datavorous/entrance-exam-dataset huggingface dataset:
Let a curve y=yx pass through the point 3,3 and the area of the region under this curve, above the x-axis and between the abscissae 3 and x>3 be yx3. If this curve also passes through the point α,610 in the first quadrant, then α is equal to _______.

Options:

The correct answer is:
6

Solution:
Given, the area of the region under this curve, above the x-axis and between the abscissae 3 and x>3 be yx3

So, ∫3xydx=yx3

⇒x3∫3xydx=y3

Now differentiating both side w.r.t x, we get

⇒x3y+3x2y3x3=3y2dydx as given ∫3xydx=yx3

⇒x4+3y2=3yxdydx

⇒3xydydx=3y2+x4

Now put y2=t,ydydx=12dtdx

⇒dtdx-2xt=23x3

Now IF=e∫-2xdx=1x2

So, solution is given by

t×IF=∫23x3×IF

⇒tx2=x23+C

Now given curve passes through 3,3 so C=-2

⇒y2x2=x23-2

So the equation of curve is 3y2=x4-6x2

Now curve also passes through α,610

So, α4-6α2=1080

⇒ α=6

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Let AD and BC be two vertical poles at A and B respectively on a horizontal ground. If AD=8m, BC=11m, AB=10m; then the distance (in meters) of a point M lying in between AB from the point A such that MD2+MC2 is minimums, is__

Options:

The correct answer is:
5

Solution:
Let AM=x and S=MD2+MC2

So, MD2+MC2=64+x2+121+(10-x)2

⇒S=285-20x+2x2

dSdx=-20+4x⇒dSdx=0⇒x=5

d2Sdx2=4=+ve

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Let a+d=c+b and ad=bc, where a,b,câR+. If the family of lines a2x+b2y+c2+d2x=0 passes through a fixed point x0,y0, then the value of x0+y0 is

Options:

The correct answer is:
0

Solution:
We have, a+d+2ad=c+b+2bc
âa+d=c+b (As, ad=bc )
â´a2+d2+2ad=c2+b2+2bcâa2+d2=c2+b2
Now, a2+d2x+b2y+c2=0
âb2+c2x+b2y+c2=0
âb2y+x+c2x+1=0
ây+x+c2b2x+1=0
which is of the form L1+Î»L2=0
So, the fixed point is -1,1=x0,y0 (Given).
Hence, x0+y0=-1+1=0.

Topic: Mathematics, Straight Lines, JEE Main
Let A denote the event that a 6-digit integer formed by 0,1,2,3,4,5,6 without repetitions, be divisible by 3 . Then probability of event A is equal to :

Options:
A 956
B 49
C 37
D 1127

The correct answer is:
49

Solution:
Total cases : 6·6·5·4·3·2

ns=6·6 !

Favourable cases :

Number divisible by 3≡ Sum of digits must be divisible by 3

Case-I

1,2,3,4,5,6

Number of ways =6 !

Case-II

0,1,2,4,5,6

Number of ways =5·5 !

Case-III

0,1,2,3,4,5

Number of ways =5·5 !

n( favourable )=6 !+2·5·5 !

P=6 !+2·5·5 !6·6 !=49

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let A denote the matrix $\left(\begin{array}{ll}0 & i \\ i & 0\end{array}\right)$, where $i^{2}=-1$, and let I denote the identity matrix $\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$. Then $\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\ldots+\mathrm{A}^{2010} \mathrm{is}-$

Options:
A $\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$
B $\left(\begin{array}{ll}0 & i \\ i & 0\end{array}\right)$
C $\left(\begin{array}{ll}1 & 1 \\ \mathrm{i} & 1\end{array}\right)$
D $\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)$

The correct answer is:
$\left(\begin{array}{ll}1 & 1 \\ \mathrm{i} & 1\end{array}\right)$

Solution:
$$
\begin{array}{l}
\mathrm{A}=\left[\begin{array}{ll}
0 & \mathrm{i} \\
\mathrm{i} & 0
\end{array}\right] ; \mathrm{A}^{2}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right] ; \quad \mathrm{A}^{3}=\left[\begin{array}{cc}
0 & -\mathrm{i} \\
-\mathrm{i} & 0
\end{array}\right] ; \quad \mathrm{I}=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \quad ; \quad \mathrm{A}^{4}=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{I} \\
\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+\ldots \ldots \mathrm{A}^{210} \\
\left(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}\right)+\mathrm{A}^{4}\left(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}\right)+\ldots+\mathrm{A}^{2008}\left(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}\right) \\
=\left(\begin{array}{ll}
1 & \mathrm{i} \\
\mathrm{i} & 1
\end{array}\right)
\end{array}
$$

Topic: Mathematics, Matrices, JEE Main
Let a die be rolled n times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is k215, then k is equal to

Options:
A 60
B 15
C 90
D 30

The correct answer is:
60

Solution:
Given that P(getting odd 7 times) =P(getting odd 9 times).

This is the case of binomial distribution with n successes and probability of success =p=36 and failure is q=1-p=12.

We know that PX=k=Cknpkqn-k

Now PX=7=PX=9

⇒C7n12712n-7=C9n12912n-9

⇒C7n=C9n

⇒Cn-7n=C9n

⇒ n = 16

⇒P(2 times even )=C2161214122

=15×4215=60215=k215

⇒k=60

Hence this is the required option.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let a differentiable function f satisfy fx+∫3xfttdt=x+1,x≥3. Then 12f8 is equal to:

Options:
A 34
B 19
C 17
D 1

The correct answer is:
17

Solution:
Given:

fx+∫3xfttdt=x+1

⇒f'x+fxx=12x+1

Put y=fx, then

dydx+yx=12x+1

So, I.F.=e∫dxx=elnx=x

Hence, solution is

xy=12∫xx+1dx

⇒xy=12∫x+1-1x+1dx

⇒xy=12∫x+1-1x+1dx

⇒xy=1223x+132-2x+1+C

⇒xy=13x+132-x+1+C

Put x=3, then f3=2

So,

6=83-2+C⇒C=163

Hence,

xy=13x+132-x+1+163

So, put x=8, then

8f8=273-3+163

⇒f8=343×8

⇒12f8=17

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let a fully charged lead storage battery contains 1.5 L of 5 M H2SO4 . What will be the concentration of H2SO4 in the battery after 2.5 ampere current is drawn from the battery for 6 hour?

Options:
A 4.626 M
B 0.1865 M
C 0.373 M
D 9.627 M

The correct answer is:
4.626 M

Solution:
Number of moles of H2SO4 before electrolysis
=MV1000
MÃVmL=5Ã1.5=7.5
Write cell rection âPb0+PbO2+4+2H2SO4â2PbSO4+2+2H2O
n=2
For 2 moles of electrons
2 moles of H2SO4 are used
Moles of electrons used =Iin AÃtin sec96500=2.5Ã6Ã360096500=0.56
Hence, moles of H2SO4 used =0.56
Remaining mol of H2SO4=7.5-0.56=6.94
Final molarity =nVÃ1000=6.941.5Ã1000Ã1000
=4.626

Topic: Chemistry, Electrochemistry, JEE Main
Let a function $f:(0, \infty) \rightarrow(0, \infty)$ be defined by

$f(x)=\left|1-\frac{1}{x}\right| .$ Then $f$ is :

Options:
A not injective but it is surjective
B injective only
C neither injective nor surjective
D None of the above

The correct answer is:
None of the above

Solution:
$f:(0, \infty) \rightarrow(0, \infty)$

$f(x)=\left|1-\frac{1}{x}\right|$ is not a function

$\because f(1)=0$ and $1 \in$ domain but $0 \notin$ co-domain

Hence, $f(x)$ is not a function.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let a function f:0,5→R be continuous, f1=3 and F be defined as:
Fx=∫1xt2gtdt, where gt=∫1tfudu.

Then for the function Fx, the point x=1 is:

Options:
A a point of local minima
B not a critical point
C a point of local maxima
D a point of inflection

The correct answer is:
a point of local minima

Solution:
Given, Fx=∫1xt2gtdt

By Leibnitz rule we get,

F'x=x2gx

⇒F'1=1.g1=0 ∵g1=0

Now F''x=2xgx+x2g'x

⇒F''x=2xgx+x2fx ∵g'x=fx

⇒F''1=0+1×3

⇒F''1=3

Fx has a local minimum at x=1.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Let a function f:1,∞→0,∞ be defined by fx=1-1x. Then f is

Options:
A injective but not surjective
B both injective as well as surjective
C not injective but it is surjective
D neither injective nor surjective

The correct answer is:
both injective as well as surjective

Solution:
fx=1-1x=x-1x=x-1x
f ′ x = 1 x 2
f ′ x >0
∴fx is injective and range of function is 0,∞Hence, injective and surjective

Topic: Mathematics, Functions, JEE Main
Let a function f:2,ââ0,â defined as fx=x-3x-2, then f is

Options:
A injective & surjective
B not injective but surjective
C injective but not surjective
D neither injective nor surjective

The correct answer is:
not injective but surjective

Solution:
f(x)=1-1x-2
∵2<x<∞
0<x-2<∞⇒∞>1x-2>0
-∞<-1x-2<0⇒-∞<1-1x-2<1
∵0≤1-1x-2<∞
∴ Range of fx∈0,∞= co-domain

Hence, f(x) is surjective
Let, fx=12⇒1-1x-2=12
1-1x-2=12, 1-1x-2=-12
1x-2=12, 1x-2=32
x=4, x=83
∴f4=f83=12
∴fx is many-one (not injective)

Topic: Mathematics, Functions, JEE Main
Let a function f:ℝ→ℝ be defined as:

fx=∫0x5-t-3dt,x>4x2+bx,x≤4

where b∈ℝ. If f is continuous at x=4, then which of the following statements is NOT true?

Options:
A f is not differentiable at x=4
B f'3+f'5=354
C f is increasing in -∞,18∪8,∞
D f has a local minima at x=18

The correct answer is:
f is increasing in -∞,18∪8,∞

Solution:
Given fx=∫0x5-t-3dt,x>4x2+bx,x≤4

Also given, fx is continuous at x=4

So limx→4-fx=limx→4+fx=f4

So 16+4b=∫032-tdt+∫348-tdt

⇒16+4b=2t-t2203+8t-t2234

⇒16+4b=15

So b=-14

Now differentiating fx we get,

f'x=5-x-3,x>42x+b,x<4

Now at x=4

LHD=2x+b=2×4-14=314

RHD=5-x-3=4

LHD≠RHD

So, option A is true

And f'3+f'5=234+3=354

So, option B is true

Now fx=x2-x4 at x≤4

f'x=2x-14

Thus the function is not increasing in the interval in x∈-∞,18

So, option C is NOT TRUE.

And function fx is also have local minima at x=18 so option D is also true.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let a function f:R→R be defined as, fx=sinx−ex if x≤0a+-x if 0<x<12x−b if x≥1

Where x is the greatest integer less than or equal to x. If f is continuous on R, then a+b is equal to:

Options:
A 4
B 3
C 2
D 5

The correct answer is:
3

Solution:
We have,

fx=sinx−ex if x≤0a+-x if 0<x<12x−b if x≥1

If fx is continuous at x=a then limx→a-fx=fa=limx→a+fx.

Continuous at x=0

f0+=f0-

⇒ a-1=0-e0

⇒ a=0

Continuous at x=1

f1+=f1-

⇒ 21-b=a+-1

⇒ b=2-a+1

⇒ b=3

∴ a+b=3

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let a function fx be continuous in an interval a,b. Let δ>0 be a very small real number. Let c∈a,b be such that fc-δ<fc and fc+δ<fc for every δ>0. Let fα-δ-fαfα+δ-fα<0 ∀α∈a,b and α≠c. Then

Options:
A fx has a local maximum at c and a local minimum at α
B fx has a local maximum at α and a local minimum at c
C fx has only one local maximum at c
D fx has only one local minimum at c

The correct answer is:
fx has only one local maximum at c

Solution:
Since, fx is continuous in a,b and fc-δ<fc & fc+δ<fc, threfore following scenario is possible

Hence, fx must have local maximum at x=c.

Also,

fa−δ−fαfα+δ−fα<0 ...1

Case 1: When

fα-δ-fα>0 and fα+δ-fα<0

⇒fα<fα-δ and fα+δ<fα

So, fx has neither maximum nor minima at x=α.

Case 2:

fα−δ−fα<0 and fα+δ−fα>0

⇒fα−δ<fα and fα−δ>fα

Here also fx has neither maxima nor minima at x=α.

Topic: Mathematics, Application of Derivatives, JEE Main
Let a function \( f:[0,5] \rightarrow R \) be continuous, \( f(1)=3 \) and \( F \) be defined as:
\( F(x)=\int_{1}^{x} t^{2} g(t) d t \), where \( g(t)=\int_{1}^{t} f(u) d u . \)
Then for the function \( F(x) \), the point \( x=1 \) is:

Options:
A a point of local minima
B not a critical point
C a point of local maxima
D a point of inflection

The correct answer is:
a point of local minima

Solution:
Given, Fx=∫1xt2gtdt

By Leibnitz rule we get,

F'x=x2gx

⇒F'1=1.g1=0 ∵g1=0

Now F''x=2xgx+x2g'x

⇒F''x=2xgx+x2fx ∵g'x=fx

⇒F''1=0+1×3

⇒F''1=3

Fx has a local minimum at x=1.

Topic: Mathematics, Application of Derivatives, JEE Main
Let a function g:0,4→R be defined as

gx=max t3-6t2+9t-30≤t≤x,0≤x≤34-x,3<x≤4

then the number of points in the interval 0, 4 where g(x) is NOT differentiable, is _________.

Options:

The correct answer is:
1

Solution:
We have,

gx=max t3-6t2+9t-30≤t≤x,0≤x≤34-x,3<x≤4

Let

fx=x3-6x2+9x-3

⇒f'(x)=3x2-12x+9

⇒f'(x)=3(x-1)(x-3)

For critical points:

f'x=0

⇒x=1, 3

And,

f"x=6x-12

Then,

f"1=6-12=-6<0(maxima)

f"3=6>0(minima)

Hence,

gx=f(x),0≤x≤11,1≤x≤34-x,3<x≤4

Hence, gx is continuous function.

g'x=3(x-1)(x-3),0<x<10,1<x<3-1,3<x<4

Clearly, g(x) is non-differentiable at x=3.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Let $A(h, k), B(1,1)$ and $C(2,1)$ be the vertices of a right angled triangle with $A C$ as its hypotenuse. If the area of the triangle is $1$, then the set of values which ' $\mathrm{k}$ ' can take is given by

Options:
A
$\{1,3\}$

B
$\{0,2\}$

C
$\{-1,3\}$

D
$\{-3,-2\}$

The correct answer is:
$\{-1,3\}$

Solution:
$\frac{1}{2} \times 1(k-1)=\pm 1$
$k-1=\pm 2$
$k=3$
$k=-1$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2007
Let a→=i^+2j^+3k^ and b→=i^+j^-k^. If c→ is a vector such that a→·c→=11,b→·a→×c→=27 and b→·c→= -3|b→|, then |a→×c→|2 is equal to

Options:

The correct answer is:
285

Solution:
Given,

a→=i^+2j^+3k^, b→=i^+j^-k^, a→·c→=11, b→·a→×c→=27 and b→·c→=-3b→

Now finding, b→×a→ we get,

b→×a→=i^j^k^11-1123=5i^-4j^+k^

Let c→=c1i^+c2j^+c3k^

Now solving, a→·c→=11 we get,

c1+2c2+3c3=11 .......1

Now solving, b→·c→=-3b→ we get,

c1+c2-c3=-33

c1+c2-c3=-3 ..........2

Now solving b→×a→·c→=27 we get,

5c1-4c2+c3=27...........3

Now on solving equation 1, 2 & 3 we get,

c→=3i^-2j^+4k^

Hence, a→×c→2=i^j^k^1233-2+42=14i^+5j^-8k^2

⇒|a→×c→|2=142+52+82=285

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\overrightarrow{\mathrm{r}}$ be anit vector along $\vec{b}+\vec{c}$. If $\vec{r} \cdot \vec{a}=3$, then $3 \lambda$ is equal to:

Options:
A 21
B 30
C 25
D 27

The correct answer is:
25

Solution:
$\begin{aligned} & \overrightarrow{\mathrm{r}}=\mathrm{k}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=3 \\ & \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{k}(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \\ & 3=\mathrm{k}(2+6-15+3-2+3 \lambda)\end{aligned}$
$3=\mathrm{k}(-6+3 \lambda)$ ...(1)
$\overrightarrow{\mathrm{r}}=\mathrm{k}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-(5-\lambda) \hat{\mathrm{k}})$
$|\vec{r}|=k \sqrt{25+4+25+\lambda^2-10 \lambda}=1$ ...(2)
$\mathrm{k}=\frac{3}{-6+3 \lambda}=\frac{1}{-2+\lambda} \quad$ put in (2)
$\begin{aligned} & 4+\lambda^2-4 \lambda=54+\lambda^2-10 \lambda \\ & 6 \lambda=50 \\ & 3 \lambda=25\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let $\quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \quad$ and $\overrightarrow{\mathbf{c}}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+(2 \lambda-1) \hat{\mathbf{k}}$. If $\overrightarrow{\mathbf{c}}$ is parallel to the plane containing $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$, then $\lambda$ is equal to

Options:
A 0
B 1
C -1
D 2

The correct answer is:
0

Solution:
Given that, $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
$\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
and $\quad \overrightarrow{\mathbf{c}}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+(2 \lambda-1) \hat{\mathbf{k}}$
Also, since $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ lies in the same plane, then $(\vec{a} \times \vec{b})$ is perpendicular vector to this plane. Given that vector $\overrightarrow{\mathbf{c}}$ is parallel to the plane containing $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, so vector $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})$ also perpendicular to the vector $\overrightarrow{\mathbf{c}}$ ie, $\left(\theta=90^{\circ}\right)$ So, $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}}$ should be equal to zero
or $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}}=0$ ...(i)
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 3 \\ 2 & 3 & -1\end{array}\right|$
$=(2-9) \hat{\mathbf{i}}+(6+1) \hat{\mathbf{j}}+(3+4) \hat{\mathbf{k}}$
$=-7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$
Then from Eq. (i)
$(-7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \cdot(\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+(2 \lambda-1) \hat{\mathbf{k}})=0$
$\Rightarrow \quad-7 \lambda+7+7(2 \lambda-1)=0$
$\Rightarrow \quad-7 \lambda+7+14 \lambda-7=0$
$\Rightarrow \quad 7 \lambda=0$
$\Rightarrow \quad \lambda=0$
Hence, the value of $\lambda$ is 0 .

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let a â = i ^ +2 j ^ +3 k ^ , b â =2 i ^ +3 j ^ + k ^ , c â = k ^ + i ^ and xâÃbâ=aâÃcâÃbâ. If xââaâ=0, then xâ is equal to
use 3=1.73)

Options:

The correct answer is:
3.46

Solution:
x→×b→=a→⋅b→c→-b→⋅c→a→=11c→-3a→
As, a→⋅b→=11 and b→⋅c→=3
Now x→×b→×a→=11c→×a→
⇒x→⋅a→b→-b→⋅a→x→=11c→×a→
⇒x→=a→×c→ as a→⋅x→=0
⇒x→=i^j^k^123101=2i^+2j^-2k^
⇒x→=23
=3.46

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=i^+2j^+3k^,b→=i^-j^+2k^ and c→=5i^-3j^+3k^, be there(three) vector. If r→ is a vector such that, r→×b→=c→×b→ and r→·a→=0, then 25r→2 is equal to

Options:
A 560
B 339
C 449
D 336

The correct answer is:
339

Solution:
Given,

r→×b→=c→×b→

⇒r→×b→-c→×b→=0

r→-c→×b→=0

Hence, we can say that r→-c→ is parallel to vector b→,

Now let , r→-c→=λb→

⇒r→=c→+λb→ ......1

∵ r→·a→=0⇒c→·a→+λb→·a→=0

⇒λ=-c→·a→b→·a→=-85

Now putting the value of λ in equation 1 we get,

r→=c→-85b→=5i^-3j^+3k^-85i^-j^+2k^

⇒5r→=17i^-7j^-k^

⇒25r→2=172+-72+-12

⇒25r→2=289+49+1=339

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let a→=i^+2j^+3k^,b→=i^-j^+2k^ and c→=x-2i^-x-3j^-k^. If c→ lies in the plane of a→ and b→, then 1x is equal to

Options:

The correct answer is:
0.75

Solution:
c→,a→,b→ are coplanar ⇒a→b→c→=0
1231-12x-2-x-3-1=0
⇒11+2x-6-2-1-2x+4+3-x+3+x-2=0
⇒2x-5-6+4x+3=0
⇒6x-8=0⇒x=43

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=i^-2j^+3k^, b→=i^+j^+k^ and c→ be a vector such that a→×(b→+c→)=0→, then the value of 3c→.a→ is equal to _______.

Options:

The correct answer is:
29

Solution:
Given, a→=i^-2j^+3k^, b→=i^+j^+k^

And a→×(b→+c→)=0→,

Let c→=xi^+yj^+zk^

So, a→×b→+c→=0→⇒a→×b→+a→×c→=0→

⇒a→×b→=-a→×c→⇒a→×b→=c→×a→

⇒i^j^k^1-23111=i^j^k^xyz1-23

⇒-5i^+2j^+3k^=i^3y+2z+j^z-3x+k^-2x-y

On comparing we get,

3y+2z=-5 .....(i)

z-3x=2 ............(ii)

-2x-y=3 ........(iii)

On solving equation (i), (ii) and (iii) we get,

x=-16, y=-83, z=32

So, c→=-16i^+-83j^+32k^

So, c→.a→=-16i^+-83j^+32k^.i^-2j^+3k^=293

So, 3c→.a→=29

Note: This question appeared in JEE Main 2022, 29th june shift 2. The question was incorrect, so it is modified to make it correct.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let $\vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}$ and

$\vec{c}=2 \hat{i}+4 \hat{j}+\left(\lambda^{2}-1\right) \hat{k}$ be coplanar vectors. Then the

non-zero vector $\vec{a} \times \vec{c}$ is:

Options:
A $-10 \hat{i}-5 \hat{j}$
B $-14 \hat{i}-5 \hat{j}$
C $-14 \hat{i}+5 \hat{j}$
D $-10 \hat{i}+5 \hat{j}$

The correct answer is:
$-10 \hat{i}+5 \hat{j}$

Solution:
$\because \bar{a}, \bar{b}$ and $\bar{c}$ are coplanar

$\therefore\left|\begin{array}{ccc}1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & \left(\lambda^{2}-1\right)\end{array}\right|=0$

$\Rightarrow \quad \lambda^{3}-\lambda-16+2\left(8-\lambda^{2}+1\right)+4(4-2 \lambda)=0$

$\Rightarrow \quad \lambda^{3}-2 \lambda^{2}-9 \lambda+18=0$

i.e. $\quad(\lambda-2)(\lambda-3)(\lambda+3)=0$

For $\lambda=2, \vec{c}=2 \hat{i}+4 \hat{j}+3 \hat{k}$

$\therefore \quad \vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3\end{array}\right|=-10 \hat{i}+5 \hat{j}$

For $\lambda=3$ or $-3, \bar{c}=2 \bar{a} \Rightarrow \bar{a} \times \bar{c}=0$ (Rejected)

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let a→=αi^+2j^-k^ and b→=-2i^+αj^+k^, where α∈R. If the area of the parallelogram whose adjacent sides are represented by the vectors a→ and b→ is 15α2+4, then the value of 2a→2+a→·b→b→2 is equal to

Options:
A 10
B 7
C 9
D 14

The correct answer is:
14

Solution:
Given,

a→=αi^+2j^-k^,b→=-2i^+αj^+k^,

Area of parallelogram =a^×b^

a^×b^=α+22+α-22+α2+42

Given a^×b^=15α2+4

So, 2α2+4+α2+42=15α2+4

⇒α2+42=13α2+4

⇒α2+4=13 ∴ α2=9

Now, 2a→2+a→·b→b→2

a→2=α2+4+1=α2+5

|b→|2=4+α2+1=α2+5

a→·b→=-2α+2α-1=-1

∴ 2a→2+a→·b→b→2

2α2+5-1α2+5=α2+5=14

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let a→=i^-2j^+k^ and b→=i^-j^+k^, be two vectors. If c→, is a vector such that b→×c→=b→×a→ and c→⋅a→=0, then c→⋅b→, is equal to.

Options:
A -32
B 12
C -12
D -1

The correct answer is:
-12

Solution:
a→×b→×c→=a→×b→×a→
⇒-a→.b→c→=a→.a→b→-a→.b→a→

Using given information a→=i^-2j^+k^ and b→=i^-j^+k^ , we can write a→.b→=4, a→.a→=6
⇒-4c→=6i^-j^+k^-4i^-2j^+k^
⇒-4c→=2i^+2j^+2k^
⇒c→=-12i^+j^+k^
⇒b→.c→=-12.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Let a→=i^+2j^+λk^, b→=3i^-5j^-λk^, a→·c→=7, 2b→·c→+43=0, a→×c→=b→×c→, then a→·b→ is equal to

Options:

The correct answer is:
8

Solution:
Given:

a→×c→=b→×c→

⇒a→×c→-b→×c→=0

⇒a→-b→×c→=0

So, a→-b→∥c→, hence

a→-b→=μc→

where, μ is any scalar.

⇒-2i^+7j^+2λk^=μc→ ....1

Taking dot product with a→, we get

⇒-2+14+2λ2=7μ

⇒7μ=2λ2+12 ...2

Taking dot product with b→ in 1, we get

-6-35-2λ2=-432μ

⇒82+4λ2=43μ ....3

Solving 2 & 3, we get

λ2=1, μ=2

So,

a→·b→=3-10-λ2=-8

⇒a→·b→=8

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let a→=i^+2j^-k^, b→=i^-j^ and c→=i^-j^-k^ be three given vectors. If r→ is a vector such that r→×a→=c→×a→ and r→·b→=0, then r→·a→ is equal to

Options:

The correct answer is:
12

Solution:
r→-c→×a→=0

⇒ r→=c→+λa→

taking dot product with b→

Now, 0=b→·c→+λa→·b→

⇒ λ=-b→·c→a→·b→=-2-1=2

So, r→·a→=a→·c→+2a→2=0+2×6=12

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Let $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$. If $\vec{a} \cdot \vec{c}=130$, then $\vec{b} \cdot \vec{c}$ is equal to _______

Options:

The correct answer is:
30

Solution:
$\begin{aligned} & (\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}) \\ & (2 \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{a}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}})=\lambda(8 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+30 \hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=130 \\ & 8 \lambda+42 \lambda+210 \lambda=130 \\ & \lambda=\frac{1}{2} \\ & \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+15 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=8+7+15=30\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Let a→=αi^+3j^-k^,b→=3i^-βj^+4k^ and c→=i^+2j^-2k^ where α,β∈R be three vectors. If the projection of a→ on c→ is 103 andb→×c→=-6i^+10j^+7k^ , then the value of α+β equal to

Options:
A 3
B 4
C 5
D 6

The correct answer is:
3

Solution:
Given a→=αi^+3j^-k^,b→=3i^-βj^+4k^ and c→=i^+2j^-2k^ where α,β∈R be three vectors.

Projection of a→ on c→ is a→·c→c→=103 (given)

⇒α+6+21+4+4=103⇒α=2

and b→×c→=i^j^k^3-β412-2=-6i^+j^+k^

⇒2β-8=-6⇒β=1

Hence α+β=3

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (29 Jun Shift 1)