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JEE Main Maths Mock Test- 2 - JEE MCQ


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25 Questions MCQ Test - JEE Main Maths Mock Test- 2

JEE Main Maths Mock Test- 2 for JEE 2024 is part of JEE preparation. The JEE Main Maths Mock Test- 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Maths Mock Test- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Maths Mock Test- 2 below.
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JEE Main Maths Mock Test- 2 - Question 1
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The orthocentre of the traingle whose vertices are (5, -2), (-1, 2) and (1,4) is

JEE Main Maths Mock Test- 2 - Question 2
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If the equation [(k(x+1)2/3)]+[(y+2)2/4]=1 represents a circle, then k=

Detailed Solution for JEE Main Maths Mock Test- 2 - Question 2
The given equation can be write as


=> 4k(x+1) ²+3(y+2) ²=12

on expanding wee get x² coefficient as 4k
and y² coefficient as 3
but in equation of circle x² coefficient is equal to y² coefficient

therefore 4k=3
=> k=3/4
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JEE Main Maths Mock Test- 2 - Question 3
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The eccentric angles of the extremities of the latus-rectum intersecting positive x-axis of the ellipse ((x2/a2) + (y2/b2) = 1) are given by

JEE Main Maths Mock Test- 2 - Question 4
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If arg (z) = θ, then arg(z̅) =
JEE Main Maths Mock Test- 2 - Question 5
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Detailed Solution for JEE Main Maths Mock Test- 2 - Question 5


Hence the answer will be 1.

JEE Main Maths Mock Test- 2 - Question 6
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If N N+ denotes the set of all positive integers and if f : NN+ → N is defined by f(n)    = the sum of positive divisors of (n)  then f (2k . 3), where k is a positive integer is
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 6

f(2k. 3) = The sum of positive divisors of 2k . 3 

JEE Main Maths Mock Test- 2 - Question 7
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If a, b, c are different and 

Detailed Solution for JEE Main Maths Mock Test- 2 - Question 7

Correct Answer : b

Explanation : A = {(a, a2, a3-1) (b, b2, b3-1) (c, c2, c3-1)}

=> {(a, a2, a3) (b, b2, b3) (c, c2, c3)} - {(a, a2, 1) (b, b2, 1) (c, c2, 1)} = 0

=> abc{(1, a, a2) (1, b, b2) (1, c, c2)} - {(a, a2, 1) (b, b2, 1) (c, c2, 1)} = 0

=> abc{(a, a2, 1) (b, b2, 1) (c, c2, 1)} - {(a, a2, 1) (b, b2 1) (c, c2, 1)} = 0

=> (abc-1){(a, a2, 1) (b, b2, 1) (c, c2, 1)} = 0

abc - 1 = 0

=> abc = 1

JEE Main Maths Mock Test- 2 - Question 8
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Detailed Solution for JEE Main Maths Mock Test- 2 - Question 8
  1. Angle Measurement Conversion:
    • Degrees to Radians: Since calculus typically operates in radians, it's essential to convert degrees to radians.
    • Conversion Formula:

      θ radians = θ° × (π/180)

  2. cos(x°) = cos(x × π/180)

  3. Apply the Chain Rule:

    The chain rule states that if you have a composite function f(g(x)), then its derivative is f'(g(x)) · g'(x).

  4. Differentiate cos(x × π/180):

    d/dx [cos(x × π/180)] = -sin(x × π/180) × (π/180)

  5. Simplify the Expression:

    d/dx [cos(x°)] = -(π/180) × sin(x°)

    Here, sin(x°) implies that the sine function takes the angle in degrees, consistent with the original function's angle measurement.

  6. Final Answer

    d/dx [cos(x°)] = -(π/180) sin(x°)

     

 

JEE Main Maths Mock Test- 2 - Question 9
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In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion (A): Angle between is acute angle

Reason (R): If is acute then is obtuse then

JEE Main Maths Mock Test- 2 - Question 10
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The point on the curve y = x2 which is nearest to (3, 0) is
JEE Main Maths Mock Test- 2 - Question 11
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The degree of the differential equation
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JEE Main Maths Mock Test- 2 - Question 12
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In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion (A): are non zero vectors then is a vector perpendicular to all the vectors a → , b → , c →
Reason (R): are perpendicular to both

JEE Main Maths Mock Test- 2 - Question 13
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If i2 = -1, then the sum i + i2 + i3 + ..... upto 1000 terms is equal to
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 13
There will equal n opposite signed terms that is 500 +ve one and 500-ve one therefore it's value comes to zero.
JEE Main Maths Mock Test- 2 - Question 14
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A parallelogram is cut by two sets of m lines parallel to the sides, the number of parallelogram thus formed is
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 14
Parallelogram is cut by two sets of m parallel lines to its sides.
then we have  2 sets of (m+2) parallel lines ( 2 lines of the parallelogram)
so parallelogram is formed by taking 2 lines from each set
 = m+2C2 * m+2C2
 = [(m+2)(m+1)/2 ]2
 this also include 1 original parallelogram
 so total number of new parallelogram formed is  =  (m + 2)2(m + 1)2/4
JEE Main Maths Mock Test- 2 - Question 15
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If A and B are two events such that 
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 15

JEE Main Maths Mock Test- 2 - Question 16
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The probability that a leap year will have exactly 52 Tuesdays is
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 16

The probability of a year being a leap year is 1/4 and being non-leap is 3/4.A leap year has 366 days or 52 weeks and 2 odd days. The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday}, Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.So there are 7 possibiliyies out of which 2 have a Sunday. So the probability of 53 Sundays in a leap year is 2/7.

So, the probability of 52 sundays is 1-2/7 = 5/7.

JEE Main Maths Mock Test- 2 - Question 17
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Product of the real roots of the equation t2x2 + ∣x∣ + 9 = 0
JEE Main Maths Mock Test- 2 - Question 18
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If A.M. between two numbers is 5 and their G.M. is 4, then their H.M. is
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 18

If x, y and z respectively represent AM, GM and HM between two numbers a and b, then
y2 = xz
Here x = 5, y = 4
then 16 = 5 x z
z = 16/5

JEE Main Maths Mock Test- 2 - Question 19
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If the coefficient of correlation between x and y is 0.28, covariance between x and y is 7.6, and the variance of x is 9, then the standard deviation of the y series is
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 19
N the given problem it is SD of x is 3: (or Variance of x is 9). As Variance = (Sx)^2. We know the relation : correlation coefficient (r) = Cov (x,y) / (Sx * Sy) so, 0.28 = 7.6 / (3 * Sy) From here we get the value of SD of Y : Sy = 9.05.
JEE Main Maths Mock Test- 2 - Question 20
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The equation line passing through the point P(1,2) whose portion cut by axes is bisected at P, is
Detailed Solution for JEE Main Maths Mock Test- 2 - Question 20

*Answer can only contain numeric values
JEE Main Maths Mock Test- 2 - Question 21
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If 2 tan2x – 5 sec x is equal to 1 for exactly 7 distinct values of X ∈ [0, nπ/2], n ∈ N, then the greatest value of n is

Detailed Solution for JEE Main Maths Mock Test- 2 - Question 21

2tan2x – 5sec x = 1
2 (sec2x – 1) – 5secx = 1
2sec2x – 5sec – 3 = 1
∴  cosx = 1/3

*Answer can only contain numeric values
JEE Main Maths Mock Test- 2 - Question 22
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If the mean deviation of the number 1, 1 + d, ... , 1 + 8d from their mean is 205, then d is equal to

Detailed Solution for JEE Main Maths Mock Test- 2 - Question 22

*Answer can only contain numeric values
JEE Main Maths Mock Test- 2 - Question 23
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The number of 5-digit numbers of the form xyzyx in which x < y is :-

Detailed Solution for JEE Main Maths Mock Test- 2 - Question 23

for z → 10 choice
for  First two x and y → 9C2 choice
Last two y and x → 1 choice
10 × 9C2 × 1= 360

*Answer can only contain numeric values
JEE Main Maths Mock Test- 2 - Question 24
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If z1 and z2 are two unimodular complex numbers that satisfy z12 + z22 = 5, then  is equal to -

Detailed Solution for JEE Main Maths Mock Test- 2 - Question 24

*Answer can only contain numeric values
JEE Main Maths Mock Test- 2 - Question 25
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The AM of 9 term is 15. If one more term is added to this series, then the A.M. becomes 16. The value of added term is :

Detailed Solution for JEE Main Maths Mock Test- 2 - Question 25

Sum of 9 term = 9 × 15 = 135
New sum when Mean is 16
= 16 × 10 = 160
New term = 160 – 135 ⇒ 25

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