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JEE Exam  >  JEE Tests  >  BITSAT Maths Test - 2 - JEE MCQ

BITSAT Maths Test - 2 - JEE MCQ


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30 Questions MCQ Test - BITSAT Maths Test - 2

BITSAT Maths Test - 2 for JEE 2024 is part of JEE preparation. The BITSAT Maths Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Maths Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Maths Test - 2 below.
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BITSAT Maths Test - 2 - Question 1
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The area (in square units) of the region enclosed by the curves y = x2 and y = x3 is

Detailed Solution for BITSAT Maths Test - 2 - Question 1
By solving equations 1and 2 we can get 0,1as values for x which can be the limits and by subtracting y=x^2 from x^3 and integrating the result within the limits 0 to 1 we can get area = 1\12
BITSAT Maths Test - 2 - Question 2
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In the expansion of (y1/6 - y-1/3)9 the term independent of y is :

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BITSAT Maths Test - 2 - Question 3
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If the two circles 2x2 + 2y2 -3x + 6y + k = 0 and x2 + y2 - 4x + 10y + 16 = 0 cut orthogonally, then the value of k is

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BITSAT Maths Test - 2 - Question 4
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Equation of the diameter of the circle x2 + y2 - 6x + 2y = 0 which passes thro' the origin is
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BITSAT Maths Test - 2 - Question 5
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If then a and b are
BITSAT Maths Test - 2 - Question 6
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The order and degree of differential equation √(dy/dx) - 4 (dy/ dx) - 7x = 0 are
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√(dy/ dx)- 4 (dy/dx) -7x = 0
or √(dy/ dx) = 4 (dy/dx) + 7x
Squaring both side
dy/ dx = [4 (dy/dx ) + 7x]2
or dy dx = 16 ( dy/dx )2 + 49x2 + 56 dy/ dx x
∴ order = 1
degree = 2

BITSAT Maths Test - 2 - Question 7
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If x dy = y(dx + y dy), y > 0 and y (1) = 1, then y (-3) is equal to
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BITSAT Maths Test - 2 - Question 8
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If x=a[(cost)+(log tan(t/2))], y=a sint, (dy/dx)=
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BITSAT Maths Test - 2 - Question 9
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Solution of the differential equation 
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BITSAT Maths Test - 2 - Question 10
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then  = f '(1)
BITSAT Maths Test - 2 - Question 11
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1/2! - 1/3! + 1/4! - 1/5! + ....... equals
BITSAT Maths Test - 2 - Question 12
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The length of the shadow of a rod inclined at 10o to the vertical towards the sun is 2.05 metres when the elevation of the sun is 38o.The length of the rod is
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BITSAT Maths Test - 2 - Question 13
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The product of the perpendicular, drawn from any point on a hyperbola to its asymptotes is
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99996222375-0-0

BITSAT Maths Test - 2 - Question 14
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The area of the region bounded by the curve y = x - x2 between x = 0 and x = 1 is
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BITSAT Maths Test - 2 - Question 15
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The area bounded by the curve x = at2, y = 2at and the X-axis is 1 ≤ t ≤ 3 is
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BITSAT Maths Test - 2 - Question 16
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The sum of first 50 terms of the series cot⁻13 + cot⁻1 7 + cot⁻1 13 + cot⁻1 21 + ... is
BITSAT Maths Test - 2 - Question 17
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The domain of the function f(x) = √(2 - 2x - x2) is
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BITSAT Maths Test - 2 - Question 18
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The difference of an integer and its cube is divisible by
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Solution:


  • Let the integer be x.

  • The difference of the integer and its cube is x - x^3.

  • For this difference to be divisible by a number, the number should divide the difference without leaving a remainder.

  • We need to find a number which divides x - x^3 without leaving a remainder.

  • Factoring out x from x - x^3 gives x(1 - x^2).

  • Further factoring 1 - x^2 gives (1 - x)(1 + x).

  • Therefore, the difference x - x^3 = x(1 - x)(1 + x).

  • The number that divides x - x^3 without leaving a remainder is the common factor of x, 1 - x, and 1 + x.

  • The common factor of x, 1 - x, and 1 + x is 1.

  • Therefore, the difference of an integer and its cube is divisible by 1, which means it is divisible by any integer including 4, 6, 10, and 9.

  • However, the smallest number among the given options that divides x - x^3 without leaving a remainder is 6.

  • Hence, the correct answer is B: 6.


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BITSAT Maths Test - 2 - Question 19
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If A and B are two matrices such that AB = B and BA = A, then A2 + B2 is equal to
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BITSAT Maths Test - 2 - Question 20
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if A is a 3 x 3 matrix and B is its adjoint matrix. If ∣B∣ = 64, then ∣A∣ =
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BITSAT Maths Test - 2 - Question 21
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The strength of a beam varies as the product of its breadth b and square of its depth d. A beam cut out of a circular log of radius r would be strong when
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BITSAT Maths Test - 2 - Question 22
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Which of the following is correct
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Explanation:


  • To compare two complex numbers, we compare their real parts first. If the real parts are equal, then we compare the imaginary parts.

  • In this case, the real parts are 5 and 6. Since 5 is less than 6, we can say that 5 + 3i is less than 6 + 4i.

  • Therefore, the correct answer is option C: 5 + 3i < 6 + 4i.


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BITSAT Maths Test - 2 - Question 23
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Equation 2x2+7xy+3y2+8x+14y+λ=0 represents a pair of straight lines, value of λ is
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BITSAT Maths Test - 2 - Question 24
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The pole of the line lx+my+n=0 w.r.t. the parabloa y2 =4ax
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BITSAT Maths Test - 2 - Question 25
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The latus rectum of the parabola y2 = 5x + 4y + 1 is
Detailed Solution for BITSAT Maths Test - 2 - Question 25

y2 = 5x + 4y + 1
or y2 - 4y = 5x + 1
or y2 - 2.2.y + (2)2 - (2)2 = 5x + 1
or (y - 2)2 = 5x + 5
or (y - 2)2 = 5(x + 1)
Length of the latus rectum = 5

BITSAT Maths Test - 2 - Question 26
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How many total words can be formed from the letters of the word INSURANCE in which vowels are always together?
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BITSAT Maths Test - 2 - Question 27
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A polygon has 44 diagonals. The number of its sides is
Detailed Solution for BITSAT Maths Test - 2 - Question 27
The number of diagonals for n sided polygon = [n(n-3)]/2 . therefore, => 44 = [n(n-3)]/2 . => n^2 - 3n - 88 = 0 . => (n+8)(n-11) =0. => n = -8 or 11. Neglect n = -8. Therefore, => n = 11 . therefore, the no. of sides = 11 Hence, correct answer is (B).
BITSAT Maths Test - 2 - Question 28
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A single letter is selected at random from the word "PROBABILITY". The probability that it is a vowel is
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BITSAT Maths Test - 2 - Question 29
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A committee consists of 9 experts from three institutions A, B and C, of which 2 are from A, 3 from B and 4 from C. If three experts resign, then the probability that they belong to different institutions is
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BITSAT Maths Test - 2 - Question 30
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If the area of a Δ A B C be λ then a2 sin 2B + b2 sin 2A is equal to
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