datavorous/entrance-exam-dataset huggingface dataset:
If an=-24n2-16n+15, then a1+a2+….+a25 is equal to:

Options:
A 51144
B 49138
C 50141
D 52147

The correct answer is:
50141

Solution:
We have,

an=-24n2-16n+15

⇒an=-22n-52n-3

⇒an=-121n-52n-32

⇒an=-121n-52-1n-32

⇒an=121n-32-1n-52

⇒∑n=125an=12∑n=1251n-32-1n-52

⇒∑n=125an=121-12-1-32+112-1-12+132-112+152-132+172-152+...+1452-1432+1472-1452

⇒∑n=125an=1223+247

⇒∑n=125an=13+147

⇒∑n=125an=50141

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If $a_n=\sqrt{7+\sqrt{7+\sqrt{7+\ldots .}}}$ having $\mathrm{n}$ radical signs then by methods of mathematical induciton which is true

Options:
A
$a_n>7 \forall n \geq 1$

B
$\mathrm{a}_{\mathrm{n}}>7 \forall \mathrm{n} \geq 1$

C
$a_n < 4 \forall n \geq 1$
D
$a_n < 3 \forall \mathrm{n} \geq 1$

The correct answer is:
$\mathrm{a}_{\mathrm{n}}>7 \forall \mathrm{n} \geq 1$

Solution:
$\mathrm{a}_1=\sqrt{7} < 7$. Leta $_{\mathrm{m}} < 7$. Then $\mathrm{a}_{\mathrm{m}}+1=\sqrt{7+\mathrm{a}_{\mathrm{m}}} \Rightarrow \mathrm{a}_{\mathrm{m}+1}^2=7+\mathrm{a}_{\mathrm{m}} < 7+7 < 14$
$\Rightarrow \mathrm{a}_{\mathrm{m}+1} < \sqrt{14} < 7 ;$ So $\mathrm{a}_{\mathrm{n}} < 7 \forall \mathrm{n} \therefore \mathrm{a}_{\mathrm{n}}>3$

Topic: Mathematics, Mathematical Induction, JEE Main, JEE Main 2002
If $\left|\mathrm{A}_{\mathrm{n} \times \mathrm{n}}\right|=3$ and $|\operatorname{adj} \mathrm{A}|=243$, what is the value of $\mathrm{n}$?

Options:
A 4
B 5
C 6
D 7

The correct answer is:
6

Solution:
As given: $\left|\mathrm{A}_{\mathrm{nxn}}\right|=3$ and $|\operatorname{adjA}|=243$
Dterminant of adjoint $\mathrm{A}$ is given by:
$|\operatorname{adj} \mathrm{A}|=\left|\mathrm{A}_{\mathrm{n} \times \mathrm{n}}\right|^{\mathrm{n}-1}$
$\Rightarrow 243=3^{n-1} \Rightarrow 3^{5}=3^{n-1} \Rightarrow n-1=5 \Rightarrow n=6$

Topic: Mathematics, Matrices, JEE Main
If a new planet is discovered rotating around Sun with the orbital radius double that of earth, then what will be its time period (in earth's days)

Options:
A $1032$
B $1023$
C $1024$
D $1043$

The correct answer is:
$1032$

Solution:
$T^2 \propto R^3 \Rightarrow\left(\frac{T_P}{T_E}\right)^2=\left(\frac{R_P}{R_E}\right)^3=\left(\frac{2 R_E}{R_E}\right)^3$
$\begin{aligned} & \frac{T_P}{T_E}=(2)^{3 / 2}=2 \sqrt{2} \\ \Rightarrow & T_P=2 \sqrt{2} \times 365=1032.37=1032 \text { days }\end{aligned}$

Topic: Physics, Gravitation, JEE Main
If a new planet is discovered rotating around the sun with the orbital radius double that of earth, then what will be its time period (in earth's days)? (2=1.4)

Options:

The correct answer is:
1032

Solution:
From Kepler's third law of planetary motion
T2âR3
Given, Rp=2Re
â´ Te2Tp2=Re3Rp3
â¹ Te2Tp2=Re32Re3
â¹ Tp=22 Te
Since, Te=365 days=1 year, we have
Tp=22Ã365 days
TP=1032 days

Topic: Physics, Gravitation, JEE Main
If a normal drawn to the parabola $y^2=4 a x$ at the point $(a, 2 a)$ meets parabola again on $\left(a t^2, 2 a t\right)$, then the value of $t$ will be

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

The correct answer is:
$-3$

Solution:
We know that
$t_2=-t_1-\frac{2}{t_1}$
Put $t_1=1$ and $t_2=t$. Hence $t=-3$.

Topic: Mathematics, Parabola, JEE Main
If a number of ellipses whose major axis is x-axis and the minor axis is y-axis be described having the same length of the major axis as 2 but a variable minor axis, then the tangents at the ends of their latus rectum pass through fixed points whose distance from the centre is equal to

Options:
A 12 units
B 1 unit
C 32 units
D 2 units

The correct answer is:
1 unit

Solution:
Since the minor axis is variable, hence e is a variable quantity

Therefore the equation of the tangent to the ellipse can be written as xaea2+yb2ab2=1
âex+y=aây-a+ex=0
This line passes through the fixed point 0,a (where a=1)

Hence, distance fro the centre of the ellipse =1

Topic: Mathematics, Ellipse, JEE Main
If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2 c \\ a & b & r\end{array}\right|=0$, then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
We have,
$\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2 c \\ a & b & r\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{lll}p & b & c \\ p & b & c \\ a & b & r\end{array}\right|+\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0$
$\Rightarrow \quad 0+\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0$
$\Rightarrow p(q r-b c)-b(a r-a c)+c(a b-a q)=0$
$\Rightarrow \quad p q r-p b c-b a r+b a c+a b c-a c q=0$
$\Rightarrow \quad p q r-p b c-b a r-a c q=-2 a b c$
$-p q r+p b c+b a r+a c q=2 a b c$
On simplifying $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2$

Topic: Mathematics, Determinants, JEE Main
If a parabola passess through the points $(-2,1)$, $(1,2)$ and $(-1,3)$ having horizontal axis, then the length of the latus rectum of that parabola is

Options:
A 5
B $\frac{5}{2}$
C $\frac{2}{5}$
D $\frac{1}{5}$

The correct answer is:
$\frac{2}{5}$

Solution:
Let the equation of parabola having horizontal axis, vertex at $(h, k)$ is
$$
(y-k)^2=4 a(x-h)
$$
$\because$ Parabola (i) passes through the points $(-2,1)$, $(1,2)$ and $(-1,3)$, so
$$
\begin{aligned}
& (k-1)^2=4 a(-2-h) \\
& \Rightarrow \quad k^2-2 k+1=-8 a-4 a h \\
& \Rightarrow \quad\left(k-2^2=4 a(1-h)\right. \\
& \Rightarrow \quad k^2-4 k+4=4 a-4 a h \\
& \text { and } \quad(3-k)^2=4 a(-1-h) \\
& \Rightarrow \quad k^2-6 k+9=-4 a-4 a h \\
&
\end{aligned}
$$
From Eqs. (ii), (iii) and (iv), we get
$$
4 a=\frac{2}{5}
$$

Topic: Mathematics, Parabola, JEE Main
If a parallelogram is cut by two sets of m lines parallel to its sides, then the number of parallelogram thus formed, is

Options:
A C2m×C2m
B 2C2m+2
C C2m+22
D None of these

The correct answer is:
C2m+22

Solution:
The two sets of m parallel lines along with two sets of two parallel lines of the given parallelogram will form two sets of m+2 parallel lines. Each parallelogram is formed by choosing two parallel lines from each of the above

∴ Total number of parallelograms

=C2m+2×C2m+2=C2m+22

Hence, option (c) is correct.

Topic: Mathematics, Permutation Combination, JEE Main
If a parallelogram is formed with two sides represented by vectors \( \vec{a} \) and \( \vec{b} \), then \( \vec{a}+\vec{b} \) represented the

Options:
A major diagonal when the angle between vectors is acute.
B minor diagonal when the angle between vectors is obtuse.
C Both of the above.
D None of the above.

The correct answer is:
Both of the above.

Solution:
If a → & b→ represents side of parallelogram, then a→+b→ is represented by diagonal of parallelogram.
Case-1, Angle between them is acute, then the addition is represented major diagonal shown as below,

Case-2, Angle between them is obtuse, then the addition is represented minor diagonal shown as below,

a→+b→→major diagonal, if angle between vectors is acute.

a→+b→→minor diagonal, if angle between vectors is obtuse.

Topic: Physics, Mathematics in Physics, JEE Main
If a particle moves from point $P(2,3,5)$ to point $Q(3,4,5)$. Its displacement vector be

Options:
A $\hat{i}+\hat{j}+10 \hat{k}$
B $\hat{i}+\hat{j}+5 \hat{k}$
C $\hat{i}+\hat{j}$
D $2 \hat{i}+4 \hat{j}+6 \hat{k}$

The correct answer is:
$\hat{i}+\hat{j}$

Solution:
Displacement vector $\vec{r}=\Delta x \hat{i}+\Delta \hat{y j}+\Delta z \hat{k}$
$=(3-2) \hat{i}+(4-3) \hat{j}+(5-5) \hat{k}=\hat{i}+\hat{j}$

Topic: Physics, Mathematics in Physics, JEE Main
If a particle moves in a circle, describing equal angles in equal intervals of time, the velocity vector:

Options:
A Remains constant.
B Changes in magnitude.
C Changes in direction.
D Changes both in magnitude and direction

The correct answer is:
Changes in direction.

Solution:
The figure clearly describes motion of the particles and it is evident that the velocity vector changes direction continuously during the motion. As the particle is covering equal angles in equal time, the magnitude remains constant while the direction changes.

Topic: Physics, Rotational Motion, JEE Main
If a particle of mass $m$ is moving with constant velocity $v$ parallel to $x$-axis in $x$-y plane as shown in fig. Its angular momentum with respect to origin at any time $t$ will be

Options:
A $m v b \hat{k}$
B $-m v b \hat{k}$
C $m v b \hat{i}$
D $m v \hat{i}$

The correct answer is:
$-m v b \hat{k}$

Solution:
We know that, Angular momentum
$\vec{L}=\vec{r} \times \vec{p}$ in terms of component becomes
$\vec{L}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z\end{array}\right|$
As motion is in $x-y$ plane $\left(z=0\right.$ and $\left.P_z=0\right)$, so $\vec{L}=\vec{k}\left(x p_y-y p_x\right)$
Here $x=v t, y=b, p_x=m v$ and $p_y=0$
$\therefore \vec{L}=\vec{k}[v t \times 0-b m v]=-m v b \hat{k}$

Topic: Physics, Mathematics in Physics, JEE Main
If a person can throw a stone to maximum height of \(\mathrm{h}\) metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

Options:
A \(\frac{\mathrm{h}}{2}\)
B \(\mathrm{h}\)
C \(2 \mathrm{~h}\)
D \(3 \mathrm{~h}\)

The correct answer is:
\(2 \mathrm{~h}\)

Solution:
Hints:
\(\mathrm{h}=\frac{\mathrm{u}^2}{2 \mathrm{~g}} \Rightarrow \mathrm{u}^2=2 \mathrm{gh}\)
Maximum horizontal distance
\(\begin{aligned}
& \mathrm{R}_{\max }=\frac{\mathrm{u}^2}{\mathrm{~g}}\left(\text { when } \theta=45^{\circ}\right) \\
& \mathrm{R}_{\max }=2 \mathrm{~h}
\end{aligned}\)

Topic: Physics, Motion In Two Dimensions, JEE Main
If a person can throw a stone to the maximum height of h metre vertically, then the maximum horizontal distance through which the stone can be thrown by the same person is

Options:
A h2
B h
C 2h
D 3h

The correct answer is:
2h

Solution:
h=u22g, u2=2gh
Maximum horizontal distance
Rmax=u2g
Rmax=2h

Topic: Physics, Motion In Two Dimensions, JEE Main
If a person with a spring balance and a body hanging from it goes up and up in an aeroplane, then the reading of the weight of the body as indicated by the spring balance will

Options:
A Go on increasing
B Go on decreasing
C First increase and then decrease
D Remain the same

The correct answer is:
First increase and then decrease

Solution:
Initially due to upward acceleration apparent weight of the body increases but then it decreases due to decrease in gravity.

Topic: Physics, Laws of Motion, JEE Main
If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ0, the graph between the temperature T of the metal and time t will be closest to :

Options:
A
B
C
D

The correct answer is:

Solution:
According to Newton's cooling law, rate of cooling is

dθdt∝θ-θ0

Here, θ=the temperature of the body

θ0=Room temperature(constant)

Let k be the proportionality constant and θi be the initial temperature.
Therefore,

dθdt=-kθ-θ0⇒∫dθθ-θ0=-k∫dt⇒lnθ-θ0=-kt⇒θ=θ0+θ-θ0e-kt

Hence, the graph is exponential.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2013 (07 Apr)
If a plane cuts off intercepts $-6,3,4$ from the co-ordinate axes, then the length of the perpendicular from the origin to the plane is

Options:
A $\frac{1}{\sqrt{61}}$
B $\frac{13}{\sqrt{61}}$
C $\frac{12}{\sqrt{29}}$
D $\frac{5}{\sqrt{41}}$

The correct answer is:
$\frac{12}{\sqrt{29}}$

Solution:
Equation is $\frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1$ or $-2 x+4 y+3 z=12$
$\therefore$ Length of perpendicular from origin
$=\frac{12}{\sqrt{4+16+9}}=\frac{12}{\sqrt{29}} .$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If a plane meets the co-ordinate axes at $A, B$ and $C$ such that the centroid of the triangle is $(1,2$, 4) then the equation of the plane is

Options:
A $x+2 y+4 z=12$
B $4 x+2 y+z=12$
C $x+2 y+4 z=3$
D $4 x+2 y+z=3$

The correct answer is:
$4 x+2 y+z=12$

Solution:
Given, plane meets the co-ordinate axes at $A(a, 0,0), B(0, b, 0) C(0,0, c)$
$\begin{aligned}
& \quad \equiv\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(1,2,4) \\
& \therefore \text { Centroid } \\
& \Rightarrow a=3, b=6, c=12
\end{aligned}$
Hence, equation of required plane is, $\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1$
$\Rightarrow 4 x+2 y+z=12 \text {. }$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If a plane meets the co-ordinate axes at \( A, B \) and \( C \) such that the centroid of the triangle is \( (1,2,4) \), then the equation of the plane is

Options:
A \( x+2 y+4 z=12 \)
B \( 4 x+2 y+z=12 \)
C \( x+2 y+4 z=3 \)
D \( 4 x+2 y+z=3 \)

The correct answer is:
\( 4 x+2 y+z=12 \)

Solution:
Given plane meets the co-ordinate axes at Aa,0,0, B0,b,0 C0,0,c

G =Centroid ≡a3,b3,c3=1, 2, 4

a=3,b=6,c=12

Hence, equation of required plane is, x3+y6+z12=1

4x+2y+z=12 .

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If a plane passes through the points (-1,k,0),(2,k,-1), (1,1,2) and is parallel to the line x-11=2y+12 =z+1-1, then the value of k2+1(k-1)(k-2) is

Options:
A 175
B 517
C 613
D 136

The correct answer is:
136

Solution:
Given,

A plane passes through the points (-1,k,0),(2,k,-1), (1,1,2) and is parallel to the line x-11=2y+12 =z+1-1,

So, on rearranging x-11=2y+12=z+1-1 we get,

⇒x-11=y+121=z+1-1

Now given point : A(-1,k,0),B(2,k,-1),C(1,1,2)

So, CA→=-2i^+(k-1)j^-2k^,

CB→=i^+(k-1)j^-3k^

And CA→×CB→=i^j^k^-2k-1-21k-1-3

⇒CA→×CB→=i^(-3k+3+2k-2)-j^(6+2)+k^(-2k+2-k+1)

⇒CA→×CB→=i^(-3k+3+2k-2)-j^(6+2)+k^(-2k+2-k+1)

⇒CA→×CB→=(1-k)i^-8j^+(3-3k)k^

The line x-11=y+121=z+1-1 is perpendicular to normal vector.

Now using the condition of perpendicular vector we get,

1(1-k)+1(-8)+(-1)(3-3k)=0

⇒1-k-8-3+3k=0

⇒2k=10⇒k=5

Hence, k2+1(k-1)(k-2)=264×3=136

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (30 Jan Shift 2)
If a plant BBCC is pollinated by pollens of plant AADD, the fruit thus obtained will have:

Options:
A AaBbCcDd genotype
B AABBCCDd genotype
C BBCC genotype
D AABBCCDD genotype

The correct answer is:
BBCC genotype

Solution:
Fruit is a characteristic feature found in flowering plants which develops from ovary after fertilization. If fruit develops without fertilization then it is called as parthenocarpic fruit. The fruit basically is formed by maturation of ovary after fertilization. Ovary cells are not involved in fertilization. The genotype of fruit will remain the same as that of the female reproductive organ. The female plant genotype given is BBCC. Thus, the fruit thus obtained will have the same genotype BBCC.
The embryo within the seed of the fruit is formed by the fusion of gametic cells. The pollen which is male gamete has genotype -AD and the female gamete has BC as the genotype. Thus, the embryo will have the genotype -AABBCCDD.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
If a point P0,-2 and Q is any point on the circle, x2+y2-5x-y+5=0, then the maximum value of PQ2 is

Options:
A 8+53
B 47+1062
C 14+53
D 25+ 62

The correct answer is:
14+53

Solution:
Given circle x2+y2-5x-y+5=0
⇒x-522+y-122=32
On circle Q=(52+32cosθ,12+32sinθ)
PQ2=52+32cos⁡θ​2+52+32sin⁡θ​2
PQ2=252+32+532cosθ​+sinθ​
=14+532cos⁡θ​+sin⁡θ​
∴Maximum value of PQ2=14+5322
=14+53.

Topic: Mathematics, Circle, JEE Main, JEE Main 2017 (08 Apr Online)
If a point $P$ denotes a complex number $z=x+i y$ in the argand plane and if $\frac{z+1}{z+i}$ is a purely real number, then the locus of $P$ is

Options:
A $x+y+1=0$
B $x^2+y^2+x+y=0$
C $x^2+y^2+2 y+1=0,(x, y) \neq(0,-1)$
D $x+y+1=0,(x, y) \neq(0,-1)$

The correct answer is:
$x+y+1=0,(x, y) \neq(0,-1)$

Solution:
Since
$$
\begin{aligned}
& \frac{z+1}{z+i}=\frac{x+i y+1}{x+i y+i}=\frac{(x+1)}{x+i(y+1)} \\
& \times \frac{x-i(y+1)}{x-i(y+1)}
\end{aligned}
$$
If $\frac{z+1}{z+i}$ is a purely real number, $z \neq-i$
$\begin{aligned} & \text { Then, } \quad I_m\left(\frac{z+1}{z+i}\right)=0, z=-i \\ & \Rightarrow \quad x y-(x+1)(y+1)=0, \quad(x, y) \neq(0,-1) \\ & \Rightarrow \quad x y-x y-x-y-1=0 \\ & \Rightarrow \quad x+y+1=0, \quad(x, y) \neq(0,-1)\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If a point $P$ denotes the complex number $z=x+$ iy in the Argand plane and if $\frac{z-(2+i)}{z+(1-2 i)}$ is purely real, then the locus of $\mathrm{P}$ is

Options:
A the line $x+3 y-5=0$ excluding the point $(-1,2)$
B the circle $x^2+y^2-x-3 y=0$ excluding the point $(-1,2)$
C the line $x+3 y-5=0$ and the circle $x^2+y^2-x-3 y$ $=0$ excluding the point $(-1,2)$
D the circle $x^2+y^2-2 x-6 y+5=0$ excluding the point $(-1,2)$

The correct answer is:
the line $x+3 y-5=0$ excluding the point $(-1,2)$

Solution:
$\frac{Z-(2+i)}{Z+(1-2 i)}$
$Z$ is purely real
$$
\begin{aligned}
& \therefore \quad \frac{(x+i y)-2-i}{x+i y+1-2 i}=\frac{(x-2)+(y-1) i}{(x+1)+(y-2) i} \\
& =\frac{[(x-2)+(y-1) i][(x+1)-(y-2) i]}{[(x+1)+(y-2 i)][(x+1)-(y-2) i]} \\
& \{(x-2)(x+1)+(y-1)(y-2)\} \\
& =\frac{+i\{(x+1)(y-1)-(x-2)(y-2)\}}{(x+1)^2+(y-2)^2} \\
& \text { Imaginary part }=0 \\
& \therefore \quad \frac{(x+1)(y-1)-(x-2)(y-2)}{(x+1)^2+(y-2)^2}=0 \\
& \therefore \quad(x+1)(y-1)-(x-2)(y-2)=0 \text { and } x \neq-1, y \neq 2 \\
& \Rightarrow x y-x+y-1-x y+2 x+2 y-4=0 \\
& \Rightarrow x+3 y-5=0 \text { and } x \neq-1, y \neq 2 \\
&
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main
If a point Pα,β,γ satisfying αβγ2108938848=000 lies on the plane 2x+4y+3z=5, then 6α+9β+7γ is equal to

Options:
A 54
B -1
C 11
D 115

The correct answer is:
11

Solution:
Given,

αβγ2108938848=000

⇒2α+9β+8γ10α+3β+4γ8α+8β+8γ=000

Now on comparing both side we get,

2α+2β+8γ=0 …1

10α+3β+4γ=0 …2

8α+8β+8γ=0 …3

Now from 1 and 3 by cross multiplication we get,

α1=β6=γ-7=k⇒α=k, β=6k, γ=-7k

Point Pα,β,γ lie on the plane 2x+4y+3z=5⇒2α+4β+3γ=5

⇒2k+24k-21k=5

⇒k=1

Hence, 6α+9β+7γ=6k+54k-49k=11k=11.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (31 Jan Shift 2)
If a point R4,y,z lies on the line segment joining the points P2,-3,4 and Q8,0,10, then the distance of R from the origin is

Options:
A 221
B 53
C 6
D 214

The correct answer is:
214

Solution:
Given point R lies on the line segment joining the points P and Q, hence direction ratios of the line
PR and PQ must be identical

⇒26=y+33=z-46
⇒y=-2  and z=6
⇒R4, -2, 6.
Hence, distance from origin =16+4+36=214

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If a polygon of $n$ sides has 275 diagonals, then $n$ is equal to

Options:
A $25$
B $35$
C $20$
D $15$

The correct answer is:
$25$

Solution:
A polygon of $n$ sides has number of diagonals
$=\frac{n(n-3)}{2}=275 \quad$ (given)
$\begin{array}{cc}\Rightarrow & n^2-3 n-550=0 \\ \Rightarrow & n^2-25 n+22 n-550=0 \\ \Rightarrow & n(n-25)+22(n-25)=0 \\ \Rightarrow & (n-25)(n+22)=0 \\ \Rightarrow & n=25,(\because n=-22 \text { is not possible })\end{array}$

Topic: Mathematics, Permutation Combination, JEE Main
If a polythene sample contains two monodisperse fractions in the ratio $2: 3$ with degree of polymerization $100$ and $200$, respectively, then its weight average molecular weight will be :

Options:
A
$4900$

B
$4600$

C
$4300$

D
$5200$

The correct answer is:
$4900$

Solution:
No Solution Available

Topic: Chemistry, Polymers, JEE Main, JEE Main 2013 (09 Apr Online)
If a progressive wave is represented as $y=2 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ where $\mathrm{x}$ is in metre and $\mathrm{t}$ is in second, then the distance travelled by the wave in $5 \mathrm{~s}$ is

Options:
A $5 \mathrm{~m}$
B $10 \mathrm{~m}$
C $25 \mathrm{~m}$
D $32 \mathrm{~m}$

The correct answer is:
$10 \mathrm{~m}$

Solution:
$\begin{array}{l}
\text { Given, } y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right) \\
=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)
\end{array}$
Comparing it with standard equation
$\begin{array}{l}
y=r \sin \frac{2 \pi}{\lambda}(v t-x) \\
=r \sin \left(\frac{2 \pi v t}{\lambda}-\frac{2 \pi x}{\lambda}\right)
\end{array}$
We have, $\frac{2 \pi \mathrm{v}}{\lambda}=\frac{\pi}{2}$ or $\mathrm{v}=\frac{\lambda}{4}$
and $\frac{2 \pi}{\lambda}=\frac{\pi}{4}$ or $\lambda=8 \mathrm{~m} \therefore \mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{s}$
So, the distance travelled by wave in $t$ second $=\mathrm{vt}=2 \times 5=10 \mathrm{~m}$

Topic: Physics, Waves and Sound, JEE Main
If Ar=1rr+113r23, then limnâââ¡âr=1nAr is equal to

Options:
A 0
B 2
C 4
D 8

The correct answer is:
2

Solution:
Ar=1rr+113r23=1r-1r+113r23
⇒limn→∞⁡∑r=1nAr

=1-12+12-13+13-14+… 13+132+133+…23
=1 131-1323=11223=3-1=2

Topic: Mathematics, Determinants, JEE Main
If a ∈ R and the equation -3(x - [x])2+2(x - [x])+a2 = 0 (where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval

Options:
A (-2, -1)
B -∞ , - 2 ∪ 2 , ∞
C - 1 , 0 ∪ 0 , 1
D (1,2)

The correct answer is:
- 1 , 0 ∪ 0 , 1

Solution:
Let x-[x]=t
∴ -3t2+2t+a2=0
As x is not an integer, t≠0 ∴ a≠0
For a root of x to exist at least one of the roots of t should be between 0 and 1.
Roots =-2±4-4-3a2-6
=-2±4+12a2-6
=-1±1+3a2-3=1∓1+3a23
As we observe, one root is surely less than zero,
i.e., 1-1+3a23
∴ For a solution to exist,
1+1+3a23<1
∴ 1+1+3a2<3
∴ 1+3a2<2
∴ 1+3a2<4
∴ 3a2<3
∴ a2<1
∴ a∈-1,0∪0,1. ∵ a≠0

Topic: Mathematics, Quadratic Equation, JEE Main
If a ∈ R and the equation -3(x - [x])2+2(x - [x])+a2 = 0 (where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval

Options:
A (-2, -1)
B -∞ , - 2 ∪ 2 , ∞
C - 1 , 0 ∪ 0 , 1
D (1,2)

The correct answer is:
- 1 , 0 ∪ 0 , 1

Solution:
Let x-[x]=t
∴ -3t2+2t+a2=0
As x is not an integer, t≠0 ∴ a≠0
For a root of x to exist at least one of the roots of t should be between 0 and 1.
Roots =-2±4-4-3a2-6
=-2±4+12a2-6
=-1±1+3a2-3=1∓1+3a23
As we observe, one root is surely less than zero,
i.e., 1-1+3a23
∴ For a solution to exist,
1+1+3a23<1
∴ 1+1+3a2<3
∴ 1+3a2<2
∴ 1+3a2<4
∴ 3a2<3
∴ a2<1
∴ a∈-1,0∪0,1. ∵ a≠0

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2014 (06 Apr)
If ar is the coefficient of x10-r in the Binomial expansion of 1+x10, then ∑r=110r3arar-12 is equal to

Options:
A 4895
B 1210
C 5445
D 3025

The correct answer is:
1210

Solution:
Since, ar is the coefficient of x10-r in the binomial expansion of 1+x10, therefore

ar=C10-r10=Cr10

ar-1=Cr-110

So,

arar-1=Cr10Cr-110

⇒arar-1=10!r!×10-r!×r-1!×11-r!10!

⇒arar-1=11-rr

Now,

∑r=110r3arar-12=∑r=110r311-rr2

⇒∑r=110r3arar-12=∑r=110r11-r2

⇒∑r=110r3arar-12=∑r=110r121+r2-22r

⇒∑r=110r3arar-12=∑r=110121r+r3-22r2

⇒∑r=110r3arar-12=121×10×112+10×1122-22×10×11×216

⇒∑r=110r3arar-12=6655+3025-8470

⇒∑r=110r3arar-12=1210

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (25 Jan Shift 1)
If a radioactive element having half-life of 30 min is undergoing beta decay, the fraction of radioactive element remains undecayed after 90 min will be :

Options:
A 18
B 116
C 14
D 12

The correct answer is:
18

Solution:
Given: Half life of radioactive element t12=30 min.

In 90 min, there will be three half lives, so n=3.

Using radioactive decay law, fraction of radioactive element remains undecayed is

NN0=12n=123

⇒NN0=18

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (29 Jan Shift 1)
If a random variable X follows the Binomial distribution B33,p such that 3PX=0=PX=1, then the value of PX=15PX=18-PX=16PX=17 is equal to

Options:
A 1320
B 1088
C 10881089
D 1201331

The correct answer is:
1320

Solution:
We know that p=success, q=failure

Given 3PX=0=PX=1

3C033p0q33=C133p1q32

3×1×q=33×p

q=11p ... i

Also p+q=1

Now from equation i, p+11p=1

12p=1

p=112, q=1112

Now PX=15PX=18-PX=16PX=17

⇒C1533p15q18C1833p18q15-C1633p16q17C1733p17q16

=q3p3-qp=qpqp2-1=111112-1

=11×120

=1320

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (24 Jun Shift 1)