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Sample BITSAT Maths Test - JEE MCQ


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45 Questions MCQ Test - Sample BITSAT Maths Test

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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




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

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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

Sample BITSAT Maths Test - Question 31

The perimeter of a triangle is 16cm. One of the sides is of length 6cm. If the area of the triangle is 12sq.cm, then the triangle is

Detailed Solution for Sample BITSAT Maths Test - Question 31

Sample BITSAT Maths Test - Question 32

The product of cube roots of -1 is equal to

Sample BITSAT Maths Test - Question 33

The sum of all 2-digit odd numbers is

Detailed Solution for Sample BITSAT Maths Test - Question 33
The two digit odd numbers
11,13,15,...., 99

a = 11, n = 45, d = 2

Sn = n/2[2a + (n-1)d}]

= 45/2 [22 + 44 × 2]

= 45/2 × 110

= 45 × 55

= 2475
Sample BITSAT Maths Test - Question 34

If the sum of first n terms of an A.P. be 3n2 - n and its common difference is 6, then its first term is

Sample BITSAT Maths Test - Question 35

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B, and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

Detailed Solution for Sample BITSAT Maths Test - Question 35

n(A)=4000

n(B)=2000

n(C)=1000

n(A∩B)=500

n(B∩C)=300

n(C∩A)=400

n(A∩B∩C)=200

 

We want to find n(A only)=n(A)–[n(A∩B)+n(A∩C)]+n(A∩B∩C)

 

n(A only)=4000–[500+400]+200=4000–700=3300

Sample BITSAT Maths Test - Question 36

If f : N x N → N is such that f (m,n) = m + n where N is the set of natural numbers, then which of the following is true?

Sample BITSAT Maths Test - Question 37

The ortho centre of triangle whose vertices are (0,0), (3,0) and (0,4) is

Sample BITSAT Maths Test - Question 38

The angle between the curves y2=x at x2=y at (1,1) is

Detailed Solution for Sample BITSAT Maths Test - Question 38




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Sample BITSAT Maths Test - Question 39

The angle between the planes 2x-y+z=6 and x+y+2z=7 is

Detailed Solution for Sample BITSAT Maths Test - Question 39

Sample BITSAT Maths Test - Question 40

The acute angle between the planes 2x-y+z=6 and x+y+2z=3 is

Sample BITSAT Maths Test - Question 41

If 2cos2x+3sinx-3=0, 0≤x≤180o, then x=

Sample BITSAT Maths Test - Question 42

The equation sinx+cosx=2 has

Sample BITSAT Maths Test - Question 43

Maximum value of cos2x+cos2y-cos2z is

Detailed Solution for Sample BITSAT Maths Test - Question 43

Max. value will occur when cosx=cosy=1 and cosz=0

Sample BITSAT Maths Test - Question 44

If 3i+4j and -5i+7j represent the sides of a triangle, then its area is

Detailed Solution for Sample BITSAT Maths Test - Question 44

Explanation:

Given sides of the triangle:

- Side 1: 3i + 4j
- Side 2: -5i + 7j

Calculating the area of the triangle:

- The area of a triangle can be found using the formula: Area = 1/2 * |(x1y2 - x2y1)|
- Let the given sides be represented as vectors A = 3i + 4j and B = -5i + 7j
- Calculate the area using the formula: Area = 1/2 * |(3 * 7 - 4 * (-5))| = 1/2 * |(21 + 20)| = 1/2 * 41 = 41/2

Therefore, the area of the triangle formed by the given sides is 41/2, which corresponds to option C.

Sample BITSAT Maths Test - Question 45

If and  are mutually perpendicular, then (a+b)2=

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