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JEE Exam  >  JEE Test  >  BITSAT Mock Tests Series & Past Year Papers 2026  >  BITSAT Mathematics Test - 5 - JEE MCQ

BITSAT Mathematics Test - 5 Free Online Test 2026


MCQ Practice Test & Solutions: BITSAT Mathematics Test - 5 (40 Questions)

You can prepare effectively for JEE BITSAT Mock Tests Series & Past Year Papers 2026 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "BITSAT Mathematics Test - 5". These 40 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 50 minutes
  • - Number of Questions: 40

Sign up on EduRev for free to attempt this test and track your preparation progress.

BITSAT Mathematics Test - 5 - Question 1
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If cos 20° = k and cos x = 2k2-1, then the possible values of x between 0° and 360° are

Detailed Solution: Question 1

BITSAT Mathematics Test - 5 - Question 2
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sin2 25° + sin265° is equal to 

Detailed Solution: Question 2

sin2 25° = sin2 65°
= sin2 25° + sin2 (90° - 25°)
= sin2 25° + cos2 25° = 1

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BITSAT Mathematics Test - 5 - Question 3
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cos 1° cos 2° cos 3°.... cos 179° is equal to 

Detailed Solution: Question 3

The given product contains the factor cos 90° = 0.

BITSAT Mathematics Test - 5 - Question 4
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Which of the following is correct ?

Detailed Solution: Question 4

cos 2 = cos 114° 35’ 30” is surely negative.

BITSAT Mathematics Test - 5 - Question 5
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Detailed Solution: Question 5

BITSAT Mathematics Test - 5 - Question 6
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The area enclosed between the curves y = x2 and x = y2 is

Detailed Solution: Question 6

The two curves must in (0,0) and (1,1).
∴ Required area lies above the curve y = x2 and below x = y2 is

BITSAT Mathematics Test - 5 - Question 7
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The differential coefficient of log (I log xl) with respect to log x is

Detailed Solution: Question 7

Let, y = log (Hog xl) and z = log x,
then 

BITSAT Mathematics Test - 5 - Question 8
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is equal to 

Detailed Solution: Question 8


= log(√2+1).

BITSAT Mathematics Test - 5 - Question 9
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is equal to

Detailed Solution: Question 9


= log (√2 + 1)

BITSAT Mathematics Test - 5 - Question 10
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Length of the tangent from (2,1) to the circle x2 + y2 + 4y + 3 = 0 is

Detailed Solution: Question 10

Required length = 

BITSAT Mathematics Test - 5 - Question 11
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If  then what is the standard deviation of all the terms?

Detailed Solution: Question 11

Let 
Therefore,

Hence, this is the required solution.

BITSAT Mathematics Test - 5 - Question 12
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Consider the given three vectors:

If are coplanar, then what is the value of

Detailed Solution: Question 12


Since, are coplanar,

BITSAT Mathematics Test - 5 - Question 13
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A person is to select an onto function from all the functions F: A → A, where A = {2, 4, 6, 8, 10, 12, 14}. Find the probability of selecting onto function.

Detailed Solution: Question 13

Total number of functions = 77
Number of onto functions = 7!
Therefore,
required probability =
Hence, this is the required solution.

BITSAT Mathematics Test - 5 - Question 14
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Consider the equations given below:
y = (1 - x)2
y = 0
x = 0
A straight line representing x = k separates the area enclosed by the above curves. Say both the areas are A1 (0 ≤ x ≤ k) and A2 (k ≤ x ≤ 1). If A1 - A2 = 1/4 , then what is the value of k?

Detailed Solution: Question 14

Here, area between 0 and k is A1, and between k and 1 is A2. Therefore,

BITSAT Mathematics Test - 5 - Question 15
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Evaluate 

Detailed Solution: Question 15

Consider the given expression:

Since is positive when 0 < x < 2, and negative when 2< x < 4.
Therefore,

Therefore,

Hence, this is the required solution.

BITSAT Mathematics Test - 5 - Question 16
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Consider the function given below:

Which of the following statements is true?

Detailed Solution: Question 16

Consider the given expression:

Differentiate both sides with respect to x.

Hence, this is the required solution.

BITSAT Mathematics Test - 5 - Question 17
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Consider the given function:
f(x) = |x - 3| + |x| + |x + 3|
The function is

Detailed Solution: Question 17

Consider the graph of the above function.

Therefore,
the function is not differentiable at x = -3, 0, 3.
Hence, this is the required solution.

BITSAT Mathematics Test - 5 - Question 18
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Evaluate:

Detailed Solution: Question 18

Consider the given expression.

Hence, this is the required solution.

BITSAT Mathematics Test - 5 - Question 19
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Points of intersection of a plane on the coordinate axes are P, Q and R. If (a, b, c) is the intersection point of the medians of ∆PQR, then what is the equation of the plane?

Detailed Solution: Question 19

We know that the intercept form of the equation of a plane is
where x1, y1 and z1 are the intercepts made by the plane on the axes.
Therefore,

Therefore, the equation of the plane is

Hence, this is the required solution.

BITSAT Mathematics Test - 5 - Question 20
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The fraction is equivalent to

Detailed Solution: Question 20

We can write the given expression as

Multiplying numerator and denominator by the conjugate of

BITSAT Mathematics Test - 5 - Question 21
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The roots of the equation x -2 - 2x - 1 = 8 are

Detailed Solution: Question 21

Given:

BITSAT Mathematics Test - 5 - Question 22
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The total number of solutions of the system of equations 5x−y=3,y2−6x2=25 are

Detailed Solution: Question 22

The given equations are
5x−y=3 ...(i)
y2−6x2=25 ...(ii)
From (i), we have
y = 5x - 3
Substituting this value of y in (ii), we get
(5x−3)2 −6x2 = 25
⇒19x2 − 30x − 16 = 0
⇒19x2−38x+8x−16=0
⇒19x (x − 2) + 8(x − 2)=0
⇒(19x + 8)(x − 2)=0

And substituting these values in (i), we get

BITSAT Mathematics Test - 5 - Question 23
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The roots of the equationare

Detailed Solution: Question 23

Given:


Hence,

BITSAT Mathematics Test - 5 - Question 24
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Express with rational denominator 

Detailed Solution: Question 24

Given, 

In order to rationalise we will first obtain the rationalising factor of

Therefore, the rationalising factor is

Now multiplying and dividing the given expression by we get

BITSAT Mathematics Test - 5 - Question 25
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Expresswith rational denominator.

Detailed Solution: Question 25

Given, 
In order to rationalise we will first obtain the rationalising factor of 

Therefore, the rationalising factor is

Now multiplying and dividing the given expression by (i), we have

BITSAT Mathematics Test - 5 - Question 26
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Detailed Solution: Question 26


Multiplying and dividing (1)by (–√3−√2), we get

Now, multiplying and dividing (2) by (√3 + √2), we get

BITSAT Mathematics Test - 5 - Question 27
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Let A(1,−1,2) A and B(2,3,−1) be two points. If a point P P divides AB AB internally in the ratio 2:3, then the position vector of P is

Detailed Solution: Question 27

If A(x1, y1, z1)  & B(x2, y2, z2) and P divides AB internally in ratio m1:m2 then



BITSAT Mathematics Test - 5 - Question 28
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If the direction cosines of a line are then

Detailed Solution: Question 28

As we know if direction cosines of a line are (l,m,n) then l2+m2+n2=1
Since direction cosines of line are 

BITSAT Mathematics Test - 5 - Question 29
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The angle between two diagonals of a cube is

Detailed Solution: Question 29

Let edge of a cube be 1 units. The diagonals of a cube are OA and BC. So, DR's of diagonals OA are (1, 1, 1) and BC are (0, -1, 1, 1), i.e., (-1, 1, 1)

Now, angle between diagonals,

BITSAT Mathematics Test - 5 - Question 30
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The angle between two diagonals of a cube is

Detailed Solution: Question 30

Let edge of a cube be 1 units. The diagonals of a cube are OA and BC. So, DR's of diagonals OA are (1, 1, 1) and BC are (0, -1, 1, 1), i.e., (-1, 1, 1)

Now, angle between diagonals,

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About this JEE Practice Test

This test contains 40 MCQs on BITSAT Mathematics Test - 5 with detailed solutions for each question. It is part of the BITSAT Mock Tests Series & Past Year Papers 2026 course on EduRev, designed to align with the current JEE syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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