datavorous/entrance-exam-dataset huggingface dataset:
If the straight line drawn through the point P 3 , 2 and inclined at an angle Ï 6 with the x - axis meets the line 3 x- 4 y + 8 = 0 at Q, then the length of PQ is

Options:

The correct answer is:
6

Solution:
Point Q can be obtained by using parametric form of a straight line, ⇒Q≡3-3r2, 2-r2

Also, point Q is on 3 x - 4 y + 8 so it will satisfy this equation, hence,
33-3r2-4 2-r2+8=0
3 - 3 2 r - 8 + 2 r + 8 = 0
⇒r2-32=-3 ⇒r=-6

Hence, the length of PQ=6

Topic: Mathematics, Straight Lines, JEE Main
If the straight line drawn through the point \( P(\sqrt{3}, 2) \) and inclined at an angle \( \frac{\pi}{6} \) with the \( x \) - axis meets the line \( \sqrt{3} x-4 y+8=0 \) at \( Q \), then the length of \( P Q \) is

Options:

The correct answer is:
6

Solution:
Point Q can be obtained by using parametric form of a straight line, ⇒Q≡3-3r2, 2-r2

Also, point Q is on 3x-4y+8 so it will satisfy this equation, hence,
33-3r2-4 2-r2+8=0
3-32r-8+2r+8=0
⇒r2-32=-3

⇒r=-6

Hence, the length of PQ=6

Topic: Mathematics, Straight Lines, JEE Main
If the straight line $x \cos \alpha+y \sin \alpha=p$ be a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then

Options:
A $a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p^2$
B $a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2$
C $a^2 \sin ^2 \alpha+b^2 \cos ^2 \alpha=p^2$
D $a^2 \sin ^2 \alpha-b^2 \cos ^2 \alpha=p^2$

The correct answer is:
$a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2$

Solution:
$x \cos \alpha+y \sin \alpha=p \Rightarrow y=-\cot \alpha . x+p \operatorname{cosec} \alpha$
It is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ Therefore, $p^2 \operatorname{cosec}^2 \alpha=a^2 \cot ^2 \alpha-b^2$ $\Rightarrow a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2$.

Topic: Mathematics, Hyperbola, JEE Main
If the straight line $x+y=1$ touches the parabola $y^2-y+x=0$, then the co-ordinates of the point of contact are

Options:
A $(1,1)$
B $\left(\frac{1}{2}, \frac{1}{2}\right)$
C $(0,1)$
D $(1,0)$

The correct answer is:
$(0,1)$

Solution:
$m$ of tangent $=-1$
Also from equation of parabola, we get gradjent at ${ }^{(h, k)}$ as the slope of parabola
$=\frac{d y}{d x}=\frac{-1}{2 y-1}=\frac{-1}{2 k-1}$
Since line and parabola touch at $(h, k)$
$\Rightarrow \frac{-1}{2 k-1}=-1 \Rightarrow-2 k+1=-1 \Rightarrow k=1$
Putting this value in $x+y=1$, we have $h=0$, so the point of contact is $(0,1)$.

Topic: Mathematics, Parabola, JEE Main
If the straight line $y=m x+c$ is parallel to the axis of the parabola $y^2=l x$ and intersects the parabola at $\left(\frac{c^2}{8}, c\right)$, then the length of the latusrectum is

Options:
A 2
B 3
C 4
D 8

The correct answer is:
8

Solution:
Here, length of latusrectum $=l$
Since, point $\left(\frac{c^2}{8}, c\right)$ is the point of intersection of parabola and the line $y=m x+c$ then it will satisfy the equations.
$\begin{array}{cc}
\Rightarrow & y^2=l x \\
\Rightarrow & (c)^2=\frac{l c^2}{8} \Rightarrow l=8
\end{array}$
$\therefore$ Length of latusrectum $(l)=8$

Topic: Mathematics, Parabola, JEE Main
If the straight line $y=m x+c$ touches the parabola $y^2-4 a x+4 a^3=0$, then $c$ is

Options:
A
$a m+\frac{a}{m}$

B
$a m-\frac{a}{m}$

C
$\frac{a}{m}+a^2 m$

D
$\frac{a}{m}-a^2 m$

The correct answer is:

$\frac{a}{m}-a^2 m$

Solution:
Solving the given equations,
$$
\begin{aligned}
&(m x+c)^2=4 a x-4 a^3 \\
& \Rightarrow m^2 x^2+2 m c \cdot x+c^2=4 a x-4 a^3 \\
& \Rightarrow m^2 x^2+(2 m c-4 a) x+c^2+4 a^3=0
\end{aligned}
$$
Since the straight line touches the parabola at a point, so, the discriminant $=0$
$$
\begin{aligned}
& \Rightarrow \quad(2 m c-4 a)^2-4 m^2\left(c^2+4 a^3\right)=0 \\
& \Rightarrow 4 m^2 c^2-16 a m c+16 a^2-4 m^2 c^2-16 a^3 m^2=0 \\
& \Rightarrow-m c+a-a^2 m^2=0 \\
& \Rightarrow m c=a-a^2 m^2 \Rightarrow c=\frac{a}{m}-a^2 m
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main
If the straight line $y=m x$ is outside the circle $x^2+y^2-20 y+90=0$, then

Options:
A $m\gt3$
B $m \lt 3$
C $|m|\gt3$
D $|m| \lt 3$

The correct answer is:
$|m| \lt 3$

Solution:
If the straight line $y=m x$ is outside the given circle then distance from centre of circle $\gt$ radius of circle
$\frac{10}{\sqrt{1+m^2}}\gt\sqrt{10}$
$\Rightarrow\left(1+m^2\right) \lt 10 \Rightarrow m^2 \lt 9 \Rightarrow|m| \lt 3$.

Topic: Mathematics, Circle, JEE Main
If the straight line y=x meets y=f(x) at P, where fx is a solution of the differential equation dydx=x2+xyx2+y2 such that f1=3, then the value of f ′ x at the point P is

Options:
A 35
B 53
C 2
D 1

The correct answer is:
1

Solution:
Note f ′ x = dy dx = 1+ y x 1+ y x 2
At point P,yx=1
So, dydxP=1+11+1=1

Topic: Mathematics, Differential Equations, JEE Main
If the straight line \(y=m x\) lies outside of the circle \(x^2+y^2-20 y+90=0\), then the value of \(m\) will satisfy

Options:
A \(m < 3\)
B \(|m| < 3\)
C \(\mathrm{m} > 3\)
D \(|\mathrm{m}| > 3\)

The correct answer is:
\(|m| < 3\)

Solution:
Hints: \(x^2+m^2 x^2-20 m x+90\)
\(\begin{aligned}
& x^2\left(1+m^2\right)-20 m x+90=0 \\
& D < 0 \\
& 400 m^2-4 \times 90\left(1+m^2\right) < 0 \\
& 40 m^2 < 360 \\
& m^2 < 9 ;|m| < 3
\end{aligned}\)

Topic: Mathematics, Circle, JEE Main
If the straight lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$ intersect at a point, then the integer $k$ is equal to

Options:
A
$-5$

B
5

C
2

D
$-2$

The correct answer is:
$-5$

Solution:
$\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$
Since lines intersect in a point
$$
\begin{aligned}
& \left|\begin{array}{ccc}
k & 2 & 3 \\
3 & k & 2 \\
1 & 1 & -2
\end{array}\right|=0 \\
& \therefore 2 k^2+5 k-25=0 \\
& k=-5,5 / 2 .
\end{aligned}
$$
Directions: Questions number 17 to 21 are Assertion-Reason type questions. Each of these questions contains two statements : Statement - 1 (Assertion) and Statement-2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2008
If the straight lines $x=1+s, y=-3-\lambda s, z=1+\lambda s$ and $x=\frac{t}{2}, y=1+t, z=2-t$ with parameters $s$ and $t$ respectively, are co-planar then $\lambda$ equals

Options:
A
$-2$

B
$-1$

C
$-\frac{1}{2}$

D
0

The correct answer is:
$-2$

Solution:
Given lines $\frac{x-1}{1}=\frac{y+s}{-\lambda}=\frac{z-1}{\lambda}=s$ and $\frac{x}{1 / 2}=\frac{y-1}{1}=\frac{z-2}{-1}=t$ are coplanar then plan passing through these lines has normal perpendicular to these lines $\Rightarrow a-b \lambda+c \lambda=0 \quad$ and $\frac{a}{2}+b-c=0$ (where $a, b, c$ are direction ratios of the normal to the plan) On solving, we get $\lambda=-2$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
If the straight lines x+2y=3, 2x+3y=5 and k2x+ky=-1 represent a triangle which is right-angled, then the values of k are k1 and k2. The value of k1+k2k1-k2 is

Options:

The correct answer is:
7

Solution:
Area of triangle≠0

12-323-5k2k1≠0⇒k2+k+1≠0⇒k∈R
This should be a right-angled triangle if
-12-k2k=-1 or -23-k2k=-1
⇒k1=-2 or k2=-3/2 ∵k≠0

Topic: Mathematics, Straight Lines, JEE Main
If the straight lines $x+3 y=4,3 x+y=4$ and $x+y$ $=0$ form a triangle, then the triangle is

Options:
A
scalene

B
equilateral triangle

C
isosceles

D
right angled isosceles

The correct answer is:
isosceles

Solution:
Let equation of $A B: x+3 y=4$
Let equation of $B C: 3 x+y=4$
Let equation of $C A: x+y=0$
Now, By solving these equations we get $A=(-2,2), B=(1,1)$ and $C=(2,-2)$
Now, $A B=\sqrt{9+1}=\sqrt{10}$,
$$
B C=\sqrt{1+9}=\sqrt{10}
$$
and $C A=\sqrt{16+16}=\sqrt{32}$
Since, length of $A B$ and $B C$ are same
therefore triangle is isosceles.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (07 May Online)
If the straight lines \( x=1+s, y=-3-\lambda s, z=1+\lambda s \) and \( x=\frac{t}{2}, y=1+t, z=2-t \) with parameters \( s \) and \( t \) respectively, are coplanar, then \( \lambda \) Equals

Options:
A \( -2 \)
B \( -1 \)
C \( \frac{-\frac{1}{2}}{2} \)
D \( 0 \)

The correct answer is:
\( -2 \)

Solution:
Given lines can be rewritten as
x-11=y+3-λ=z-1λ=s
and x-01=y-12=z-2-2=t2

As we know that for coplaner lines determinant formed by elements of direction ratio of lines must be zero.
∴ 1-0-3-11-21-λλ12-2=0
⇒ 1-4-11-λλ12-2=0
⇒ 12λ-2λ+4-2-λ-12+λ=0
⇒ -8-4λ-2-λ=0
⇒ λ=-2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the sum 3 1 2 + 5 1 2 + 2 2 + 7 1 2 + 2 2 + 3 2 + ..... + up to 20 terms is equal to k 2 1 , then k is equal to

Options:
A 240
B 120
C 60
D 180

The correct answer is:
120

Solution:
S20=312+512+22+712+22+32+....+ up to 20 terms
=Î£r=1202r+112+22+32+.... r2
= Î£ r = 1 20 2r + 1 r 6 r + 1 2r + 1
= 6 Î£ r = 1 20 1 r r + 1
=6Î£r=1201r-1r+1
= 6 1 1 - 1 2 + 1 2 - 1 3 + 1 3 - 1 4 + .... + 1 20 - 1 21
= 6 1 - 1 21
= 6 Ã 20 21
= 1 20 21
= k 21
k=120

Topic: Mathematics, Sequences and Series, JEE Main
If the sum 312+512+22+712+22+32+.....+ up to 20 terms is equal to k21, then k is equal to

Options:
A 240
B 120
C 60
D 180

The correct answer is:
120

Solution:
S20=312+512+22+712+22+32+....+upto 20terms
=Σr=1202r+112+22+32+.... r2
=Σr=1202r+1r6r+12r+1
=6Σr=1201rr+1
=6Σr=1201r-1r+1
=611-12+12-13+13-14+....+120-121
=61-121
=6×2021
=12021
=k21
k=120

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (09 Apr Online)
If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

Options:
A 7
B 92
C 3
D 14

The correct answer is:
7

Solution:
Let us consider four positive consecutive terms of a G.P. as follow:

a,ar,ar2,ar3a,r>0

Product =a4r6=1296

⇒a2r3=36

⇒a=6r3/2

Sum =a+ar+ar2+ar3=126

⇒1r3/2+rr3/2+r2r3/2+r3r3/2=1266

⇒r-3/2+r3/2+r1/2+r-1/2=21

Let r1/2+r-1/2=A

⇒r-3/2+r3/2=(r12+r-12)3-3(r12+r-12)

=A3-3A

Therefore,

A3-3A+A=21

⇒A3-2A=21

⇒A=3

∴ r+1r=3

⇒r+1=3r

On squaring both sides:

r2+2r+1=9r

⇒r2-7r+1=0

Here, sum of roots=7

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (31 Jan Shift 1)
If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th term is:

Options:
A -25
B -35
C 25
D -36

The correct answer is:
-25

Solution:
Let, the first 3 terms of the A.P. are a-d, a, a+d

⇒sum=3a=33⇒a=11

And product a-d a a+d=aa2-d2=1155

⇒11112-d2=1155

⇒121-d2=115511=105

⇒d2=121-105=16

⇒d=±4

For a=11, d=4, the A.P. is 7, 11, 15,…

The nth term of an A.P., with first term A and common difference d is Tn=A+n-1d

So, T11=7+10×4=47

For a=11, d=-4, the A.P. is 15, 11, 7,…

So, T11=15+10-4=-25.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

Options:
A 33232
B 33229
C 33228
D 33227

The correct answer is:
33228

Solution:
Let p=probability of success and q=probability of failure,

Now given,

Sum np+npq=24 ...i

Product np×npq=128 ...ii

Also p=1-q ....iii

Solving i , iiand iii we get,

p=12,q=12,n=32

Now, PX=1+PX=2

=C132pq31+C232p2q30

=33228

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (25 Jul Shift 1)
If the sum of \( 5^{\text {th }} \) and \( 6^{t h} \) terms is zero in the expansion of \( (x-2 y)^{10} \), then the value of \( \frac{x}{y} \) is equal to

Options:

The correct answer is:
2.4

Solution:
T5+T6=0

⇒10C4x6-2y4+10C5x5-2y5=0

⇒10!4! 6!x6+16y4=10!5! 5!x5×32y5

⇒56x=2y

⇒xy=125=2.4

Topic: Mathematics, Binomial Theorem, JEE Main
If the sum of a certain number of terms of the A.P. $25,22,19, \ldots$ is 116 then the last term is

Options:
A 0
B 2
C 4
D 6

The correct answer is:
4

Solution:
$a=25, d=22-25=-3 \text {. }$

Let $n$ be the no. of terms

$\begin{array}{l}

\text { Sum }=116 ; \\

\text { Sum }=\frac{n}{2}[2 a+(n-1) d] \\

116=\frac{n}{2}[50+(n-1)(-3)] \\

\text { or } 232=n[50-3 n+3]=n[53-3 n]=-3 n^{2}+53 n \\

\Rightarrow 3 n^{2}-53+232=0 \\

\Rightarrow(n-8)(3 n-29)=0 \\

\Rightarrow n=8 \text { or } n=\frac{29}{3}, \\

n \neq \frac{29}{3} \\

\therefore n=8 \\

\therefore \text { Now, } \mathrm{T}_{8}=a+(8-1) d=25+7 \times(-3)=25-21=4 \\

\therefore \text { Last term }=4

\end{array}$

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of all the solutions of
tan-12x1-x2+cot-11-x22x=π3, -1<x<1,x≠0, is α-43, then α is equal to _____ .

Options:

The correct answer is:
2

Solution:
Given,

tan-12x1-x2+cot-11-x22x=π3 where x∈-1,1

Case 1: When x<0, then

tan-12x1-x2+cot-11-x22x=π3

⇒2tan-1x+π+2tan-1x=π3

⇒4tan-1x=-π+π3

⇒4tan-1x=-2π3

⇒tan-1x=-π6

⇒x=-13

Case 2: When x≥0,

tan-12x1-x2+cot-11-x22x=π3

⇒2tan-1x+2tan-1x=π3

⇒tan-1x=π12

⇒x=2-3

So, sum of solution will be

-13+2-3=2-43

Now, on comparing with α-43, we get

α=2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2023 (25 Jan Shift 1)
If the sum of an infinite GP, a,ar,ar2,ar3,… is 15 and the sum of the squares of its each term is 150, then the sum of ar,2ar4,ar6,… is:

Options:
A 252
B 92
C 12
D 52

The correct answer is:
12

Solution:
Given:

a+ar+ar2+ar3+......∞=15

⇒a1-r=15 ....1

And,

a2+a2r2+a2r4+a2r6+......∞=150

⇒a21-r2=150

⇒a1+ra1-r=150

⇒a(1+r)=10 ...2

Solving equation (1) and (2), we get

r=15⇒a=12

Now,

ar2+ar4+ar6......∞

=ar21-r2=12×1251-125=12

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (26 Aug Shift 1)
If the sum of an infinite GP be 9 and sum of first two terms be 5 then their common ratio is …..

Options:
A 13
B 3
C 23
D 32

The correct answer is:
23

Solution:
We have,
Sum of infinite GP, a1-r=9
⇒a+ar=5
⇒a1+r=5
⇒91-r1+r=5
⇒1-r2=59⇒r2=49⇒r=23

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of first 11 terms of an A.P. ,a1,a2,a3…… is 0a1≠0 then the sum of the A.P a1,a3,a5,…..a23 is ka1 where k is equal to

Options:
A -12110
B 12110
C 725
D -725

The correct answer is:
-725

Solution:
Let a1 be the first term and d be the common difference of A.P.

∴S11=0 ⇒1122a1+10d=0

⇒ a1+5d=0

Now a1+a3+a5+….+a23=1222a1+12-12d =12a1+11d
=12a1-11a15
=-72a15

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (02 Sep Shift 2)
If the sum of first $n$ terms of an $\mathrm{AP}$ is $\mathrm{cn}^2$, then the sum of squares of these $n$ terms is

Options:
A
$\frac{n\left(4 n^2-1\right) c^2}{6}$

B
$\frac{n\left(4 n^2+1\right) c^2}{3}$

C
$\frac{n\left(4 n^2-1\right) c^2}{3}$

D
$\frac{n\left(4 n^2+1\right) c^2}{6}$

The correct answer is:
$\frac{n\left(4 n^2-1\right) c^2}{3}$

Solution:
Let $S_n=c^2$
$$
\begin{aligned}
& S_{n-1}=c(n-1)^2=c n^2+c-2 c n \\
& \therefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}\right) \\
& T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \\
& \therefore \text { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \\
& =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \\
& =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3} \\
& =\frac{n c^2\left(4 n^2-1\right)}{3}
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of mean and variance of a binomial distribution for 5 trials is 1.8 , then probability of a success is

Options:
A 0.2
B 0.6
C 0.4
D 0.8

The correct answer is:
0.2

Solution:
We have $n p+n p q=1.8$, where $n=5$
$$
\begin{aligned}
& \therefore \mathrm{np}(1+\mathrm{q})=1.8 \Rightarrow 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \\
& \therefore 5 \mathrm{p}(2-\mathrm{p})=1.8 \Rightarrow 10 \mathrm{p}-5 \mathrm{p}^2=1.8 \text { i.e. } \\
& 5 \mathrm{p}^2-10 \mathrm{p}+1.8=0 \Rightarrow 5 \mathrm{p}^2-9 \mathrm{p}-\mathrm{p}+1.8=0 \\
& \therefore 5 \mathrm{p}(\mathrm{p}-1.8)-1(\mathrm{p}-1.8)=0 \Rightarrow(5 \mathrm{p}-1)(\mathrm{p}-1.8)=0 \\
& \therefore \mathrm{p}=\frac{1}{5}, 1.8 \text { but } \mathrm{p} \leq 1 \Rightarrow \mathrm{p}=\frac{1}{5}=0.2
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
If the sum of mean and variance of a Binomial Distribution is $\frac{15}{2}$ for 10 trials, then the variance is

Options:
A $1.5$
B $2.5$
C $4.5$
D $3.5$

The correct answer is:
$2.5$

Solution:
$\begin{array}{ll} & \text { Mean }+ \text { variance }=\frac{15}{2} \\ & \Rightarrow n p+n p q=\frac{15}{2} \\ & \Rightarrow n p+n p(1-p)=\frac{15}{2} \quad \ldots[\because p+q=1] \\ & \Rightarrow n\left(2 p-p^2\right)=\frac{15}{2} \\ & \Rightarrow 2 p-p^2=\frac{15}{2 \times 10} \quad \ldots[n=10] \\ & \Rightarrow 4 p^2-8 p+3=0 \\ & \Rightarrow(2 p-3)(2 p-1)=0 \\ & \Rightarrow p=\frac{3}{2} \text { or } p=\frac{1}{2} \\ & p=\frac{1}{2}...[0 < p < 1] \\ & q=1-\frac{1}{2}=\frac{1}{2} \\ & \text { Variance }=n p q=10 \times \frac{1}{2} \times \frac{1}{2}=2.5\end{array}$

Topic: Mathematics, Probability, JEE Main
If the sum of $n$ terms of an A.P. is $2 n^2+5 n$, then the $n^{\text {th }}$ term will be

Options:
A $4 n+3$
B $4 n+5$
C $4 n+6$
D $4 n+7$

The correct answer is:
$4 n+3$

Solution:
Given that $S_n=2 n^2+5 n$
Putting $n=1,2,3, \ldots \ldots \ldots . ., S_1=2 \times 1+5 \times 1=7$
So, $T_1=S_1=a=7, T_2=S_2-S_1=18-7=11$
$T_3=S_3-S_2=33-18=15$
Therefore series is $7,11,15, \ldots . . . .$.
Now, $n^{\text {th }}$ term $=a+(n-1) d=7+(n-1) 4=4 n+3$
Aliter : As we know $T_n=S_n-S_{n-1}$
$=\left(2 n^2+5 n\right)-\left\{2(n-1)^2+5(n-1)\right\}$
$=2 n^2+5 n-2 n^2+4 n-2-5 n+5=4 n+3$.

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of $n$ terms of an A.P. is $n A+n^2 B$, where $A, B$ are constants, then its common difference will be

Options:
A $A-B$
B $A+B$
C $2 A$
D $2 B$

The correct answer is:
$2 B$

Solution:
Given that $S_n=n A+n^2 B$
Putting $n=1,2,3, \ldots \ldots . . . . .$. , we get
$S_1=A+B, S_2=2 A+4 B, S_3=3 A+9 B$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
Therefore $T_1=S_1=A+B, T_2=S_2-S_1=A+3 B$,
$T_3=S_3-S_2=A+5 B$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
Hence the sequence is $(A+B),(A+3 B),(A+5 B)$
Here $a=A+B$ and common difference $d=2 B$ $\ldots\ldots. . . . .$

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of solutions of the system of equations 2sin2θ-cos2θ=0 and 2cos2θ+3sinθ=0 in the interval 0,2π is kπ, then k is equal to _______.

Options:

The correct answer is:
3

Solution:
Given,

2sin2θ-cos2θ=0

⇒2sin2θ-1-2sin2θ=0

⇒sin2θ=122

⇒θ=π6,5π6,7π6,11π6

Also given, 2cos2θ+3sinθ=0

⇒2sin2θ-3sinθ-2=0

⇒ sinθ=-12

⇒θ=7π6,11π6

So, the common solution is

θ=7π6,11π6

Sum =7π+11π6=3π=kπ

⇒K=3

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (26 Jul Shift 2)
If the sum of squares of all real values of α, for which the lines 2x-y+3=0, 6x+3y+1=0 and αx+2y-2=0 do not form a triangle is p, then the greatest integer less than or equal to p is ________.

Options:

The correct answer is:
32

Solution:
Given lines are 2x-y+3=0, 6x+3y+1=0 and αx+2y-2=0.

These lines will not form a triangle if αx+2y-2=0 is concurrent with 2x-y+3=0 and 6x+3y+1=0 or parallel to either of them.

Case-1: Concurrent lines

⇒2-13631α2-2=0

⇒2-6-2+1-12-α+312-3α=0

⇒-16-12-α+36-9α=0

⇒-10α+8=0

⇒α=45

Case-2: Parallel lines

-α6=-23, -α2=-2-1

⇒α=4, α=-4

⇒P=16+16+1625

[P]=32+1625=32

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (27 Jan Shift 2)
If the sum of squares of distances of a point from the planes x+y+z=0, x-z=0 and x-2y+z=0 is p2 then locus of the point is

Options:
A x2+z2=p2
B x2+2xy+y2+z2=p2
C x+y+z=p2
D x2+y2+z2=p2

The correct answer is:
x2+y2+z2=p2

Solution:
Let distances of a point Px,y,z from the planes x+y+z=0, x-z=0 and x-2y+z=0 are x+y+z3, x-z2 and x-2y+z6 respectively, then the sum of the squares of distances, is as

⇒x+y+z32+x-z22+x-2y+z62=p2

⇒2x+y+z2+3x-z2+x-2y+z2=6p2

⇒ 2x2+2y2+2z2+4xy+4yz+4zx+3x2+3z2-6xz+x2+4y2+z2-4xy-4yz+2xz=6p2

⇒6x2+6y+6z2=6p2

⇒x2+y2+z2=p2

Hence, option (d) is correct.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the sum of squares of the deviations from the mean of the data $x_i,(i=1,2, \ldots, n)$ is $n^2$, where $\bar{x}$ is the mean of $x_1$ 's, then the sum of squares of $x_i$ 's is

Options:
A $4 n x^{-2}$
B $3 \mathrm{nx}^{-2}$
C $12 \mathrm{nx}^{-2}$
D $2 n \mathrm{x}^{-2}$

The correct answer is:
$2 n \mathrm{x}^{-2}$

Solution:
Given, Sum of squares of deviations from the mean of data $x_i=n \bar{x}^2$
$\begin{aligned} & \Rightarrow \sum_{i=1}^n\left(x_i-\bar{x}\right)^2=n \bar{x}^2 \\ & \Rightarrow \sum_{i=1}^n x_i^2+n \bar{x}^2-2 \bar{x} \sum_{i=1}^n x_i=n \bar{x}^2 \\ & \Rightarrow \sum_{i=1}^n x_i^2-2 \bar{x} \cdot n \bar{x}=0 \quad\left\{\because \bar{x}=\frac{\sum x_i}{n}\right\} \\ & \Rightarrow \sum_{i=1}^n x_i^2=2 n \bar{x}^2\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If the sum of the coefficients in the expansion of 1+3xn lies between 4000 and 10000, then the value of the greatest coefficient must be

Options:
A 3954
B 6342
C 4806
D 1458

The correct answer is:
1458

Solution:
Putting x=1 to get the sum of the coefficients, we get,
4000<4n<10000⇒n=6
45=210=1032
The greatest coefficient is the greatest term in the expansion of 1+3x6 when x=1.
For 1+3x6, m=3x6+13x+1=3×74=5.25
⇒ Greatest term is Tm+1=T5+1
=6C53x5=6×35×x5
Greatest coefficient =6×35=1458

Topic: Mathematics, Binomial Theorem, JEE Main
If the sum of the coefficients in the expansion of $(a+b)^n$ is 4096 , then the greatest coefficient in the expansion is

Options:
A
1594

B
792

C
924

D
2924

The correct answer is:
924

Solution:
We have $2^n=4096=2^{12} \Rightarrow n=12$; So middle term $=t_7 ; t_7=t_{6+1}={ }^{12} C_6=\frac{12 !}{6 ! 6 !}=924$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2002
If the sum of the coefficients in the expansion of (x+y)n is 4096, then the greatest coefficient in the expansion is _____.

Options:

The correct answer is:
924

Solution:
Sum of coefficients of the expression (x+y)n is 2n

2n=4096=212

⇒n=12

Now greatest binomial coefficient of the expression(x+y)n is the middle term's coefficient so C6n=C612=924

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (01 Sep Shift 2)
If the sum of the coefficients of all even powers of x in the product 1+x+x2+…+x2n1-x+x2-x3+…+x2n is 61, then n is equal to

Options:

The correct answer is:
30

Solution:
Let 1-x+x2..1+x+x2…=a0+a1×a2x2+…
Put x=1,
12n+1=a0+a1+a2+..a2n …2
Put x=-1
2n+1×1=a0-a1+a2+..a2n …2
From 1+2
4n+2=2a0+a2+…
=2×61
⇒2n+1=61⇒n=30

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If the sum of the co-efficients of all the positive even powers of x in the binomial expansion of 2x3+3x10 is 510-β·39, then β is equal to _____.

Options:

The correct answer is:
83

Solution:
Given,

Binomial expression 2x3+3x10, so its nth term is given by,

Tr+1=Cr102x310-r3xr

=Cr10210-r3rx30-4r

Now putting r=0,1,2,.....7 and adding all term we get

2x3+3x10=C01021030x30+C1102931x26..............C7102337x2+C81022381x2....

2x3+3x10-C81022381x2.......=C01021030x30+C1102931x26..............C7102337x2

Put x=1 both side we get,

510-3960+20+3=C01021030+C1102931..............C7102337

510-3983=sum of even coefficients

So β=83

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (25 Jun Shift 2)
If the sum of the coefficients of all the positive powers of x, in the binomial expansion of xn+2x57 is 939, then the sum of all the possible integral values of n is

Options:

The correct answer is:
57

Solution:
Plotting the table for coefficient of xn+2x57 and there cumulative sum :

Coefficient	Commulative sum

x7n→C07	1

x6n-5→2×C17	1+14

x5n-10→22×C27	1+14+84

x4n-15→23×C37	1+14+84+280

x3n-20→24×C47	1+4+84+280+560=939

x2n-25→25×C57

So, 3n-20≥0 ∩ 2n-25<0 ∩ n∈I

∴ 7≤n≤12

Sum =7+8+9+10+11+12=57

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (27 Jun Shift 2)
If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :

Options:
A 30
B 51
C 50
D 31

The correct answer is:
31

Solution:
Let the observations be x1, x2, ..., x50.
Given x1-30+x2-30+.....+x50-30=50
⇒ x1+x2+.....+x50=50+30×50=1550
Now, Mean =x1+x2+....+x5050=155050=31.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (12 Jan Shift 1)