datavorous/entrance-exam-dataset huggingface dataset:
Let the foci of the ellipse $\frac{x^{2}}{9}+y^{2}=1$ subtend a right angle at a point $P$. Then, the locus of $P$ is

Options:
A $x^{2}+y^{2}=1$
B $x^{2}+y^{2}=2$
C $x^{2}+y^{2}=4$
D $x^{2}+y^{2}=8$

The correct answer is:
$x^{2}+y^{2}=8$

Solution:
$\frac{x^{2}}{9}+\frac{y^{2}}{1}=1 \Rightarrow e=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$
0) ie, $(\pm 2 \sqrt{2}, 0)$ Two foci are $(\pm a e,$ Let $P(h, k)$ by any point on the ellipse
$\therefore$
$\frac{k-0}{h-2 \sqrt{2}} \times \frac{k-0}{h+2 \sqrt{2}}=-1$
[From given condition]
$\Rightarrow \quad h^{2}-8=-k^{2}$
$\Rightarrow \quad x^{2}+y^{2}=8$

Topic: Mathematics, Ellipse, JEE Main
Let the foci of the ellipse x216+y27=1 and the hyperbola x2144-y2α=125 coincide. Then the length of the latus rectum of the hyperbola is:

Options:
A 329
B 185
C 274
D 2710

The correct answer is:
2710

Solution:
Given equation of ellipse x216+y27=1

Now finding eccentricity =1-716=34

So, foci ≡±ae,0≡±3,0

Now, hyperbola: x214425-y2α25=1

Eccentricity will be =1+α144=112144+α

Foci ≡±ae,0≡±125·112144+α,0

Given foci coincide then 3=15144+α⇒α=81

Hence, hyperbola is x21252-y2952=1

Length of latus rectum =2b2a=2·8125125=2710

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Let the focus(S) of a parabola divides its one of the focal chords PQ in the ratio 2:1. If the tangent at Q cuts the directrix at R such that RQ=6, then the distance (in units) of the focus from the tangent at P is

Options:

The correct answer is:
4

Solution:
Let, the foot of the perpendicular from the focus on the tangent PR is A which lies on the tangent at the vertex.
Also, â PRQ=90o
Now, in ÎPRQ & ÎPAS,â P is common & â R=â A
Then, ÎPRQ is similar to ÎPAS
âPSPQ=ASRQ=23âAS=23Ã6=4
â The distance of the focus from the tangent at P=4 units

Topic: Mathematics, Parabola, JEE Main
Let the focus S of the parabola y2=8x lie on the focal chord PQ of the same parabola. If the length QS=3 units, then the ratio of length PQ to the length of the latus rectum of the parabola is

Options:
A 25
B 45
C 54
D 98

The correct answer is:
98

Solution:
For the focal chord PQ, we know that,

PS,4,SQ are in H.P.
â1PS,14,13 are in A.P.
â1PS=24-13=16âPS=6
âPQ=6+3=9
âPQlength of latus rectum=98

Topic: Mathematics, Parabola, JEE Main
Let the focus S of the parabola y2=8x lies on the focal chord PQ of the same parabola. If PS=6, then the square of the slope of the chord PQ is

Options:
A 25
B 45
C 54
D 94

The correct answer is:
45

Solution:
We know that, PS, 4, SQ are in H.P.
⇒16,14,1SQ are in A.P.
⇒1SQ=24-16=13⇒SQ=3
⇒PQ=6+3=9=4cosec2θ (where, tan⁡θ is the slope of PQ )
⇒1+cot2⁡θ=94⇒cot2⁡θ=54
⇒tan2⁡θ=45

Topic: Mathematics, Parabola, JEE Main
Let the foot of perpendicular from a point P(1,2,-1) to the straight line L:x1=y0=z-1 be N. Let a line be drawn from P parallel to the plane x+y+2z=0 which meets L at point Q. If α is the acute angle between the lines PN and PQ, then cosα is equal to

.

Options:
A 15
B 32
C 13
D 123

The correct answer is:
13

Solution:
Assume, x1=y0=z-1=λ

So, coordinate of any point on the line is λ,0,-λ.

As N is the foot of the perpendicular,

PN→·i^-k^=0

λ-1i^+0-2j^+-λ+1k^·i^-k^=0

⇒λ-1i^+0-2j^+-λ+1k^·i^-k^=0

⇒λ-1×1+-2×0+-λ+1×-1=0

⇒λ-1+0+λ-1=0

⇒2λ-1=0

⇒λ=1

Hence, the coordinate of N=(1,0,-1)

Now,

Assume, the coordinate of Q=μ,0,-μ

So, PQ→·i^+j^+2k^=0

μ-1i^+0-2j^+-μ+1k^·i^+j^+2k^=0

μ-1×1+0-2×1+-μ+1×2=0

μ-1+-2+-2μ+2=0

-μ-1=0

⇒μ=-1

So, the coordinate of Q=(-1,0,1)

Hence, PN→=2j^ and PQ→=2i^+2j^-2k^

⇒cosα=PN→·PQ→PN→PQ→

=2j^·2i^+2j^-2k^2222+22+22=13

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let the foot of perpendicular from the point A4,3,1 on the plane P: x-y+2z+3=0 be N. If B(5,α,β),α,β∈ℤ is a point on plane P such that the area of the triangle ABN is 32, then α2+β2+αβ is equal to _____________

Options:

The correct answer is:
7

Solution:
Given,

The foot of perpendicular from the point A4,3,1 on the plane P: x-y+2z+3=0 be N

So, by foot of perpendicular formula we get,

x1-41=y1-3-1=z1-12=-4-3+2+312+12+22

⇒N≡x1,y1,z1=(3,4,-1)

And B5,α,β,α,β∈ℤ is a point on plane P such that the area of the triangle ABN is 32,

So putting the value of point B5,α,β in plane we get,

5-α+2β+3=0

⇒α-2β=8

And given area of triangle ABN is 32, we get

12AN×BN=32

⇒AN×BN=62

⇒66×BN=62

⇒BN=23

⇒4+α-42+β+12=12

⇒α-42+β+12=8

⇒5β2+18β+9=0 as α-2β=8

⇒β=-3 as β≠-35

So, β=-3, α=2

Hence, α2+β2+αβ=32+22-6=7

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let the foot of perpendicular of the point P3, –2, –9 on the plane passing through the points –1, –2, –3, 9, 3, 4, 9, –2, 1 be Qα, β, γ. Then the distance Q from the origin is

Options:
A 42
B 38
C 35
D 29

The correct answer is:
42

Solution:
Given that three points lie on a plane,

AC→=10i^+4k^

AB→=10i^+5j^+7k^

AC→×AB→=i^j^k^10041057

⇒AC→×AB→=-20i^-30j^+50k^

⇒AC→×AB→=2i^+3j^-5k^

Equation of plane will be,

⇒2x+3y-5z=d

And the plane passing through, –1, –2, –3

⇒-2-6+15=d

⇒d=7

∴ 2x+3y-5z=7

Now finding the foot of perpendicular we get,

x-32=y+23=z+9-5=-3838

x=1, y=-5, z=-4

Q1, –5, –4

Distance from origin =1+25+16=42

Hence this is the required option.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let the foot of the perpendicular from the point 1,2,4 on the line x+24=y-12=z+13 be P. Then the distance of P from the plane 3x+4y+12z+23=0 is

Options:
A 5013
B $\frac{63}{13}$
C 6513
D 4

The correct answer is:
6513

Solution:
Let P be foot of perpendicular of point Q1,2,4

Let x+24=y-12=z+13=λ

So x=4λ-2, y=2λ+1,z=3λ-1

Then coordinates of point P will be 4λ-2, 2λ+1,3λ-1

Now direction ratios of QP

=4λ-2-1, 2λ+1-2, 3λ-1-4

=4λ-3,2λ-1,3λ-5

and D.R's of line will be 4,2,3

Now PQ and line are perpendicular

so, 44λ-3+22λ-1+33λ-5=0

⇒ λ=1

Putting the value of λ in P we get P≡2,3,2

Now distance of point P2,3,2 from plane3x+4y+12z+23=0 will be =3×2+4×3+12×2+239+16+144=6513

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let the function ,f:-7,0→R be continuous on -7,0 and differentiable on -7,0. If f-7=-3 and f'x≤2 for all x∈-7,0, then for all such functions f, f-1+f0 lies in the interval

Options:
A -∞,20
B -3,11
C -∞,11
D -6,20

The correct answer is:
-∞,20

Solution:
Lets use LMVT for x∈-7,-1
f-1-f-7-1+7≤2
f-1+36≤2⇒f-1≤9
Lets use LMVT for x∈-7,0
f0-f-70+7≤2
f0+37≤2⇒f0≤11 ∴f0+f-1≤20

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Let, the function F be defined as Fx=∫1xettdt, x>0, then the value of the integral ∫1xett+adt, where a>0, is

Options:
A eaFx-F1+a
B e-aFx+a-Fa
C eaFx+a-F1+a
D e-aFx+a-F1+a

The correct answer is:
e-aFx+a-F1+a

Solution:
Given Fx=∫1xettdt

Let, I=∫1xett+adt

Let, t+a=y ⇒dt=dy
Also, t=1 ⇒y=1+a and t=x ⇒y=x+a

∴ I=∫1+ax+aey-aydy

⇒I=e-a∫1+ax+aeyydy

Using ∫abfxdx=∫abftdt, we get

I=e-a∫1+ax+aettdt

⇒I=e-a∫11+aettdt+∫1+ax+aettdt-∫11+aettdt

Using ∫abfxdx+∫bcfxdx=∫acfxdx, a<c<b,

⇒I=e-a∫1x+aettdt-∫11+aettdt

Using, the given relation, we get

I=e-aFx+a-F1+a.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2014 (19 Apr Online)
Let the function F be defined as F x⁡ = ∫ 1 x e t t dt , x⁡ > 0 , then the value of the integral ∫ 1 x⁡ e t t + a dt , where a>0, is

Options:
A e a F x⁡ - F 1 + a
B e - a F x⁡ + a - F a
C e a F x⁡ + a - F 1 + a
D e - a F x⁡ + a - F 1 + a

The correct answer is:
e - a F x⁡ + a - F 1 + a

Solution:
F x = â« 1 x e t t dt
I = â« 1 x e t t + a dt
Let, t + a = y â dt = dy
Also, t = 1 â y = 1 + a and t = x â y = x + a
â´ I = â« 1 + a x + a e y - a y dy
= e - a â« 1 + a x + a e y y dy
= e - a â« 1 + a x + a e t t dt
= e - a F xâ¡ + a - F 1 + a

Topic: Mathematics, Definite Integration, JEE Main
Let the function $f$ be defined by $f(x)=\frac{2 x+1}{1-3 x}$, then $f^{-1}(x)$ is

Options:
A $\frac{x-1}{3 x+2}$
B $\frac{3 x+2}{x-1}$
C $\frac{x+1}{3 x-2}$
D $\frac{2 x+1}{1-3 x}$

The correct answer is:
$\frac{x-1}{3 x+2}$

Solution:
$\begin{aligned} & \text { Let } y=f(x) \Rightarrow y=\frac{2 x+1}{1-3 x} \Rightarrow y-3 x y=2 x+1 \\ & \Rightarrow x=\frac{y-1}{3 y+2} \Rightarrow f^{-1}(y)=\frac{y-1}{3 y+2} \Rightarrow f^{-1}(x)=\frac{x-1}{3 x+2}\end{aligned}$

Topic: Mathematics, Functions, JEE Main
Let the function $f$ be defined by the equation $f(x)=\left\{\begin{array}{ll}3 x & \text { if } 0 \leq x \leq 1 \\ 5-3 x & \text { if } 1 \lt x \leq 2^{\prime}\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 1} f(x)=f(1)$
B $\lim _{x \rightarrow 1} f(x)=3$
C $\lim _{x \rightarrow 1} f(x)=2$
D $\lim _{x \rightarrow 1} f(x)$ does not exist

The correct answer is:
$\lim _{x \rightarrow 1} f(x)$ does not exist

Solution:
$\begin{aligned} & =\lim _{x \rightarrow 1-0} f(x)=\lim _{h \rightarrow 0}(1-h)=\lim _{h \rightarrow 0} 3(1-h) \\ & =\lim _{h \rightarrow 0}(3-3 h)=3-3.0=3 \\ & =\lim _{x \rightarrow 1+0} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[5-3(1+h)] \\ & =\lim _{h \rightarrow 0}(2-3 h)=2-3.0=2 \\ & \text { Hence } \lim _{x \rightarrow 1} f(x) \text { does not exist. } \\ & \end{aligned}$

Topic: Mathematics, Limits, JEE Main
Let the function $f: R \rightarrow R$ be defined by $f(x)=2 x+\sin x, x \in R$. Then $f$ is

Options:
A One-to-one and onto
B One-to-one but not onto
C Onto but not one-to-one
D Neither one-to-one nor onto

The correct answer is:
One-to-one and onto

Solution:
$f^{\prime}(x)=2+\cos x\gt0$. So, $f(x)$ is strictly monotonic increasing so, $f(x)$ is one-to-one and onto.

Topic: Mathematics, Functions, JEE Main
Let the function f(x)=2x3+(2p-7)x2+3(2p-9)x-6 have a maxima for some value of x<0 and a minima for some value of x>0. Then, the set of all values of p is

Options:
A 92,∞
B 0,92
C -∞,92
D -92,92

The correct answer is:
-∞,92

Solution:
Given,f(x)=2x3+(2p-7)x2+3(2p-9)x-6

On taking differentiation w.r.t x, we get

f'(x)=6x2+2(2p-7)x+3(2p-9)

⇒f'(0)=6(0)2+2(2p-7)(0)+3(2p-9)

⇒f'(0)=3(2p-9)

Here, given that for x<0 the function has maxima and for x>0 the function has minima so the function is decreasing function about x=0,

Hence f'(0)<0

So, 3(2p-9)<0

⇒p<92

Therefore, p∈-∞,92

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let the function fx=loge1+5x-loge1+αxxif x≠010if x=0 be continuous at x=0. Then α is equal to

Options:
A 10
B -10
C 5
D -5

The correct answer is:
-5

Solution:
fx=ln1+5x-ln1+αxx x≠010 x=0

For fx to be continuous

limx→0fx=f0

limx→0ln1+5x-ln1+αxx=10

Using expansion of ln1+x=x-x22+x33- ...., we get

limx→05x-25x22+……-αx-α2x22+……x=10

i.e. 5-α=10⇒α=-5

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Let the function fx=x+1. The number of values of xâ-2,2 for which fx-3, fx-1 and fx+1 are in the arithmetic progression is

Options:
A 0
B 1
C 2
D Infinite

The correct answer is:
2

Solution:
fx=x+1âfx-3=x-2
fx-1=x, fx+1=x+2
Given fx-3,fx-1,f(x+1) are in A.P.
â2x=x-2+x+2
Case 1: x<-2
-x+2-x-2=-2xâ0=0
âx<-2
Case 2: -2â¤x<0
-x+2+x+2=-2xâx=-2
âx=-2
Case 3: 0â¤x<2
-x+2+x+2=2xâx=2 â no solution
Case 4: xâ¥2
x-2+x+2=2xâ0=0
âxâ¥2
Taking union, we get xâ¤-2 or xâ¥2
But. xâ-2,2
Taking intersection, we get,

x=-2 and 2

Topic: Mathematics, Sequences and Series, JEE Main
Let the function fx=x2sinâ¡1x âxâ 0 is continuous at x=0. Then, the value of the function at x=0 is

Options:
A 0
B -1
C 1
D indeterminate

The correct answer is:
0

Solution:
For fx to be continuous at x=0, we must have f0=limxâ0â¡fx
âf0=limxâ0â¡x2sinâ¡1x=0Ãa finite quantity =0
Hence, f0=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let the functions f:R→R and g:R→R be defined as :

fx=x+2,x<0x2,x≥0 and gx=x3,x<13x-2,x≥1

Then, the number of points in R where fogx is NOT differentiable is equal to :

Options:
A 3
B 1
C 0
D 2

The correct answer is:
1

Solution:
fgx=gx+2, gx<0gx2, gx≥0

=x3+2,x<0x6,x∈[0,1)3x-22,x∈[1,∞)

fogx'=3x2,x<06x5,x∈[0,1)23x-2×3,x∈[1,∞)

At x=0

L.H.L.≠R.H.L. (Discontinuous)

At x=1

L.H.L.=R.H.L. (continuous)

L.H.D.=6=R.H.D.

⇒fogx is differentiable for all x∈R-0

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let the hyperbola H:x2a2-y2=1 and the ellipse E:3x2+4y2=12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH and eE are the eccentricities of H and E respectively, then the value of 12eH2+eE2 is equal to _____.

Options:

The correct answer is:
42

Solution:
Given

E≡3x2+4y2=12 ⇒ x24+y23=1

Now, eE=1-b2a2=1-34=12

Length of L.R. =2b2a=2×32=3

Now H≡ x2a2-y21=1

Length of L.R.=2b2a=2×1a

Given length of L.R. of E and H are equal

So 2a=3 ⇒ a=23

Now eH=b2a2+1=1232+1 ⇒ eH=94+1=134

So 12eH2+eE2=12134+14=12144=14×3=42

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Let the hyperbola H:x2a2-y2 b2=1 pass through the point 22,-22. A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?

Options:
A 23,32
B 33,-62
C 3,-6
D 36,62

The correct answer is:
33,-62

Solution:
Given,

H: x2a2-y2 b2=1

So coordinates of foci will be : Sae,0,S'−ae,0

Now foot of directrix of parabola will be -ae,0

Also focus of parabola is which is same as focus of H will be ae, 0

Now, semi latus rectum of parabola =SS'=2ae

Given, 4ae=e2b2a

⇒b2=2a2 ...1

Also given, 22,-22 lies on H: x2a2-y2 b2=1

⇒222a2-222 b2=1

⇒1a2-1 b2=18 ...2

Now from equation 1 & 2 we get,

a2=4, b2=8

∵b2=a2e2-1

∴ e=3

So, the equation of parabola is y2=4×aex⇒y2=83x

So, only 33,-62 will satisfy the parabola y2=83x

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let the image of the point (1,0,7) in the line x1=y-12=z-23 be the point (α,β,γ). Then which one of the following points lies on the line passing through (α,β,γ) and making angles 2π3 and 3π4 with y-axis and z-axis respectively and an acute angle with x-axis?

Options:
A (1,-2,1+2)
B (1,2,1-2)
C (3,4,3-22)
D (3,-4,3+22)

The correct answer is:
(3,4,3-22)

Solution:
Given:

L1≡x1=y-12=z-23=λ

⇒M(λ,1+2λ,2+3λ)

⇒PM→=λ-1i^+1+2λj^+3λ-5k^

So, PM→ is perpendicular to line L1

⇒PM→·b→=0

⇒PM→.b→=λ-1i^+1+2λj^+3λ-5k^.i^+2j^+3k^

⇒λ-1+4λ+2+9λ-15=0

⇒14λ=14

⇒λ=1

⇒M=(1,3,5)

Now, M is midpoint of P→ and Q→

⇒1,3,5≡1+α2,0+β2,7+γ2

⇒1+α2=1, β2=3, 7+γ2=5

⇒α=1, β=6, γ=3

Now, required line having direction cosine l,m,n, where given m=cos-2π3=-12 & n=cos3π4=-12

Also, l2+m2+n2=1

⇒l2+-122+-122=1

⇒l2=14

⇒l=12

[Line make acute angle with x-axis]

So, the equation of line passing through 1,6,3 will be given by,

r→=(i^+6j^+3k^)+μ12i^-12j^-12k^

Option 3 3,4,3-22 satisfying for μ=4

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let the image of the point (1,0,7) in the line x1=y-12=z-23 be the point (α,β,γ). Then which one of the following points lies on the line passing through (α,β,γ) and making angles 2π3 and 3π4 with y-axis and z-axis respectively and an acute angle with x-axis?

Options:
A (1,-2,1+2)
B (1,2,1-2)
C (3,4,3-22)
D (3,-4,3+22)

The correct answer is:
(3,4,3-22)

Solution:
Given:

L1≡x1=y-12=z-23=λ

⇒M(λ,1+2λ,2+3λ)

⇒PM→=λ-1i^+1+2λj^+3λ-5k^

So, PM→ is perpendicular to line L1

⇒PM→·b→=0

⇒PM→.b→=λ-1i^+1+2λj^+3λ-5k^.i^+2j^+3k^

⇒λ-1+4λ+2+9λ-15=0

⇒14λ=14

⇒λ=1

⇒M=(1,3,5)

Now, M is midpoint of P→ and Q→

⇒1,3,5≡1+α2,0+β2,7+γ2

⇒1+α2=1, β2=3, 7+γ2=5

⇒α=1, β=6, γ=3

Now, required line having direction cosine l,m,n, where given m=cos-2π3=-12 & n=cos3π4=-12

Also, l2+m2+n2=1

⇒l2+-122+-122=1

⇒l2=14

⇒l=12

[Line make acute angle with x-axis]

So, the equation of line passing through 1,6,3 will be given by,

r→=(i^+6j^+3k^)+μ12i^-12j^-12k^

Option 3 3,4,3-22 satisfying for μ=4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Let the image of the point53,53,83 in the plane x- 2y+z–2=0 be P. If the distance of the point Q(6,-2,α),α>0, from P is 13, then α is equal to_______.

Options:

The correct answer is:
15

Solution:
The image of the point 53,53,83 w.r.t plane will be,

x-531=y-53-2=2-831=-21×53-2×53+83-212+22+12

⇒x-531=y-53-2=2-831=-2(-1)6

⇒P(x, y, z)=(2,1,3)

But PQ=13

42+32+(α-3)2=169

|α-3|=12

α=15 Since α>0

Hence this is the required answer.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let the image of the point P1,2,3 in the line L:x-63=y-12=z-23 be Q. let Rα,β,γ be a point that divides internally the line segment PQ in the ratio 1:3. Then the value of 22α+β+γ is equal to

Options:

The correct answer is:
125

Solution:
Plotting the rough graph of the given data we have,

Let M be the mid-point of PQ

∴ M=3λ+6,2λ+1,3λ+2

Now, PM→=3λ+5i^+2λ-1j^+3λ-1k^

∵ PM→⊥3i^+2j^+3k^

∴ 33λ+5+22λ-1+33λ-1=0

λ=-511

∴ M5111,111,711

Since R is mid-point of PM

So by using midpoint formula in P & M we get,

R=6222,2322,4022, now comparing with Rα,β,γ

We get, α=6222, β=2322 & γ=4022

So, 22α+β+γ=125

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let the image of the point P(1, 2, 3) in the plane 2x–y+z=9 be Q. If the coordinates of the point R are (6, 10, 7), then the square of the area of the triangle PQR is _______.

Options:

The correct answer is:
594

Solution:
Given,

The image of the point P(1, 2, 3) in the plane 2x–y+z=9 be Q,

The coordinates of the point R are (6, 10, 7),

Now plotting the diagram we get,

R lies on plane as 6,10,7 satisfy the plane equation 2x–y+z=9

Now length of PR=52+82+42=105

Now finding the angle between PR & PM(which is normal vector to plane) we get,

cosθ=PR→·PM→PR→·PM→

⇒cosθ=(5i^+8j^+4k^)(2i^-j^+k^)1056

⇒cosθ=6630

Now Area(∆PQR)=2area(∆PMR)

⇒∆PQR=2·12·PRsinθ·PRcosθ

⇒∆PQR=2·12(PR)2sinθ cosθ

⇒∆PQR=105·6630·594630

⇒∆PQR=594

Hence, area of the square will be 594

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let the image of the point P2,-1,3 in the plane x+2y-z=0 be Q. Then the distance of the plane 3x+2y+z+29=0 from the point Q is

Options:
A 2227
B 2427
C 214
D 314

The correct answer is:
314

Solution:
We have,

Equation of line PM is

x-21=y+12=z-3-1=λ

Any point on the line is

M≡λ+2,2λ-1,-λ+3

Point M lies on the plane, so

λ+2+22λ-1-3-λ=0

⇒λ=12

Point M≡52,0,52

For Image Qα,β,γ, we have

α+22=52,β-12=0

γ+32=52

So, Q≡3,1,2

Distance of the point Q from the plane 3x+3y+z+29=0 is

d=33+21+2+2932+22+12

⇒d=4214=314 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let the images of the point A2,3 about the lines y=x and y=mx are P and Q respectively. If the line PQ passes through the origin, then m is equal to

Options:
A 32
B 23
C -1
D 5

The correct answer is:
-1

Solution:
Reflection of A2,3 about line y=x is 3,2=P
Equation of the line PQ is the same as line OP
â equation of the line PQ is y=23x and the equation of the line OA is y=32x.
Now, y=mx & y=x are the angle bisectors of the lines 2x-3y=0 and 3x-2y=0
Now, angle bisectors of 2x-3y=0 and 3x-2y=0 are y=x and y=-x
âm=-1

Topic: Mathematics, Straight Lines, JEE Main
Let the incentre of ΔABC is Ι2,5. If A=1,13 and B=-4,1, then the coordinates of C are

Options:
A 1,10
B 10,1
C 8,2
D 9,3

The correct answer is:
10,1

Solution:
Slope of AB is 13-11--4=125
Slope of BI is 5-12--4=23
Slope of AI is 13-51-2=-8
Let, the slopes of BC and AC are m1 & m2 respectively
âm2--81+m2-8=-8-1251+-8125
m2+891=1-8m2-57
8Ã91-57=8Ã57-91m2âm2=-43
Now, m1-231+m123=23-1251+23Ã125
â3m1-239=3+2m1-26
â9m1-6=-6-4m1âm1=0
â The equation of AC is y-13=-43x-1 and the equation of BC is y=1
âC is 10,1

Topic: Mathematics, Straight Lines, JEE Main
Let the incentre of ΔABC is I2,5. If A=1,13 and B=-4,1, then the sum of the slopes of sides AC and BC is

Options:
A 34
B -43
C 43
D -34

The correct answer is:
-43

Solution:
Slope of AB is 13-11--4=125
Slope of BI is 5-12--4=23
Slope of AI is 13-51-2=-8
Let, the slopes of BC and AC are m1 and m2 respectively
tan⁡A2=tan⁡A2⇒m2--81+m2-8=-8-1251+-8125
m2+891=1-8m252
⇒m2=-43
Now tan⁡B2=tan⁡B2⇒m1-231+m123=23-1251+23×125
⇒3m1-239=3+2m1-26
⇒m1=0
⇒m1+m2=-43

Topic: Mathematics, Straight Lines, JEE Main
Let the inductance and resistance be denoted by 'L' and 'R' respectively. The dimensions of $\left(\frac{\mathrm{L}}{\mathrm{R}}\right)$ are

Options:
A $\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{1}\right]$
B $\left[\mathrm{L}^{0} \mathrm{M}^{0} \mathrm{~T}^{0}\right]$
C $\left[\mathrm{L}^{0} \mathrm{M}^{1} \mathrm{~T}^{0}\right]$
D $\left[\mathrm{L}^{0} \mathrm{M}^{0} \mathrm{~T}^{1}\right]$

The correct answer is:
$\left[\mathrm{L}^{0} \mathrm{M}^{0} \mathrm{~T}^{1}\right]$

Solution:
We know that;
Dimensional formula of $\mathrm{L}=\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}$
Dimensional formula for $\mathrm{R}=\mathrm{R}=\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}$
$\frac{\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{ML}^{2}}{\mathrm{~T}^{2} \mathrm{~A}^{2}} \times \frac{\mathrm{T}^{2} \mathrm{~A}^{2}}{\mathrm{ML}^{2}}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}$

Topic: Physics, Units and Dimensions, JEE Main
Let the initial concentration of P is 1 M, the concentration of P after 33.33 sec is equal to?

Options:
A 1e
B 2e
C 1e2
D e

The correct answer is:
1e

Solution:
keff=k1+k2

Pt=1×e-kt
=e-3×10-2×33.3
=e-0.99=1e

Topic: Chemistry, Chemical Kinetics, JEE Main
Let the integral I=∫2020x+sin-1⁡2020x1-20202xdx=K22020sin-1⁡2020x+λ (where, λ is the constant of integration), then the vaue of 2020K is

Options:
A 2020
B 2019
C e
D 1e

The correct answer is:
e

Solution:
Let, sin-1⁡2020x=t
⇒2020xln⁡20201-20202xdx=dt
⇒I=∫2020tln⁡2020dt
=2020tln20202+λ
=2020sin-1⁡2020xln20202+λ
∴K=1ln2020=log2020e
∴2020K=2020log2020⁡e=e

Topic: Mathematics, Indefinite Integration, JEE Main
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5}$, is _______

Options:

The correct answer is:
0

Solution:
$\begin{aligned}
& 2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5} \\
& \Rightarrow \quad \pi+\cos ^{-1} x=\frac{2 \pi}{5} \\
& \Rightarrow \quad \cos ^{-1} x=\frac{-3 \pi}{5}
\end{aligned}$
Not possible
Ans. 0

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let the latus rectum of the hyperbola x29-y2b2=1 subtend an angle of π3 at the centre of the hyperbola. If b2 is equal to l m(1+n), where l and m are co-prime numbers, then l2+m2+n2 is equal to __________.

Options:

The correct answer is:
182

Solution:
Given,

Equation of hyperbola x29-y2b2=1

And latusrectum LR subtends 60° at centre

Now, plotting the diagram we get,

Now, from above diagram we get Aae,b2a & Bae,-b2a

⇒tan30°=b2aae=b2a2e=13

⇒e=3 b29 as a2=9

Also, e2=1+b29

⇒1+b29=3 b481

⇒b4=3b2+27

⇒b4-3b2-27=0

⇒b2=3+1172 {ignoring the negative sign as it is a square function}

⇒b2=32(1+13)

Hence, on comparing with lm1+n we get,

⇒l=3, m=2, n=13

⇒l2+m2+n2=182

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Let the latus rectum of the parabola y2=4x be the common chord to the circles C1 and C2 each of them having radius 25. Then, the distance between the centres of the circles C1 and C2 is :

Options:
A 12
B 8
C 85
D 45

The correct answer is:
8

Solution:
Co-ordinates of focus of parabola will be S=a, 0=1, 0.

Now, length of latus-rectum =4.

So, length of semi latus-rectum =2.

Also, ∠LSC1=90°

Now, radius of circle LC1=25

Now, applying pythagoras theorem in △LSC1,

LC12=C1S2+LS2

⇒C1S2=LC12-LS2=252-22=16

⇒C1S=16=4

∴C1C2=2C1S=2×4=8.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (03 Sep Shift 2)