JEE Exam  >  JEE Tests  >  Sample BITSAT Maths Test - JEE MCQ

Sample BITSAT Maths Test - JEE MCQ


Test Description

30 Questions MCQ Test - Sample BITSAT Maths Test

Sample BITSAT Maths Test for JEE 2024 is part of JEE preparation. The Sample BITSAT Maths Test questions and answers have been prepared according to the JEE exam syllabus.The Sample BITSAT Maths Test MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Sample BITSAT Maths Test below.
Solutions of Sample BITSAT Maths Test questions in English are available as part of our course for JEE & Sample BITSAT Maths Test solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Sample BITSAT Maths Test | 45 questions in 55 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
Sample BITSAT Maths Test - Question 1

What is the area under the curve y = |x| + | x - 1| between x = 0 and x = 1 ?

Detailed Solution for Sample BITSAT Maths Test - Question 1

|x| for x≥0
=x and |x−1| for x≤1
=−(x−1),
so, ∫10(|x|+|x−|1∣)=required area
a=∫10xdx−∫10(x−1)dx
=[x22]10−[x22−x]10=12−(12−1)=1sq. units

Sample BITSAT Maths Test - Question 2

The greatest term in the expansion of (3 + 2x)9, when x = 1, is

Detailed Solution for Sample BITSAT Maths Test - Question 2




1 Crore+ students have signed up on EduRev. Have you? Download the App
Sample BITSAT Maths Test - Question 3

The coordinates of the pole of the line lx+my+n=0 with respect to the circle x2+y2=1 are

Detailed Solution for Sample BITSAT Maths Test - Question 3

ANSWER :- d

Solution :- Let (x1,y1) be the pole of the line lx+my+n=0 with respect to the hyperbola x2a2−y2b2=1. Then, the equation of the polar is

xx1/ + yy1/ = 1………(i)

Since, (x1,y1) is the pole of the line lx+my+n=0. So, the polar of (x1,y1) is also the line

lx+my+n=0...(ii)

clearly, (i) and (ii) represent the same line. Therefore,

x1(l) = y1(m) = 1/(−n)

⇒x1 = −l/n,  y1 = -m/n.

Hence, the pole of the given line with respect to the given hyperbola is

(−l/n, -m/n)

Sample BITSAT Maths Test - Question 4

If the line 2x - y + k = 0 is a diameter of the circle x2 + y2 + 6x -6y + 5 =0, then k is equal to

Detailed Solution for Sample BITSAT Maths Test - Question 4
If the given line 2x-y+k=0 is the diameter of the circle then it passes through the centre of the given circle. Which on comparing we get as (-3,3).On substituting this in the line equation we get 2(-3)- 3+k=0=> -9+k=0 K=9.
Sample BITSAT Maths Test - Question 5

If z = i log(2 - √3), then cos z =

Detailed Solution for Sample BITSAT Maths Test - Question 5


Sample BITSAT Maths Test - Question 6

The differential equation of the family of lines passing through the origin is

Detailed Solution for Sample BITSAT Maths Test - Question 6

The equation of line passing through the origin is y = mx , when m is constant
Diffrence w.r.t x

Sample BITSAT Maths Test - Question 7

Which of the following is a solution of the differential equation

Detailed Solution for Sample BITSAT Maths Test - Question 7

Sample BITSAT Maths Test - Question 8

If y' = x-y/x+y, then its solution is

Detailed Solution for Sample BITSAT Maths Test - Question 8

dy/dx= x−y/x+y

Put,y=vx

⟹dy/dx=v+x(dv/dx)

⟹v+xdv/dx=1−v/1+v

⟹xdv/dx = 1−2v−v^2/(1+v)

⟹∫v+1/(v+1)^2−2dv=−∫1/xdx

⟹1/2ln⁡|[(v+1)^2−2]| = 2ln⁡|c1/x|

⟹x^2(v^2+2v−1)=C

Where C = 2ln⁡|c1/x|

Since,v=y/x, we get

⟹ y2+2xy−x2=C

Sample BITSAT Maths Test - Question 9

Detailed Solution for Sample BITSAT Maths Test - Question 9

 

Sample BITSAT Maths Test - Question 10

(d/dx)[tan⁻1((sinx+cosx)/(cosx-sinx))]

Detailed Solution for Sample BITSAT Maths Test - Question 10

Sample BITSAT Maths Test - Question 11

Value of 1 + log x + (log x)2/2! + (log x)3/3! + ..... ∞ is

Sample BITSAT Maths Test - Question 12

The angle of elevation of a cloud from a point h mt above the surface of a lake is θ and the angle of depression of its reflection in the lake is φ . The height of the cloud is

Sample BITSAT Maths Test - Question 13

The eccentricity of the conjugate hyperbola of the hyperbola x2 - 3y2 = 1 is

Detailed Solution for Sample BITSAT Maths Test - Question 13

​​​​​​

Sample BITSAT Maths Test - Question 14

Which of the following functions is a solution of the differential equation (dy/dx)2 - x (dy/dx) + y = 0?

Detailed Solution for Sample BITSAT Maths Test - Question 14

Sample BITSAT Maths Test - Question 15

The solution of the differential equation (dy/dx) = (y/x) + (φ (y/x)/φ' (y/x)) is

Sample BITSAT Maths Test - Question 16

tan⁻1(1/4)+ tan⁻1(2/9) is equal to

Sample BITSAT Maths Test - Question 17

f(x) = ||x| - 1| is not differentiable at

Sample BITSAT Maths Test - Question 18

For every n ∈ N, 23n-7n-1 is divisible by

Detailed Solution for Sample BITSAT Maths Test - Question 18

Sample BITSAT Maths Test - Question 19

If A, B are two square matrices such that AB = A and BA = B, then

Sample BITSAT Maths Test - Question 20

For a square matrix A, it is given that AA' = I, then A is a

Sample BITSAT Maths Test - Question 21

The maximum value of xy subject to x+y=8 is

Detailed Solution for Sample BITSAT Maths Test - Question 21
Because x and y should have value 4 so that 4+4=8 and 4*4=16
Sample BITSAT Maths Test - Question 22

The real value of α for which the expression 1-i sin α/1+2 i sin α is purely real is

Detailed Solution for Sample BITSAT Maths Test - Question 22
The real value of α for which the expression (1-i sin α) / (1+2 i sin α) is purely 
Sample BITSAT Maths Test - Question 23

The angle between lines xy=0 is

Detailed Solution for Sample BITSAT Maths Test - Question 23

For xy = 0
The lines are: x = 0 & y = 0 which are the Y and X axis respectively which are perpendicular.

Sample BITSAT Maths Test - Question 24

The focus of the parabola (y-2)2=20(x+3) is

Sample BITSAT Maths Test - Question 25

The equation of the normal to the curve x2 = 4y at (1, 2) is

Detailed Solution for Sample BITSAT Maths Test - Question 25

Sample BITSAT Maths Test - Question 26

Two finite sets have m and n elements, the total number of subsets of the first set is 56 more than the total number of subsets of the second. The value of m and n are respectively

Detailed Solution for Sample BITSAT Maths Test - Question 26

Let A denote the first set and B denote the second set
We have, n(A) = 2m and n(B) = 2n
As per the question, we have
n(A) = 56 + n(B)
⇒ n(A) - n(B) = 56
⇒ 2m - 2n = 56
⇒ 2n (2m - n - 1)
⇒ 2n (2m - n - 1) = 8 x 7
⇒ 2n = 8 = 23 or (2m - n - 1) = 7
⇒ n = 3 or 2m - n = 8 = 23 = 26 - 3
⇒ n = 3 or m - n = 3
⇒ n = 3 or m = 6
Hence, the required values of m and n are 6 and 3 respectively

Sample BITSAT Maths Test - Question 27

In how many ways can the letters of the word ARRANGE be arranged so that R's are never together?

Detailed Solution for Sample BITSAT Maths Test - Question 27

Reqd. ways = = 1260 - 360 = 900

Sample BITSAT Maths Test - Question 28

A and B are events such that P(A ∪ B) = 3/4, P(A ∩ B) = 1/4, P(A̅)= 2/3, then P(A̅ ∩ B) is

Sample BITSAT Maths Test - Question 29

The probability that a number selected at random from the set of numbers {1,2,3,....,100} is a cube is

Sample BITSAT Maths Test - Question 30

In a equilateral triangle r : R : r1 is

Detailed Solution for Sample BITSAT Maths Test - Question 30

A = B = C = 60o
r : R : r1 = 4R sin (A/2) sin (B/2) sin (C/2) : R : 4R sin (A/2) cos (B/2) cos (C/2)
= 4 (1/2) (1/2) (1/2) : 1 : 4 (1/2) (√3 /2) (√3 /2) = (1/2) : 1 : (3/2) = 1 : 2 : 3

View more questions
Information about Sample BITSAT Maths Test Page
In this test you can find the Exam questions for Sample BITSAT Maths Test solved & explained in the simplest way possible. Besides giving Questions and answers for Sample BITSAT Maths Test, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE