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This mock test of JEE Main Mathematics Mock - 2 for JEE helps you for every JEE entrance exam.
This contains 25 Multiple Choice Questions for JEE JEE Main Mathematics Mock - 2 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The orthocentre of the traingle whose vertices are (5, -2), (-1, 2) and (1,4) is

Solution:

QUESTION: 2

If the equation [(k(x+1)^{2}/3)]+[(y+2)^{2}/4]=1 represents a circle, then k=

Solution:
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

=> 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

QUESTION: 3

The eccentric angles of the extremities of the latus-rectum intersecting positive x-axis of the ellipse ((x^{2}/a^{2}) + (y^{2}/b^{2}) = 1) are given by

Solution:

QUESTION: 4

If arg (z) = θ, then arg(z̅) =

Solution:

QUESTION: 5

Solution:

QUESTION: 6

If N^{ N+} denotes the set of all positive integers and if f : N^{N+} → N is defined by f(n) = the sum of positive divisors of (n) then f (2^{k} . 3), where *k* is a positive integer is

Solution:

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

QUESTION: 7

If *a, b, c * are different and

Solution:

**Correct Answer : b**

**Explanation : **A = {(a, a^{2}, a^{3}-1) (b, b^{2}, b^{3}-1) (c, c^{2}, c^{3}-1)}

=> {(a, a^{2}, a^{3}) (b, b^{2}, b^{3}) (c, c^{2}, c^{3})} - {(a, a^{2}, 1) (b, b^{2}, 1) (c, c^{2}, 1)} = 0

=> abc{(1, a, a^{2}) (1, b, b^{2}) (1, c, c^{2})} - {(a, a^{2}, 1) (b, b^{2}, 1) (c, c^{2}, 1)} = 0

=> abc{(a, a^{2}, 1) (b, b^{2}, 1) (c, c^{2}, 1)} - {(a, a^{2}, 1) (b, b^{2} 1) (c, c^{2}, 1)} = 0

=> (abc-1){(a, a^{2}, 1) (b, b^{2}, 1) (c, c^{2}, 1)} = 0

abc - 1 = 0

=> abc = 1

QUESTION: 8

Solution:

QUESTION: 9

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

Solution:

QUESTION: 10

The point on the curve *y = x*^{2} which is nearest to (3, 0) is

Solution:

QUESTION: 11

The degree of the differential equation

Solution:

QUESTION: 12

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

Solution:

QUESTION: 13

If i^{2} = -1, then the sum i + i^{2} + i^{3} + ..... upto 1000 terms is equal to

Solution:
There will equal n opposite signed terms that is 500 +ve one and 500-ve one therefore it's value comes to zero.

QUESTION: 14

A parallelogram is cut by two sets of m lines parallel to the sides, the number of parallelogram thus formed is

Solution:

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

QUESTION: 15

If *A* and *B* are two events such that

Solution:

QUESTION: 16

The probability that a leap year will have exactly 52 Tuesdays is

Solution:

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.

QUESTION: 17

Product of the real roots of the equation t^{2}x^{2} + ∣x∣ + 9 = 0

Solution:

QUESTION: 18

If A.M. between two numbers is 5 and their G.M. is 4, then their H.M. is

Solution:

If x, y and z respectively represent AM, GM and HM between two numbers a and b, then

y^{2} = xz

Here x = 5, y = 4

then 16 = 5 x z

z = 16/5

QUESTION: 19

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

Solution:
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.

QUESTION: 20

The equation line passing through the point P(1,2) whose portion cut by axes is bisected at P, is

Solution:

*Answer can only contain numeric values

QUESTION: 21

If 2 tan^{2}x – 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

Solution:

2tan^{2}x – 5sec x = 1

2 (sec^{2}x – 1) – 5secx = 1

2sec^{2}x – 5sec – 3 = 1

∴ cosx = 1/3

*Answer can only contain numeric values

QUESTION: 22

If the mean deviation of the number 1, 1 + d, ... , 1 + 8d from their mean is 205, then d is equal to

Solution:

*Answer can only contain numeric values

QUESTION: 23

The number of 5-digit numbers of the form xyzyx in which x < y is :-

Solution:

for z → 10 choice

for First two x and y → ^{9}C_{2} choice

Last two y and x → 1 choice

10 × ^{9}C_{2} × 1= 360

*Answer can only contain numeric values

QUESTION: 24

If z_{1} and z_{2} are two unimodular complex numbers that satisfy z_{1}^{2} + z_{2}^{2} = 5, then is equal to -

Solution:

*Answer can only contain numeric values

QUESTION: 25

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 :

Solution:

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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