datavorous/entrance-exam-dataset huggingface dataset:
Let the position vectors of the vertices A, B and C of a triangle be 2i^+2j^+k^, i^+2j^+2k^ and 2i^+j^+2k^ respectively. Let l1,l2 and l3 be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides AB,BC and CA respectively, then l12+l22+l32 equals :

Options:
A 15
B 12
C 14
D 13

The correct answer is:
12

Solution:
Given: A2,2,1, B1,2,2 and C2,1,2 are vertices of ∆ABC.

⇒AB=2-12+2-22+1-22=2

⇒BC=1-22+2-12+2-22=2

⇒CA=2-22+2-12+1-22=2

So, ∆ABC is equilateral.

Therefore, orthocentre and centroid will be same.

⇒G≡2+1+23,2+2+13,1+2+23

⇒G≡53,53,53

Also, mid point of AB is D32,2,32

⇒l1=32-532+2-532+32-532

⇒l1=136+19+136

⇒l1=16

Similarly, l2=l3=16

⇒l12+l22+l32=16+16+16

⇒l12+l22+l32=12

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Let the position vectors of two points $A$ and $B$ be $\mathbf{a}+\mathbf{b}+\mathbf{c}$ and $\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}$, respectively. If the points $P$ and $Q$ divide $A B$ in the ratio $1: 3$ internally and externally respectively, then 3|AB $\mid=$

Options:
A $4|P Q|$
B $3|\mathrm{PQ}|$
C $\frac{1}{2}|\mathrm{PQ}|$
D $2 \mid \mathrm{PQ}$

The correct answer is:
$4|P Q|$

Solution:
We have,
$$
\begin{aligned}
\mathbf{O A} & =\mathbf{a}+\mathbf{b}+\mathbf{c}, \mathbf{O B}=\mathbf{a}-2 \mathbf{b}+3 \mathbf{c} \\
\mathbf{A B} & =(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})-(\mathbf{a}+\mathbf{b}+\mathbf{c}) \\
& =-3 \mathbf{b}+2 \mathbf{c}
\end{aligned}
$$
Point $P$ divide $\mathbf{A B}$ in the ratio $1: 3$ internally,
$$
\therefore \mathbf{O P}=\frac{\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}+3 \mathbf{a}+3 \mathbf{b}+3 \mathbf{c}}{\mathrm{l}+3}=\frac{4 \mathbf{a}+\mathbf{b}+6 \mathbf{c}}{4}
$$
Point $Q$ divides $\mathbf{A B}$ in the ratio $1: 3$ externally,
$$
\begin{aligned}
& \therefore \quad \text { OQ }=\frac{\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}-3 \mathbf{a}-3 \mathbf{b}-3 \mathbf{c}}{1-3}=\frac{2 \mathbf{a}+5 \mathbf{b}}{2} \\
& \quad \mathbf{P Q}=\left(\frac{2 \mathbf{a}+5 \mathbf{b}}{2}\right)-\left(\frac{4 \mathbf{a}+\mathbf{b}+6 \mathbf{c}}{4}\right)=\frac{9 \mathbf{b}-6 \mathbf{c}}{4} \\
& |\mathbf{A B}|=\sqrt{9 \mathbf{b}^2+4 \mathbf{c}^2} \text { and }|\mathbf{P Q}|=\frac{1}{4} \sqrt{81 \mathbf{b}^2+36 \mathbf{c}^2} \\
& |\mathbf{P Q}|=\frac{1}{4} \sqrt{9\left(9 \mathbf{b}^2+4 \mathbf{c}^2\right)} \Rightarrow|\mathbf{P Q}|=\frac{3}{4} \sqrt{9 \mathbf{b}^2+4 \mathbf{c}^2} \\
& |\mathbf{P Q}|=\frac{3}{4}|\mathbf{A B}| \Rightarrow 4|\mathbf{P Q}|=3|\mathbf{A B}|
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
Let the positive integers be written in the form :
If the $k^{\text {th }}$ row contains exactly $k$ numbers for every natural number $k$, then the row in which the number 5310 will be, is _______

Options:

The correct answer is:
103

Solution:
$\begin{aligned} & \mathrm{S}=1+2+4+7+\ldots \ldots+\mathrm{T}_{\mathrm{n}} \\ & \mathrm{S}=1+2+4+\ldots \ldots \\ & \mathrm{Tn}=1+1+2+3+\ldots \ldots+\left(\mathrm{T}_{\mathrm{n}}-\mathrm{T}_{\mathrm{n}-1}\right) \\ & \mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{n}-1}{2}\right)[2+(\mathrm{n}-2) \times 1] \\ & \mathrm{T}_{\mathrm{n}}=1+1+\frac{\mathrm{n}(\mathrm{n}-1)}{2} \\ & \mathrm{n}=100 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{100 \times 99}{2}=4950+1 \\ & \mathrm{n}=101 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{101 \times 100}{2}=5050+1=5051 \\ & \mathrm{n}=102 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{102 \times 101}{2}=5151+1=5152 \\ & \mathrm{n}=103 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{103 \times 102}{2}=5254 \\ & \mathrm{n}=104 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{104 \times 103}{2}=5357\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Let the positive numbers a1, a2, a3, a4 and a5 be in a G.P. Let their mean and variance be 3110 and mn respectively, where m and n are co-prime. If the mean of their reciprocals is 3110 and a3+a4+a5=14, then m+n is equal to ____________.

Options:

The correct answer is:
211

Solution:
5 positive numbers a1, a2,..., a5 are in geometric progression. So, let the numbers be ar2, ar, a, ar, ar2

According to question,

ar2+ar+a+ar+ar25=3110 ...1

And,

r2a+ra+1a+1ar+1ar25=3140

⇒1ar2+r+1+1r+1r25=3140 ...2

By 1 & 2, we get

a2=4 ⇒a=2

So,

r2+r+1+1r+1r2=314

⇒r+1r2+r+1r=274+2

⇒t2+t=354

where, t=r+1r

4t2+4t-35=0

⇒t=52⇒r=2

So, numbers are 12, 1, 2, 4, 8

Variance is

=14+1+4+16+645-12+1+2+4+852

=18625=mn

m+n=211

Hence this is the required option.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let the probability of getting head for a biased coin be 14. It is tossed repeatedly until a head appears. Let N be the number of tosses required. If the probability that the equation 64x2+5Nx+1=0 has no real root is pq, where p and q are co-prime, then q-p is equal to..........

Options:

The correct answer is:
27

Solution:
Given a quadratic equation 64x2+5nx+1=0 has no real roots, so

D<0

⇒25N2-4×64<0

⇒N<165⇒N=1,2,3

And, For a biased coin, the probability of getting head is P=14, and probability of getting tail will be Q=34

So, required probability is

14+34·14+342·14+343·14=pq

⇒141-3431-34=pq

⇒1-343=pq

⇒1-2764=pq

⇒pq=3764

q-p=27

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let the quadratic curve passing through the point -1,0 and touching the line y=x at 1,1 be y=fx. Then the x-intercept of the normal to the curve at the point α,α+1 in the first quadrant is

Options:

The correct answer is:
11

Solution:
Given,

The quadratic curve passing through the point -1,0 and touching the line y=x at 1,1 be y=fx,

Now let the curve be y=ax2+bx+c

Passes through -1,0

⇒a-b+c=0 ....i

Passes through 1, 1

⇒ a+b+c = 1 .....…ii

Now given tangent to y=x at 1,1,

So, dydx=1

⇒ 2a+b = 1 ....…iii

Solving all above three equations we get,

a=14, b=12 and c=14

Hence, the equation of curve is y=x24+x2+14

Now α,α+1 lies on curve

α+1=α24+α2+14

⇒(α-3)(α+1)=0

⇒α=3
Point in first quadrant is 3,4

Now finding slope of tangent at 3,4 we get,

dydx3,4=x2+123,4=2

⇒ Slope of normal =-12

Hence, equation of normal will be, y-4=-12(x-3)

So, x- intercept = 11

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let the radius of the circle touching the parabola y2=x at 1,1 and having the directrix of y2=x as its normal is equal to k5 units, then k is equal to

Options:

The correct answer is:
1.25

Solution:
Equation of normal at P is
y-1=-2x-1
y+2x=3
at x=-14, y=3+12=72
⇒ radius =1+142+1-722=554

⇒554=k5
⇒k=54=1.25

Topic: Mathematics, Parabola, JEE Main
Let the range of the function $f(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R}$ be $[a, b]$. If $\alpha$ and $\beta$ are respectively the A.M. and the G.M. of $a$ and $b$, then $\frac{\alpha}{\beta}$ is equal to

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

The correct answer is:
$\sqrt{2}$

Solution:
$\begin{aligned} & f(x) \frac{1}{2+\sin 3 x+\cos 3 x} \\ & {\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right]} \\ & \frac{\alpha}{\beta}=\frac{a+b}{2 \sqrt{a b}}=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right) \\ & =\frac{1}{2}\left(\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}+\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}\right) \\ & =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{2 \times \sqrt{2}}=\sqrt{2}\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of 24+134n, in the increasing powers of 134 be 64:1. If the sixth term from the beginning is α34, then α is equal to _______.

Options:

The correct answer is:
84

Solution:
Given,

The ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of 24+134n in the increasing powers of 134 be 64:1

Now according to given data we have,

T5Tn-3=C4n21/4n-43-1/44Cn-4n21/443-1/4n-4=641

⇒2n-843n-84=61/4

⇒6n-8=6

⇒n-8=1⇒n=9

So, T6=C5921/443-1/45=8434

Now comparing with T6=α34,

We get α=84

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let the refractive index of a denser medium with respect to rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90o. Angle A is given by

Options:
A tan-1⁡sin⁡θC
B 1tan-1⁡sin⁡θC
C cos-1⁡sin⁡θC
D 1cos-1⁡sin⁡θC

The correct answer is:
tan-1⁡sin⁡θC

Solution:
In accordance with Snell's law, μ=μRμD=sin⁡icsin⁡90o

here ic is the angle of incidence.
μRμD=sin⁡ic
μ=μRμD=sin⁡Asin⁡r
=sinAsin90°-A=sinAcosA
μRμD=tan⁡A.
The critical angle is related to the angle A as tan⁡A=sin⁡θC
A=tan-1⁡sin⁡θC.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2017 (08 Apr Online)
Let the relation $R_{1}$ be defined on $R$ as $a R_{1}b$ if $1+a b > 0 .$ Then

Options:
A $R_{1}$ is reflexive only.
B $R_{1}$ is equivalence relation.
C $R_{1}$ is reflexive and transitive but not symmetric
D $R_{1}$ is reflexive and symmetric but not transitive.

The correct answer is:
$R_{1}$ is reflexive and symmetric but not transitive.

Solution:
We observe the following properties:
Reflexivity : Let a be an arbitrary element of $R$, Then, $a \in R$ $\Rightarrow 1+a \cdot a=1+a^{2}>0 \quad\left[\because a^{2}>0\right.$ for all $\left.a \in R\right]$
$\Rightarrow \quad(a, a) \in R_{1}$
$\left[\right.$ By def. of $\left.R_{1}\right]$
Thus, $(a, a) \in R_{1}$ for all $a \in R$. So, $R_{1}$ is reflexive on $R$.
Symmetry : Let $(a, b) \in R$ Then, $(a, b) \in R_{1}$
$\Rightarrow \quad 1+a b>0$
$[\because a b=b a$ for all $]$
$\Rightarrow \quad 1+b a>0 \Rightarrow(b, a) \in R_{1}$
Thus, $(a, b) \in R_{1} \neq(b, a) \in R,$ for all $a, b \in R$
So, $R_{1}$ is symmetric on $R$.
Transitivity : we observe that $\left(1, \frac{1}{2}\right) \in R_{1}$ and $\left(\frac{1}{2},-1\right) \in R_{1}$ but (1,-1)$\notin R_{1}$ because
$1+1 \times(-1)=0 \gg 0$
So, $R_{1}$ is not transitive on $R$.

Topic: Mathematics, Sets and Relations, JEE Main
Let the relation $R$ is defined in $N$ by $a R b$, if $3 a+2 b=27$ then $R$ is

Options:
A $\{(1,12)(3,9)(5,6)(7,3)\}$
B $\left\{\left(0, \frac{27}{2}\right)(1,12)(3,9)(5,6)(7,3)\right\}$
C $\{(1,12)(3,9)(5,6)(7,3)(9,0)\}$
D $\{(2,1)(9,3)(6,5)(3,7)\}$

The correct answer is:
$\{(1,12)(3,9)(5,6)(7,3)\}$

Solution:
Given relation is $3 a+2 b=27$, where $a, b \in N$
$$
\begin{aligned}
& 3 a+2 b=27 \\
& \Rightarrow \quad b=\frac{27-3 a}{2} \\
& a=1 \Rightarrow b=\frac{27-3}{2}=\frac{24}{2}=12 \\
& a=3 \Rightarrow b=\frac{27-9}{2}=\frac{18}{2}=9 \\
& a=5 \Rightarrow b=\frac{27-15}{2}=\frac{12}{2}=6 \\
& a=7 \Rightarrow b=\frac{27-21}{2}=\frac{6}{2}=3
\end{aligned}
$$
Hence, we can see that all the elements of option (a) satisfies the given relation. Hence, option (a) is correct.

Topic: Mathematics, Sets and Relations, JEE Main
Let the set C=(x,y)∣x2-2y=2023,x,y∈ℕ. Then ∑(x,y)∈C(x+y) is equal to _______.

Options:

The correct answer is:
46

Solution:
Given

x2-2y=2023

⇒x2=2y+2023

Now, if y=1 we get,

⇒x2=2025

⇒x=45

And there is no other value which will satisfy the above equation,

Hence, x=45 & y=1

So, ∑(x,y)∈C(x+y)=45+1=46

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let the set of all values of $p$, for which $f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right)+2(2-p) x+7$ does not have any critical point, be the interval $(a, b)$. Then $16 a b$ is equal to _______

Options:

The correct answer is:
252

Solution:
$\begin{aligned} & f(x)=-\left(p^2-6 p+8\right) \cos 4 n+2(2-p) n+7 \\ & f^1(x)=+4\left(p^2-6 p+8\right) \sin 4 x+(4-2 p) \neq 0 \\ & \sin 4 x \neq \frac{2 p-4}{4(p-4)(p-2)}\end{aligned}$
$\begin{aligned}
& \sin 4 x \neq \frac{2(p-2)}{4(p-4)(p-2)} \\
& p \neq 2 \\
& \quad \sin 4 x \neq \frac{1}{2(p-4)} \\
& \Rightarrow\left|\frac{1}{2(p-4)}\right|>1
\end{aligned}$
on solving we get
$\therefore \mathrm{p} \in\left(\frac{7}{2}, \frac{9}{2}\right)$
Hence $\mathrm{a}=\frac{7}{2}, \mathrm{~b}=\frac{9}{2}$
$\therefore 16 \mathrm{ab}=252$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let the set $S=\{2,4,8,16, \ldots, 512\}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\mathrm{S}$ and $\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{A} \cap \mathrm{C}=\phi$. The maximum number of such possible partitions of $S$ is equal to:

Options:
A 1680
B 1640
C 1520
D 1710

The correct answer is:
1680

Solution:
$\frac{9 !}{(3 ! 3 ! 3 !) 3 !} \times 3 !$

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let the sets A=2,4,6,8â¦ and B=3,6,9,12â¦ such that nA=200 and nB=250. If nAâªB=k, then k100 is equal to

Options:

The correct answer is:
3.84

Solution:
Let, the last element of set A is, t200=2+200-12=400

and the last element of set B is t250=3+250-13=750

Common elements are →6,12,18…

We have to find the common elements upto 400
tn=6+n-16⇒400=6+6n-6 ⇒n=4006=66.6…
t66=6+66-16=396
Number of common elements =66

nA∪B=nA+nB-nA∩B
=200+250-66=384⇒k=384
⇒k100=3.84

Topic: Mathematics, Sequences and Series, JEE Main
Let the sets A and B denote the domain and range respectively of the function fx=1x-x, where x denotes the smallest integer greater than or equal to x. Then among the statements

S1: A∩B=1, ∞-ℕ and

S2: A∪B=1, ∞

Options:
A Only (S2) is true
B Only (S1) is true
C Neither S1 nor S2 is true
D Both S1 and S2 are true

The correct answer is:
Neither S1 nor S2 is true

Solution:
Given,

fx=1x-x

We know that, x=x+x where x is fractional part of x whose value is 0<x<1,

So, x-x=-x

Now putting the value in given function we get,

fx=1x-x

⇒fx=1-x

Now we know that 1gx is defined when gx>0, but -x∈-1,0

Hence, -x is not defined

Hence, Domain and Range will be=ϕ

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Let the shortest distance between the lines L:x-5-2=y-λ0=z+λ1,λ≥0 and L1: x+1=y-1=4-z be 26. If (α,β,γ) lies on L, then which of the following is NOT possible?

Options:
A α+2γ=24
B 2α+γ=7
C 2α-γ=9
D α-2γ=19

The correct answer is:
α+2γ=24

Solution:
Given equations of lines are

L: x-5-2=y-λ0=z+λ1=k

L1: x+11=y-11=z-4-1

Here,

b1→×b2→=i^j^k^-20111-1

⇒b1→×b2→=-i^-j^-2k^

And,

a2→-a1→=6i^+λ-1j^+-λ-4k^

∵ Shortest distance is 26.

⇒26=-6-λ+1+2λ+81+1+4

⇒|λ+3|=12⇒λ=9,-15

⇒α=-2k+5, γ=k-λ where k∈R

So,

α+2γ=5-2λ=-13, 35

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let the six numbers a1, a2, ..., a6 be in A.P. and a1+a3=10.If the mean of these six numbers is 192 and their variance is σ2, then 8σ2 is equal to

Options:
A 220
B 210
C 200
D 105

The correct answer is:
210

Solution:
Let a1=a and common difference be d.

Now,

a1+a3=10

⇒2a+2d=10

⇒a+d=5 ...i

Now,

Given mean x¯=192

⇒a+a+d+a+2d+...+a+5d6=192

⇒6a+15d6=192

⇒2a+5d=19 ...ii

Solving i & ii, we get

a=2, d=3

So, A.P is 2, 5, 8, 11, 14, 17.

Now variance is given by,

σ2=16∑i=16xi-x¯2

⇒σ2=4+25+64+121+196+2896-1922

⇒σ2=6996-3614

⇒σ2=1054

⇒8σ2=210

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let the sixth term in the binomial expansion of 2log210-3x+2(x-2)log235m powers of 2x-2log23, be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is _____ .

Options:

The correct answer is:
4

Solution:
Given,

Binomial expression,

2log210-3x+2(x-2)log235m

10-3x+3(x-2)5m

Now, T6=C5m10-3xm-52·3x-2=21 …1

Also given,

C1m,C2m,C3m are in A.P.

So, 2·C2m=C1m+C3m

⇒2×m!2!m-2!=m+m!3!m-3!

Solving for m, we get m=2 , 7 and m=2 (rejected), so m=7

Put in equation 1

21·10-3x3x9=21

⇒10-3x3x=9×1

⇒3x=30,32

⇒x=0, 2

Sum of the squares of all possible values of x=4.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let the slope of the tangent to a curve y=fx at x,y be given by 2tanxcosx-y. if the curve passes through the point π4,0, then the value of ∫0π2ydx is equal to

Options:
A 2-2+π2
B 2-π2
C 2+2+π2
D 2+π2

The correct answer is:
2-π2

Solution:
Given,

dydx=2tanxcosx-2tanx·y

dydx+2tanxy=2sinx

Now solving linear differential equation by finding Integrating factor =e∫2tanxdx=1cos2x

Solution is given by,

y1cos2x=∫2sinxcos2xdx

⇒ysec2x=2cosx+C

⇒y=2cosx+Ccos2x

Passes through π4,0

⇒0=2+C2⇒C=-22

So, fx=2cosx-22cos2x: Required curve

Now, ∫0π2ydx=2∫0π2cosxdx-22∫0π2cos2xdx

=2sinx0π2-22x2+sin2x40π2

=2-π2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let the solubility of an aqueous solution of $\mathrm{Mg}(\mathrm{OH})_2$ be $x$ then its $\mathrm{k}_{\text {sp }}$ is

Options:
A
$4 x^3$

B
$108 x^5$

C
$27 x^4$

D
$9 x$

The correct answer is:
$4 x^3$

Solution:

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2002
Let the solution curve of the differential equation xdydx-y=y2+16x2,y1=3 be y=yx. Then y2 is equal to

Options:
A 15
B 11
C 14
D 17

The correct answer is:
15

Solution:
Given,

xdydx-y=y2+16x2

⇒ dydx=y2+16x2+yx

Let y=xt⇒dydx=xdtdx+t

⇒xdtdx+t=t+t2+16

⇒dtt2+16=dxx

Now integrating both side we get,

⇒∫dtt2+16=∫dxx

⇒ lnt+t2+16=lnx+lnc

⇒ yx+y2+16x2x=cx

∵ y1=3 so ⇒ 31+32+16×121=c×1⇒c=8

So solution becomes,

yx+y2+16x2x=8x

Now y2=y2+y2+16×222=8×2

⇒y+y2+64=32

⇒y=15

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Let the solution curve of the differential equation xdy=x2+y2+ydx,x>0, intersect the line x=1 at y=0 and the line x=2 at y=α. Then the value of α is

Options:
A 12
B 32
C -32
D 52

The correct answer is:
32

Solution:
Given,

xdy=x2+y2+ydx

⇒xdy-ydx=x2+y2dx

⇒xdy-ydxx2=1+y2x2·dxx

⇒dyx1+yx2=dxx

Now integrating both side we get,

⇒∫dyx1+yx2=∫dxx

⇒lnyx+yx2+1=lnx+lnc

⇒y+y2+x2x=cx

⇒y+y2+x2=cx2

Now given when x=1,y=0⇒0+1=c⇒c=1

So equation of curve is y+x2+y2=x2

Now at x=2,y=α

So putting the value in curve we get, 2+4+α2=4

⇒4+α2=16+α2=8α

⇒α=32

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Let the solution curve y=yx of the differential equation 1+e2xdydx+y=1 pass through the point 0,π2. Then, limx→∞exyx is equal to

Options:
A π4
B 3π4
C π2
D 3π2

The correct answer is:
3π4

Solution:
Given,

1+e2xdydx+y=1

Now on rearranging we get,

⇒dydx+y=11+e2x

We can see it is a linear differential equation,

So integrating factor is e∫1·dx=ex

So solution will be

y·IF=∫11+e2x×IF dx

⇒yex=∫11+e2x×ex dx

⇒y·ex=tan-1ex+c

Now as curve is passing through 0,π2 so

⇒c=π4

Now calculating the limit limx→∞y·ex we get,

⇒limx→∞y·ex=limx→∞tan-1ex+π4=π2+π4=3π4

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let the solution curve y=yx of the differential equation 4+x2dy-2xx2+3y+4dx=0 pass through the origin. Then y2 is equal to _____.

Options:

The correct answer is:
12

Solution:
Given 4+x2dy-2xx2+3y+4dx=0

x2+4dydx=2x3+6xy+8x

x2+4dydx-6xy=2x3+8x

dydx-6xx2+4y=2x3+8xx2+4

This is of the form of linear differential equation

I.F. =e-∫6xx2+4dx=e-3logex2+4

=elogex2+4-3=1x2+43

So the solution of the differential equation will be

y.1x2+43=∫2x3+8xx2+43x2+4dx

yx2+43=∫2xx2+4x2+43x2+4dx

Let x2+4=t, 2xdx=dt

So yx2+43=∫dtt3

yx2+43=-12x2+42+C

Since the curve passes through origin 0,0

0=-12×16+C⇒C=132

i.e. yx2+43=-12x2+42+132

y=-x2+42+x2+4332

Hence y2=-82+8×8×832=12

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let the solution $y=y(x)$ of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}-y=1+4 \sin x$ satisfy $y(\pi)=1$. Then $y\left(\frac{\pi}{2}\right)+10$ is equal to ______

Options:

The correct answer is:
7

Solution:
$\begin{aligned} & y e^{-x}=\int\left(e^{-x}+4 e^{-x} \sin x\right) d x \\ & y e^{-x}=-e^{-x}-2\left(e^{-x} \sin x e^{-x} \cos x\right)+C \\ & y=-1-2(\sin x+\cos x)+c e^x \\ & \because y(\pi)=1 \Rightarrow c=0 \\ & y(\pi / 2)=-1-2=-3 \\ & \text { Ans }=10-3=7\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$ and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then, $b$ equals to

Options:
A
$\frac{3}{4}$

B
$\frac{1}{2}$

C
$\frac{1}{3}$

D
$\frac{1}{4}$

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

Solution:
Here, area between 0 to $b$ is $R_1$ and $b$ to
$$
\begin{aligned}
& 1 \text { is } R_2 \\
& \therefore \int_0^b(1-x)^2 d x-\int_b^1(1-x)^2 d x=\frac{1}{4} \\
& \Rightarrow \quad\left(\frac{(1-x)^3}{-3}\right)_0^b-\left(\frac{(1-x)^3}{-3}\right)_b^1=\frac{1}{4} \\
& \Rightarrow-\frac{1}{3}\left\{(1-b)^3-1\right\}+\frac{1}{3}\left\{0-(1-b)^3\right\} \\
& =\frac{1}{4}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad-\frac{2}{3}(1-b)^3=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \\
& \Rightarrow \quad(1-b)^3=\frac{1}{8} \\
& \Rightarrow \quad(1-b)=\frac{1}{2} \Rightarrow b=\frac{1}{2} \\
&
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
Let the sum ∑ n=1 9 1 n( n+1 )( n+2 ) , written in the rational form be pq (where p and q are co-prime), then the value of q-p10 is, (where [.] is the greatest integer function)

Options:

The correct answer is:
8

Solution:
ân=1912 n+2-nnn+1 n+2
=12 11.2-12.3+12.3-13.4+â¦+19Ã10-110Ã11
=1212-1110=1255-1110=27110
â q-p=110-27=83

Topic: Mathematics, Sequences and Series, JEE Main
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5. Let the sum of its first five terms be 9825. Then the sum of the first 21 terms of an AP, whose first term is 10ar, nth term is an and the common difference is 10ar2, is equal to

Options:
A 21a11
B 22a11
C 15a16
D 14a16

The correct answer is:
21a11

Solution:
For the infinite G.P.

S∞=a1-r=5

Also a+ar+ar2+ar3+ar4=5

⇒a1-r51-r=9825

i.e. 1-r5=98125⇒r5=3353⇒r=3535

Now S21=21220ar+21-110ar2

=2110ar+100ar2

=21·a11

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, ...., an be 50n+n(n-7)2A, where A is a constant. If d is the common difference of this A.P., then the ordered pair d, a50 is equal to

Options:
A 50,Â 50+46A
B A,Â 50+45A
C 50, 50+45A
D A,Â 50+46A

The correct answer is:
A, 50+46A

Solution:
Given sum of n terms is Sn=50n+nn-72A …(1)

⇒Sn-1=50(n-1)+(n-1)n-82A …2

Subtracting (1) and (2), we get

Sn-Sn-1=50n-50n-1+nn-72A-n-1n-82A

⇒Sn-Sn-1=50n-n+1+A2nn-7-n-1n-8

⇒Sn-Sn-1=50+A2n2-7n-n2+9n-8

⇒Tn=Sn-Sn-1=50+A(n-4)

Hence, T1=50-3A and T2=50-2A

⇒d=T2-T1=A and T50=50+46A.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Let the sum of the first three terms of an A.P. be 39 and the sum of its last four terms be 178. If the first term of this A.P. is 10, then the median of the A.P. is :

Options:
A 26.5
B 29.5
C 28
D 31

The correct answer is:
29.5

Solution:
Let the A.P. be a, a+d,a+2d⋯⋯,l-3d, l-2d, l-d,l
Where a is the first term and l is the last term,
Sum of 1st, 3terms is, 39.
30+3d=39 as a=10 (Given)
d=93=3
Sum of last 4 terms is178.
4l-6d=178
l=49
10, 13, 16, 19.......46, 49
Let number of terms are, n then nthterm is10 + (n - 1) 3 = 49
⇒n = 14
So the median of the series would be mean of 7th and 8th term
10+6⋅3+10+7⋅32
=592 =29.5
Alternate way
The median would be mean of 10 and 49, That is 29.5.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2015 (10 Apr Online)
Let the sum \( \sum_{n=1}^{9} \frac{1}{n(n+1)(n+2)} \), written in the rational form be \( \frac{p}{q} \) (where \( p \) and \( q \) are co-prime), then the value of \( \left[\frac{q-p}{10}\right] \) is, (where [.] is the greatest integer function)

Options:

The correct answer is:
8

Solution:
ân=1912 n+2-nnn+1 n+2
=12 11.2-12.3+12.3-13.4+â¦+19Ã10-110Ã11
=1212-1110=1255-1110=27110
â q-p=110-27=83

Topic: Mathematics, Sequences and Series, JEE Main