JEE Main Maths Mock Test- 1 - JEE MCQ

# JEE Main Maths Mock Test- 1 - JEE MCQ

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## 25 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Maths Mock Test- 1

JEE Main Maths Mock Test- 1 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Maths Mock Test- 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Maths Mock Test- 1 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- 1 below.
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JEE Main Maths Mock Test- 1 - Question 1

### 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):If Cr is the coefficient of xr in the expansion of (1 + x)20 Reason(R) : Cr = Cn − r for any positive integer n

JEE Main Maths Mock Test- 1 - Question 2

### 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) :The inverse of does not exist. Reason(R) :The matrix is non singular.

JEE Main Maths Mock Test- 1 - Question 3

### If the line 3x-4y=λ touches the circle x2+y2-4x-8y-5=0, λ  can have the values

JEE Main Maths Mock Test- 1 - Question 4

The area (in square units) bounded by the curves y2 = 4x and x2 = 4y in the plane is

JEE Main Maths Mock Test- 1 - Question 5

The length of the tangent from (0,0) to the circle 2x2 + 2y2 + 7x -7 y + 5 = 0 is

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 5

We have

S=2(x2+y2)+7x−7y+5=0

=x2+y2+7/2x−7/2y+5/2=0

Length of tangent from (0,0) =√S1

=√5/2

Hence, the required answer is √5/2.

JEE Main Maths Mock Test- 1 - Question 6

If sin θ is real, then θ =

JEE Main Maths Mock Test- 1 - Question 7
If A, B, C are represented by 3 + 4i, 5 - 2i, -1 + 16i respectively, then A, B, C are
JEE Main Maths Mock Test- 1 - Question 8

The differential equation which represents the family of plane curves y=exp. (cx) is

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 8

y = ecx
dy/dx = c. ecx
y' = cy

JEE Main Maths Mock Test- 1 - Question 9

If sin y = x sin (a + y), then (dy/dx) =

JEE Main Maths Mock Test- 1 - Question 10

The fundamental period of the function f(x) = 2 cos 1/3(x - π) is

JEE Main Maths Mock Test- 1 - Question 11
Which of the following is not a statement ?
JEE Main Maths Mock Test- 1 - Question 12
In the following question, a Statement-1 is given followed by a corresponding Statement-2 just below it. Read the statements carefully and mark the correct answer-
Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5.
Statement-1:
The parametric equations of the line of intersection of the given planes are
x = 3 + 14t, y = 1 + 2t, z = 15t.
Statement-2:
The vector 14î+2ĵ+15k̂ is parallel to the line of intersection of given planes
JEE Main Maths Mock Test- 1 - Question 13

The value of a for which the system of equations
a3x+(a+1)3y+(a+2)3z = 0
ax+(a+1)y+(a+2)z = 0
x+y+z = 0
has a non-zero solution, is

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 13

The system of equation has a non-zero solution

JEE Main Maths Mock Test- 1 - Question 14

In the following question, a Statement-1 is given followed by a corresponding Statement-2 just below it. Read the statements carefully and mark the correct answer-
Tangents are drawn from the point (17,7) to the circle x2+y2=169.
Statement-1:
The tangents are mutually perpendicular.
Statement-2:
The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2+y2=338.

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 14

Clearly, m1m2 = - 1.
Hence, the two tangents arc mutually perpendicular.
Statement 1 is true.
Now, the locus of the point of intersection of two mutually perpendicular tangents to the circle x2 + y2 = r2 is the director circle i.e. the circle x2 +y2 = 2r2
For the given circle r = 13. ..
Its director circle is x2 + y2 = 338.
Hence, statement 2 is true and  a cogect explanation of statement as the point (17, 7) lies on the director circle of the circle (i).

JEE Main Maths Mock Test- 1 - Question 15

If nC12=nC8, then n=

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 15

JEE Main Maths Mock Test- 1 - Question 16

The pole of the line 2x + 3y − 4 = 0 with respect to the parabola y2 = 4 x is

JEE Main Maths Mock Test- 1 - Question 17

The chance of getting a doublet with 2 dice is

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 17

Total outcomes = 36
Doublet are 6 (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)

Probability of getting doublet = 6/36
= 1/6

JEE Main Maths Mock Test- 1 - Question 18

If covariance (x, y) = 0, then ρ(x, y) equals

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 18

Covariance is a quantitative measure of the extent to which the deviation of one variable from its mean matches the deviation of the other from its mean. It is a mathematical relationship that is defined as:

Cov(X,Y) = E[(X − E[X])(Y − E[Y])]

Correlation between two random variables, ρ(X,Y) is the covariance of the two variables normalized by the variance of each variable. This normalization cancels the units out and normalizes the measure so that it is always in the range [0, 1]:

JEE Main Maths Mock Test- 1 - Question 19

The two opposite vertices of a square on xy-plane are A(-1,1) and B(5,3), the equation of other diagonal (not passing through A and B) is

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 19

Given: Here,AB is the diagonal of square.
The vertices of a square A
Let the mid−point of AB be EThen coordinates of E are
Therefore equation of other diagonal is

JEE Main Maths Mock Test- 1 - Question 20

If the normal to the curve y=f(x) at the point (3,4) makes an angle 3π/4 with the positive x-axis, then f'(3)

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 20

Given y = f(x)

differentiating w.r.t x

y' = f'(x) which is the slope of the tangent

Hence the slope of the normal is - 1/f'(x) = 3pi/4 = -1

therefore f'(x) = 1

Hence f'(3) = 1

*Answer can only contain numeric values
JEE Main Maths Mock Test- 1 - Question 21

Assume e-4/5 = 2/5. If x, y satisfy, y = ex and the minimum value of (x2 + y2) is expressed in the form of m/n then (2m - n)/5 equals (where m & n are coprime natural numbers)

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 21

OP2 = x2 + y2
y = ex, y' = ex,

*Answer can only contain numeric values
JEE Main Maths Mock Test- 1 - Question 22

Let ƒ(x) be non-constant thrice differentiable function defined on (–∞, ∞) such that ƒ(x) = ƒ(6 – x) and ƒ'(0) = 0 = ƒ'(2) = ƒ'(5). If 'n' is the minimum number of roots of (ƒ"(x))2 + ƒ'(x)ƒ"'(x) = 0 in the interval x ∈ [0, 6] then sum of digits of n equals

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 22

ƒ(x) = ƒ(6 – x)
⇒ ƒ'(x) = –ƒ'(6 – x) .... (1)
put x = 0, 2, 5
ƒ'(0) = ƒ'(6) = ƒ'(2) = ƒ'(4) = ƒ'(5) = ƒ'(1) = 0
and from equation (1) we get ƒ'(3) = –ƒ'(3)
⇒ ƒ'(3) = 0
So ƒ'(x) = 0 has minimum 7 roots in
x ∈ [0, 6] ⇒ ƒ"(x) has min 6 roots in x ∈ [0,6]
h(x) = ƒ'(x).ƒ"(x)
h'(x) = (ƒ"(x))2 + ƒ'(x) ƒ"'(x)
h(x) = 0 has 13 roots in x ∈ [0, 6]
h'(x) = 0 has 12 roots in x ∈ [0, 6]

*Answer can only contain numeric values
JEE Main Maths Mock Test- 1 - Question 23

If a, b, c, x, y, z are non-zero real numbers and  then the value of (a3 + b3 + c3 + abc) equals

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 23

x2(y + z)y2(z + x)z2(x + y) = a3b3c3 = x3y3z3
⇒ (x + y) (y + z) (z + x) = xyz
⇒ x2(y + z) + y2(z + y) + z2(x + y) + xyz = 0
⇒ a3 + b3 + c3 + abc = 0

*Answer can only contain numeric values
JEE Main Maths Mock Test- 1 - Question 24

If the co-ordinate of the vertex of the parabola whose parametric equation is x = t2 – t + 1 and y = t2 + t + 1, t ∈ R is (a, b) then (2a + 4b) equals

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 24

x = t2 – t + 1    .... (1)
y = t2 + t + 1    .... (2)
y – x = 2t  &  x + y = 2(t2 + 1)
________on elminating 't' we get
⇒ (x + y – 2) = 2(y - x)/22
(x – y)2 = 2(x + y – 2)
Axis : x – y = 0
Tangent at vertex : x + y – 2 = 0
Vertex : (1, 1) = (x, y)

*Answer can only contain numeric values
JEE Main Maths Mock Test- 1 - Question 25

Detailed Solution for JEE Main Maths Mock Test- 1 - Question 25

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