Sample JEE Main Mathematics Mock Test - JEE MCQ

Let coordinates of the pole be (h,k) then equation of the polar of y2=4ax is ky=2a(x+h)

y=(2a/k)x+(2ah)/K

Since line (1) is touching the ellipse

x²/a²+y²/b²=1

4a²b²/k²=a².4a²/K²+b²

Required locus is 4a²x²=4a⁴+b²y²

=> 4a²x²−b²y²=4a⁴

Hence A is the correct answer.

(3+w+3w²)⁴ = [(1+w+w²)+2w²+2]⁴= 2⁴(w²+1)=16(-w)⁴ = 16w⁴= 16w

If you draw a graph it will be clear let us consider 4 coordinates (4,0),(-4,0),(0,-4),(0,4) if you join this points you will get a square.this square can again be divided into 4 symmetric triangles.therefore the area = 4 times area of one triangle

area of triangle = 1/2bh=1/2*4*4=8

therefore total area = 4*8=32

Sine function is a periodic function. Its value repeat after an interval of 2π. Since 2π/3 is out of its inverse principle range (which is from -π/2 to π/2) so we will write sin 2π/3 = sin π/3. Now its in the principle range. Therefore sin⁻1(sin 2π/3) = sin⁻1(sin π/3) = π/3

We have, dy/dx=(x−y)²
Let (x−y)=v. Then,
1−dy/dx=dv/dx⇒dy/dx=1−dv/dx
∴ dy/dx=(x−y)²
⇒ 1−dv/dx=v²
⇒ 1−v²=dv/dx
⇒ dx=1/1−v²dv
⇒ 2∫dx=2∫1/1−v²dv
⇒ 2x=log(1+v/1−v)+logC
⇒ C(1+v/1−v)=e^2x
⇒C(x−y+1/y−x+1)=e^2x⇒C(x−y+1)=e^2x(y−x+1)
Taking C = 1, we find that option (a) is correct.

For unique solution where a, b, c, d {0, 1}

total cases = 16
Favorable cases = 6 (either ad =1, bc = 0 or ad = 0, bc =1).

Probability that system of equation has unique solution is 

and system of equations has either unique solution or infinite solutions so that probability for system to have a solution is 1.

Solution :- A = i^+ j^ and B = j^+k^

A = (2)^1/2 , B = (2)^1/2

​θ=(cos^−1)(A.B/AB)=(cos^−1)(1/2).

or, θ=(cos^−1)(cosπ/3) = π/3

Therefore, both Assertion and Reason are correct and Reason is the correct explanation for Assertion.

Since, B is a subset of A so B ∩ A = B
now, P(A|B) = P(A ∩ B)/P(B) = P(B)/P(B) = 1

A.M. ,G.M., H.M. between two numbers are 144/15 , 15 , 12.

We know that A.M > G.M > H.M

So, the order is 

Therefore, A.M  = 15 ,    G.M = 12 ,    H.M = 144/15

Then, 

H.M, G.M and A.M = 144/15 ,12, 15

Standard deviation (SD) is the most commonly used measure of dispersion. It is a measure of spread of data about the mean. SD is the square root of sum of squared deviation from the mean divided by the number of observations.

Given, 
y = 3x + 1 , slope of it {m1} = 3 
2y = x + 3 , slope of it {m2} = 1/2
angle between y = 3x + 1 and y = mx + 4 is ∅
then , tan∅ = |3 - m|/| 1 + 3m| ____(1)

it is given that angle between the lines 2y = x + 3 and y = mx + 4 is also ∅ .
then, tan∅ = |1/2 - m|/| 1 +m/2|________(2)
from equations (1) and (2), 
|3 - m|/|1 + 3m| = | 1/2 - m|/| 1 + m/2| 
|3 - m|/|1 + 3m| = |1 - 2m|/|2 + m| 
taking positive sign , 
(3 - m)(2 + m) = (1 - 2m)( 1 + 3m) 
6 + 3m -2m -m^2 = 1 + 3m -2m - 6m^2
5m^2 = -5 
m^2 = -1 it's not possible .
taking negative sign, 
(3 - m)(2 + m) = -(1 - 2m)(1 + 3m)
6 + 3m - 2m - m^2 = -1 - 3m + 2m + 6m^2
7m^2 - 2m -7 = 0
m = { 2 ± √(4 + 49 × 4)}/14 
= { 2 ± 2√50}/14 
= { 1 ± √50}/7 
= {1 ± 5√2}/7 
hence, m = { 1 ± 5√2}/7

Mode = 3 median –2 mean      ...(1)
Given mean – mode = 63
⇒ Mode = mean –63               ...(2)
from (1) & (2)
Mean –63 = 3 median –2 mean
3 mean –3median = 63
(mean – median) = 21

3,1,1 & 2,2,1 → two Methods only

(3 - even)         or         (2 – odd and 1 - even)
³C₃ × 3!            +          ³C₂ × ³C₁ × 3! = 60
(Total numbers).