datavorous/entrance-exam-dataset huggingface dataset:
Let α>0, be the smallest number such that the expansion of x23+2x330 has a term βx-α,β∈N. Then α is equal to _____ .

Options:

The correct answer is:
2

Solution:
General term of the binomial expansion of x23+2x330 is

Tr+1=Cr30x2/330-r2x3r

⇒Tr+1=Cr30·2r·x60-11r3

So, for term of the form βx-α, we have

60-11r3<0

⇒11r>60

⇒r>6011⇒r=6

So,

T7=C630·26x-2

We have also observed β=C63026 is a natural number.

∴α=2

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let ε0 denote the dimensional formula of the permittivity of vacuum. If M= mass, L= length, T= time and A= electric current, then dimensions of permittivity is given as M p L q T r A s . Find the value of p-q+rs

Options:

The correct answer is:
3

Solution:
From Coulomb's law,
F=q1q24πε0r2
∴ε0=q1q24πFr2=(A1T1)(A1T1)[M1L1T-2][L2]=M-1L-3T4A2
∴p=-1,q=-3,r=4,s=2
∴p-q+rs=-1-(-3)+42=62=3

Topic: Physics, Units and Dimensions, JEE Main
Let ∈ 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then :

Options:
A ∈ 0 = M -1 L 2 T -1 A - 2
B ∈ 0 = M -1 L 2 T -1 A
C ∈ 0 = M -1 L - 3 T 2 A
D ∈ 0 = M -1 L - 3 T 4 A 2

The correct answer is:
∈ 0 = M -1 L - 3 T 4 A 2

Solution:
F = 1 4 π ε 0 q 1 q 2 R 2
ε 0 = q 1 q 2 4 π FR 2
Hence, ε 0 = C 2 N . m 2 = AT 2 ML T - 2 . L 2 = M -1 L - 3 T 4 A 2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2013 (07 Apr)
Let α>0. If ∫0αxx+α-xdx=16+20215 then α is equal to :

Options:
A 2
B 22
C 4
D 2

The correct answer is:
2

Solution:
Given,

∫0αxx+α-xdx=16+20215....(1)

Rationalise the denominator in L.H.S we get,

∫0αx(α+x+x)αdx=1α∫0αxα+xdx+∫0αx3/2dx

Put α+x=t2⇒dx=2tdt

Limit when x=0→t=α, x=α→t=2α

1α∫α2αt2-α2t2dt+2α3/25

=1α2t55-2αt33α2α+2α3/25

=4α2α+10αα15

From equation 1

∫0αxx+α-xdx=16+20215

⇒4α2α+10αα15=16+20215

On comparing both sides we get,

α=2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let $0 \leq \mathrm{r} \leq \mathrm{n}$. If ${ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}:{ }^n \mathrm{C}_{\mathrm{r}}:{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}=55: 35: 21$, then $2 \mathrm{n}+5 \mathrm{r}$ is equal to:

Options:
A 50
B 62
C 55
D 60

The correct answer is:
50

Solution:
$\begin{aligned} & \frac{{ }^{n+1} C_r}{{ }^n C_r}=\frac{55}{35} \\ & \frac{(n+1) !}{(r+1) !(n-r)} ! \frac{r !(n-r) !}{n !}=\frac{11}{7} \\ & \frac{(n+1)}{r+1}=\frac{11}{7}\end{aligned}$
$\begin{aligned} & 7 \mathrm{n}=4+11 \mathrm{r} \\ & \frac{{ }^n C_r}{{ }^{n-1} C_{r-1}}=\frac{35}{21} \\ & \frac{n !}{r !(n-r) !}=\frac{(r-1) !(n-r) !}{(n-1) !}=\frac{5}{3} \\ & \frac{n}{r}=\frac{5}{3} \\ & 3 n=5 r \\ & \text { By solving } r=6 \quad n=10 \\ & 2 n+5 r=50\end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Let 0<z<y<x be three real numbers such that 1x,1y,1z are in an arithmetic progression and x,2y,z are in a geometric progression. If xy+yz+zx=32xyz, then 3(x+y+z)2 is equal to

Options:

The correct answer is:
150

Solution:
Given that 1x,1y,1z are in AP and x,2y,z are in GP.

As given, 2y=1x+1y .......i
Also, 2y2=xz .......ii

Also given that xy+yz+zx=32xyz
⇒1x+1y+1z=32 .....iii
From (i) and (iii) we get 3y=32

y=2 .....iv

Now from (ii) xz=4 .......v

Now using (ii), (iv) and (v)

⇒x+z=42

Hence 3(x+y+z)2=3(2+42)2

=150

Therefore, this is the required answer.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let Ι1=∫01ex2dx and Ι2=∫012x2ex2dx, then the value of Ι1+Ι2 is equal to

Options:
A 1
B 2
C e
D e2

The correct answer is:
e

Solution:
Î2=â«012x2ex2dx
=â«012xex2xdx
Using integration by parts,
Î2=xex201-â«01ex2dx
Î2+Î1=e

Topic: Mathematics, Definite Integration, JEE Main
Let Ι1=∫01ln⁡xx2+4x+1dx and Ι2=∫1∞ln⁡xx2+4x+1dx, then

Options:
A Î1=Î2
B Î1>Î2
C Î1+Î2=0
D Ι1=2Ι2

The correct answer is:
Ι1=Ι2

Solution:
In Ι2 substitute x=1t⇒dx=-dtt2
So, Ι2=-∫10ln⁡tt2+4t+1⋅t2-dtt2=∫01-ln⁡tt2+4t+1dt=Ι1 as ln⁡t=-ln⁡t in 0,1

Topic: Mathematics, Definite Integration, JEE Main
Let Ι1=∫1π2dt1+t6 and Ι2=∫0π2xcos⁡xdx1+xsinx+cos⁡x6, then

Options:
A 2Ι1=Ι2
B Ι1=2Ι2
C Ι1=Ι2
D Ι1=Ι2=0

The correct answer is:
Ι1=Ι2

Solution:
In Î2,
let, xsinx+cosâ¡x=t
âxcosx+sinâ¡x-sinâ¡xdx=dt or xcosxdx=dt
â´Î2=â«1Ï2dt1+t6=Î1
Also, Î1 & Î2 are both positive as
11+t6>0 âtâ1,Ï2

Topic: Mathematics, Definite Integration, JEE Main
Let $\alpha_{1}, \alpha_{2}$ and $\beta_{1}, \beta_{2}$ be the roots of $a x^{2}+b x+c$ $=0$ and $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0$ respectively. If the system of equations $\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}=0$ and $\beta_{1} \mathrm{y}+\beta_{2} \mathrm{z}=0$ has a non-trivial solution, then

Options:
A $\frac{b^{2}}{q^{2}}=\frac{a c}{p r}$
B $\frac{\mathrm{c}^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{ab}}{\mathrm{pq}}$
C $\frac{a^{2}}{p^{2}}=\frac{b c}{q r}$
D None of these

The correct answer is:
$\frac{b^{2}}{q^{2}}=\frac{a c}{p r}$

Solution:
Since $\alpha_{1}, \alpha_{2}$ and $\beta_{1}, \beta_{2}$ are the roots of ax $^{2}$ $+b x+c=0$ and $p x^{2}+q x+r=0$ respectively

therefore

$$

\alpha_{1}+\alpha_{2}=\frac{-b}{a}, \alpha_{1} \alpha_{2}=\frac{c}{a}

$$

and $\beta_{1}+\beta_{2}=\frac{-q}{p}, \beta_{1} \beta_{2}=\frac{r}{p}$

Since the given system of equation has a non-trivial solution

$$

\therefore\left|\begin{array}{ll}

\alpha_{1} & \alpha_{2} \\

\beta_{1} & \beta_{2}

\end{array}\right|=0 \text { i.e. } \alpha_{1} \beta_{2}-\alpha_{2} \beta_{1}=0

$$

or $\quad \frac{\alpha_{1}}{\beta_{1}}=\frac{\alpha_{2}}{\beta_{2}}=\frac{\alpha_{1}+\alpha_{2}}{\beta_{1}+\beta_{2}}=\sqrt{\frac{\alpha_{1} \alpha_{2}}{\beta_{1} \beta_{2}}}$

$\Rightarrow \frac{\mathrm{pb}}{\mathrm{qa}}=\sqrt{\frac{\mathrm{pc}}{\mathrm{ra}}} \Rightarrow \frac{\mathrm{b}^{2}}{\mathrm{q}^{2}}=\frac{\mathrm{ac}}{\mathrm{pr}}$

Topic: Mathematics, Determinants, JEE Main
Let $\omega_1, \omega_2$ and $\omega_3$ be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If $x_1, x_2$ and $x_3$ are their respective angular distances in 1 minute then the factor which remains constant $(k)$ is

Options:
A $\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=k$
B $\omega_1 x_1=\omega_2 x_2=\omega_3 x_3=k$
C $\omega_1 x_1^2=\omega_2 x_2^2=\omega_3 x_3^2=k$
D $\omega_1^2 x_1=\omega_2^2 x_2=\omega_3^2 x_3=k$

The correct answer is:
$\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=k$

Solution:
$\omega_1=\frac{2 \pi}{60} ; \quad x_1=\frac{2 \pi}{60} \times 60=2 \pi$
$\omega_2=\frac{2 \pi}{3600} ; \quad x_2=\frac{2 \pi}{3600} \times 60=\frac{2 \pi}{60}$
$\omega_3=\frac{2 \pi}{3600 \times 12} ; \quad x_3=\frac{2 \pi}{3600 \times 12} \times 60=\frac{2 \pi}{720}$
$\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=\frac{1}{60}=k$

Topic: Physics, Oscillations, NEET, NEET 2024 (Re-NEET)
Let $(1,2)$ be the focus and $x+y+1=0$ be the directrix of a hyperbola $\mathrm{H}$. If $\sqrt{3}$ is the eccentricity of $\mathrm{H}$, then its equation is

Options:
A $x^2-6 x y+y^2-14 x-22 y+17=0$
B $x^2-6 x y+y^2+10 x+14 y-7=0$
C $x^2+6 x y+y^2-14 x-22 y+17=0$
D $x^2+6 x y+y^2+10 x+14 y-7=0$

The correct answer is:
$x^2+6 x y+y^2+10 x+14 y-7=0$

Solution:
Given equation of directrix is $x+y+1=0$ and Focus is $(1,2)=5$
Let $P(x, y)$ be any point on hyperbola $\mathrm{H}$
Let $\mathrm{PM}$ be the length of the perpendicular from $\mathrm{P}$ to the directrix.
then
$\frac{P S}{P M}=\sqrt{3} \Rightarrow P S^2=3 P M^2$
$\begin{aligned} & \Rightarrow(x-1)^2+(y-2)^2=3\left|\frac{x+y+1}{\sqrt{2}}\right|^2 \\ & \begin{aligned} \Rightarrow 2\left(x^2+1-2 x+y^2+4-4 y\right)=3\left(x^2+y^2+1\right. & +2 x y \\ & +2 y+2 x) \\ \Rightarrow 2 x^2+2-4 x+2 y^2+8-8 y=3 x^2+3 y^2+3 & +6 x y +6 y+6 x\\ \Rightarrow x^2+y^2+6 x y+10 x+14 y-7=0 &\end{aligned}\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
Let λ1, λ2 be the values of λ for which the points 52,1, λ and -2, 0, 1 are at equal distance from the plane 2x+3y-6z+7 If λ1>λ2

then the distance of the point λ1-λ2,λ2,λ1 from the line x-51=y-12=z+72 is ______

Options:

The correct answer is:
9

Solution:
Given,

λ1, λ2 be the values of λ for which the points 52,1, λ and -2, 0, 1 are at equal distance from the plane 2x+3y-6z+7,

So, by using distance formula of a point from a plane we get,

-4+0-6+722+32+62=5+3-6λ+722+32+62

⇒-4+0-6+77=5+3-6λ+77

⇒37=15-6λ7

⇒λ=2 or 3

⇒λ1=3, λ2=2 as λ1> λ2

So, λ1-λ2, λ2, λ1=(1,2,3)

Now finding the distance of point 1,2,3 from the line x-51=y-12=z+72=λ we get,

Now form diagram we can see that, PM→ is perpendicular to given line,

So, by perpendicular condition we get,

PM→·i^+2j^+2k^=0

⇒λ+4+22λ-1+22λ-10=0

⇒9λ=18 or λ=2

Hence, the point M will be M7,5,-3

So, the distance PM→=62+32+62=9

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let 116,a and b be in G.P. and 1a,1 b,6 be in A.P., where a,b>0. Then 72(a+b) is equal to _______ .

Options:

The correct answer is:
14

Solution:
a2=b16⇒1b=116a2

2b=1a+6

18a2=1a+6

1a2-8a-48=0

1a=12,-4⇒a=112,-14

a=112,a>0

b=16a2=19

⇒72(a+b)=6+8=14

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let α→=1ai^+4bj^+bk^ and β→=bi^+aj^+1bk^ ∀a,b>0, then the maximum value of 126+α→⋅β→ is

Options:
A 1211
B 2
C 1
D 109

The correct answer is:
1211

Solution:
Î±ââÎ²â=ba+4ab+1
âµba+4abâ¥4
1+ba+4abâ¥5
6+Î±ââÎ²ââ¥11⇒6+Î±ââÎ²â12â¥1112
⇒126+Î±ââÎ²ââ¤1211

Topic: Mathematics, Vector Algebra, JEE Main
Let $\rho_{1}$ and $\rho_{2}$ be two equivalence relations defined on a non-void set $S$. Then

Options:
A both $\rho_{1} \cap \rho_{2}$ and $\rho_{1} \cup \rho_{2}$ are equivalence relations
B $\rho_{1} \cap \rho_{2}$ is equivalence relation but $\rho_{1} \cup \rho_{2}$ is not so.
C $\rho_{1} \cup \rho_{2}$ is equivalence relation but $\rho_{1} \cap \rho_{2}$ is not so
D neither $\rho_{1} \cap \rho_{2}$ nor $\rho_{1} \cup \rho_{2}$ is equivalence relation.

The correct answer is:
$\rho_{1} \cap \rho_{2}$ is equivalence relation but $\rho_{1} \cup \rho_{2}$ is not so.

Solution:
Hint:
Union of two transitive may or may not be transitive

Topic: Mathematics, Sets and Relations, JEE Main
Let $\omega \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $\left[\begin{array}{ccc}1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1\end{array}\right]$, where each of $a, b$ and $c$ is either $\omega$ or $\omega^2$. Then, the number of distinct matrices in the set $S$ is

Options:
A
2

B
6

C
4

D
8

The correct answer is:
2

Solution:
$|A| \neq 0$, as non-singular.
$$
\begin{aligned}
& \therefore \quad\left|\begin{array}{ccc}
1 & a & b \\
\omega & 1 & c \\
\omega^2 & \omega & 1
\end{array}\right| \neq 0 \\
& \Rightarrow \quad 1(1-c \omega)-a\left(\omega-c \omega^2\right) \\
& +b\left(\omega^2-\omega^2\right) \neq 0 \\
& \Rightarrow \quad 1-c \omega-a \omega+a c \omega^2 \neq 0 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad(1-c \omega)(1-a \omega) \neq 0 \\
& \Rightarrow \quad a \neq \frac{1}{\omega}, c \neq \frac{1}{\omega} \Rightarrow a=\omega, c=\omega
\end{aligned}
$$
and $b \in\left\{\omega, \omega^2\right\} \Rightarrow 2$ solutions

Topic: Mathematics, Matrices, JEE Main
Let $\theta_1$ be the angle between two lines $2 x+3 y+$ $c_1=0$ and $-x+5 y+c_2=0$ and $\theta_2$ be the angle between two lines $2 x+3 y+c_1=0$ and $-x+5 y+$ $c_3=0$, where $c_1, c_2, c_3$ are any real numbers :
Statement-1: If $c_2$ and $c_3$ are proportional, then $\theta_1=\theta_2$.
Statement-2: $\theta_1=\theta_2$ for all $c_2$ and $c_3$.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation of Statement-1.

B
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation of Statement-1.

C
Statement-1 is false; Statement- 2 is true.

D
Statement-1 is true; Statement- 2 is false.

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation of Statement-1.

Solution:
Two lines $-x+5 y+c_2=0$ and $-x+5 y+c_3$ $=0$ are parallel to each other. Hence statement-1 is true, statement-2 is true and statement-2 is the correct explanation of statement-1.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (23 Apr Online)
Let 1 be the directrix of the parabola $9 y^2+12 y+9 x-14=0$ and $l_1$ be the line passing through the vertex of this parabola and the origin. If $(\mathrm{h}, \mathrm{k})$ is the point of intersection of $l$ and $l_1$, then $\mathrm{h}+\mathrm{k}=$

Options:
A $-9 / 2$
B $3 / 2$
C $-3 / 4$
D $9 / 4$

The correct answer is:
$-9 / 2$

Solution:
Given: $9 y^2+12 y+9 x-14=0$
$\Rightarrow\left(y+\frac{2}{3}\right)^2=-4 \times \frac{1}{4}(x-2)$
Whose vertex is $\left(-\frac{2}{3}, 2\right)$ Equation of line passing through $\left(-\frac{2}{3}, 2\right)$ and $(0,0)$ :
$(y-0)=\frac{2-0}{-\frac{2}{3}-0}(x-0) \Rightarrow y=-3 x$
Now, the equation of directrix of parabola (i) is
$x=\frac{1}{4}+2 \Rightarrow x=\frac{9}{4}$ ...(iii)
Solving $\mathrm{eq}^{\mathrm{n}}$ (ii) \& (iii), we get
$x=\frac{9}{4}$ and $y=-\frac{27}{4} \quad \therefore h=\frac{9}{4} \& k=-\frac{27}{4}$
So, $h+k=\frac{9}{4}-\frac{27}{4}=\frac{-18}{4}=\frac{-9}{2}$

Topic: Mathematics, Parabola, JEE Main
Let α=-1+i32. If a=1+α∑k=0100α2k and b=∑k=0100α3k, then a and b, are the roots of the quadratic equation.

Options:
A x2+101x+100=0
B x2-102x+101=0
C x2-101x+100=0
D x2+102x+101=0

The correct answer is:
x2-102x+101=0

Solution:
Given,

α=-1+i32

a=1+α∑k=0100α2k and

b=∑k=0100α3k

Now

α=ω ; ω=-1+i32

Using this a and b can be written as
a=1+ω1+ω2+ω4+…ω198+ω200
=1+ω1-ω21011-ω2=1+ω1-ω1-ω2=1

Similarly,

b=1+ω3+ω6+…+ω300=101
So, the required quadratic equation is

x2-a+bx+ab=0

⇒x2-102x+101=0.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Let ω ω≠1 is a cube root of unity, such that 1+ω28=a+bω where a,b∈R, then a+b is equal to

Options:
A 1
B 3
C 0
D 2

The correct answer is:
2

Solution:
-ω8=a+bω
⇒ω2=a+bω………i
⇒-12-i32=a+b-12+i32
Comparing real and imaginary parts
a=-1,b=-1⇒a+b=2

Topic: Mathematics, Complex Number, JEE Main
Let ∫1-ln⁡xx2dx=fx, for all positive x. If fe=1e, then f2+f4 is equal to

Options:
A ln⁡4
B ln⁡2
C 1
D 0

The correct answer is:
ln⁡2

Solution:
fx=∫1x2dx-∫ln⁡x⋅1x2dx
=-1x-ln⁡x-1x+∫1x2dx
=-1x+ln⁡xx+1x+c=ln⁡xx+c
Now, fe=1e⇒c=0

Hence, f2+f4=ln⁡22+ln⁡44=ln2

Topic: Mathematics, Indefinite Integration, JEE Main
Let 1x1, 1x2,…,1xn(xi≠0 for i=1, 2,…., n) be in A.P. such that x1=4 and x21=20 . If n is the least positive integer for which xn>50, then ∑i=1n1xi is equal to

Options:
A 3
B 18
C 134
D 138

The correct answer is:
134

Solution:
Given 1x1,1x2,………,1xn are in A.P. and x1=4 and x21=20.

⇒a=14

Given, x21=20.

⇒14+20·d=120

⇒d=−1100
Given, xn>50.

⇒1xn<150

⇒14- n-1100< 150⇒n>24

∵ n=25

Now, ∑i=1251xi=2522×14-1100×24=134

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (16 Apr Online)
Let 1+x+2x220=a0+a1x+a2x2+…+a40x40, then a1+a3+a5+…+a37 is equal to

Options:
A 220220-21
B 219220-21
C 219220+21
D 220220+21

The correct answer is:
219220-21

Solution:
Given 1+x+2x220=a0+a1x+…+a40x40

Put x=1 we get,

a0+a1+a2+…+a40=420

Put x=-1 we get,

a0-a1+a2+….+a40=220

⇒a1+a3+….+a39=420-2202

⇒a1+a3+…+a37=239-219-a39

Here a39=20!(2)19×119!=20×219

⇒a1+a3+…+a37=219220-1-20

=219220-21

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Let $(1+x)^{10}=\sum_{r=0}^{10} c_{r} x^{r}$ and $(1+x)^{7}=\sum_{r=0}^{7} d_{r} x^{r}$
If $P=\sum_{r=0}^{5} c_{2}$ and $Q=\sum_{r=0}^{3} d_{2 r+1},$ then $\frac{P}{Q}$ is equal
to

Options:
A 4
B 8
C 16
D 32

The correct answer is:
8

Solution:
$P=\sum_{r=0}^{5} C_{2 r}$
$={ }^{10} C_{0}+{ }^{10} C_{2}+\quad+{ }^{10} C_{10}=\frac{2^{0}}{2}=2^{9}$
$Q=\sum_{r=0}^{3} d_{2 r+1}=d_{1}+d_{3}+d_{5}+d_{7}$
$={ }^{7} C_{1}+{ }^{7} C_{3}+{ }^{7} C_{5}+{ }^{7} C_{7}=\frac{2^{7}}{2}=2^{6}$
$\therefore \frac{P}{\Omega}=\frac{2^{9}}{3^{6}}=2^{3}=8$

Topic: Mathematics, Binomial Theorem, JEE Main
Let $\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}$
$+\ldots+a_{18} x^{18} \cdot$ Then

Options:
A $a_{0}+a_{2}+\ldots+a_{18}=a_{1}+a_{3}+\ldots+a_{17}$
B $a_{0}+a_{2}+\ldots+a_{18}$ is even
C $a_{0}+a_{2}+\ldots+a_{18}$ is divisible by 9
D $a_{0}+a_{2}+\ldots+a_{18}$ is divisible by 3 but not by 9

The correct answer is:
$a_{0}+a_{2}+\ldots+a_{18}$ is even

Solution:
$\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}+ \dots+a_{18}x^{18}$
$$
Put $x=-1,$ we get
$$
\begin{array}{l}
(1-1+1)^{9}=a_{0}-a_{1}+a_{2}+\ldots+a_{18} \\
\Rightarrow 1=a_{0}-a_{1}+a_{2}+\ldots+a_{18}
\end{array}
$$
Put $x=1,$ we get
$$
(1+1+1)^{9}=a_{0}+a_{1}+a_{2}+\ldots+a_{14}
$$
$\Rightarrow 3^{9}=a_{0}+a_{1}+a_{2}+\ldots+a_{18}$
On adding Eqs. (i) and (ii), we get
$$
\begin{array}{c}
3^{9}+1=2\left(a_{0}+a_{2}+\ldots+a_{10}\right) \\
a_{0}+a_{2}+a_{4}+\ldots+a_{18}=\frac{3^{0}+1}{2} \\
=\frac{19683+1}{2} \\
=\frac{19684}{2}
\end{array}
$$
$=9842$ which is even number.

Topic: Mathematics, Binomial Theorem, JEE Main
Let $\left(1+x+x^{2}\right)^{2014}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\ldots . .+a_{s+28} x^{4028}$ and let
$$
\begin{array}{l}
A=a_{0}-a_{3}+a_{6}-\ldots \ldots+a_{4026} \\
B=a_{1}-a_{4}+a_{7}-\ldots \ldots .-a_{4027} \\
C=a_{2}-a_{5}+a_{8}-\ldots \ldots .+a_{4028}
\end{array}
$$
Then

Options:
A $|A|=|B|>|C|$
B $|A|=|B| < |C|$
C $|A|=|C|>|B|$
D $|A|=|C| < |B|$

The correct answer is:
$|A|=|C|>|B|$

Solution:

Topic: Mathematics, Binomial Theorem, JEE Main
Let 21012-10-12. If adj(adjadj2A)=(16)n, then n is equal to

Options:
A 8
B 10
C 9
D 12

The correct answer is:
10

Solution:
Given that A=21012-10-12 and adjadjadj2A=16n.

We know that adjA=An-1 and adjadjA=An-12

Similarly, adjadjadjA=An-13 where n is the order of the square matrix.

Let us find A.

A=22×2--1-1-11×2-0-1+01×-1-0×2

=23-2+0=4

⇒2An-13=2A3-13

=2A8=283A8

⇒224×48=16n

⇒166×164=16n

⇒1610=16n

⇒n=10

Therefore, the value of n is 10.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let λ,2,1 be a point on the plane which passes through the point 4,-2,2. If the plane is perpendicular to the line joining the points -2,-21,29 and -1,-16,23, then λ112-4λ11-4 is equal to ________.

Options:

The correct answer is:
8

Solution:
AB→.PQ→=0

⇒i^+5j^-6k^.4-λi^-4j^+k^=0

⇒4-λ-20-6=0

⇒λ=-22

⇒λ112-4λ11-4=4+8-4=8

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Let $\left(-2-\frac{1}{3} i\right)^{3}=\frac{x+i y}{27}(i=\sqrt{-1}),$ where $x$ and $y$ are real numbers then $\mathrm{y}-\mathrm{x}$ equals

Options:
A 91
B -85
C 85
D -91

The correct answer is:
91

Solution:
$-(6+i)^{3}=x+i y$

$\Rightarrow \quad-\left[216+i^{3}+18 i(6+i)\right]=x+i y$

$\Rightarrow \quad-[216-i+108 i-18]=x+i y$

$\Rightarrow \quad-216+i-108 i+18=x+i y$

$\Rightarrow-198-107 i=x+i y$

$\Rightarrow \quad x=-198, y=-107$

$\Rightarrow \quad y-x=-107+198=91$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let 2a+2b+c=0, l1 and l2 are straight lines of the family ax+by+c=0 which are at 1 unit distance from the point 1,1, then the area (in sq. units) bounded by l1,l2 and coordinate axes is

Options:

The correct answer is:
4

Solution:
All the lines of the family of ax+by+c=0 pass from the point 2,2.
The line passing through 2,2 and at unit distance from 1,1 is x=2 and y=2
â Required Area =2Ã2=4 sq. units

Topic: Mathematics, Straight Lines, JEE Main
Let 2a+2b+c=0, then the equation of the straight line ax+by+c=0 which is farthest from the point 1,1 is

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

The correct answer is:
y+x=4

Solution:
All the lines ax+by+c=0 pass through the point 2,2.
The line passing through 2,2 & farthest from 1,1 is perpendicular to the line joining 1,1 & 2,2
Now, the slope of the line joining 1,1 & 2,2 is 2-12-1=1
⇒ The slope of the required line is -1
⇒ Required line is y-2x-2=-1
⇒y+x=4

Topic: Mathematics, Straight Lines, JEE Main
Let α→=λ-2 a→+b→ and β→=4λ-2 a→+3b→, be two given vectors where vectors a→ and b→ are non-collinear. The value of λ for which vectors α→ and β→ are collinear, is:

Options:
A -4
B -3
C 4
D 3

The correct answer is:
-4

Solution:
∵α→, β→ are collinear then, λ-24λ-2=13⇒λ=-4.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let 2aâ=bâÃcâ+2bâ where aâ, bâ and câ are three unit vectors, then sum of all possible values of 3aâ+4bâ+5câ is

Options:
A 10
B 12
C 14
D 16

The correct answer is:
14

Solution:
2a→=b→×c→+2b→
Taking dot with b→
2a→⋅b→=2b→2⇒a→⋅b→=1
⇒a→=b→ also b→×c→=0 (from given equation)
⇒b→=c→ or b→=-c→
So, if b→=c→, then 3a→+3b→+5c→=12
or if b→=-c→, then 3a→+4b→+5c→=2

Topic: Mathematics, Vector Algebra, JEE Main
Let 2nd, 8th and 44th, terms of a non-constant A.P. be respectively the 1st, 2nd and 3rd terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

Options:
A 980
B 960
C 990
D 970

The correct answer is:
970

Solution:
Given: 2nd, 8th, 44th terms of A.P are 1st, 2nd & 3rd terms respectively of G.P.

And first term of A.P. is 1 and let d be common difference

So, 1+d, 1+7d, 1+43d are 1st, 2nd & 3rd term of G.P.

⇒1+7d1+d=1+43d1+7d

⇒1+7d2=1+43d1+d

⇒1+49d2+14d=1+43d2+44d

⇒6d2=30d

⇒d=5

Now finding the sum of first 20 terms of AP,

⇒S20=102×1+19×5

⇒S20=970

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let 2 planes are being contained by the vectors αi^+3j^-k^, i^+α-1j^+2k^ and 3i^+5j^+2k^. If the angle between these 2 planes is θ, then the value of cos2⁡θ is equal to

Options:
A 1517
B 289717
C 2892151
D 172151

The correct answer is:
2892151

Solution:
All given vectors are coplaner
α3-11α-12352=0
⇒α2α-2-10-32-6-15-3α+3=0
⇒2α2-12α+12-8+3α=0
⇒2α2-9α+4=0
⇒α=4,12
A normal vector to the plane is i^j^k^1α-12352
=i^2α-12-j^-4+k^8-3α
If α=4, then a normal vector to one of the plane is n1→=-4i^+4j^-4k^=-4i^-j^+k^
If α=12, then a normal vector to the other plane is n2→=-11i^+4j^+132k^
cos⁡θ=n1→·n2→n1→n2→=-11-4+1323121+16+1694
⇒cos⁡θ=-173717
⇒cos2⁡θ=2892151

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let 2x2+3x+410=∑r=020arxr, then the value of $\frac{a_{8}}{a_{12}},$ is

Options:
A 2
B 4
C 8
D 16

The correct answer is:
4

Solution:
Given, 2x2+3x+410=∑r=020arxr

Replacing x by 2x, we get

8x2+6x+410=∑r=020ar2xr

⇒2102x2+3x+410=∑r=020ar2rx20-r

⇒ 210∑r=020arxr=∑r=020ar2rx20-r

Comparing ccefficient of x8 both sides, we get

210a8=a12212⇒a8a12=212210=4

Topic: Mathematics, Binomial Theorem, JEE Main
Let 2x+ay+6z=8, x+2y+bz=5 and x+y+3z=4 be three equations. If these 3 equations are consistent, then

Options:
A b=3, a≠2
B a=2, bâ 3
C a≠2, b≠3
D a≠2, b=4

The correct answer is:
a=2, bâ 3

Solution:
From Cramerâs rule,
for consistent system â³=â³1=â³2=â³3=0
â³= 2 a 6 1 2 b 1 1 3 =2 6âb âa 3âb +6 1â2
=12-2b-3a+ab-6
=6-2b-3a+ab
=a-2b-3
â³1=8a652b413=86-b-a15-4b+65-8
=48-8b-15a+4ab-18
=30-15a-8b+4ab
=(4b-15)(a-2)
â³2=28615b143=215-4b-83-b+64-5
=30-8b-24+8b-6=0
â³3=2a8125114=28-5-a4-5+81-2=a-2
For consistent system a=2

Topic: Mathematics, Determinants, JEE Main
Let 3, 6, 9, 12, … upto 78 terms and 5, 9, 13, 17, … upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ______.

Options:

The correct answer is:
2223

Solution:
Given,

3,6,9,⋯ upto 78 terms

⇒t78=3+77×3=234

5,9,13,⋯ upto 59 terms

⇒t59=5+58×4=237

Common difference of common terms =LCM3,4=12

First common term is 9 and last common term is 225

So series will be 9,21,33,⋯,225 ⇒n=19

S=n2a+l=1929+225=2223

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let 3, 7,11, 15,..,403 and 2, 5, 8, 11,...,404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to_________

Options:

The correct answer is:
6699

Solution:
Given AP are: 3, 7,11, 15,..,403 and 2, 5, 8, 11,...,404

a1=3, a2=2, d1=4, d2=3

So, for the common AP,

a=11, d=LCM3,4=12

Also, an≤403

⇒11+n-112≤403

⇒-1+12n≤403

⇒n≤40412

⇒n≤33.6

⇒n=33

⇒Sn=3322×11+32×12

⇒Sn=6699

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (01 Feb Shift 1)