datavorous/entrance-exam-dataset huggingface dataset:
Let there be $\mathrm{n}$ resistors $\mathrm{R}_1 \ldots \mathrm{R}_{\mathrm{n}}$, with $\mathrm{R}_{\max }=\max \left(\mathrm{R}_1 \ldots . \mathrm{R}_{\mathrm{n}}\right)$ and $R_{\min }=\min \left\{R_1 \ldots R_n\right\}$. Show that when they are connected in parallel, the resultant resistance $R_p=R_{\text {min }}$ and when they are connected in series, the resultant resistance $\mathrm{R}_{\mathrm{s}}>\mathrm{R}_{\max }$. Interpret the result physically.

Options:

None

Solution:
Let $R_{\min }$ and $R_{\max }$ are the minimum and maximum resistances.
When all resistors are connected in parallel, the resultant resistance $R_p$ is
$$
\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_1}+\ldots+\frac{1}{\mathrm{R}_{\mathrm{n}}}
$$
On multiplying both sides by $R_{\min }$ we get,
$$
\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\mathrm{p}}}=\frac{\mathrm{R}_{\min }}{\mathrm{R}_1}+\frac{\mathrm{R}_{\min }}{\mathrm{R}_2}+\ldots+\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\mathrm{n}}}
$$
Here, in RHS, there exist one term $\frac{R_{\text {min }}}{R_{\text {min }}}=1$ and other terms are positive, so we have
$$
\frac{\mathrm{R}_{\min }}{\mathrm{R}_p}=\frac{\mathrm{R}_{\min }}{\mathrm{R}_1}+\frac{\mathrm{R}_{\min }}{\mathrm{R}_2}+\ldots+\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\mathrm{n}}}>1
$$
i.e., the resultant resistance $\left(R_p < R_{\text {min }}\right)$.
So, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in combination of resistors.
Now, in series combination, the equivalent resistant is
$$
\mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\ldots+\mathrm{R}_{\mathrm{n}}
$$
Here, in RHS, there exist one term having resistance $\mathrm{R}_{\max }$
So, we get
$$
\begin{aligned}
\mathrm{R}_{\mathrm{s}} &=\mathrm{R}_1+\ldots+\mathrm{R}_{\max } \ldots+\ldots+\mathrm{R}_{\mathrm{n}} \\
\mathrm{R}_{\mathrm{s}} &=\mathrm{R}_1+\ldots \mathrm{R}_{\max }+\mathrm{R}_{\mathrm{n}} \\
&=\mathrm{R}_{\max }+\ldots\left(\mathrm{R}_1+\ldots+\mathrm{R}_{\mathrm{n}}\right.\\
\mathrm{R}_{\mathrm{s}} & \geq \mathrm{R}_{\max } \\
\mathrm{R}_{\mathrm{s}} &=\mathrm{R}_{\max }\left(\mathrm{R}_1+\ldots \mathrm{R}_{\mathrm{n}}\right)
\end{aligned}
$$
So, in series combination, the equivalent resistance is always greater than the maximum resistance $\left(\mathrm{R}_{\max }\right)$ available in combination of resistors. Physical interpretation
In Fig. (b), $R_{\text {min }}$ provides an equivalent route as in Fig (a) for current. But in addition there are $(\mathrm{n}-1)$ routes by the remaining $(\mathrm{n}-1)$ resistors. Current in Fig (b) is greater than current in Fig. (a). Effective resistance in Fig. (b) $ < \mathrm{R}_{\text {min. }}$. Second circuit evidently affords a greater resistance.
In the above figure, $R_{\max }$ provides an equivalent route as in Fig. (c) for current. Current in Fig.(d) < current in Fig. (c). Effective resistance in Fig. (d) $>\mathrm{R}_{\max }$. Second circuit evidently affords a greater resistance.

Topic: Physics, Current Electricity
Let there be three independent events E1,E2 and E3. The probability that only E1 occurs is α only E2 occurs is β and only E3 occurs is γ. Let p'' denote the probability of none of events occurs that satisfies the equations (α-2β)p=αβ and (β-3γ)p=2βγ. All the
given probabilities are assumed to lie in the interval 0, 1.

Then, Probability of occurrence of E1 Probability of occurrence of E3 is equal to ________.

Options:

The correct answer is:
6

Solution:
Let PE1=P1;PE2=P2;PE3=P3

PE1∩E¯2∩E¯3=α=P11-P21-P3 …1

PE¯1∩E2∩E¯3=β=1-P1P21-P3 …2

PE¯1∩E¯2∩E3=γ=1-P11-P2P3 …3

PE¯1∩E¯2∩E¯3=p=1-P11-P21-P3 …4

Given that, (α-2β)p=αβ

⇒P11-P21-P3-21-P1P21-P3p=P1P2

1-P11-P21-P32

⇒P11-P2-21-P1P2=P1P2

⇒P1-P1P2-2P2+2P1P2=P1P2

⇒P1=2P2 …1

and similarly, (β-3γ)P=2βγ

P2=3P3 …2

So, P1=6P3⇒P1P3=6

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Let there be two parabolas y2=4x and y2=-8x. Then the locus of the mid-points of the intercepts between the parabolas made on the lines parallel to the common axis is

Options:
A y2=16x
B x2=16y
C y2=-8x
D x2=8y

The correct answer is:
y2=16x

Solution:
Let line parallel to common axis be y=h. Then coordinates of A and B are
h24,h and -h28,h respectively.
If P≡α,β, then

α=12h24-h28, β=h

⇒2α=h2411-12=h28
∴ Locus of P is x=y216

Topic: Mathematics, Parabola, JEE Main
Let three positive numbers a, b, c are in geometric progression, such that a, b+8, c are in arithmetic progression and a, b+8, c+64 are in geometric progression. If the arithmetic mean of a,b,c is k, then 313k is equal to

Options:

The correct answer is:
4

Solution:
Here, b2=ac,2b+8=a+c,
b+82=ac+64
⇒b2+64+16b=ac+64a
⇒b+4=4a⇒b=4a-4
2b+8=a+c⇒24a+4=a+c
⇒c=7a+8
b2=ac⇒4a-42=a7a+8
⇒16a2+16-32a=7a2+8a
⇒9a2-40a+16=0
⇒9a-4a-4=0⇒a=49,4
But for a=49⇒b<0
So, a=4⇒b=12 and c=36
⇒k=a+b+c3=523
⇒3k13=313Ã523=4

Topic: Mathematics, Sequences and Series, JEE Main
Let three positive numbers \( a, b, c \) are in geometric progression, such that \( a, b+8, c \) are in arithmetic progression and \( a, b+8, c+64 \) are in geometric progression. If the arithmetic mean of \( a, b, c \) is \( k \), then \( \frac{3}{13} k \) is equal to

Options:

The correct answer is:
4

Solution:
Here, b2=ac,2b+8=a+c,

b+82=ac+64

⇒b2+64+16b=ac+64a

⇒b+4=4a⇒b=4a-4

2b+8=a+c⇒24a+4=a+c

⇒c=7a+8

b2=ac⇒4a-42=a7a+8

⇒16a2+16-32a=7a2+8a

⇒9a2-40a+16=0

⇒9a-4a-4=0⇒a=49,4

But for a=49⇒b<0

So, a=4⇒b=12 and c=36

⇒k=a+b+c3=523

⇒3k13=313×523=4

Topic: Mathematics, Sequences and Series, JEE Main
Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression. If $a>10$ and the arithmetic mean of $a, b$ and $c$ is 8, then the cube of the geometric mean of $a, b$ and $c$ is

Options:
A 128
B 316
C 120
D 312

The correct answer is:
120

Solution:
$\begin{aligned} & 2 b=a+c, b^2=(a+1)(c+3) \\ & \frac{a+b+c}{3}=8 \rightarrow b=8, a+c=16 \\ & 64=(a+1)(19-a)=19+18 a-a^2 \\ & a^2-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a>10) \\ & a=15, c=1, b=8 \\ & \left((a b c)^{1 / 3}\right)^3=a b c=120\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Let three vectors a→,b→ and c→ be such that c→ is coplanar with a→ and b→,a→·c→=7 and b→ is perpendicular to c→, where a→=-i^+j^+k^ and b→=2i^+k^, then the value of 2a→+b→+c→2 is

Options:

The correct answer is:
75

Solution:
Let c→=λb→×a→×b→

=λb→·b→a→-b→·a→b→

=λ5-i^+j^+k^+2i^+k^

=λ-3i^+5j^+6k^

c→·a→=7⇒3λ+5λ+6λ=7

λ=12

∴ 2-32-1+2i^+52+1j^+3+1+1k^2

=214+494+25=25+50=75

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Let triangle ABC is right-angled at A. The circle with centre A and radius AB cuts BC and AC internally at D and E respectively. If BD=20 units and DC=16 units, then the length AB is equal to

Options:
A 621 units
B 610 units
C 300 units
D 18 units

The correct answer is:
610 units

Solution:
b2+r2=362…i
Also CD⋅CB=CE⋅CX
16⋅36=b-rb+r
∴b2-r2=16⋅36
From i and ii
2b2=3636+16=36⋅52
b2=36⋅26⇒b=626 units

and r=362-3626=610 units

Topic: Mathematics, Circle, JEE Main
Let truth values of \( p \) be \( F \) and \( q \) be \( T \). Then, truth value of \( \sim(\sim p \vee q) \) is

Options:
A \( T \)
B \( F \)
C Either \( T \) or \( F \)
D Neither \( T \) nor \( F \)

The correct answer is:
\( F \)

Solution:
Given that truth values of p is F & q is T.

~p = T.

We know that p∨q is true if at least one of them is true.

⇒~p∨q = T

⇒~~p∨q = F .

Topic: Mathematics, Mathematical Reasoning, JEE Main
Let two circles having radii r1 and r2 are orthogonal to each other. If the length of their common chord is k times the square root of harmonic mean between the squares of their radii, then k4 is equal to

Options:
A 13
B 7
C 4
D 2

The correct answer is:
4

Solution:
Area of ΔABC
=12r1r2=12l2r12+r22
⇒l=2r1r2r12+r22
⇒l=22r12r22r12+r22⇒k=2
⇒k4=4

Topic: Mathematics, Circle, JEE Main
Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is not true?

Options:
A E1 and E3 are independent
B E1, E2 and E3 are independent
C E1 and E2 are independent
D E2 and E3 are independent

The correct answer is:
E1, E2 and E3 are independent

Solution:
E1→A show up 4
E2→B shows up 2
E3→ Sum is odd (i.e., even + odd or odd + even)
PE1=66.6=16
PE2=66.6=16
PE3=3.36.6+3.36.6=24=12
PE1∩E2=16.6=PE1 . PE2
⇒ E1 & E2 are independent
PE1∩E3=1.36.6=PE1 . PE3
⇒ E1 & E3 are independent
PE2∩E3=1.36.6=112=PE2 . PE3
⇒ E2 & E3 are independent
PE1∩E2∩E3=0 ≠P(E1)×P(E2)×P(E3)

Topic: Mathematics, Probability, JEE Main, JEE Main 2016 (03 Apr)
Let two numbers have an arithmetic mean 9 and geometric mean 4, then these numbers are the roots of the quadratic equation

Options:
A x 2 + 1 8 x - 1 6 = 0
B x 2 - 1 8 x + 1 6 = 0
C x 2 + 1 8 x + 1 6 = 0
D x 2 - 1 8 x - 1 6 = 0

The correct answer is:
x 2 - 1 8 x + 1 6 = 0

Solution:
Let the two numbers be Î± , Î²
â´ Î± + Î² 2 = 9 and Î± Î² = 4 â Î± + Î² = 18, Î± Î² = 16
â´ Required equation is
x2-Î±+Î²x+Î±Î²=0
â x2-18x+16=0

Topic: Mathematics, Quadratic Equation, JEE Main
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation

Options:
A
$x^2+18 x+16=0$

B
$x^2-18 x-16=0$

C
$x^2+18 x-16=0$

D
$x^2-18 x+16=0$

The correct answer is:
$x^2-18 x+16=0$

Solution:
Let numbers be $a, b \Rightarrow a+b=18, \sqrt{a b}=4 \quad \Rightarrow a b=16, a$ and $b$ are roots of the equation
$\Rightarrow \quad x^2-18 x+16=0$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2004
Let two points be A1,-1 and B0,2. If a point Px', y' be such that the area of ∆PAB=5 sq. units and it lies on the line 3x+y-4λ=0, then a value of λ is

Options:
A 4
B 3
C 1
D -3

The correct answer is:
3

Solution:
∆=120211-11x'y'1
⇒-21-x'+y'+x'=±10
⇒-2+2x'+y'+x'=±10
⇒3x'+y'=12 Or 3x'+y'=-8 ...i

x', y' lies on the line 3x+y-4λ=0,

So, 3x'+y'-4λ=0...ii

Solving i and ii, we get,
λ=3, -2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let two sides of a rectangle of area 20 sq. units are along lines xây=0 and x+y=2, then the locus of the point of intersection of diagonals is a

Options:
A pair of ellipse
B pair of straight lines
C pair of hyperbola having eccentricity 2 and 23
D pair of hyperbola each having eccentricity 2

The correct answer is:
pair of hyperbola each having eccentricity 2

Solution:
Let diagonals AC & BD intersect at Fh,k
â Area of quad DEFG is 1420=5
Now distance of h,k from x+y=2 and xây=0 is h+k-22 & h-k2 respectively.
â h+k-22 h-k2=5
â (h-1)+(k-1)h-1-k-1=10
â (h-1)2-(k-1)2=10
â (x-1)2-(y-1)2=10 or x-12-y-12=-10
â Both are hyperbola with eccentricity = 2

Topic: Mathematics, Straight Lines, JEE Main
Let two vertices of a triangle ABC be 2, 4, 6 and 0,-2,-5, and its centroid be 2,1,-1. If the image of the third vertex in the plane x+2y+4z=11 is α, β, γ, then αβ+βγ+γα is equal to

Options:
A 70
B 76
C 74
D 72

The correct answer is:
74

Solution:
Given,

Two vertices of a triangle ABC be 2, 4, 6 and 0,-2,-5, and its centroid be 2,1,-1

Now, let the vertex 'C' be a, b, c and using centroid formula we get,

2=2+0+a3⇒a=4

1=4-2+b3⇒b=1

-1=6-5+c3⇒c=-4

So, the third vertex will be, C4, 1, -4

Now finding image of C in x+2y+4z=11 we get,

α-41=β-12=γ+44=-24+2-16-1116+4+1

⇒α-41=2⇒α=6, β-12=2⇒β=5 and γ+44=2⇒γ=4

Now find the required answer we get,

αβ+βγ+αγ=30+20+24=74

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let u→ be a vector coplanar with the vectors a→=2i^+3j^-k^ and b→= j^+k^ . If u→ is perpendicular to a→ and u→⋅b→=24, then u→2 is equal to:

Options:
A 84
B 336
C 315
D 256

The correct answer is:
336

Solution:
As u → =λ[ a → ×( a → × b → ) ]
u → =λ[ [ a → . b → ] a → −| a → | 2 b → ]
u → =λ[ 2 a → −14 b → ]
u → . b → =2λ[ a → .  b → −7| b → | 2 ]
=2λ[ 2−14 ]
24=−24λ
λ=−1
u → =14 b → −2 a →
u→=-4i^+8j^+16k^
| u → | 2 = 4 2 + 8 2 + 16 2
=336

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2018 (08 Apr)
Let U→=i^+j^, V→=i^-j^ and W→=3i^+5j^+3k^. If n^ is a unit vector such that U→⋅n^=0 and V→⋅n^=0, then W→⋅n^ is equal to

Options:

The correct answer is:
3

Solution:
âµ n^ is perpendicular to both Uâ and Vâ
Hence, n^=Î»UâÃVâ
n^=Â±Î»-2k^
Since n^ is unit vector hence n^=1
â2Î»=1
âÎ»=12
Hence, n^=Â±k^
So, Wâân^=3

Topic: Mathematics, Vector Algebra, JEE Main
Let $\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{w}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If $\hat{\mathrm{n}}$ is a unit vector such that $\vec{u} \cdot \hat{n}=0$ and $\overrightarrow{\mathrm{v}} \cdot \hat{\mathrm{n}}=0$, then $|\overrightarrow{\mathrm{w}} \cdot \hat{\mathrm{n}}|$ is equal to

Options:
A
3

B
0

C
1

D
2

The correct answer is:
3

Solution:
Since $\overrightarrow{\mathrm{n}}$ is perpendicular $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}, \overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}$
$\hat{\mathrm{n}}=\frac{\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 1 & 0 \\ 1 & -1 & 0\end{array}\right|}{\sqrt{2} \times \sqrt{2}}=\frac{-2 \hat{\mathrm{k}}}{2}=-\hat{\mathrm{k}}$
$|\vec{\omega} \cdot \hat{n}|=|(\mathrm{i}+2 \mathrm{j}+3 \mathrm{k}) \cdot(-\mathrm{k})|=|-3|=3$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
Let u â, v â and w â are vectors such that u â+ v â+w â=0 â. If u â=3, v â=4 and w â=5 , then the value of u â.v â+v â.w â+w â.u â is

Options:
A -25
B -27
C 28
D 25

The correct answer is:
-25

Solution:
Given,
u →+ v →+w →=0 →
If u →= 3; v →= 4 and w →= 5
now u→+v→+w→2=0
⇒ u→2+v→2+w→2+2u→.v→+v→.w→+w→.u→=0
∴ 9+16+25+2u→.v→+v→.w→+w→.u→=0
∴ u→.v→+v→.w→+w→.u→=-25

Topic: Mathematics, Vector Algebra, JEE Main
Let $\bar{u}, \bar{v}, \bar{w}$ be such that $|\bar{u}|=1,|\bar{v}|=2,|\bar{w}|=3$. If the projection $\bar{v}$ along $\bar{u}$ is equal to that of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$ and $\overline{\mathrm{v}}, \overline{\mathrm{w}}$ are perpendicular to each other then $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|$ equals

Options:
A
2

B
$\sqrt{7}$

C
$\sqrt{14}$

D
14

The correct answer is:
$\sqrt{14}$

Solution:
Projection of $\bar{v}$ along $\bar{u}$ and $\bar{w}$ along $\bar{u}$ is $\frac{\bar{v} \cdot \bar{u}}{|\bar{u}|}$ and $\frac{\bar{w} \cdot \bar{u}}{|\bar{u}|}$ respectively According to question $\frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} \Rightarrow \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}$. and $\overline{\mathrm{v}} \cdot \overline{\mathrm{w}}=0$
$$
|\bar{u}-\bar{v}+\bar{w}|^2=|\bar{u}|^2+|\bar{v}|^2+|\bar{w}|^2-2 \bar{u} \cdot \bar{v}+2 \bar{u} \cdot \bar{w}-2 \bar{v} \cdot \bar{w}=14 \Rightarrow|\bar{u}-\bar{v}+\bar{w}|=\sqrt{14}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2004
Let us consider a curve, y=fx passing through the point -2, 2 and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x)+xf'(x)=x2. Then

Options:
A x3-3xf(x)-4=0
B x2+2xf(x)-12=0
C x3+xf(x)+12=0
D x2+2xf(x)+4=0

The correct answer is:
x3-3xf(x)-4=0

Solution:
y+xdydx=x2

dydx+yx=x

I.F.=e∫1xdx=x

Solution of differential equation,

y·x=∫x·x dx

xy=x33+C

Passes through -2, 2, So,

-12=-8+3C

C=-43

∴ 3xy=x3-4

i.e 3x·fx=x3-4

x3-3x fx-4=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Let V1→=i^+aj^+k^, V2→=j^+ak^ and V3→=ai^+k^, ∀a>0. If V1→ V2→ V3→ is minimum, then the value of a is

Options:
A 3
B 3
C 13
D 13

The correct answer is:
13

Solution:
V1→ V2→ V3→=1a101aa01=a3-a+1=fa (say)

So, for minima
f'a=3a2−1=0

Hence, for minima a=13 a>0

Topic: Mathematics, Vector Algebra, JEE Main
Let V1= variance of 13,16,19..........103 and V2= variance of 20,26,32..........200. Then V1:V2 is

Options:
A 1:2
B 1:1
C 4:9
D 1:4

The correct answer is:
1:4

Solution:
V1= variance of 13,16,19..........103
= variance of 3,6,9..........93
=9 of variance of 1,2,3..........31
V2= variance of 20,26,32..........200
= variance of 6,12,18..........186
=36 of variance of 1,2,3..........31
V1V2=14

Topic: Mathematics, Statistics, JEE Main
Let $\mathbf{v}=2 \mathbf{i}+\mathbf{j}-\mathbf{k}$ and $\mathbf{w}=\mathbf{i}+3 \mathbf{k}$. If $\mathbf{u}$ is any unit vector, then the m$\sqrt{59}$
aximum value of the scalar triple product $[\mathbf{u} \mathbf{v} \mathbf{w}]$ is

Options:
A 1
B $\sqrt{10}+\sqrt{6}$
C $\sqrt{59}$
D $\sqrt{60}$

The correct answer is:
$\sqrt{59}$

Solution:
$\begin{aligned} & \text { Since, }[\mathbf{u} \mathbf{v} \mathbf{w}]=\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \\ & \Rightarrow[\mathbf{u} \mathbf{v} \mathbf{w}] \leq|\mathbf{u} \| \mathbf{v} \times \mathbf{w}| \\ & \Rightarrow[\mathbf{u} \mathbf{v} \mathbf{w}] \leq|\mathbf{v} \times \mathbf{w}| \\ & \text { Now, } \mathbf{v} \times \mathbf{w}=\left|\begin{array}{rrr}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|=3 \mathbf{i}-7 \mathbf{j}-\mathbf{k} \\ & \therefore \quad \mathbf{v} \times \mathbf{w}=\sqrt{9+49+1}=\sqrt{59}\end{aligned}$
$\therefore \quad[\mathbf{u} \mathbf{v} \mathbf{w}] \leq \sqrt{59}$

Topic: Mathematics, Vector Algebra, JEE Main
Let VâÎ¸=cosâ¡Î¸+secâ¡Î¸a^+cosâ¡Î¸-secÎ¸b^, where a^ and b^ are unit vectors and the angle between a^ and b^ is 60o, then the minimum value of Vâ4 is equal to

Options:

The correct answer is:
12

Solution:
V→2=cos⁡θ+sec⁡θa^+cos⁡θ-sec⁡θb^2
=cos⁡θ+sec⁡θ2+cos⁡θ-sec⁡θ2+(cos⁡θ-sec⁡θ)(cos⁡θ+sec⁡θ)
=3cos2⁡θ+sec2⁡θ≥23
⇒ least value of V→2=23
⇒ least value of V→4=12

Topic: Mathematics, Vector Algebra, JEE Main
Let \( \vec{a}=(1,1,-1), \vec{b}=(5,-3,-3) \) and \( \vec{c}=(3,-1,2) \). If \( \vec{r} \) is collinear with \( \vec{c} \) and has length of \( \frac{|\vec{a}+\vec{b}|}{2} \), then
\( \vec{r} \) equals

Options:
A \( \pm 3 \vec{c} \)
B \( \pm \frac{3}{2} \vec{c} \)
C \( \pm \vec{c} \)
D \( \pm \frac{2}{3} \vec{c} \)

The correct answer is:
\( \pm \vec{c} \)

Solution:
Since r→ is collinear with c→, we can write,

r→=kc→

r→=kc→

Given,

r→=a→+b→2

∴a→+b→2=kc→

6i^−2j^−4k^=2k3i^−j^+2k^

56=2k14

∴k=±1

∴r→=±c→

Topic: Mathematics, Vector Algebra, JEE Main
Let \( \vec{a}=\hat{i}-2 \hat{j}+\hat{k} \) and \( \vec{b}=\hat{i}-\hat{j}+\hat{k} \), be two vectors. If \( \vec{c} \), is a vector such that \( \vec{b} \times \vec{c}=\vec{b} \times \vec{a} \) and \( \vec{c} \cdot \vec{a}=0 \)
then \( \vec{c} \cdot \vec{b} \), is equal to.

Options:
A \( -\frac{3}{2} \)
B \( \frac{1}{2} \)
C \( -\frac{1}{2} \)
D \( -1 \)

The correct answer is:
\( -\frac{1}{2} \)

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
Let \( \vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k} \) and \( \vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k} \)
If the vectors \( \vec{c} \) lies in the plane of \( \vec{a} \) and \( \vec{b} \), then \( x \) equals

Options:
A \( -4 \)
B \( -2 \)
C \( 0 \)
D \( 1 \)

The correct answer is:
\( -2 \)

Solution:
a→=i^+j^+k^ , b→=i^-j^+2k^ and c→=xi^+(x-2)j^-k^

If the vector lies On the same Plane then , we have
a→ b→ c→ =0

1111-12xx-2-1=0
⇒ 1(1 - 2x+ 4) -1 (-1 - 2x) + 1 (x - 2 + x) = 0
⇒ 5 - 2x +1 + 2x + 2x - 2 = 0
⇒ x = - 2.

Topic: Mathematics, Vector Algebra, JEE Main
Let w1 be the point obtained by the rotation of z1=5+4i about the origin through a right angle in the anticlockwise direction, and w2 be the point obtained by the rotation of z2=3+5i about the origin through a right angle in the clockwise direction. Then the principal argument w1-w2 is equal to

Options:
A π-tan-189
B -π+tan-1335
C -π+tan-189
D π-tan-1335

The correct answer is:
π-tan-189

Solution:
Given,

Two complex numbers w1 & w2

w1=3+5i and w2=5+4i are both rotated by 90° with respect to origin anticlock-wise and clockwise respectively,

So by concept of rotation we get,

w3=iw1=i3+5i=-5+3i

And w4=-iw2=-i5+4i=4-5i

Now principal argument of w3-w4=-9+8i will be π-tan-189 {as complex number is in second quadrant so principal argument is given by π-tan-1yx}

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let W denote the words in the English dictionary. Define the relation R by :
$R=\{(x, y) \in W \times W \mid$ the words $x$ and $y$ have at least one letter in common $\}$. Then $R$ is

Options:
A
not reflexive, symmetric and transitive

B
reflexive, symmetric and not transitive

C
reflexive, symmetric and transitive

D
reflexive, not symmetric and transitive

The correct answer is:
reflexive, symmetric and not transitive

Solution:
Clearly $(\mathrm{x}, \mathrm{x}) \in \mathrm{R} \forall \mathrm{x} \in \mathrm{W}$. So, $\mathrm{R}$ is reflexive.
Let $(x, y) \in R$, then $(y, x) \in R$ as $x$ and $y$ have at least one letter in common. So, $R$ is symmetric.
But $R$ is not transitive for example
Let $x=$ DELHI, $y=$ DWARKA and $z=$ PARK
then $(x, y) \in R$ and $(y, z) \in R$ but $(x, z) \notin R$.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2006
Let W denotes the words in the English dictionary. If the relation R is given by

R={x, yâWÃW: the word x and y have at least one letter in common}, then R is

Options:
A reflexive, symmetric and not transitive
B reflexive, symmetric and transitive
C reflexive, not symmetric and transitive
D not reflexive, symmetric and transitive

The correct answer is:
reflexive, symmetric and not transitive

Solution:
Given, R={(x,y)âWÃW: the word x and y have atleast one letter in common}
Let, W={cat, toy, you,â¦}
Clearly, R is reflexive and symmetric but not transitive.
[Since, catRtoy, toyRyouâ catRyou]

Topic: Mathematics, Sets and Relations, JEE Main
Let wIm w≠0 be a complex number. Then, the set of all complex numbers z satisfying the equation w-w¯z=k1-z, for some real number k, is

Options:
A z: z≠1
B z: z=1, z≠1
C z: z=z¯
D z: z=1

The correct answer is:
z: z=1, z≠1

Solution:
w∈C, Im w≠0 ⇒w=a+ib:b≠0 (suppose)
w-w¯ z=k1-z
If z=x+iy
⇒a+ib-a-ibx+iy=k{1-x+iy}
⇒a+ib-ax+iay-ibx+by=k-kx-iky
Equating real and imaginary parts
a-ax-by=k-kx …(1)
and b-ay+bx=-ky …2
From 1 k-ax-1=by
2 k-ay=b(x+1)
Dividing the two equations, we get,

x-1y=yx+1
⇒x2-1=y2
⇒x2+y2=1
⇒z=1

z≠1+i0 (∴ if z=1 then from original equation w=w ⇒a+ib=a-ib⇒b=0)

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2014 (09 Apr Online)
Let w=zz¯+k1z+k2iz+λ(1+i),k1,k2∈ℝ. . Let Re(w)=0 be the circle C of radius 1 in the first quadrant touching the line y=1 and the y-axis. If the curve Im(w)=0 intersects C at A and B, then 30(AB)2 is equal to _______.

Options:

The correct answer is:
24

Solution:
Given,

w=zz¯+k1z+k2iz+λ(1+i),k1,k2∈ℝ. .

And Re(w)=0,

So, taking z=x+iy & z¯=x-iy

We get,

w=x2+y2+k1x+iy+k2ix+iy+λ(1+i)

⇒Rew=0⇒x2+y2+k1x-k2y+λ=0

And, Im(w)=0⇒k1y+k2x+λ=0

Now given Re(w)=0 circle is touching y=1 and y-axis, with radius 1, so we have equation of circle

x2+y2-2x-4y+4=0

So, on comparing with Re(w)=0 we get,

k1=-2,k2=4,λ=4

So, Im(w)=0⇒-2y+4x+4=0

And given -2 y+4 x+4=0 intersects the circle,

So, solving x2+y2-2x-4y+4=0 and -2 y+4 x+4=0 we get,

Intersecting point as A0,2 and B25,145

Hence, by distance formula we get,

30(AB)2=300-252+2-1452=24

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (13 Apr Shift 1)