datavorous/entrance-exam-dataset huggingface dataset:
If the value of the integral ∫012x21-x232dx is k6, then k is equal to:

Options:
A 23+π
B 23-π
C 32+π
D 32-π

The correct answer is:
23-π

Solution:
k6=∫012x21-x232dx

Let x=sinθ⇒dx=cosθdθ

⇒k6=∫0π6sin2 θ1-sin2 θ32cosθdθ⇒k6=∫0π6sin2 θcos3 θcosθdθ

⇒k6=∫0π6tan2θdθ=∫0π6sec2θ-1dθ⇒k6=tanθ-θ0π6=13-π6=23-π6

⇒k=23-π

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If the value of the integral Î=â«01dxx+1-x2 is equal to Ïk, then the value of k is equal to

Options:

The correct answer is:
4

Solution:
Let, x=sin⁡θ, ∀θ∈0,π2
⇒dx=cos⁡θdθ
So, Ι=∫0π/2cos⁡θdθsin⁡θ+cos⁡θ
Applying a+b-x property and adding, we get
2Ι=∫0π/2cos⁡θ+sin⁡θsin⁡θ+cos⁡θdθ
=∫0π/21dθ=θ0π2
⇒2Ι=π2
⇒Ι=π4∴k=4

Topic: Mathematics, Definite Integration, JEE Main
If the value of the integral ∫0π2cotθ21-cos4θdθ is equal to λ, then the value of 3λ is equal to

Options:

The correct answer is:
6.25

Solution:
Let I=∫0π2cotθ21-cosθ1+cosθ+cos2θ+cos3θdθ
So, I=∫0π2cosθ2sinθ2·2sin2θ21+cosθ+cos2⁡θ+cos3⁡θdθ

=∫0π2sin⁡θ1+cosθ+cos2θ+cos3θdθ
Let cos⁡θ=t

⇒I=∫011+t+t2+t3dt=1+12+13+14=2512

Hence, λ=2512⇒3λ=254=6.25

Topic: Mathematics, Definite Integration, JEE Main
If the value of the integral $\int_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$ is $\frac{2}{\pi}$. Then, a value of $\alpha$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $\frac{\pi}{4}$
D $\frac{\pi}{2}$

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

Solution:
Let $I=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^x} d x$ ...(I)
$I=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^{-x}} d x$
$\left(u \operatorname{sing} \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right)$ ...(II)
Add (1) and (II)
$\begin{aligned}
& 2 I=\int_{-1}^{+1} \cos (\alpha x) d x=2 \int_0^1 \cos (\alpha x) d x \\
& I=\frac{\sin \alpha}{\alpha}=\frac{2}{\pi}(\text { given }) \\
& \therefore \alpha=\frac{\pi}{2}
\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (04 Apr Shift 2)
If the value of the integral∫-π2π2x2cosx1+πx+1+sin2x1+esinx2023dx=π4π+a-2, then the value of a is

Options:
A 3
B -32
C 2
D 32

The correct answer is:
3

Solution:
Given,

∫-π2π2x2cosx1+πx+1+sin2x1+esin2023xdx=π4π+α-4

Now let, I1=∫-π2π2x2cosx1+πxdx .......i

Now using the property ∫abfxdx=∫abfa+b-xdx we get,

⇒I1=∫-π2π2x2cosx1+π-xdx ........ii

Now adding above equations we get,

⇒2I1=2∫0π2x2cosxdx as ∫-aafxdx=2∫0afxdx, if fx is even

⇒I1=x2sinx0π2-∫0π22x·sinxdx

⇒I1=π24--2xcosx0π2+∫0π22cosxdx

⇒I1=π24-2

Similarly solving, I2=∫-π2π21+sin2x1+esin2023xdx

⇒I2=∫0π21+sin2xdx

⇒I2=∫0π21+1-cos2x2dx

⇒I2=12∫0π23-cos2xdx

⇒I2=123x-sin2x20π2

⇒I2=3π4

⇒I1+I2=π24-2+3π4

⇒π4π+3-2=π4π+α-2

⇒α=3

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If the value of the integral Ι=∫π4π3maxsinx,tan⁡xdx is equal to ln⁡k, then the value of k2 is equal to

Options:

The correct answer is:
2

Solution:
For x∈π4,π3, tan⁡x>sin⁡x (as cos⁡x<1 )
∴maxtan⁡x,sin⁡x=tan⁡x
Thus, Ι=∫π4π3tan⁡xdx
=-ln⁡cos⁡xπ4π3
=-ln⁡12+ln⁡12
=ln⁡2

Topic: Mathematics, Definite Integration, JEE Main
If the value of the integral I=∫02πsgnexdx is equal to kπ, then the smallest prime number greater than 2k is (where, sgnx represents the signum function of x)

Options:
A 3
B 5
C 7
D 11

The correct answer is:
5

Solution:
As ex>0 ∀x∈R
∴sgnex=1
⇒I=∫02π1dx=2π
⇒k=2
∴2k=4
Hence, the required prime number =5

Topic: Mathematics, Definite Integration, JEE Main
If the value of the limit limn→∞⁡110+210+…+n10n11 is equal to K, then the value of 12K is equal to (where, . represents the greatest integer function)

Options:
A 4
B 5
C 10
D 11

The correct answer is:
5

Solution:
The given limit is
K=limn→∞∑r=1nr10n11
=limn→∞∑r=1nrn101n
=∫01x10dx=x111110
=111
⇒12K=112
⇒12K=112=5.5=5

Topic: Mathematics, Definite Integration, JEE Main
If the value of the sum 29C300+28C301+27C302+.â¦â¦.+1C3028+0âC3029-C3030 is equal to Kâ232, then the value of K is equal to

Options:
A 7
B 14
C 52
D 72

The correct answer is:
72

Solution:
S=29C300+28C301+27C302+.…….+1C3028+0⋅C3029-C3030
Also,
S=-C300+0⋅C301+1C302+.…….+27C3028+28C3029+29C3030
Adding we get,
2S=28C300+C301+C302+.…..+C3028+C3029+C3030
⇒S=14⋅230=144232=72⋅232
⇒K=72

Topic: Mathematics, Binomial Theorem, JEE Main
If the values $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots \ldots, \frac{1}{n}$ occur at frequencies 1 , $2,3,4,5, \ldots, n$ in a distribution, then the mean is

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

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

Solution:
Mean $=\frac{1 \cdot 1+\frac{1}{2} \cdot 2+\frac{1}{3} \cdot 3+\frac{1}{4} \cdot 4+\frac{1}{5} \cdot 5+\ldots . .+\frac{1}{n} n}{1+2+3+\ldots . .+n}$
$=\frac{1+1+1+1+\ldots .+1}{\frac{n(n+1)}{2}}=\frac{n}{\frac{n(n+1)}{2}}=\frac{2}{n+1}$.

Topic: Mathematics, Statistics, JEE Main
If the values of $\Lambda_{\infty}$ of $\mathrm{NH}_4 \mathrm{Cl}, \mathrm{NaOH}$ and $\mathrm{NaCl}$ are 130, 217 and $109 \mathrm{ohm}^{-1} \mathrm{~cm}^2$ equiv $^{-1}$ respectively, the $\Lambda_{\infty}$ of $\mathrm{NH}_4 \mathrm{OH}$ in ohm ${ }^{-1} \mathrm{~cm}^2$ equiv $^{-1}$ is

Options:
A 238
B 196
C 22
D 456

The correct answer is:
238

Solution:
On adding Eq. (i) and (ii) and subtracting the Eq. (iii)
$\begin{aligned}
\Lambda_{\infty} \mathrm{NH}_4^{+}+\Lambda_{\infty} \mathrm{OH}^{-} & =130+217-109 \\
\Lambda_{\infty} \mathrm{NH}_4 \mathrm{OH} & =347-109 \\
& =238 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \text { equiv }^{-1}
\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main
If the Vant Hoff factor for 0.1MBaNO32 solution is 2.74, the degree of dissociation is

Options:
A 0.87
B 0.74
C 0.91
D 87

The correct answer is:
0.87

Solution:
BaNO32⇌Ba2++2NO3-;n=3
α=i-1n-1=2.74-13-1=0.87

Topic: Chemistry, Solutions, JEE Main
If the variable line 3x+4y=α lies between the two circles (x-1)2+(y-1)2=1 and (x-9)2+(y-1)2=4, without intercepting a chord on either circle, then the sum of all the integral values of α is

Options:

The correct answer is:
165

Solution:
Given line 3x + 4y = α

Given circles (x-1)2+(y-1)2=1 & (x-9)2+(y-1)2=4

According to the given information centres of both the circles must lie on opposite side of line.

L11 . L22 < 0

⇒ 3 +4 -α 27 + 4 - α <0

α ∈ 7, 31 .

Line is neither touching nor intersecting that means perpendicular distance from centre to line must more than or equal to radius.

⇒ 3+4-α5 ≥ 1 & 27+4-α5 ≥ 2

⇒ α ∈ (-∞, 2] ∪ [12,∞) & α ∈ (-∞, 21] ∪ [41,∞)

⇒ α ∈ 12, 21

Sum of integer values of α = 12 + 13+ --21=165.

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If the variance σ2 of the data xi0156101217fi3232633 is k then the value of k is ______ {where . denotes the greatest integer funciton}

Options:

The correct answer is:
29

Solution:
Finding the fixi & fixi2 we get,

Now, from above table we get, ∑fixi=176

Now, finding x¯=∑fixi∑fi=17622=8

Now, finding variance, we get,

σ2=1N∑fixi-2(x¯)2

⇒σ2=122×2048-(8)2

⇒σ2= 93.090964

⇒σ2= 29.0909

⇒σ2= 29.0909=29

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If the variance of 10 natural numbers 1, 1, 1, …, 1, k is less than 10, then the maximum possible value of k is ___________.

Options:

The correct answer is:
11

Solution:
σ2=Σx2n-Σxn2

=9+k210-9+k102<10

90+10k2-81-k2-18k<1000

9k2-18k-991<0

k2-2k<9919

k-12<10009

-10103<k-1<10103

k<10103+1

k≤11

Maximum value of k is 11.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If the variance of first n even natural numbers is 133, then the value of n is equal to

Options:
A 19
B 24
C 21
D 20

The correct answer is:
20

Solution:
Var2,4,6.........2n=133
⇒4Var1,2,3.........n=133
⇒Var1,2,3.........n=1334
⇒12+22+.......n2n-1+2+3+.......nn2=1334
⇒n+12n+16-n+124=1334
⇒n+122n+13-n+12=1334
⇒n+12n-16=1334
⇒n2-1=399⇒n=20

Topic: Mathematics, Statistics, JEE Main
If the variance of observations $x_1, x_2, \ldots \ldots . x_n$ is $\sigma^2$, then the variance of $a x_1, a x_2 \ldots \ldots ., a x_n, \alpha \neq 0$ is

Options:
A $\sigma^2$
B $a \sigma^2$
C $a^2 \sigma^2$
D $\frac{\sigma^2}{a^2}$

The correct answer is:
$a^2 \sigma^2$

Solution:
Let $\bar{x}$ be the mean of $x_1, x_2, x_3, \cdots, x_n$ and a be a nonzero real number.
Then, $\bar{x}=\frac{1}{n}\left(x_1+x_2+x_3+\cdots+x_n\right)$
Let $y_i=a x_i$ for each $i=1,2,3, \ldots, n$. Then,
$\begin{aligned} & \bar{y}=\frac{1}{n}\left(y_1+y_2+y_3+\cdots+y_n\right) \\ & =\frac{1}{n}\left(a x_1+a x_2+a x_3+\cdots+a x_n\right)=\text { a. } \frac{1}{n}\left(x_1+x_2+x_3+\cdots x_n\right)=a \bar{x}\end{aligned}$
Thus, $\bar{y}=a \bar{x}$
Now, the variance of new observations is given by
variance $(\mathrm{y})=\sigma_1^2$
$=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2$
$=\frac{1}{n} \sum_{i=1}^n\left(a x_i-a \bar{x}\right)^2\left[\because \quad y_i=a x_i\right.$ for each $i$ and $\left.\bar{y}=a \bar{x}\right]$
$=a^2 \cdot \frac{1}{n} \cdot \sum_{\mathrm{i}=1}^{\mathrm{n}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2$
$\begin{aligned} & =\mathrm{a}^2 .\{\text { variance }(\mathrm{x})\}=\mathrm{a}^2 \sigma^2 \\ & \therefore \text { new variance }=\mathrm{a}^2 \sigma^2 \\ & \text { Remark } \sigma_1=\sqrt{\mathrm{a}^2 \sigma^2}=|\mathrm{a}| \cdot \sigma\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If the variance of the data $2,4,5,6,8,17$ is 23.33 , then the variance of $4,8,10,12,16,34$ will be

Options:
A 93.32
B 25.33
C 23.33
D 48.66

The correct answer is:
93.32

Solution:
We find that each element of new data is 2 times of each element in old data.
Variance of old data $=23.33$
$\therefore$ variance of new data $=(23.33)(2)^2=93.32$

Topic: Mathematics, Statistics, JEE Main
If the variance of the data 12,14,18,19,21,36 is λ, then the value of 3λ is equal to

Options:

The correct answer is:
181

Solution:
x-=1206=20
The square of deviations about mean are 64,36,4,1,1,256
Variance, Ï2=3626=1813=Î»

Hence, 3Î»=181

Topic: Mathematics, Statistics, JEE Main
If the variance of the distribution
\begin{array}{cccccccc}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 4 & 8 & 11 & 17 & 20 & 24 & 32 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1
\end{array}
is 45.8 , then the variance of the distribution.
\begin{array}{llllllll}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 10 & 18 & 24 & 36 & 42 & 50 & 66 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1 \\
\hline
\end{array}

Options:
A 93.6
B $\sqrt{93.9}$
C 183.2
D $\sqrt{183.2}$

The correct answer is:
183.2

Solution:
Mean of given observation $\left(x_i\right)$ is
$$
\begin{array}{r}
\frac{(4 \times 3)+(8 \times 5)+(11 \times 9)+(17 \times 5)+(20 \times 4)}{+(24 \times 3)+(32 \times 1)} \\
=\frac{420}{30}=14
\end{array}
$$
So,
$$
\begin{aligned}
\mathrm{ce} & =k \frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30} \\
& =45.8 \text { (given) }
\end{aligned}
$$
Now, mean of given observation $\left(y_i\right)$ is
$$
\begin{array}{r}
\begin{array}{r}
(10 \times 3)+(18 \times 5)+(24 \times 9)+(36 \times 5)+(42 \times 4) \\
+(50 \times 3)+(66 \times 1)
\end{array} \\
\begin{aligned}
30 & \frac{900}{30}=30
\end{aligned}
\end{array}
$$
So, variance
$$
\begin{aligned}
& =k \frac{20^2+12^2+6^2+6^2+12^2+20^2+36^2}{30} \\
& =4 \times k\left(\frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30}\right) \\
& =4 \times 45.8=183.2
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
If the variance of the first 50 odd natural numbers is V1 and the variance of next 50 odd natural numbers is V2, then V1+V2 is equal to

Options:

The correct answer is:
1666

Solution:
Note that V1=V2 (Variance is independent of shifting of orgin)
Now V1=12+32+…+99250-1+3+..+99502
=15050×101×993-2500
=3333-2500=833
Hence, V1+V2=1666

Topic: Mathematics, Statistics, JEE Main
If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then the value of m+n is equal to

Options:

The correct answer is:
18

Solution:
var1,2,…n=10⇒12+22+…+n2n-1+2+…+nn2=10

⇒n+12n+16-n+122=10

⇒n2-1=120⇒n=11

var2,4,6….,2m=16⇒var1,2….,m=4

⇒m2-1=48⇒m=7⇒m+n=18

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then the value of m+n is equal to

Options:

The correct answer is:
18

Solution:
var1,2,…n=10⇒12+22+…+n2n-1+2+…+nn2=10
⇒n+12n+16-n+122=10
⇒n2-1=120⇒n=11
var2,4,6….,2m=16⇒var1,2….,m=4
⇒m2-1=48⇒m=7⇒m+n=18

Topic: Mathematics, Statistics, JEE Main
If the variance of the following data :

6, 8, 10, 12, 14, 16, 18, 20, 22, 24 is K, then the value of K11 is

Options:

The correct answer is:
3

Solution:
The given data is :
6, 8, 10, 12, 14, 16, 18, 20, 22, 24
∴ x ¯ = 6 + 8 + 1 0 + 1 2 + 1 4 + 1 6 + 1 8 + 2 0 + 2 2 + 2 4 1 0
= 1 5 0 1 0 = 1 5
The table is as below :

x i	x i - x ¯	x i - x ¯ 2

6	- 9	81

8	- 7	49

10	- 5	25

12	- 3	9

14	- 1	1

16	1	1

18	3	9

20	5	25

22	7	49

24	9	81

Total	330

Hence, Variance σ 2 = Σ x i - x ¯ 2 n = 3 3 0 1 0 = 3 3 . â

Topic: Mathematics, Statistics, JEE Main
If the variance of the frequency distribution
$\begin{array}{|c|c|c|c|c|c|c|}
\hline x & c & 2 c & 3 c & 4 c & 5 c & 6 c \\
\hline f & 2 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}$
is 160, then the value of $c \in N$ is

Options:
A 7
B 8
C 5
D 6

The correct answer is:
7

Solution:
$\begin{array}{|c|c|c|c|c|c|c|}
\hline \mathrm{x} & \mathrm{C} & 2 \mathrm{C} & 3 \mathrm{C} & 4 \mathrm{C} & 5 \mathrm{C} & 6 \mathrm{C} \\
\hline \mathrm{f} & 2 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}$
$\bar{x}=\frac{(2+2+3+4+5+6) C}{7}=\frac{22 C}{7}$
$\begin{aligned} & \operatorname{Var}(x)=\frac{\mathrm{c}^2\left(2+2^2+3^2+4^2+5^2+6^2\right)}{7} \\ & -\left(\frac{22 \mathrm{c}}{7}\right)^2 \\ & =\frac{92 \mathrm{c}^2}{7}-\mathrm{c}^2 \times \frac{484}{49} \\ & =\frac{(644-484) \mathrm{c}^2}{49}=\frac{160 \mathrm{c}^2}{49} \\ & 160=\frac{160 \times \mathrm{c}^2}{49} \Rightarrow \mathrm{c}=7\end{aligned}$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
If the variance of the frequency distribution

xi	2	3	4	5	6	7	8

Frequency fi	3	6	16	α	9	5	6

is 3, then α is equal to

Options:

The correct answer is:
5

Solution:
We have,

xi	fi	di=xi-5	fidi2	fidi

2	3	-3	27	-9

3	6	-2	24	-12

4	16	-1	16	-16

5	α	0	0	0

6	9	1	9	9

7	5	2	20	10

8	6	3	54	18

σx2=σd2=∑fidi2∑fi-∑fidi∑fi2

⇒3=15045+α-0

⇒150=135+3α

⇒3α=15⇒α=5

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (31 Jan Shift 1)
If the variance of the numbers $2,3,11$ and $x$ is $\frac{49}{4}$, then the values of $x$ are

Options:
A $6, \frac{14}{3}$
B $4, \frac{13}{5}$
C $6, \frac{16}{3}$
D $6, \frac{14}{5}$

The correct answer is:
$6, \frac{14}{3}$

Solution:
Mean of given number is $\frac{2+3+11+x}{4}$ i.e. $\frac{16+x}{4}$
$$
\begin{aligned}
& \text { Variance }=\frac{1}{\mathrm{n}} \sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2 \\
& \therefore \frac{49}{4}=\frac{1}{4}\left[\left(\frac{16+\mathrm{x}}{4}-2\right)^2+\left(\frac{16+\mathrm{x}}{4}-3\right)^2+\left(\frac{16+\mathrm{x}}{4}-11\right)^2+\left(\frac{16+\mathrm{x}}{4}-\mathrm{x}\right)^2\right] \\
& \therefore 49=\frac{(8+\mathrm{x})^2}{16}+\frac{(4+\mathrm{x})^2}{16}+\frac{(\mathrm{x}-28)^2}{16}+\frac{(16-3 \mathrm{x})^2}{16} \\
& \therefore(49)(16)=\left(64+\mathrm{x}^2+16 \mathrm{x}\right)+\left(16+\mathrm{x}^2+8 \mathrm{x}\right)
\end{aligned}
$$
$$
\begin{aligned}
& +\left(\mathrm{x}^2-56 \mathrm{x}+784\right)+\left(256+9 \mathrm{x}^2-96 \mathrm{x}\right) \\
& \therefore 784=12 \mathrm{x}^2-128 \mathrm{x}+80+784+256 \\
& \therefore 12 \mathrm{x}^2-128 \mathrm{x}+336=0 \\
& \Rightarrow 3 \mathrm{x}^2-32 \mathrm{x}+84=0 \\
& \therefore \mathrm{x}=\frac{32 \pm \sqrt{(32)^2-(4)(3)(84)}}{2(3)}=\frac{32 \pm \sqrt{1024-1008}}{6}=\frac{32 \pm 4}{6} \\
& \therefore \mathrm{x}=\frac{36}{6} \text { or } \mathrm{x}=\frac{28}{6} \\
& \Rightarrow \mathrm{x}=6, \frac{14}{3}
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
If the variance of the observations x1,x2,……,xn is σ2, then the variance of αx1,αx2,……,αxn,α≠0 is

Options:
A σ2
B ασ2
C α2σ2
D σ2α2

The correct answer is:
α2σ2

Solution:
Variance =1nΣ(x-x¯)2=σ2

New variance =1nΣ(αx-αx¯)2

=α21nΣ(x-x¯)2=α2σ2

Topic: Mathematics, Statistics, JEE Main
If the variance of the terms in an increasing A.P. b1b2,b3,……..,b11 is 90 then the common difference of this A.P. is

Options:

The correct answer is:
3

Solution:
Givenb1,b2,b3,…..b11, are in A.P.

∴ Variance of b1,b2,…..b11=Variance of (0,d,2d,…..10d)=90

⇒ ∑i=111bi211-∑i=111bi112=90

⇒ d211·10.11.216-d2·5511·5511=90

⇒ 35d2-25d2=90

⇒ d2=9 ⇒d=±3

It is an increasing A.P. ⇒d is positive.

⇒ d=3

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (02 Sep Shift 2)
If the variate of a distribution takes the values 12,22,32,...n2 with frequencies n,n-1,n-2,...3,2,1 respectively, then the mean value of the distribution is

Options:
A nn+23
B nn+1n+26
C n+23
D n+1n+26

The correct answer is:
n+1n+26

Solution:
xi:12 22 32 42...n-12n2fi:n n-1n-2n-3... 21
∴∑i=1nxifi=12⋅n+22n-1+32n-2+...+n-12x+n2⋅1
=∑r=1nn+1r2-r3
=n+1∑r=1nr2-∑r=1nr3

=nn+122n+16-n2n+124
nn+121222n+1-3n=nn+12n+212
Also, Σfi=1+2+....+n=nn+12
Mean =ΣfixiΣfi=nn+12n+212×2nn+1

=n+1n+26

Topic: Mathematics, Statistics, JEE Main
If the vector b→=3j^+4k^ is written as the sum of a vector b1→, parallel to a→= i^+ j^ and a vector b→2, perpendicular to a→, then b→1×b→2 is equal to :

Options:
A 6i^-6j^+92k^
B -3i^+3j^-9k^
C -6i^+6j^-92k^
D 3i^-3j^+9k^

The correct answer is:
6i^-6j^+92k^

Solution:
b→1=b→ ⋅ a→a^2
=3j^+4k^⋅i^+ j^2 i^+ j^2
=3i^+ j^2×2=3i^+ j^2
b→1+b→2= b→
b→2= b→- b→1
=3j^+4k^-32i^+ j^
b→2= -32i^+32j^+4k^
b→1×b→2=i^j^k^32320-32324
⇒i^6- j^6+ k^94+94=6i^-6j^+92k^

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2017 (09 Apr Online)
If the vectors $2 \mathbf{i}-3 \mathbf{j}, \mathbf{i}+\mathbf{j}-\mathbf{k}$ and $3 \mathbf{i}-\mathbf{k}$ form three concurrent edges of a parallelopiped, then the volume of the parallelopiped is

Options:
A 8
B 10
C 4
D 14

The correct answer is:
4

Solution:
Here,
$\begin{aligned}
& \overrightarrow{O A}=2 \mathbf{i}-3 \mathbf{j}=\mathbf{a} \quad \text { (say) } \\
& \overrightarrow{O B}=\mathbf{i}+\mathbf{j}-\mathbf{k}=\mathbf{b} \quad \text { (say) }
\end{aligned}$
and
$\overrightarrow{O C}=3 \mathbf{i}-\mathbf{k}=\mathbf{c} \quad \text { (say) }$
Hence volume is $[\mathbf{a} \mathbf{b} \mathbf{c}]=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=\left|\begin{array}{ccc}2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1\end{array}\right|=4$

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors $2 \mathbf{i}-\mathbf{j}+\mathbf{k}, \mathbf{i}+2 \mathbf{j}-3 \mathbf{k}$ and $3 \mathbf{i}+\lambda \mathbf{j}+5 \mathbf{k}$ be coplanar, then $\lambda=$

Options:
A -1
B -2
C -3
D -4

The correct answer is:
-4

Solution:
If the given vectors are coplanar, then their scalar triple product is zero.
$\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & -3 \\
3 & \lambda & 5
\end{array}\right|=0 \Rightarrow \lambda=-4$

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors $6 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k}, 2 \mathbf{i}+3 \mathbf{j}-6 \mathbf{k}$ and $3 \mathbf{i}+6 \mathbf{j}-2 \mathbf{k}$ form a triangle, then it is

Options:
A Right angled
B Obtuse angled
C Equilateral
D Isosceles

The correct answer is:
Obtuse angled

Solution:
$\overrightarrow{A B}=\text { Position vector of } \vec{B} \text { - Position vector of } \vec{A}=(2 \mathbf{i}+3 \mathbf{j}-6 \mathbf{k})-(6 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})=-4 \mathbf{i}+5 \mathbf{j}-9 \mathbf{k}$
$\Rightarrow|\overrightarrow{A B}|=\sqrt{16+25+81}=\sqrt{122}, \overrightarrow{B C}=\mathbf{i}+3 \mathbf{j}+4 \mathbf{k}$
$\Rightarrow|\overrightarrow{B C}|=\sqrt{1+9+16}=\sqrt{26} \text { and } \overrightarrow{A C}=-3 \mathbf{i}+8 \mathbf{j}-5 \mathbf{k}$
$\Rightarrow|\overrightarrow{A C}|=\sqrt{98}$
$\text {Therefore, } A B^2=122, B C^2=26 \text { and } A C^2=98 .$
$\Rightarrow A B^2+B C^2=26+122=148$
Therefore, $A B^2=122, B C^2=26$ and $A C^2=98$.
Since $A C^2 \lt A B^2+B C^2$, therefore $\triangle A B C$ is an obtuse-angled triangle.

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors $\mathbf{a}$ and $\mathbf{b}$ are mutually perpendicular, then $\mathbf{a} \times\{\mathbf{a} \times\{\mathbf{a} \times(\mathbf{a} \times \mathbf{b})\}\}$ is equal to

Options:
A $|\mathbf{a}|^2 \mathbf{b}$
B $|\mathbf{a}|^3 \mathbf{b}$
C $|\mathbf{a}|^4 \mathbf{b}$
D None of these

The correct answer is:
$|\mathbf{a}|^4 \mathbf{b}$

Solution:
$\quad \mathbf{a} \times[\mathbf{a} \times\{\mathbf{a} \times(\mathbf{a} \times \mathbf{b})\}]=\mathbf{a} \times[\mathbf{a} \times\{\mathbf{a} \times \mathbf{a b} \hat{\mathbf{n}}\}]$
$=\mathbf{a} \times\left[\mathbf{a} \times a^2 \mathbf{b}\right]=\mathbf{a} \times a^3 b \hat{\mathbf{n}}=|\mathbf{a}|^4 \mathbf{b}$.

Topic: Mathematics, Vector Algebra, JEE Main
If the vectors $\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}},-2 \overrightarrow{\mathbf{a}}+3 \overrightarrow{\mathbf{b}}-4 \overrightarrow{\mathbf{c}}$
and $\overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{b}}+5 \overrightarrow{\mathbf{c}}$ are coplanar, then the value of $\lambda$ is

Options:
A 2
B $-1$
C 1
D $-2$

The correct answer is:
$-2$

Solution:
Since, the given three vectors are coplanar, therefore one of them should be expressible as a linear combination of the remaining two ie, there exist two scalars $x$ and $y$ such that
$$
\begin{aligned}
\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=x(-2 \overrightarrow{\mathbf{a}}+3&\overrightarrow{\mathbf{b}}-4 \overrightarrow{\mathbf{c}}) \\
+& y(\overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{b}}+5 \overrightarrow{\mathbf{c}})
\end{aligned}
$$
On comparing the coefficient of $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ on both sides, we get
$$
-2 x+y=1 ; 3 x-3 y=\lambda
$$
and $-4 x+5 y=3$
On solving first and third equations, we get
$$
x=-\frac{1}{3}, y=\frac{1}{3}
$$
Since, the vectors are coplanar, therefore these values of $x$ and $y$, also satisfy the second equation ie, $-1-1=\lambda$
$$
\therefore \quad \lambda=-2
$$

Topic: Mathematics, Vector Algebra, MHT CET, MHT CET 2007
If the vectors $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+a \hat{\mathbf{j}}+a^{2} \hat{\mathbf{k}}$,
$\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+b \hat{\mathbf{j}}+b^{2} \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+c \hat{\mathbf{j}}+c^{2} \hat{\mathbf{k}}$ are three
non-coplanar vectors and $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$,
then the value of $a b c$ is

Options:
A 0
B 1
C 2
D $-1$

The correct answer is:
$-1$

Solution:
Since, $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$, and $\overrightarrow{\mathbf{c}}$ are non-coplanar vectors,
therefore $[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \neq 0$
$\Rightarrow \Delta=\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right| \neq 0$
$\Rightarrow$
$\Delta \neq 0$
$$
\begin{array}{l}
\text { Now, } & \left(\begin{array}{lll}
a & a^{2} & 1+a^{3} \\
b & b^{2} & 1+b^{3} \\
c & c^{2} & 1+c^{3}
\end{array} \mid=0\right. \\
\Rightarrow\left|\begin{array}{lll}
a & a^{2} & 1 \\
b & b^{2} & 1 \\
c & c^{2} & 1
\end{array}\right|+\left|\begin{array}{lll}
a & a^{2} & a^{3} \\
b & b^{2} & b^{3} \\
c & c^{2} & c^{3}
\end{array}\right|=0 \\
\Rightarrow & \quad \Delta(1+a b c)=0
\end{array}
$$
$\Rightarrow$
$$
a b c=-1
$$

Topic: Mathematics, Determinants, JEE Main
If the vectors $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mu \hat{\mathrm{k}}$ are mutually orthogonal, then $(\lambda, \mu)=$

Options:
A
$(2,-3)$

B
$(-2,3)$

C
$(3,-2)$

D
$(-3,2)$

The correct answer is:
$(-3,2)$

Solution:
$$
\begin{aligned}
& \vec{a} \cdot \vec{b}=0, \quad \vec{b} \cdot \vec{c}=0, \quad \vec{c} \cdot \vec{a}=0 \\
& \Rightarrow 2 \lambda+4+\mu=0 \quad \lambda-1+2 \mu=0 \\
&
\end{aligned}
$$
Solving we get: $\lambda=-3, \mu=2$

Topic: Mathematics, Vector Algebra, JEE Main