datavorous/entrance-exam-dataset huggingface dataset:
Let the system of equations x+2y+3z=5,2x+3y+z=9,4x+3y+λz=μ have infinite number of solutions. Then λ+2μ is equal to:

Options:
A 28
B 17
C 22
D 15

The correct answer is:
17

Solution:
Given: x+2y+3z=5, 2x+3y+z=9 and 4x+3y+λz=μ.

For infinite solutions, Δ=Δ1=Δ2=Δ3=0.

⇒Δ=12323143λ=0

⇒3λ-3-4λ+8+18-36=0

⇒λ=−13

Now, finding

⇒Δ1=523931μ3−13=0

⇒μ=15

Now, finding

⇒Δ2=1532914μλ

⇒Δ2=153291415−13

⇒Δ2=0

And

⇒Δ3=12523943μ

⇒Δ3=1252394315

⇒Δ3=0

For λ=−13, μ=15 system of equation has infinite solutions.

⇒λ+2μ=17.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Let the system of linear equations
4x+λy+2z=0
2x-y+z=0
μx+2y+3z=0, λ, μ∈R
has a non-trivial solution. Then which of the following is true?

Options:
A μ=6, λ∈R
B λ=2, μ∈R
C λ=3, μ∈R
D μ=-6, λ∈R

The correct answer is:
μ=6, λ∈R

Solution:
For non-trivial solution

4λ22-11μ23=0

⇒-20-6λ+λμ+8+2μ=0

⇒2μ-6λ+λμ=12

when μ=6, 12-6λ+6λ=12

which is satisfied by all λ

When λ=2, 4μ=24

When λ=3, 5μ=30

When μ=-6, -12-6λ-6λ=12

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let the system of linear equations

–x+2y-9z=7

-x+3y+7z=9

-2x+y+5z=8

-3x+y+13z=λ

has a unique solution x=α, y=β, z=γ. Then the distance of the point α, β, γ from the plane 2x-2y+z=λ is

Options:
A 11
B 7
C 9
D 13

The correct answer is:
7

Solution:
Given,

System of linear equation,

–x+2y-9z=7 .......i

–x+3y+7z=9 .......ii

–2x+y+5z=8 ......iii

–3x+y+13z=λ ......iv

Now solving, equations i, ii & iii we get,

x=–3, y=2, z=0

Now substituting in equation iv we get,

3×3+2=λ

⇒λ=11

Now finding the distance of Point –3, 2, 0 from the plane 2x-2y+z=11 we get,

d=-6-4-1122+22+1=213=7

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let the system of linear equations x+2y+z=2, αx+3y-z=α,-αx+y+2z=-α be inconsistent. Then α is equal to

Options:
A 52
B -52
C 72
D -72

The correct answer is:
-72

Solution:
Given, the system of linear equations x+2y+z=2, αx+3y-z=α, and -αx+y+2z=-α

Now for system to be inconsistent, ∆=0 & either of ∆1,∆2 or ∆3 ≠0

So, Δ=12123-1-212

=6+y-22α-α+1α+3α

=7-2α+4α

=7+2α

Δ=0⇒α=-72

Now calculating, Δ1=221α3-1-α12

=14+2α

Where α=-72 so, ∆1=14-7=7

Δ1≠0, hence system is inconsistent at α=-72

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Let the tangent at any point P on a curve passing through the points 1,1 and 110,100, intersect positive x-axis and y-axis at the points A and B respectively. If P A: P B=1: k and y=yx is the solution of the differential equation edydx=kx+k2,y0=k, then 4y1-5loge3 is equal to _______________

Options:

The correct answer is:
5

Solution:
Given,

The tangent at any point P on a curve passing through the points 1,1 and 110,100, intersect positive x-axis and y-axis at the points A and B respectively,

And P A: P B=1: k and y=yx is the solution of the differential equation edydx=kx+k2,y0=k,

Now on plotting the diagram we get,

Equation of tangent at Px, y is :

Y-y=dydxX-x

Coordinate of A=x-ydxdy,0

Coordinate of B=0,y-xdydx

∴ x,y=kx-kydxdyk+1,y-xdydxk+1

So, y=y-xdydxk+1

⇒yk+1=y-xdydx

⇒ky=-xdydx

⇒kdxx=-dyy

Now integrating both side, we get

kln|x|+ln|y|=lnC ......i

Now given equation i passes through 1, 1 and 110,100

So, C = 1 and k = 2

Now putting the value of k, we get

edydx=kx+k2

⇒edydx=2x+1 as k=2

⇒dydx=ln2x+1

Integrating the above equation we get,

y=122x+1ln2x+1-1+C

And the above equation passes through 0, 2

⇒C=52

So, 2y=2x+1ln2x+1-1+5

⇒2y1=3ln3-1+5

⇒2y1=3ln3+2

Hence, 4y1-5loge3=23ln3+2-5ln3

⇒4y1-5loge3=6ln3+4-5ln3

⇒4y1-5loge3=4+ln3≈5

Note: This question was bonus in Jee Mains 2023 April session.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let the tangent drawn to the parabola y2=24x at the point α,β is perpendicular to the line 2x+2y=5. Then the normal to the hyperbola x2α2-y2β2=1 at the point α+4,β+4 does NOT pass through the point:

Options:
A 25,10
B 20,12
C 30,8
D 15,13

The correct answer is:
15,13

Solution:
Given, the tangent drawn to the parabola y2=24x at the point α,β is perpendicular to the line 2x+2y=5.

So, tangent at α,β has slope 1

And α,β lies on y2=24x so β2=24α

Equation of tangent will be yβ=12x+α so its slope will be 12β=1, so β=12

⇒α=6,β=12

∴ α+4,β+4=10,16

Now finding normal at 10,16 to x236-y2144=1,

First finding slope of the tangent 2x36-2y144dydx=0

⇒dydx10,16=144×1036×16=52, so slope of the normal will be -25

Now equation of normal will be y-16=-25x-10

⇒2x+5y=100

Now satisfying all option one by one we can see 15,13 will not satisfy the given line.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Let the tangent to the circle x2+y2=25 at the point R(3,4) meet x -axis and y-axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r2 is equal to

Options:
A 52964
B 12572
C 62572
D 58566

The correct answer is:
62572

Solution:
The equation of tangent to a circle x2+y2=a2 at a point x1, y1 is xx1+yy1=a2.

Thus, the tangent to circle x2+y2=25 at R3, 4 is 3x+4y=25.

To find the coordinates of the point P on the x-axis, put y=0, to get 3x=25

⇒x=253

Thus, the coordinates of the point P are 253, 0.

Similarly, to find the coordinates of the point Q on the y-axis, put x=0, to get 4y=25

⇒y=254

Thus, the coordinates of the point Q are 0, 254.

The incentre of a triangle with vertices x1, y1, x2, y2 and x3, y3 is ax1+bx2+cx3a+b+c, ay1+by2+cy3a+b+c, where a, b, c units are respectively the lengths of the sides opposite to vertices with coordinates x1, y1, x2, y2 & x3, y3.

In ∆OPQ, we have OP=253, OQ=254 and PQ=253-02+0-2542=12512 units.

Thus, the incentre of ∆OPQ=0+0+254×25325, 0+0+254×25325

=2512, 2512.

The point 2512, 2512 is the centre of the circle of radius r which passes through origin O, thus r=2512-02+2512-02

∴ r2=225122=2×625144=62572.

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let the tangent to the curve x2+2x-4y+9=0 at the point P1,3 on it meet the y-axis at A. Let the line passing through P and parallel to the line x-3y=6 meet the parabola y2=4x at B. If B lies on the line 2x-3y=8, then AB2 is equal to _______.

Options:

The correct answer is:
292

Solution:
Given,

The tangent to the curve x2+2x-4y+9=0 at the point P1,3 on it meet the y-axis at A,

So, tangent to circle at point P1,3 is given by,

x+x+1-2y+3+9=0

⇒2x-2y=-4

So, on y-axis, A0,2

And the line passing through P and parallel to the line x-3y=6 meet the parabola y2=4x at B,

So, the equation of line through P will be y-3=13x-1

⇒3y=x+8

Now finding intersection of y2=4x & 3y=x+8 we get,

B=4,4 & 16,8,

Also B lies on the line 2x-3y=8, so 4,4 will not satisfy,

Hence, B=16,8

So, AB2=0-162+2-82=256+36=292

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let the tangent to the parabola S:y2=2x at the point P2,2 meet the x-axis at Q and normal at it meet the parabola S at the point R. Then the area (in sq. units) of the triangle PQR is equal to:

Options:
A 252
B 352
C 152
D 25

The correct answer is:
252

Solution:
We have,

y2=2x

Tangent at P2,2 is

y2=212x+2

⇒ 2y=x+2

It meets x-axis at y=0 i.e., x=-2

∴ Q≡-2,0

Normal at P is

⇒ y-2=-2x-2

⇒ y=6-2x

∴ Solving with y2=2x, we get

6-2x2=2x

⇒ 4x2+36-24x=2x

⇒ 4x2-26x+36=0

⇒ 2x2-13x+18=0

⇒ x=13±169-1444=13±54

⇒ x=92, 12

∴ R≡92,-3

∴ ArΔPQR=12221-20192-31

⇒ ArΔPQR=122·3-2-2-92+6

⇒ ArΔPQR=1212+13
⇒ ArΔPQR=252 sq.units

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let the tangent to the parabola y2=12x at the point 3, α be perpendicular to the line 2x+2y=3. Then the square of distance of the point 6, -4 from the normal to the hyperbola α2x2-9y2=9α2 at its point α-1, α+2 is equal to .............

Options:

The correct answer is:
116

Solution:
Given,

The tangent to the parabola y2=12x at the point 3, α be perpendicular to the line 2x+2y=3

So, Slope of tangent =1=6α⇒α=6
So, equation of hyperbola will be,

36x2-9y2=324

⇒x29-y236=1
Now equation of tangent at 5,8 will be,

5x9-8y36=1

⇒5x-2y=9
So, slope of normal=-25
Now finding, equation of normal y-8=-25x-5
⇒5y-40=-2x+10

⇒5y+2x=50

Now finding,distance from 6, -4, we get,
=12-20-5029=5829

=229

Hence, 2292=116

Topic: Mathematics, Hyperbola, JEE Main
Let the tangent to the parabola y2=12x at the point 3, α be perpendicular to the line 2x+2y=3. Then the square of distance of the point 6, -4 from the normal to the hyperbola α2x2-9y2=9α2 at its point α-1, α+2 is equal to .............

Options:

The correct answer is:
116

Solution:
Given,

The tangent to the parabola y2=12x at the point 3, α be perpendicular to the line 2x+2y=3

So, Slope of tangent =1=6α⇒α=6
So, equation of hyperbola will be,

36x2-9y2=324

⇒x29-y236=1
Now equation of tangent at 5,8 will be,

5x9-8y36=1

⇒5x-2y=9
So, slope of normal=-25
Now finding, equation of normal y-8=-25x-5
⇒5y-40=-2x+10

⇒5y+2x=50

Now finding,distance from 6, -4, we get,
=12-20-5029=5829

=229

Hence, 2292=116

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let the tangents at points P and R on the parabola y=x2 intersects at T. Tangent at point Q (lies in between the points P and R ) on the parabola intersect PT and RT at A and B respectively. The value of TATP+TBTR is

Options:
A 12
B 1
C 2
D 4

The correct answer is:
1

Solution:
Let, P=t1,t12, Q=t2,t22 & R=t3,t32
⇒TATP=t1+t32-t1+t22t1+t32-t1=t3-t2t3-t1 and

TBTR=t1-t2t1-t3=t2-t1t3-t1
⇒TATP+TBTR=t3-t2t3-t1+t2-t1t3-t1=t3-t1t3-t1=1

Topic: Mathematics, Parabola, JEE Main
Let the tangents at the points A(4,-11) and B(8,-5) on the circle x2+y2-3x+10y-15=0, intersect at the point C. Then the radius of the circle, whose centre is C and the line joining A and B is its tangent, is equal to

Options:
A 334
B 213
C 13
D 2133

The correct answer is:
2133

Solution:
Given,

The tangents at the points A(4,-11) and B(8,-5) on the circle x2+y2-3x+10y-15=0, intersect at the point C.

So, equation of tangent at A(4,-11) on circle will be

⇒4x-11y-3x+42+10y-112-15=0

⇒5x-12y-152=0 .........1

And equation of tangent at B(8,-5) on circle is

⇒8x-5y-3x+82+10y-52-15=0

⇒13x-104=0⇒x=8 .........2

Now putting the value of x=8 in equation 1 we get, y=-283

So, the point C will be C8,-283

Now equation of tangent AB will be,

y+11=-5+118-4x-4

⇒3x-2y-34=0

r=3×8+2×283-3413=-26313=2133

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let the tangents at two points A and B on the circle x2+y2-4x+3=0 meet at origin O0,0. Then the area of the triangle of OAB is

Options:
A 332
B 334
C 323
D 343

The correct answer is:
334

Solution:
Given the tangents at two points A and B on the circle x2+y2-4x+3=0 meet at origin O0,0,

Plotting the diagram we get,

Now circle C:x-22+y2=1

Equation of chord AB is given by T=0 i.e x×0+y×0-4x+02+3=0

⇒AB:2x=3

Now OA=OB=02+02-4×0+3=3

So, AM=OA2-OM2=3-94=34=32

Now area of triangle OAB=122AMOM

=334 sq. units

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let the tangents drawn from the origin to the circle, x2+y2-8x-4y+16=0 touch it at the points A and B . Then AB2 is equal to

Options:
A 525
B 565
C 645
D 325

The correct answer is:
645

Solution:
L=S1=16=4
R=16+4-16=2
Length of chord of contact =2LRL2+R2=2×4×216+4=1620

Topic: Mathematics, Circle, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Let the tangents drawn to the circle, x2+y2=16 from the point P0,h meet the x-axis at points A and B. If the area of ΔAPB is minimum, then positive value of h is:

Options:
A 42
B 32
C 43
D 33

The correct answer is:
42

Solution:
Let equation of tangent be xcos⁡ α+ysin⁡ α=4
Now, its points of intersection with co-ordinate axis are
A4cosα,0, P0,4sinα
Now, ar∆PAB=2ar∆PAO
=2×12OA×OP
=4sin⁡α . 4cos⁡α=32sin⁡2α≥32
which is maximum if, α=π4
∴ P0,4sin⁡π4
∴ h=42

Topic: Mathematics, Circle, JEE Main, JEE Main 2015 (10 Apr Online)
Let the tangents PQ and PR are drawn to y2=4ax from any point P on the line x+4a=0. The angle subtended by the chord of contact QR at the vertex of the parabola y2=4ax is

Options:
A π4
B π3
C π2
D π6

The correct answer is:
π2

Solution:
Let, Q=at12,2at1 and R=at22,2at2, then P=at1t2,at1+t2 which lies on x+4a=0
âat1t2=-4aât1t2=-4
Now, the vertex of the parabola y2=4ax is the origin O0,0
âslope of OQ=2t1 & slope of OR=2t2
â2t1Ã2t2=4t1t2=4-4=-1
âOQ is perpendicular to OR
ââ QOR=90â=Ï2

Topic: Mathematics, Parabola, JEE Main
Let the tangents to the parabola y2=4ax drawn from point P have slope m1 and m2. If m1m2=2, then the locus of point P is

Options:
A x=a
B x=a2
C x+a=0
D x=2a

The correct answer is:
x=a2

Solution:
Let co-ordinates of P be h,k and equation of the tangent is y=mx+am
Now, hm2-km+a=0

m1m2=ah=2 (given)
Hence, the locus is x=a2

Topic: Mathematics, Parabola, JEE Main
Let the variable line y=kx+h is tangent to the hyperbola x24-y29=1. If the locus of Ph,k is a conic, then which of the following statement is false about this conic?

Options:
A Vertices of conic lies on y-axis
B The eccentricity of conic is greater than the eccentricity of given hyperbola.
C The centre of conic is at the origin
D The conic is an ellipse

The correct answer is:
The conic is an ellipse

Solution:
Equation of the tangent y=kx+h is compared with y=mx±4m2-9
⇒1=mk=±4m2-9h
i.e. m=k and h=4m2-9
⇒h=4k2-9
⇒4y2-x2=9
⇒y2322-x232=1
⇒32=322e2-1
⇒e2=5⇒e=5>1

Hence, the conic is not an ellipse

Topic: Mathematics, Hyperbola, JEE Main
Let the vectors 2+a+bi^+a+2b+cj^-b+ck^, 1+bi^+2bj^-bk^ and 2+bi^+2bj^+1-bk^, ∀a,b,c∈R be co-planar. Then which of the following is true?

Options:
A 2b=a+c
B 3c=a+b
C a=b+2c
D 2a=b+c

The correct answer is:
2b=a+c

Solution:
If the vectors are co-planar,

a+b+2a+2b+c-b-cb+12b-bb+22b1-b=0

Now R1↔R1-R2, R3↔R3-R2

So a+1a+c-cb+12b-b101=0

⇒(a+1)2b-(a+c)(2b+1)-c(-2b)=0

⇒2ab+2b-2ab-a-2bc-c+2bc=0

⇒2b-a-c=0⇒2b=a+c

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let the vectors a→, b→ and c→ are perpendicular to b→+c→, c→+a→ and a→+b→ respectively. If a→+b→=6, |b→+c→|=8 and c→+a→=10, then the value of |a→+b→+c→| is equal to

Options:
A 55
B 50
C 102
D 10

The correct answer is:
10

Solution:
Since, a→⋅b→+c→=0
⟹a→⋅b→+a→⋅c→=0
Similarly, b→⋅c→+b→⋅a→=0
& c→⋅a→+c→⋅b→=0
∴ a→⋅b→+a→⋅c→+b→⋅c→=0 ...(i)
Given, |a→+b→|=6
⟹|a→|2+|b→|2+2a→⋅b→=36 ....(ii)
Similarly, |b→|2+|c→|2+2b→⋅c→=64 ...(iii)
& |c→|2+|a→|2+2c→⋅a→=100 ....(iv)
On adding Eqs. (ii),(iii) and (iv), we get
2a→|2+2b→|2+2|c→|2+2a→⋅b→+b→⋅c→+c→⋅a→=200
⟹|a→|2+|b→|2+|c→|2=100...(v)[from Eqs. (i)]
Now, |a→+b→+c→|2=a→2+|b→|2+|c→|2+2a→⋅b→+b→⋅c→+c→⋅a→
⟹|a→+b→+c→|2=100 [from Eqs. (i) and (v)]
⟹|a→+b→+c→| =10

Topic: Mathematics, Vector Algebra, JEE Main
Let the vectors a→,b→,c→ be such that |a→|=2,|b→|=4 and |c→|=4. If the projection of b→ on a→ is equal to the projection of c→ on a→ and b→ is perpendicular to c→, then the value of |a→+b→-c→| is …

Options:

The correct answer is:
6

Solution:
b→.a→=c→.a→
a→+b→-c→2=|a→|2+|b→|2+|c→|2+2a→·b→-b→·c→-a→·c→
=4+16+16+2(a→.b→-0-a→.b→)=36
⇒|a→+b→-c→|=6

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Let the vectors a→, b→, c→ represent three coterminous edges of a parallelopiped of volume V. Then the volume of the parallelopiped, whose coterminous edges are represented by a→, b→+c→ and a→+2b→+3c→ is equal to

Options:
A 2V
B 6V
C V
D 3V

The correct answer is:
V

Solution:
We know that,

Volume of parallelepiped whose edges determined by vectors a→, b→, c→ is a→b→c→.

Now it is given that V is volume of parallelepiped a→, b→, c→, so required volume of parallelepiped with new edges will be

=a→b→+c→a→+2b→+3c→

=100011123a→b→c→

=13-2-0+0a→b→c→

=a→b→c→

=V

Hence this is the correct option.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Let the vectors u1→=i^+j^+ak^, u2→=i^+bj^+k^, and u3→=ci^+j^+k^ be coplanar. If the vectors v1→=a+bi^+cj^+ck^, v2→=ai^+b+cj^+ak^ and v→3=bi^+bj^+c+ak^ are also coplanar, then 6a+b+c is equal to

Options:
A 0
B 4
C 12
D 6

The correct answer is:
12

Solution:
Given: u→1=i^+j^+ak^, u→2=i^+bj^+k^ and u→3=ci^+j^+k^ are coplanar.

Also given that,v→1=a+bi^+cj^+ck^, v→2=ai^+b+cj^+ak^and v→3=bi^+bj^+c+ak^ are coplanar.

Now, using the condition of coplanar we get,

⇒11a1b1c11=0

Expanding the determinant along R1.

⇒b-1-1-c+a1-bc=0

⇒a+b+c=2+abc .....i

Again using the coplanar condition we get,

⇒a+bccab+cabbc+a=0

Apply row transformations, R3→R3-R1+R2

⇒a+bccab+ca-2a-2c0=0

Expand the determinant along R1.

⇒a+b0+2ac-c0+2a2+c-2ac+2ab+c=0

⇒2a2c+2abc-2a2c-2ac2+2abc+2ac2=0

⇒abc=0

∴ a+b+c=2 (From eq i)

∴ 6a+b+c=12

Hence this is the correct option.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let the vertices of a triangle are A=-3+2sinâ¡Î¸,4+2cosâ¡Î¸, B=-3+2cosâ¡Î¸,4-2sinâ¡Î¸ and C=-3-2sinâ¡Î¸,4-2cosâ¡Î¸, then the distance between the centroid and the circumcentre of ÎABC is

Options:
A 23 units
B 32 units
C 12 units
D 13 units

The correct answer is:
23 units

Solution:
A,B,C lie on the circle x+32+y-42=22
â circumcentre of ÎABC is -3,4=S & the centroid of ÎABC is -3+23cosâ¡Î¸,4-23sinâ¡Î¸=G
Distance between S & G is 232cos2Î¸+232sin2Î¸=23 units

Topic: Mathematics, Circle, JEE Main
Let the volume of a parallelepiped whose coterminous edges are given by u→=i^+j^+λk^, v→=i^+j^+3k^ and w→=2i^+j^+k^ be 1 cu. unit. If θ be the angle between the edges u→ and w→, then the value of cos⁡θ can be

Options:
A 766
B 763
C 57
D 533

The correct answer is:
763

Solution:
±1=11λ113211 ⇒=-λ+3=±1⇒λ=2 or λ=4
For λ=4
cosθ=2+1+4618=763

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let the volume of a parallelepiped whose coterminous edges are given by uâ=i^+j^+Î»k^, vâ=i^+j^+3k^ and wâ=2i^+j^+k^ be 1 cu. unit. If Î¸ be the angle between the edges uâ and wâ, then the value of cosâ¡Î¸ can be

Options:
A 766
B 763
C 57
D 533

The correct answer is:
763

Solution:
Â±1=11Î»113211â⇒=-Î»+3=Â±1⇒Î»=2âorâÎ»=4
For Î»=4
cosÎ¸=2+1+4618=763

Topic: Mathematics, Vector Algebra, JEE Main
Let the volume of a parallelepiped whose coterminous edges are given by \( \vec{u}=\hat{i}+\hat{j}+\lambda \hat{k}, \vec{v}=\hat{i}+\hat{j}+3 \hat{k} \) and \( \vec{w}=2 \hat{i}+\hat{j}+\hat{k} \) be \( 1 \) cu. unit. If \( \theta \) be the angle between the edges \( \vec{u} \) and \( \vec{w} \), then the value of \( \cos \theta \) can be

Options:
A \( \frac{1}{33} \)
B \( \frac{7}{6 \sqrt{3}} \)
C \( \frac{3}{2} \)
D \( \frac{5}{3 \sqrt{3}} \)

The correct answer is:
\( \frac{7}{6 \sqrt{3}} \)

Solution:
±1=11λ113211 ⇒=-λ+3=±1⇒λ=2 or λ=4
For λ=4
cosθ=2+1+4618=763

Topic: Mathematics, Vector Algebra, JEE Main
Let the $x-z$ plane be the boundary between two transparent media. Medium 1 in $z \geq 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z < 0$ has a refractive index of $\sqrt{3}$. A ray of light in medium 1 given by the vector $\vec{A}=6 \sqrt{3} \hat{i}+8 \sqrt{3} \hat{j}-10 \hat{k}$ is incident on the plane of separation. The angle of refraction in medium 2 is

Options:
A
$45^{\circ}$

B
$60^{\circ}$

C
$75^{\circ}$

D
$30^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
Normal to the plane is $z$-axis
$$
\begin{aligned}
& \cos \theta_1=\frac{A_z}{A}=\frac{10}{20}=\frac{1}{2}, \theta_1=60 \\
& \mu_1 \sin \theta_1=\mu_2 \sin \theta_2 \Rightarrow \sqrt{2} \times \frac{\sqrt{3}}{2}=\sqrt{3} \sin \theta_2 \Rightarrow \theta_2=45^{\circ}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2011
Let there are 4 sections of 25 students each in a coaching class. Now, out of 150 students 100 are to be selected randomly and enrolled in these sections. Then, the probability that the students A and B (both present in 150 students) are selected and placed in the same section, is

Options:
A 1013
B 16149
C 233
D 5149

The correct answer is:
16149

Solution:
Probability that A is selected =100150=23
Now, probability that B is in same section =24149
Then, the required probability =23Ã24149=16149

Topic: Mathematics, Probability, JEE Main
Let there are exactly two points on the ellipse x2a2+y2b2=1 whose distances from 0,0 are equal to a22+b2. Then, the eccentricity of the ellipse is equal to

Options:
A 12
B 122
C 12
D None of these

The correct answer is:
12

Solution:
Since, a22+b2>b
Hence, major axis =2a & minor axis =2b and a22+b2=aâa22+b2=a2
âb2=a22âb2a2=12
â1-e2=12âe2=12=e=12

Topic: Mathematics, Ellipse, JEE Main
Let there be a spherical symmetric charge distribution with charge density varying as pr=p0rR upto r=R and Ïr=0 for r>R, where r is the distance from the origin. The electric field at on a distance r r<R from the origin is given by -

Options:
A ρ0r24ε0R
B ρ0r4ε0R
C ρ0r4ε0R
D ρ0r2ε0R

The correct answer is:
ρ0r24ε0R

Solution:
â®Eââ. dsâ=qinÎµ0 ....(1)
q in =â«0rÏ4Ïr2dr=â«0rÏ0rR4Ïr2dr=4ÏÏ0Rr44
=ÏÏ0r4R , from (1)
E.4Ï r2 =ÏÏ0r4Îµ0R â´ E =Ï0r24Îµ0R

Topic: Physics, Electrostatics, JEE Main
Let there be a spherically symmetric charge distribution with charge density varying as $\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=R$, and $\rho(r)=0$ for $r>R$, where $r$ is the distance from the origin. The electric field at a distance $r(r < R)$ from the origin is given by

Options:
A
$\frac{4 \pi \rho_0 r}{3 \varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)$

B
$\frac{\rho_0 r}{4 \varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)$

C
$\frac{4 \rho_0 r}{3 \varepsilon_0}\left(\frac{5}{4}-\frac{r}{R}\right)$

D
$\frac{\rho_0 r}{3 \varepsilon_0}\left(\frac{5}{4}-\frac{r}{R}\right)$

The correct answer is:
$\frac{\rho_0 r}{4 \varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)$

Solution:
Apply shell theorem the total charge upto distance $r$ can be calculated as followed $d q=4 \pi r^2 \cdot d r . \rho$
$$
\begin{aligned}
& =4 \pi r^2 \cdot d r \cdot \rho_0\left[\frac{5}{4}-\frac{r}{R}\right] \\
& =4 \pi \rho_0\left[\frac{5}{4} r^2 d r-\frac{r^3}{R} d r\right] \\
& \int d q=q=4 \pi \rho_0 \int_0^r\left(\frac{5}{4} r^2 d r-\frac{r^3}{R} d r\right) \\
& =4 \pi \rho_0\left[\frac{5}{4} \frac{r^3}{3}-\frac{1}{R} \frac{r^4}{4}\right] \\
& E=\frac{k q}{r^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{1}{r^2} \cdot 4 \pi \rho_0\left[\frac{5}{4}\left(\frac{r^3}{3}\right)-\frac{r^4}{4 R}\right] \\
& E=\frac{\rho_0 r}{4 \varepsilon_0}\left[\frac{5}{3}-\frac{r}{R}\right] \\
&
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2010