datavorous/entrance-exam-dataset huggingface dataset:
Let x→ and y→ are 2 non-zero and non-collinear vectors, then the largest value of k such that the non-zero vectors k2-5k+6x→+k-3y→ and 2x→+5y→ are collinear is

Options:
A 3
B 6
C 125
D -1

The correct answer is:
125

Solution:
Let, aâ=k2-5k+6xâ+k-3yâ
and bâ=2xâ+5yâ
âµaâ and bâ are collinear vectors
âk2-5k+62=k-35
âk=3,125

But, for k=3, aâ=0 (not possible)

Topic: Mathematics, Vector Algebra, JEE Main
Let $X$ and $Y$ are two events such that $P(X \cup Y=) P X \cap(Y . \quad)$
Statement 1: $P\left(X \cap Y^{\prime}=\dot{P} X^{\prime} \cap(Y=0 \quad)\right.$
Statement 2: $P(X) P Y \in 2) P X \cap Y(\quad)$

Options:
A
Statement 1 is false, Statement 2 is true.

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

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1.

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

Solution:
Let $X$ and $Y$ be two events such that
$$
P(X \cup Y)=P(X \cap Y)
$$
We know
$$
\begin{aligned}
& P(X \cup Y)=P(X)+P(Y)-P(X \cap Y) \\
& P(X \cap Y)=P(X)+P(Y)-P(X \cap Y) \\
& \text { (from }(1) \\
& \Rightarrow P(X)+P(Y)=2 P(X \cap Y)
\end{aligned}
$$
Hence, Statement $-2$ is true.
Now, $P\left(X \cap Y^{\prime}\right)=P(X)-P(X \cap Y)$
and $P\left(X^{\prime} \cap Y\right)=P(Y)-P(X \cap Y)$
This implies statement-1 is also true.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2012 (07 May Online)
Let x and y be distinct integers where 1≤x≤25 and 1≤y≤25. Then, the number of ways of choosing x and y, such that x+y is divisible by 5 , is _____ .

Options:

The correct answer is:
120

Solution:
Given:

1≤x,y≤25

And, x,y are distinct integers.

Let

x+y=5k, where k∈N

So, we have

x	y	Number of ways

5λ i.e, 5, 10, 15, 20, 25	5λ i.e, 5, 10, 15, 20, 25	Since, x and y are distinct integers, so we cannot pair 5,5, 10,10,...25,25	20

5λ+1 i.e., 1, 6, 11, 16, 21	5λ+4 i.e., 4, 9, 14, 19, 14	25

5λ+2 i.e, 2,7, 12, 17, 22	5λ+3 i.e., 3, 8, 13, 18, 23	25

5λ+3 i.e., 3, 8, 13, 18, 23	5λ+2 i.e, 2, 7, 12, 17, 22	25

5λ+4 i.e., 4, 9, 14, 19, 14	5λ+1 i.e, 1, 6, 11, 16, 21	25

Total number of ways =120

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let $X$ and $Y$ be subsets of $R$, the set of all real numbers. The function $f: X \rightarrow Y$ defined by $f(x)=x^2$ for $x \in X$ is one-one but not onto if (Here $R^{+}$is the set of all positive real numbers)

Options:
A $X=Y=R^{+}$
B $X=R, Y=R^{+}$
C $X=R^{+}, Y=R$
D $X=Y=R$

The correct answer is:
$X=R^{+}, Y=R$

Solution:
$\begin{array}{l} f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1^2=x_2^2 \Rightarrow x_1=x_2,\left[\text { if } X=R^{+}\right] \\ \Rightarrow f \text { is one-one. Since } R_f=R^{+} \subseteq R=Y ; \therefore f \text { is not onto }\end{array}$

Topic: Mathematics, Functions, JEE Main
Let X be a binomially distributed random variable with mean 4 and variance 43. Then 54 PX≤2 is equal to

Options:
A 7327
B 14627
C 14681
D 12681

The correct answer is:
14627

Solution:
Mean np=4

Variance npq=43

i.e. n=6, p=23, q=13

Now 54PX≤2=54PX=2+PX=1+PX=0

=54C26232134+C16231135+C06230136

=5415×436+6×236+136=2×337336=14627

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Let X be a random variable having binomial distribution B7,p. If PX=3=5PX=4, then the sum of the mean and the variance of X is

Options:
A 10516
B 7736
C 3136
D 3536

The correct answer is:
7736

Solution:
Let probability of success is p and failure is q
Given n=7,

also , PX=3=5 PX=4 ⇒C37p3q4=5·C47p4q3

⇒q=5p ...i

We know that p+q=1⇒6p=1 (from i)

∴ p=16 and q=56

∴ Mean =np=7·16=76

and Variance =npq=7·16·56=3536

∴ Sum of mean and variance =76+3536 =7736

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Let X be a set containing 10 elements and PX be its power set. If A and B are picked up at random from PX, with replacement, then the probability that A and B have equal number of elements is:

Options:
A 210-1220
B 20C10220
C 20C10210
D 210-1210

The correct answer is:
20C10220

Solution:
The power set of X will contain 210 elements where each element is subset of A in which
10C0 will contain 0 element.
10C1 will contain 1 element.
10C2 will contain 2 elements.
⋮
⋮
10C10 will contain 10 elements.
So total numbers of ways in which we can select two sets with replacement is
And favorable cases would be 10C0⋅10C0+10C1.10C1+ ……+10C10.10C10=20C10
Hence, Probability would be = 20C10220

Topic: Mathematics, Probability, JEE Main, JEE Main 2015 (10 Apr Online)
Let x→ be a vector in the plane containing vectors a→=2i^-j^+k^ and b→=i^+2j^-k^. If the vector x→ is perpendicular to (3i^+2j^-k^) and its projection on a→ is 1762, then the value of x→2 is equal to _______.

Options:

The correct answer is:
486

Solution:
A plane containing two vectors can be expressed as a linear combination of the vectors.

Hence, let x→=λa→+μb→ (λ and μ are scalars).

⇒x→=i^(2λ+μ)+j^(2μ-λ)+k^(λ-μ)

Since x→ is perpendicular to 3i^+2j^-k^,

⇒x→·3i^+2j^-k^=0

⇒i^(2λ+μ)+j^(2μ-λ)+k^(λ-μ)·3i^+2j^-k^=0

⇒3(2λ+μ)+2(2μ-λ)-(λ-μ)=0

⇒3λ+8μ=0 ...1

Also, the projection of x→ on a→ is 1762

⇒x→·a→|a→|=1762

⇒i^(2λ+μ)+j^(2μ-λ)+k^(λ-μ)·2i^-j^+k^22+12+12=1762

⇒2(2λ+μ)-(2μ-λ)+(λ-μ)6=1762

⇒6λ-μ=51 ...2

On solving the equations (1) and (2), we get λ=8, μ=-3.

Thus, x→=13i^-14j^+11k^

⇒x→=132+-142+112

⇒|x→|2=486.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let x be the greatest integer less than or equal to x, for a real number x. Then the equation x2=x+1 has

Options:
A two solutions
B one solution
C no solution
D more than two solutions

The correct answer is:
no solution

Solution:
From the graph it is clear that the equation has solution

Topic: Mathematics, Functions, JEE Main
Let ϕ(x) be the inverse of the function f(x) and f′x=11+x5, then d dx ϕ x is equal to

Options:
A 11+ϕx5
B 11+fx5
C 1+ϕx5
D 1+f(x)

The correct answer is:
1+ϕx5

Solution:
We have, Ïx=f-1(x)
âx=f[Ïx]
On differentiating both sides w.r.t. x, we get
1=f'Ïx.Ï'(x)
âÏ'x=1f'Ïx ...(i)
Since, fâ²x=11+x5 (given)
f'Ïx=11+Ïx5
From Equation (i),
Ï'x=1f'[Ïx]=1+Ïx5

Topic: Mathematics, Differentiation, JEE Main
Let x denote the fractional part of x and fx=cos−11−x2sin−11−xx−x3,x≠0. If L and R respectively denotes the left hand limit and the right hand limit of fx at x=0, then 32π2L2+R2 is equal to __________.

Options:

The correct answer is:
18

Solution:
Given: fx=cos−11−x2sin−11−xx−x3

⇒RHL=limx→0+fx=limh→0f0+h

⇒RHL=limh→0fh

⇒RHL=limh→0cos−11−h2sin−11−hh1−h2

⇒RHL=limh→0cos−11−h2hsin−111

Let cos−11−h2=θ⇒cosθ=1−h2

⇒RHL=π2limθ→0θ1−cosθ

⇒RHL=π2limθ→011−cosθθ2

⇒RHL=π2112

⇒RHL=π2

⇒R=π2

Now finding left hand limit,

⇒LHL=limx→0−fx

⇒LHL=limh→0f−h

⇒LHL=limh→0cos−11−−h2sin−11−−h−h−−h3

⇒LHL=limh→0cos−11−−h+12sin−11−−h+1−h+1−−h+13

⇒LHL=limh→0cos−1−h2+2hsin−1h1−h1−1−h2

⇒LHL=limh→0π2sin−1h1−1−h2

⇒LHL=π2limh→0sin−1h−h2+2h

⇒LHL=π2limh→0sin−1hh1−h+2

⇒LHL=π4

⇒L=π4

⇒32π2L2+R2=32π2π22+π216

⇒32π2L2+R2=16+2

⇒32π2L2+R2=18

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Let x denote the greatest integer function and fx=max1+x+x, 2+x, x+2x, 0≤x≤2, where f is not continuous and n be the number of points in 0, 2, where f is not differentiable. Then m+n2+2 is equal to

Options:
A 2
B 11
C 6
D 3

The correct answer is:
3

Solution:
Given,

fx=max1+x+x, 2+x, x+2x, 0≤x≤2

⇒fx=maxx+1, x+2, x0≤x<1maxx+2, x+2, x+21≤x<2max5, 4, 6x=2

⇒fx=x+2,0≤x<1x+2,1≤x<26,x=2

⇒f(x)=x+20≤x<26x=2

So, f is not continuous at x=2 as f2-=4 & f2=6

And f is differentiable in 0, 2 as x+2 is a linear function,

∴ m=1, n=0

⇒m+n2+2=1+2=3

Hence this is the required option.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let $[\mathrm{x}]$ denote the greatest integer less than or equal to $\mathrm{X}$. Then :

$\lim _{x \rightarrow 0} \frac{\tan \left(\pi \sin ^{2} x\right)+(|\mathrm{x}|-\sin (x[x]))^{2}}{x^{2}}$

Options:
A does not exist
B equals $\pi$
C equals $\pi+1$
D equals 0

The correct answer is:
does not exist

Solution:
RHL is, $\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+(x-0)^{2}}{x^{2}}$

$=\lim _{x \rightarrow 0^{+}}\left(\frac{\tan \left(\pi \sin ^{2} x\right.}{x^{2}}+1\right)=1+\pi$

And LHL is, $\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+(-x+\sin x)^{2}}{x^{2}}$

$=\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+x^{2}+\sin ^{2} x-2 x \sin x}{x^{2}}$

$=\pi+1+1-2=\pi$

Since, LHL $\neq$ RHL Hence, limit does not exist.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let x denote the greatest integer less than or equal to x. Then, the values of x∈R satisfying the equation ex2+ex+1-3=0 lie in the interval:

Options:
A 0, 1e
B loge2, loge3
C 1, e
D 0, loge2

The correct answer is:
0, loge2

Solution:
Given equation is ex2+ex+1-3=0

We know that x+I=x+I, I∈Z, where ·, represents the greatest integer function.

⇒ex2+ex+1-3=0

Let, ex=t

⇒t2+t-2=0

⇒t+2t-1=0

⇒t=-2, 1

But ex=-2, (Not possible because we know that ex>0)

Hence, ex=1 and also, we know that if x=1, ⇒1≤x<2

∴ 1≤ex<2

And, now by using the definition of logarithm, we have, if ex=a, then logea=x

⇒ loge1≤x<loge2

⇒ 0≤x<loge2

⇒ x∈0, loge2.

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let $[x]$ denote the greatest integer not more than $x$. If $A$ and $B$ are the domains of the functions $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ and $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ respectively, then

Options:
A $A \cup B=R$
B $A \cap B=\phi$
C $A-B=(-\infty, 0)$
D $B-A=(0, \infty)$

The correct answer is:
$A \cap B=\phi$

Solution:
Given function, $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ will define, if
$|x|>x \Rightarrow x < 0 \Rightarrow x \in(-\infty, 0)$ $\ldots(i)$
and the function, $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ will define, if
$|x|+x>0 \Rightarrow x>0 \Rightarrow x \in(0, \infty)$ $\ldots$ (ii)
$\therefore \quad A=\{x \mid x \in \mathbf{R}$ and $x < 0\}$
and $B=\{x \mid x \in \mathbf{R}$ and $x>0\}$
$\therefore \quad A \cap B=\phi$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 1)
Let [x] denote the greatest integer ≤x. Consider the function f(x)=maxx2,1+[x]. Then the value of the integral ∫02f(x)dx is :

Options:
A 5+423
B 8+423
C 1+523
D 4+523

The correct answer is:
5+423

Solution:
Plotting the diagram of the given function f(x)=maxx2,1+[x] we get,

Now when x∈[0,1)⇒fx=1

When x∈[1,2)⇒fx=2

And when x∈2,2⇒fx=x2

So, from above diagram area can be calculated as,

A=∫011.dx+∫122dx+∫22x2dx

⇒A=1+22-2+83-223

⇒A=53+423

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let x denote the greatest integer ≤x, where x∈R. If the domain of the real valued function fx=x-2x-3 is -∞,a∪b,c∪4,∞,a<b<c, then the value of a+b+c is:

Options:
A 8
B 1
C -2
D -3

The correct answer is:
-2

Solution:
We have,

fx=x-2x-3

For domain,

x-2x-3≥0

Case I: When |[x]|-2≥0 and x-3>0, then

∴ x∈-∞,-3∪4,∞ ...1

Case II: When x-2≤0 and x-3<0, then

∴ x∈-2,3 ...2

So, from 1 and 2, we get

Domain of function is

-∞,-3∪-2,3∪4,∞

∴ a+b+c=-3+-2+3=-2

∵ a<b<c

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions from the set A to the set A×B. Then :

Options:
A y=273x
B 2y=273x
C 2y=91x
D y=91x

The correct answer is:
2y=91x

Solution:
Given, nA=3,nB=5

⇒nA×B=15

x=C35×3!=5×4×3=60

y=C315×3!=15×14×13=30×91 2y=60×91

∴ 2y=91x

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Let $[x]$ denotes the greatest integer less than or equal to $x$. If $f(x)=[x \sin \pi x]$, then $f(x)$ is

Options:
A Continuous at $x=0$
B Continuous in $(-1,0)$
C Differentiable in $(-1,1)$
D All the above

The correct answer is:
All the above

Solution:
Here, when $-1 \leq x \leq 1,0 \leq x \sin \pi x \lt 1$
$\Rightarrow f(x)=[x \sin \pi x]=0 \text { for }-1 \leq x \leq 1 \text {, }$
i.e., $f(x)$ is constant function (equal to zero) in $[-1,1]$. $\Rightarrow f(x)$ is differentiable in $(-1,1)$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let xk+yk=ak,a,k>0 and dydx+yx13=0, then k is

Options:
A 32
B 43
C 23
D 13

The correct answer is:
23

Solution:
Differentiating the given equation with respect to x

k.xk-1+k.yk-1dydx=0

⇒dydx=-xyk-1

⇒dydx+xyk-1=0

⇒k-1=-13

⇒k=1-13=23

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Let x¯, M and σ2 be respectively the mean, mode and variance of n observations x1, x2,...., xn and di=-xi-a, i=1, 2,...., n, where a is any number.

Statement I: Variance of d1, d2,..., dn is σ 2 .

Statement II: Mean and mode of d1, d2,...., dn are -x¯-a and -M-a, respectively.

Options:
A Statement I and Statement II are both true
B Statement I and Statement II are both false
C Statement I is true and Statement II is false
D Statement I is false and Statement II is true

The correct answer is:
Statement I and Statement II are both true

Solution:
Mean of d1, d2, d3... dn

=∑i=1n -xi-a n

=-∑i=1n xi n+ -a n∑i=1n1

=-x+ -a nn

=-x→-a

And, the mode of d1, d2, d3... dn will be -M-a

Also, we know that the variance σ2 is unaffected by change of origin, hence it remains same.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2014 (19 Apr Online)
Let x=mn(m,n are co-prime natural numbers) be a solution of the equation cos2sin-1x=19 and let α,β(α>β) be the roots of the equation mx2-nx-m+n=0. Then the point (α,β) lies on the line

Options:
A 3x+2y=2
B 5x-8y=-9
C 3x-2y=-2
D 5x+8y=9

The correct answer is:
5x+8y=9

Solution:
Given: cos2sin-1x=19

Let, sin-1x=θ

⇒cos2θ=19

⇒1-2sin2θ=19

⇒89=2sin2θ

⇒sinθ=±23

⇒x=±23

It is given that m and n are co-prime natural numbers,

⇒m=2, n=3

Also, mx2-nx-m+n=0.

⇒2x2-3x-2+3=0

⇒2x2-3x+1=0

⇒2x2-2x-x+1=0

⇒2xx-1-x-1=0

⇒2x-1x-1=0

⇒x=1,12

⇒α=1, β=12

Out of the given options, 1,12 lines on 5x+8y=9.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let $X_{n}=\{1,2,3, \ldots . n)$ and let a subset $A$ of $X_{n}$ be chosen so that every pair of elements of A differ by at least 3. (For example, if $\mathrm{n}=5, \mathrm{~A}$ can be $\varnothing,\{2\}$ or $\{1,5\}$ among others). When $n=10$, let the probability that $1 \in A$ be p and let the probability that $2 \in A$ be $\mathrm{q}$. Then

Options:
A $p>q$ and $p-q=\frac{1}{6}$
B $p < q$ and $q-p=\frac{1}{6}$
C $p>q$ and $p-q=\frac{1}{10}$
D $p < q$ and $q-p=\frac{1}{10}$

The correct answer is:
$p>q$ and $p-q=\frac{1}{10}$

Solution:
When $\mathrm{n}=10$, Let $\mathrm{A}_{\mathrm{r}}$ be no. of ways of selecting $\mathrm{r}$ numbers
No. of selection of $\mathrm{A}$ is
$=\mathrm{n}\left(\mathrm{A}_{0}\right)+\mathrm{n}\left(\mathrm{A}_{1}\right)+\mathrm{n}\left(\mathrm{A}_{2}\right)+\mathrm{n}\left(\mathrm{A}_{3}\right)+\mathrm{n}\left(\mathrm{A}_{4}\right)$ $=1+10+(7+6+5+\ldots+1)+\{(4+3+2+1)+(3+2+1)+(2+1)+1\}+1$ $=11+\frac{7.8}{2}+10+6+3+1+1=60$
$\mathrm{N}(\mathrm{p})=\mathrm{n}$ (no. of ways 1 is selected $)=1+7+4+3+2+1+1=19$
$\mathrm{N}(\mathrm{q})=\mathrm{n}$ (no. of ways 2 is selected $)=1+6+3+2+1=13$
So. $\mathrm{p}=\frac{19}{60}$ and $\mathrm{q}=\frac{13}{60} \quad \Rightarrow \mathrm{p}-\mathrm{q}=\frac{1}{10}$

Topic: Mathematics, Permutation Combination, JEE Main
Let Xn denote the mean of first n natural numbers, then the mean of X1,X2,…,X100 is

Options:
A 25
B 50
C 25.5
D 25.75

The correct answer is:
25.75

Solution:
Xn=∑r=1nrn=nn+12n=n+12
∑n=1100Xnn=100×1012+100200
=1014+12=25.75

Topic: Mathematics, Statistics, JEE Main
Let x=x(t) and y=y(t) be solutions of the differential equations dxdt+ax=0 and dydt+by=0 respectively, a,b∈R. Given that x(0)=2 ; y(0)=1 and 3 y(1)=2 x(1), the value of t, for which x(t)=y(t), is :

Options:
A log232
B log43
C log34
D log432

The correct answer is:
log432

Solution:
Given: dxdt+ax=0

⇒dxx=-adt

Integrating both sides,

⇒∫dxx=-a∫dt

⇒log|x|=-at+c

At t=0,x=2

⇒log2=0+c

⇒logx=-at+log2

⇒logx-log2=-at

⇒logx2=-at

⇒x2=e-at

⇒x=2e-at ...(i)

It is also given that, dydt+by=0

⇒dyy=-bdt

⇒log|y|=-bt+λ

At, t=0,y=1

⇒0=0+λ

⇒y=e-bt ...(ii)

According to question,

3 y(1)=2 x(1)

⇒3e-b=22e-a

⇒ea-b=43

For xt=yt

⇒2e-at=e-bt

⇒2=e(a-b)t

⇒2=43t

⇒log2=t×log43

⇒log432=t

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Let x, y and z be the respective sum of the first n terms, the next n terms and the next n terms of a geometric progression, then x, y, z are in

Options:
A arithmetic progression
B geometric progression
C harmonic progression
D None of these

The correct answer is:
geometric progression

Solution:
Let A be the 1st term and R be the common ratio of the given G.P.
Sum of the first n terms of the G.P. is x =A(1-Rn)1-R. The next n terms form a G.P. series of n terms with the first term as the n+1th term of the given G.P. is given by tn+1=ARn . Sum of the next n terms of the G.P. is y â y =ARn(1-Rn)1-R and
z=AR2n1-Rn1-R â y2=zx
â´ x, y, z are in G.P.

Topic: Mathematics, Sequences and Series, JEE Main
Let $(x, y)$ be a variable point on the curve $4 x^{2}+9 y^{2}-8 x-36 y+15=0$. Then $\min \left(x^{2}-2 x+y^{2}-4 y+5\right)+\max \left(x^{2}-2 x+y^{2}-4 y+5\right)$ is -

Options:
A $\frac{325}{36}$
B $\frac{36}{325}$
C $\frac{13}{25}$
D $\frac{25}{13}$

The correct answer is:
$\frac{325}{36}$

Solution:
$\begin{array}{l}
4 x^{2}+9 y^{2}-8 x-36 y+15=0 \\
4\left(x^{2}-2 x\right)+9\left(y^{2}-4 y\right)=-15 \\
4\left(x^{2}-2 x+1\right)+9\left(y^{2}-4 y+4=-15+4+36\right. \\
4(x-1)^{2}+9(y-2)^{2}=25 \\
\frac{(x-1)^{2}}{\left(\frac{5}{2}\right)^{2}}+\frac{(y-2)^{2}}{\left(\frac{5}{3}\right)^{2}}=1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .(1) \\
x^{2}-2 x+y^{2}-4 y+5 \\
(x-1)^{2}+(y-2)^{2}
\end{array}$
$\min$ of $\left((x-1)^{2}+(y-2)^{2}\right)=\frac{25}{9}$
$\max$ of $\left((x-1)^{2}+(y-2)^{2}\right)=\frac{25}{4}$
$=\frac{25}{9}+\frac{25}{4}=\frac{325}{36}$

Topic: Mathematics, Ellipse, JEE Main
Let $x, y$ be positive real numbers and $m, n$ positive integers.

The maximum value of the expression $\frac{x^{\mathrm{m}} y^{\mathrm{n}}}{\left(1+x^{2 \mathrm{~m}}\right)\left(1+y^{2 \mathrm{n}}\right)}$

is :

Options:
A 1
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{m+n}{6 m n}$

The correct answer is:
$\frac{1}{4}$

Solution:
$A=\frac{x^{m} y^{n}}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)}=\frac{1}{\left(x^{-m}+x^{m}\right)\left(y^{-n}+y^{n}\right)}$

$\frac{x^{m}+y^{-m}}{2} \geq\left(x^{n+1}, x^{-m}\right)^{\frac{1}{2}} \Rightarrow x^{n+}+x^{-n} \geq 2$

In the same way, $y^{n}+y^{n} \geq 2$

Then $,\left(x^{n}+x^{-n}\right)\left(y^{-n}+y^{n}\right) \geq 4$

$\Rightarrow \frac{1}{\left(x^{\text {"1 }}+x^{-m}\right)\left(y^{-n}+y^{n}\right)} \leq \frac{1}{4}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let $\alpha|x|=|y| \mathrm{e}^{x y-\beta}, \alpha, \beta \in \mathbf{N}$ be the solution of the differential equation $x \mathrm{~d} y-y \mathrm{~d} x+x y(x \mathrm{~d} y+y \mathrm{~d} x)=0$, $y(1)=2$. Then $\alpha+\beta$ is equal to ________

Options:

The correct answer is:
4

Solution:
$\begin{aligned}
& a|x|=|y| e^{y x-\beta}, a, b \in N \\
& x d y-y d x+x y(x d y+y d x)=0 \\
& \frac{d y}{y}-\frac{d x}{x}+(x d y+y d x)=0 \\
& \ell n|y|-\ell n|x|+x y=c \\
& y(1)=2 \\
& \ell n|2|-0+2=c \\
& c=2+\ell n 2 \\
& \ell n|y|-\ell n|x|+x y=2+\ell n 2 \\
& \ell n|x|=\ell n\left|\frac{y}{2}\right|-2+x y \\
& |x|=\left|\frac{y}{2}\right| e^{x y-2} \\
& 2|x|=|y| e^{x y-2} \\
& \alpha=2 \quad \beta=2 \quad \alpha+\beta=4
\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let x×y=x2+y3 and x×1×1=x ×1×1. Then a value of 2sin-1x4+x2-2x4+x2+2 is

Options:
A π4
B π3
C π6
D π

The correct answer is:
π3

Solution:
Given x×y=x2+y3

Now x×1× 1=x×1×1

x2+13×1=x×12+13

x2+12+13=x2+23

x4+1+2x2+1=x2+8

x4+x2−6=0

Let x2=t

t2+t-6=0

t+3t−2=0

t≠−3 so t=2 or x2=2

Now 2sin-1x4+x2-2x4+x2+2=2sin-14+2-24+2+2
=2sin-112=π6×2=π3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Let x,y,z>1 and A=1logxylogxzlogyx2logyzlogzxlogzy3. Then adjadjA2 is equal to

Options:
A 64
B 28
C 48
D 24

The correct answer is:
28

Solution:
Given:

A=1logxylogxzlogyx2logyzlogzxlogzy3

⇒A=1logylogxlogzlogxlogxlogy2logzlogylogxlogzlogylogz3

⇒A=1logxlogylogzlogxlogylogzlogx2logylogzlogxlogy3logz

⇒A=1logx·logy·logzlogxlogylogzlogx2logylogzlogxlogy3logz

⇒A=logx·logy·logzlogx·logy·logz111121113

⇒A=100110102

⇒A=2

Now,

adjadjA2=adjA22=A24=A8=28

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let αx+βy+γz=1 be the equation of a plane passing through the point 3,-2,5 and perpendicular to the line joining the points 1,2,3 and -2,3,5. Then the value of α β y is equal to _____ .

Options:

The correct answer is:
6

Solution:
Given,

αx+βy+γz=1 be the equation of a plane passing through the point 3,-2,5 and perpendicular to the line joining the points 1,2,3 and -2,3,5,

Now, normal vector of plane =1--2, 2-3, 3-5≡3,-1,-2

So, the equation of Plane will be, 3x-y-2z+λ=0

Now given, point 3,-2,5 satisfies the plane

3×3+2-10+λ=0

⇒λ=-1

Hence, the equation of plane will be, 3x-y-2z=1

Now comparing with given equation of plane, we get α β y=6

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let x, y, z be positive real numbers such that x+y+z=12 and x3y4z5=0.16003. Then x3+y3+z3 is equal to

Options:
A 342
B 216
C 258
D 270

The correct answer is:
216

Solution:
x+y+z=12

A.M.≥G.M.

3x3+4y4+5z512 ≥ x33 y44 z5512

1≥1

⇒A.M.=G.M.

⇒All the numbers are equal.

∴x3=y4=z5=k

x=3k; y=4k; z=5k

x+y+z=12

⇒3k+4k+5k=12

⇒k=1

∴ x=3; y=4; z=5

Therefore, x3+y3+z3=216.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2016 (09 Apr Online)
Let $x, y, z$ be positive reals. Which of the following implies $x=y=z$ ?

Options:
A $x^{3}+y^{3}+z^{3}=3 x y z$
B $x^{3}+y^{2} z+y z^{2}=3 x y z$
C $x^{3}+y^{2} z+z^{2} x=3 x y z$
D $\quad(x+y+z)^{3}=27 x y z$

The correct answer is:
$x^{3}+y^{2} z+y z^{2}=3 x y z$

Solution:
For option (3) if $x=z=1$ and $y=2$ then option (3) is right.
So by option (3) we can't say $x=y=z$. Remaining options implies $x=y=z$

Topic: Mathematics, Basic of Mathematics, KVPY, KVPY 2015 (SA)
Let $x, y, z$ be three non-negative integers such that $x+y+z=10$. The maximum possible value of $x y z+$ $x y+y z+z x$ is

Options:
A $52$
B $64$
C $69$
D $73$

The correct answer is:
$69$

Solution:
Taking three no's.
$x+1, y+1, z+1$
$\mathrm{AM} \geq \mathrm{GM}$.
$\frac{(x+1)+(y+1)+(z+1)}{3} \geq\{(x+1)(y+1)(z+1)\}^{1 / 3}$
$\left(\frac{13}{3}\right)^{3} \geq x y z+x y+y z+z x+11$
$\left(\frac{13}{3}\right)^{3}-11 \geq x y z+x y+y z+z x$
equality hold when $x=y=z$ but $x+y+z=0$ and $x, y, z$ are integers.
So maximum value when any two of $x, y, z$ are equal to $3$ and third is equal to $4$.

Topic: Mathematics, Sequences and Series, JEE Main
Let \( x_{1}, x_{2}, \ldots \ldots \ldots \ldots x_{100} \) are \( 100 \) observations such that \( \sum x_{i}=0, \sum_{1 \leq i < j \leq 100}\left|x_{i} x_{j}\right|=80000 \& \) mean deviation from their mean is \( 5 \), then their standard deviation is

Options:

The correct answer is:
30

Solution:
x-=∑xi100=0 &
∑xi-x-100=5⇒∑xi=500
⇒∑xi2+2 ∑1≤i<j≤100xixj=5002
⇒∑xi2 100=5002-2∑xixj100=2500-1600
S.D. =∑xi-x-2100=900=30

Topic: Mathematics, Statistics, JEE Main