to

Options:
A 0
B -3α
C 3α
D 4α

The correct answer is:
-3α

Solution:
Slope of PQ=β3-α3β-α=dydxP
⇒β2+α2+βα=3α2
⇒β3+βα-2α2=0
⇒β+2αβ-α=0
⇒β=-2α as β≠α
⇒2β+α=-4α+α
=-3α

Topic: Mathematics, Application of Derivatives, JEE Main
If a tangent having slope 2 of the ellipse x2a2+y2b2=1 is normal to the circle x2+y2+4x+1=0, then the value of 4a2+b2 is equal to

Options:
A 4
B 2
C 16
D 8

The correct answer is:
16

Solution:
Equation of tangent is y=2x±4a2+b2
This is normal to the circle x2+y2+4x+1=0
This tangent passes through -2,0
0=-4±4a2+b2⇒4a2+b2=16

Topic: Mathematics, Ellipse, JEE Main
If a tangent of slope 2 of the ellipse x2a2+y21=1 passes through the point -2,0, then the value of a2 is equal to

Options:

The correct answer is:
3.75

Solution:
A tangent of slope 2 to the ellipse x2a2+y21=1 is
y=2x±4a2+1

Since, it passes through -2,0
⇒0=2-2±4a2+1
⇒4a2+1=16⇒a2=154=3.75

Topic: Mathematics, Ellipse, JEE Main
If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is:

Options:
A x2+y2–16x2y2=0
B x2+y2â4x2y2=0
C x2+y2â2xy=0
D x2+y2â2x2y2=0

The correct answer is:
x2+y2–4x2y2=0

Solution:
Any point on the circle x2+y2=r2 is rcosθ, rsinθ and the equation of the tangent at this point is xcosθ+ysinθ=r.

Thus, any point on circle x2+y2=1 is cosθ, sinθ and the equation of tangent at this point is xcosθ+ysinθ=1.

To find the point where this line intersects the x-axis, put y=0,⇒x=1cosθ

⇒P=1cosθ, 0

And, to find the point where this line intersects the y-axis, put x=0,

⇒y=1sinθ

⇒Q=0, 1sinθ

The mid-point of a line segment joining the points x1, y1 and x2, y2 is x1+x22, y1+y22

Let, midpoint of PQ is h, k

∴ h, k=12cosθ, 12sinθ

⇒ h=12cosθ & k=12sinθ

⇒cosθ=12h & sinθ=12k

∵ sin2θ+cos2θ=1

⇒ 14h2+14k2=1

⇒1h2+1k2=4

To get the locus of the required point, replace h, k by x, y

∴ Locus is 1x2+1y2=4

⇒x2+y2–4x2y2=0.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If a tangent to the ellipse x2+4y2=4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point.

Options:
A 3,0
B 2,0
C 1,1
D -1,1

The correct answer is:
3,0

Solution:
We have, x24+y21=1

Equation of tangent is cosθx+2sinθy=2

B-2,1+cosθsinθ,C2,1-cosθsinθ

≡B-2,cotθ2,C2,tanθ2

Equation of required circle will be

x+2x-2+y-cotθ2y-tanθ2=0

⇒x2-4+y2-tanθ2+cotθ2y+1=0

Hence, 3,0 satisfies the above equation.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If a thin film of thickness \( t \) and refractive index \( \mu \) is placed in the path of light coming from a source \( S \), then increase in length of optical path is

Options:
A \( \mu \mathrm{t} \)
B \( \mu / \mathrm{t} \)
C \( (\mu-1) \mathrm{t} \)
D None of these

The correct answer is:
\( (\mu-1) \mathrm{t} \)

Solution:
Increase in optical path =μt-t=μ-1t

Topic: Physics, Wave Optics, JEE Main
If a torch is used in place of monochromatic light in Young's experiment what will happens?

Options:
A Fringe will appear for a moment then, it will disappear
B Fringes will occur as from monochromatic light
C Only bright fringes will appear
D No fringes will appear

The correct answer is:
No fringes will appear

Solution:
No fringe will appear, when we used torch in place of monochromatic light because waves reaching at any point on the screen do not have a constant phase difference, as phase difference due to different colours changes randomly. Therefore, maxima and minima would also change their positions randomly and in quick succession. This will result in general illumination on the screen.

Topic: Physics, Wave Optics, JEE Main
If a tower subtends equal angles at four points P, Q, R and S that lie in a plane containing the foot of the tower, then which of the following statements is always true (here, the tower is perpendicular to the plane containing the points P, Q, R, S )

Options:
A ∠PQS=∠PRS
B ∠PQR+∠PSR=180o
C ∠PQS=90o⇒∠PRS=90o
D PQRS+PSRQ=PRQS

The correct answer is:
∠PQS=90o⇒∠PRS=90o

Solution:
Let, top of the tower is T, foot of the tower is F, height of the tower is h & â TPF=Î¸
Because TPF is a right-angled Î

so, PF=hcotâ¡Î¸,

similarly, QF=RF=SF=hcotâ¡Î¸
â points P,Q,R,S lie on the same circle
â points P,Q,R,S are vertices of a cyclic quadrilateral but may or may not be consecutive vertices in the same order
Hence, options A,B & D may or may not be correct.
But, option C is always correct

Topic: Mathematics, Heights and Distances, JEE Main
If a transverse pulse is created at the topmost point of a uniform rope suspended vertically, then

Options:
A speed of pulse remains constant.
B speed of the pulse decreases with constant rate as pulse moves downward.
C speed of the pulse decreases with increasing rate as pulse moves downward.
D speed of the pulse increases with constant rate as pulse moves downward.

The correct answer is:
speed of the pulse decreases with constant rate as pulse moves downward.

Solution:
Tension at P=mLgx=Î¼gx

v=TÎ¼=gx

a=dvdt=dgxdt

a = g2xdxdt=g2xv=g2

Thus with decrease in x, speed of pulse will also decrease but at constant rate.

Topic: Physics, Waves and Sound, JEE Main
If a tuning fork of frequency (f0) 340 Hz and tolerance Â±1% is used in the resonance column method for determining the speed of sound. If the first and the second resonance are measured at lâ¡ 1 = 2 4.0 cm and lâ¡ 2 = 74.70 cm, then the permissible error in speed of sound is

Options:
A 1.2%
B 1.8%
C 1%
D 0.8%

The correct answer is:
1.2%

Solution:
V = 2 f 0 l 2 - l 1
let l 2 - l 1 = x
l 1 = 24.0 + Â¯ 0.1 cm
l 2 = 74.70 + Â¯ 0.01 cm
l 2 - l 1 = 50.70 + Â¯ 0.11 cm
Î x x = 0.11 50.70
Î x % = 0.11 50.70 Ã 100
= 0.22%
Î V % = Î f 0 % + Î x %
= 1.22 %
= 1.2 %

Topic: Physics, Experimental Physics, JEE Main
If a unit vector a→ makes angles π3 with i^, π4 with j^ and θ∈0,π with k^, then a value of θ is:

Options:
A 5π6
B 5Ï12
C Ï4
D 2Ï3

The correct answer is:
2π3

Solution:
The unit vector a→ makes angle π3, π4 and θ with i^, j^ & k^ respectively.

Hence, the direction cosines of the vector a→ are cosα=cosπ3=12, cosβ=cosπ4=12 and cosγ=cosθ

For a line having direction cosines as cosα, cosβ, cosγ

⇒cos2α+cos2β+cos2γ=1

⇒14+12+cos2θ=1

⇒cos2θ=1-34=14

⇒cos2θ=±12

⇒θ=π3 or 2π3.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If a variable line 3x+4y-λ=0 is such that the two circles x2+y2-2x-2y+1=0 and x2+y2-18x-2y+78=0 are on its opposite sides, then the set of all values of λ is the interval :

Options:
A 13, 23
B 23, 31
C 12, 21
D 2, 17

The correct answer is:
12, 21

Solution:
Given circles x2+y2-2x-2y+1=0⇒x-12+y-12=12 and

x2+y2-18x-2y+78=0⇒x-92+y-12=22

Thus, centers of two circles are 1, 1 & 9, 1 and radii are 1 & 2 respectively.

Since, center of circles are on the opposite side of line 3x+4y-λ=0.

⇒3+4-λ27+4-λ<0

⇒7-λ31-λ<0

⇒λ∈7, 31 …….i
Since, circles should not intersect given line.

Hence, 7-λ32+42≥1   &   31-λ32+42≥2

⇒7-λ≥5  & 31-λ≥10

⇒λ≤2 or  λ≥12 …….ii

and λ≤21 or λ≥41 …….iii

Thus, λ∈i∩ii∩iii

⇒λ∈12, 21

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (12 Jan Shift 1)
If a variable line drawn through the intersection of the lines x3+y4=1 and x4+y3=1 , meets the coordinate axes at A and B, A ≠B,then the locus of the midpoint of AB is:

Options:
A 7xy=6x+y
B 4x+y2-28x+y+49=0
C 6xy=7x+y
D 14x+y2-97x+y+168=0

The correct answer is:
7xy=6x+y

Solution:
L1 :4x+3y-12=0
L2:3x+4y-12=0
L1+λL2=0
⇒4x+3y-12+λ3x+4y-12=0
⇒x4+3λ+y3+4λ-121+λ=0
Point A121+λ4+3λ, 0
Point B0,121+λ3+4λ
Mid point ⇒h=61+λ4+3λ .......i
k=61+λ3+4λ .......ii
Eliminate λ from equations i and ii then
6h+k= 7hk
Hence, locus is
6x+y= 7xy

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2016 (09 Apr Online)
If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha}+\frac{y}{\beta}=1$ and $\frac{x}{\beta}+\frac{y}{\alpha}=1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $A B$ is

Options:
A $\alpha \beta(x+y)=x y(\alpha+\beta)$
B $\alpha \beta(x+y)=2 x y(\alpha+\beta)$
C $(\alpha+\beta)(x+y)=2 \alpha \beta x y$
D None of these

The correct answer is:
$\alpha \beta(x+y)=2 x y(\alpha+\beta)$

Solution:
The equation of a line passing through the intersection of straight lines $\frac{x}{\alpha}+\frac{y}{\beta}=1$ and $\frac{x}{\beta}+\frac{y}{\alpha}=1$ is
$\begin{aligned}
& \left(\frac{x}{\alpha}+\frac{y}{\beta}-1\right)+\lambda\left(\frac{x}{\beta}+\frac{y}{\alpha}-1\right)=0 \\
& x\left(\frac{1}{\alpha}+\frac{\lambda}{\beta}\right)+y\left(\frac{1}{\beta}+\frac{\lambda}{\alpha}\right)-\lambda-1=0
\end{aligned}$
or
This meets the axes at
$A\left(\frac{\lambda+1}{\frac{1}{\alpha}+\frac{\lambda}{\beta}}, 0\right) \text { and } B\left(0, \frac{\lambda+1}{\frac{1}{\beta}+\frac{\lambda}{\alpha}}\right)$
Let $(h, k)$ be the mid point of $A B$,
$h=\frac{1}{2} \cdot \frac{\lambda+1}{\frac{1}{\alpha}+\frac{\lambda}{\beta}}, k=\frac{1}{2} \cdot \frac{\lambda+1}{\frac{1}{\beta}+\frac{\lambda}{\alpha}}$
Eliminating $\lambda$ from these two, we get
$2 h k(\alpha+\beta)=\alpha \beta(h+k)$
$\therefore$ The locus of $(h, k)$ is $2 x y(\alpha+\beta)=\alpha \beta(x+y)$

Topic: Mathematics, Straight Lines, JEE Main
If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B & C, then the locus of the centroid of ΔABC is

Options:
A 1x2+1y2+1z2=1
B 1x2+1y2+1z2=3
C 1x2+1y2+1z2=9
D 1x2+1y2+1z2=19

The correct answer is:
1x2+1y2+1z2=1

Solution:
Let centroid be h, k, l
x-intercept=3h

y-intercept=3k

and z-intercept=3l
Now, the equation of plane is x3h+y3k+z3l=1.
Distance from 0, 0, 0
⇒119h2+19k2+19l2=3
⇒1x2+1y2+1z2=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2017 (09 Apr Online)
If a variable takes the discrete values $\alpha-4, \alpha-\frac{7}{2}, \alpha-\frac{5}{2}, \alpha-3, \alpha-2, \alpha+\frac{1}{2}, \alpha-\frac{1}{2}, \alpha+5(\alpha\gt0)$, then the median is

Options:
A $\alpha-\frac{5}{4}$
B $\alpha-\frac{1}{2}$
C $\alpha-2$
D $\alpha+\frac{5}{4}$

The correct answer is:
$\alpha-\frac{5}{4}$

Solution:
Arrange the data as
$\alpha-\frac{7}{2}, \alpha-3, \alpha-\frac{5}{2}, \alpha-2, \alpha-\frac{1}{2}, \alpha+\frac{1}{2}, \alpha+4, \alpha+5$
Median $=\frac{1}{2}\left[\right.$ value of $4^{\text {th }}$ item + value of $5^{\text {th }}$ item $]$
$\therefore$ Median $=\frac{\alpha-2+\alpha-\frac{1}{2}}{2}=\frac{2 \alpha-\frac{5}{2}}{2}=\alpha-\frac{5}{4}$.

Topic: Mathematics, Statistics, JEE Main
If a vector $2 \hat{i}+3 \hat{j}+8 \hat{k}$ is perpendicular to the vector $4 \hat{j}-4 \hat{i}+\alpha \hat{k}$ then the value of $\alpha$ is:

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

The correct answer is:
$- \frac{1}{2}$

Solution:
$\vec{a}=2 \hat{i}+3 \hat{i}+8 \hat{k}, b=4 \hat{j}+3 \hat{i}+\alpha \hat{k}$
$\begin{aligned}
& \vec{a} \cdot \vec{b}=0, \text { if } a^{-1} \cdot b^{-1} \\
& \therefore(2 \hat{i}+3 \hat{j}+8 \hat{k}) \cdot(-4 i+4 j+\alpha k)=\frac{1}{2} \\
& \Rightarrow \alpha=-\frac{1}{2}
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
If a vertex of a triangle is $(1,1)$ and the mid-points of two sides through this vertex are $(-1,2)$ and $(3,2)$, then the centroid of the triangle is

Options:
A
$\left(-1, \frac{7}{3}\right)$

B
$\left(\frac{-1}{3}, \frac{7}{3}\right)$

C
$\left(1, \frac{7}{3}\right)$

D
$\left(\frac{1}{3}, \frac{7}{3}\right)$

The correct answer is:
$\left(1, \frac{7}{3}\right)$

Solution:
Vertex of triangle is $(1,1)$ and midpoint of sides through this vertex is $(-1,2)$ and $(3,2)$ $\Rightarrow$ vertex $B$ and $C$ come out to be $(-3,3)$ and $(5,3)$
$\therefore$ centroid is $\frac{1-3+5}{3}, \frac{1+3+3}{3}$ $\Rightarrow(1,7 / 3)$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2005
If a wave gets refracted into a denser medium, then which of the following is true?

Options:
A Wavelength, speed and frequency decreases.
B Wavelength increases, speed decreases and frequency remains constant.
C Wavelength and speed decreases but frequency remains constant.
D Wavelength, speed and frequency increases.

The correct answer is:
Wavelength and speed decreases but frequency remains constant.

Solution:
In refraction from one medium to another, the speed and wavelength gets changed. The frequency remains unchanged.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (27 Jun Shift 2)
If a wire is stretched to make it $0.1 \%$ longer, its resistance will :

Options:
A
increase by $0.2 \%$

B
decrease by $0.2 \%$

C
decrease by $0.05 \%$

D
increases by $0.05 \%$

The correct answer is:
increase by $0.2 \%$

Solution:
$\mathrm{R} \propto \ell^2$ (for a given volume)
$$
\Rightarrow \frac{\Delta \mathrm{R}}{\mathrm{R}} \%=\frac{2 \Delta \ell}{\ell} \%
$$
Thus when wire is stretched by $0.1 \%$ resistance increases by $0.2 \%$

Topic: Physics, Mathematics in Physics, JEE Main
If a wire is stretched to make it $0.1 \%$ longer, its resistance will :

Options:
A
increase by $0.2 \%$

B
decrease by $0.2 \%$

C
decrease by $0.05 \%$

D
increases by $0.05 \%$

The correct answer is:
increase by $0.2 \%$

Solution:
$\mathrm{R} \propto \ell^2$ (for a given volume)
$$
\Rightarrow \frac{\Delta \mathrm{R}}{\mathrm{R}} \%=\frac{2 \Delta \ell}{\ell} \%
$$
Thus when wire is stretched by $0.1 \%$ resistance increases by $0.2 \%$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2011
If $A=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right|$ and $B=\left|\begin{array}{cc}x & 1 \\ 1 & x\end{array}\right|$, then $\frac{d A}{d x}$ is equal to

Options:
A $3 B+1$
B $3 B$
C $-3 B$
D $1-3 B$

The correct answer is:
$3 B$

Solution:
$\begin{array}{ll}
\text { Given, } & A=\left|\begin{array}{lll}
x & 1 & 1 \\
1 & x & 1 \\
1 & 1 & x
\end{array}\right| \\
\Rightarrow & A=x\left(x^{2}-1\right)-(x-1)+(1-x) \\
\Rightarrow & A=x^{3}-x-x+1+1-x \\
\Rightarrow & A=x^{3}-3 x+2
\end{array}$
On differentiating w.r.t. $x$, we get
$\frac{d A}{d x}=3 x^{2}-3..(i)$
Also, given $\quad B=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|$
$\begin{array}{cc}
\Rightarrow & B=x^{2}-1 \\
\Rightarrow & 3 B=3 x^{2}-3...(ii)
\end{array}$
From Eqs. (i) and (ii), we get
$\frac{d A}{d x}=3 B$

Topic: Mathematics, Determinants, JEE Main
If $A(x)=\left|\begin{array}{ccc}1 & 2 & 3 \\ x+1 & 2 x+1 & 3 x+1 \\ x^2+1 & 2 x^2+1 & 3 x^2+1\end{array}\right|$,
then $\int_0^1 A(x) d x$ equals

Options:
A 0
B 1
C 2
D 4

The correct answer is:
0

Solution:
$A(x)=\left|\begin{array}{ccc}1 & 2 & 3 \\ x+1 & 2 x+1 & 3 x+1 \\ x^2+1 & 2 x^2+1 & 3 x^2+1\end{array}\right|$
Apply $C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_2$, we get
$$
\begin{aligned}
A(x) & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
x+1 & x & x \\
x^2+1 & x^2 & x^2
\end{array}\right| \\
& =0 \quad\left[\because C_2 \text { and } C_3 \text { are identical }\right] \\
\therefore \quad \int_0^1 A(x) d x & =\int_0^1 0 d x=0
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
If $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$, then $\frac{d y}{d x}=$

Options:
A $-\frac{a x+h y+g}{h x-b y+f}$
B $\frac{a x+h y+g}{h x-b y+f}$
C $\frac{a x-h y-g}{h x-b y-f}$
D None of these

The correct answer is:
$-\frac{a x+h y+g}{h x-b y+f}$

Solution:
$a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$
Differentiating w.r.t. x of y, we get
$\begin{aligned} & 2 a x+2 h\left(y+x \frac{d y}{d x}\right)+2 b y \frac{d y}{d x}+2 g+2 f \frac{d y}{d x}=0 \\ & \therefore \frac{d y}{d x}(2 h x+2 b y+2 f)=-(2 a x+2 h y+2 g) \\ & \text { or } \quad \frac{d y}{d x}=-\frac{(a x+h y+g)}{(h x+b y+f)} .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If ax+13y+bz+c=0 is a plane through the line of intersection of 2x+3y-z+1=0, x+y-2z+3=0 and is perpendicular to the plane 3x-y-2z=4, then the value of 2a+3b+4c is equal to

Options:
A -12
B 12
C 10
D -10

The correct answer is:
-10

Solution:
Equation of the plane through the line of intersection of given 2 planes is
2x+3y-z+1+Î»x+y-2z+3=0
2+Î»x+3+Î»y-1+2Î»z+1+3Î»=0
âµ it is perpendicular to 3x-y-2z=4
6+3Î»-3-Î»+4Î»+2=0
6Î»+5âÎ»=-56
7x+13y+4z-9=0
On comparing a=7,b=4,c=-9

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If $\mathbf{a}+x \mathbf{b}+y \mathbf{c}=\mathbf{0}$, $\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}=6(\mathbf{b} \times \mathbf{c})$, then the locus of the point $(x, y)$ is

Options:
A $x^2+y^2=1$
B $x+y-5=0$
C $2 x+6 y=5$
D $x+y+6=0$

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

Solution:
Given,
$$
\mathbf{a}+x \mathbf{b}+y \mathbf{c}=\mathbf{0}
$$
As we know that, if $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$
Then, $\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq 0$
Now, if $\mathbf{a}+x \mathbf{b}+y \mathbf{c}=0$
Let $x(\mathbf{a} \times \mathbf{b})=x y(\mathbf{b} \times \mathbf{c})=y(\mathbf{c} \times \mathbf{a})=\mathbf{p}$
$\Rightarrow \mathbf{a} \times \mathbf{b}=\frac{\mathbf{p}}{x}, \mathbf{b} \times \mathbf{c}=\frac{\mathbf{p}}{x y}$ and $\mathbf{c} \times \mathbf{a}=\frac{\mathbf{p}}{y}$
$\therefore \mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}=\frac{\mathbf{p}}{x}+\frac{\mathbf{p}}{x y}+\frac{\mathbf{p}}{y}$
$$
=\mathbf{p}\left(\frac{x+y+1}{x y}\right)
$$
$$
\begin{aligned}
& =\left(\frac{x+y+1}{x y}\right) \times x y \times(\mathbf{b} \times \mathbf{c}) \\
& =(x+y+1)(\mathbf{b} \times \mathbf{c})
\end{aligned}
$$
Comparing it with $6(\mathbf{b} \times \mathbf{c})$,
$$
\begin{aligned}
& x+y+1=6 \\
& \therefore \quad x+y=5 \\
& \Rightarrow \quad x+y-5=0 \\
&
\end{aligned}
$$
Hence, option (b) satisfies it.

Topic: Mathematics, Vector Algebra, JEE Main
If a+x=b+y=c+z+1, where a, b, c, x, y, z are non-zero distinct real numbers, thenxa+yx+ayb+yy+bzc+yz+c is equal to :

Options:
A y(b – a)
B y (a – b)
C 0
D y(a – c)

The correct answer is:
y (a – b)

Solution:
Given x+a=y+b=z+c+1

Now xa+ya+xyb+yb+yzc+yc+z=xa+yayb+ybzc+ycC3→C3-C1

=xyayybzycC2→C2-C3

=yx1ay1bz1c

R2→R2-R1 and R3→R3-R1

yx1ay-x0b-az-x0c-a=yx1aa-b0-(a-b)z-x0c-a=y(a-b)x1a10-1z-x0c-a=-y(a-b)(c-a+z-x)=y(a-b)

Topic: Mathematics, Determinants, JEE Main
If a+x=b+y=c+z+1, where a, b, c, x, y, z are non-zero distinct real numbers, thenxa+yx+ayb+yy+bzc+yz+c is equal to :

Options:
A y(b – a)
B y (a – b)
C 0
D y(a – c)

The correct answer is:
y (a – b)

Solution:
Given x+a=y+b=z+c+1

Now xa+ya+xyb+yb+yzc+yc+z=xa+yayb+ybzc+ycC3→C3-C1

=xyayybzycC2→C2-C3

=yx1ay1bz1c

R2→R2-R1 and R3→R3-R1

yx1ay-x0b-az-x0c-a=yx1aa-b0-(a-b)z-x0c-a=y(a-b)x1a10-1z-x0c-a=-y(a-b)(c-a+z-x)=y(a-b)

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If a→=xi^+yj^+zk^, b→=ı^+j^+k^ where x,y,z∈0,3,-3,6,-6, then the number of non-zero vectors a→ such that a→·b→=0 are

Options:
A 20
B 18
C 15
D 12

The correct answer is:
18

Solution:
a→·b→=x+y+z=0
If x=3,y=-3,z=0, i.e. 3!=6
If x=6,y=-6,z=0, i.e.3!=6
If x=6,y=-3,z=-3, i.e.3!2!=3
If x=-6,y=3,z=3, i.e.=3!2!=3
Total number of vectors =18

Topic: Mathematics, Vector Algebra, JEE Main
If $\Delta=\left|\begin{array}{lll}a & b & c \\ x & y & z \\ p & q & r\end{array}\right|$, then $\left|\begin{array}{lll}k a & k b & k c \\ k x & k y & k z \\ k p & k q & k r\end{array}\right|=$

Options:
A $\Delta$
B $k \Delta$
C $3 k \Delta$
D $k^3 \Delta$

The correct answer is:
$k^3 \Delta$

Solution:
$\left|\begin{array}{lll}k a & k b & k c \\ k x & k y & k z \\ k p & k q & k r\end{array}\right|=k^3\left|\begin{array}{lll}a & b & c \\ x & y & z \\ p & q & r\end{array}\right|=k^3 \Delta$

Topic: Mathematics, Determinants, JEE Main
If A=x∈R:x2-5x+6=0, then nA= ________

Options:
A 2
B 0
C 1
D 4

The correct answer is:
4

Solution:
Given: A=xâR:x2-5x+6=0
x2-5x+6=0⇒x2-5x+6=0
x=2,3⇒x=Â±2,Â±3
Hence, nA=4

Topic: Mathematics, Sets and Relations, JEE Main