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BITSAT Mathematics Test - 2 - JEE MCQ


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30 Questions MCQ Test - BITSAT Mathematics Test - 2

BITSAT Mathematics Test - 2 for JEE 2024 is part of JEE preparation. The BITSAT Mathematics Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Mathematics Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Mathematics Test - 2 below.
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BITSAT Mathematics Test - 2 - Question 1

The number of surjections from A = {1,2, ...n), n > 2 onto B = (a,b) is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 1

BITSAT Mathematics Test - 2 - Question 2

The number of bijective functions from set A to itself when A contains 106 elements is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 2

Total number of bijection from set of n elements to itself = n!

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BITSAT Mathematics Test - 2 - Question 3

The locus of point z satisfying Re when k is a non-real real number is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 3

Let 

BITSAT Mathematics Test - 2 - Question 4

If a1 a2, a3 are in G.P. with common ratio r, then value of a3 > 4a2 - 3a1 holds if

Detailed Solution for BITSAT Mathematics Test - 2 - Question 4

BITSAT Mathematics Test - 2 - Question 5

If a∈ z, ( x - a ) (x - 10) + 1 = 0 has integral roots, then values of a are

Detailed Solution for BITSAT Mathematics Test - 2 - Question 5

(x - a) (x - 10) + 1 = 0
∴ (x - a) (x - 10) = -1
∴ x - a = 1
and x - 1 0 = - 1 ,
or x - a = 1 and x - 10 = 1
∴ a = 8 ora = 12

BITSAT Mathematics Test - 2 - Question 6

 The value of a for which (1 - 2a) x- 6ax - 1 = 0 and ax2 -x + 1 = 0 have atleast one root, in common are

Detailed Solution for BITSAT Mathematics Test - 2 - Question 6

a = 0 or a = 1/2, the equation becomes linear So 
Hence, only answer is, 

BITSAT Mathematics Test - 2 - Question 7

The number of ways in which one or more balls can be selected out of 10 white, 9 green and 7 blues balls, is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 7

Number of ways = (10 + 1) (9 + 1) (7 + 1) -1
= 879

BITSAT Mathematics Test - 2 - Question 8

If sum of coefficient of (a + b)n is 4096, then greatest coefficient is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 8

BITSAT Mathematics Test - 2 - Question 9

If A is singular, then A [adj A] is matrix

Detailed Solution for BITSAT Mathematics Test - 2 - Question 9

​​​​

BITSAT Mathematics Test - 2 - Question 10


then n equals

Detailed Solution for BITSAT Mathematics Test - 2 - Question 10

Degree of the determinant is
n + (n + 2) + (n + 3) = 3n + 5 
and on R .H.S., degree = 2
3n + 5 = 2
⇒ n = -1

BITSAT Mathematics Test - 2 - Question 11

A person mistakenly calculated the mean and the median of a sample data of 200 items as 100 and 104, respectively. The maximum value of the individual data was 200. When the data was rechecked, it was found that the value of the maximum sample data was 220. The values of true mean and true median are

Detailed Solution for BITSAT Mathematics Test - 2 - Question 11

Given mean:

And the median will remain same, i.e. 104.
Hence, this is the required solution.

BITSAT Mathematics Test - 2 - Question 12

Consider the following expression:

The number of values of x which satisfy the given expression is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 12

Here,


Therefore, 3 such values of x are possible.
Hence, this is the required solution.

BITSAT Mathematics Test - 2 - Question 13

If P and Q are two sets such that  and   for same set A, then which of the following options is correct?

Detailed Solution for BITSAT Mathematics Test - 2 - Question 13

Given:

Again, consider 

From both the above results,
P = Q
Hence, this is the required solution.

BITSAT Mathematics Test - 2 - Question 14

Evaluate:

Detailed Solution for BITSAT Mathematics Test - 2 - Question 14

Here,

Hence, this is the required solution.

BITSAT Mathematics Test - 2 - Question 15

Find the maximum value of 15 cosA + 8 sinA.

Detailed Solution for BITSAT Mathematics Test - 2 - Question 15

As we know that maximum value of

According to the question,

Thus,
Maximum value of the given expression is 17.
Hence, this is required solution.

BITSAT Mathematics Test - 2 - Question 16

The axis of a parabola is along the abscissa. Its vertex is at origin and it passes through a point P(2, 3). The equation of the parabola is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 16

Given, vertex = (0, 0), point = P(2, 3) and axis is along X axis.
Since point (2, 3) lies in the first quadrant,
Therefore, equation of the parabola will be of the form y2 = 4ax , which passes through P(2, 3).
Therefore,
Therefore, required equation is

Hence, this is the required solution.

BITSAT Mathematics Test - 2 - Question 17

A canonical plastic bottle whose height is 21 m and radius of base is 7 m is being filled with milk at a uniform rate of 5/3 m3/min. When the milk level is 6 m, find the rate at which the level of the milk in the bottle is rising.

Detailed Solution for BITSAT Mathematics Test - 2 - Question 17

Here,

Let V be the volume of the milk, then

Putting the value of r,

Differentiate on both sides with respect to t,

Given, the value of h = 6m and 
Put these values.

Thus,

Hence, this is required solution.

BITSAT Mathematics Test - 2 - Question 18

Consider the given expression:

Differentiate y with respect to x.

Detailed Solution for BITSAT Mathematics Test - 2 - Question 18

Given:

Let 
So,

And,

Thus,

Hence, this is required solution.

BITSAT Mathematics Test - 2 - Question 19

Find the area (unit2) bounded by the two curves y = 6 + 5x - 2x2 y = 2x + 3.

Detailed Solution for BITSAT Mathematics Test - 2 - Question 19




BITSAT Mathematics Test - 2 - Question 20

If 0 < P(X) < 1, 0 < P(Y) < 1 and , then Which of the following is correct?

Detailed Solution for BITSAT Mathematics Test - 2 - Question 20

Here,

It means X and Y are independent events, so X' and Y' are also independent. Therefore,

Or,

Hence, this is the required solution.

BITSAT Mathematics Test - 2 - Question 21

If A and B are independent events of a random experiments such that

Detailed Solution for BITSAT Mathematics Test - 2 - Question 21

Since, A & B are independent events.

BITSAT Mathematics Test - 2 - Question 22

The mean of n items is If these n items are successively increased by 2, 22, 23, …, 2n, then the new mean is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 22

New mean

BITSAT Mathematics Test - 2 - Question 23

If are three non-coplanar vectors and are vectors defined by the relations then the value of

Detailed Solution for BITSAT Mathematics Test - 2 - Question 23

∴ Given expression is 1 + 1 + 1 = 3

BITSAT Mathematics Test - 2 - Question 24

The fourth term of equal to 200, then the value of x satisfying this is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 24

Since, fourth term of

Taking logarithm on both side

BITSAT Mathematics Test - 2 - Question 25

The value of

Detailed Solution for BITSAT Mathematics Test - 2 - Question 25

BITSAT Mathematics Test - 2 - Question 26

If ln(x + y) = 2xy, then y'(0) is equal to

Detailed Solution for BITSAT Mathematics Test - 2 - Question 26

Given equation is ln(x + y) = 2xy …(1)

ln x + y = 2xy …1

For x = 0

ln(0 + y) = 2.0.y = 0

⇒ lny = 0

⇒ y = 1

Now, differentiating (1), we get,

At point (0,1), we get,

BITSAT Mathematics Test - 2 - Question 27

The value of

Detailed Solution for BITSAT Mathematics Test - 2 - Question 27

Given limit can be written as,

Using L'Hospital' rule,

BITSAT Mathematics Test - 2 - Question 28

Two vertices of a triangle are (3,−2) and (−2, 3) and its orthocentre is (−6, 1). The coordinates of its third vertex are-

Detailed Solution for BITSAT Mathematics Test - 2 - Question 28

Let the third vertex be A(α,β)

Using the diagram, OA⊥BC

⇒ Slope of OA × Slope BC = −1

Solving Equations(i)i and (ii)ii, we get

α = −1, β = 6

∴ The third vertex is (−1, 6)

BITSAT Mathematics Test - 2 - Question 29

The equation of a circle C1 is x2 + y2 − 4x − 2y − 11 = 0. Another circle C2 of radius 1 unit rolls on the outer surface of the circle C1. Then the equation of the locus of the centre of C2 is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 29

The centre and radius of a circle x2 + y2 + 2gx + 2fy + c = 0 are (−g, −f) and

Hence, the centre of x2 + y2 − 4x −2y − 11 = 0 is A(2, 1) and the radius

If P(α, β) be the centre of C2 of radius r2 = 1

We know that, if two circles with centres A and P and radii r1 and r2 touches each other externally, then distance between their centres AP is equal to the sum of their radii i.e. AP = r1 + r2

The locus is obtained by replacing (α, β) by (x, y)

Hence, the locus is x2 + y2 − 4x − 2y − 20 = 0

BITSAT Mathematics Test - 2 - Question 30

The number of real values of k for which the lines and are intersecting is

Detailed Solution for BITSAT Mathematics Test - 2 - Question 30

Any point on the first line is (4r+k, 2r+1,r−1), and any point on the second line is (r′+k+1 ,−r′,2r′+1) for some values of r and r'. The lines are intersecting if these two points coincide i.e

4r + k = r′ + k + 1, 2r + 1 = −r′, r − 1 = 2r′ + 1 for some r and r'

⇒ 4r − r′ = 1, 2r + r′ = −1, r − 2r′ = 2

Now, 4r − r′ = 1, 2r + r′ = −1 ⇒ r = 0, r′ = −1 which satisfy r − 2 r′ = 2.

⇒ The given lines are intersecting for all real values of k.

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