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Mathematics Test - 29

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Mathematics Test - 29
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Weekly Quiz Competition
  • Question 1
    3 / -1

    The area lying in the first quadrant and bounded by the circle x2 + y2 = 4, the line x =√3 y and the x-axis is

    Solution

  • Question 2
    3 / -1

    A missile fired from the ground level rises x meters vertically upwards in t seconds, where x = 100t - 25/2 t2. The maximum height reached is

    Solution

  • Question 3
    3 / -1

    The area of the figure bounded by the curves y = | x - 1 | and y = 3 - | x | is

    Solution

  • Question 4
    3 / -1

    If a parallelepiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of diagonal of the parallelopiped is:

    Solution

    According to question, if we draw the parallelepiped in 3D space.

    Points \(\mathrm{P}(5,8,10)\) and \(\mathrm{Q}(3,6,8)\) appear to be opposite vertices of parallelepiped.

    \(\therefore\) Length of PQ gives the length of diagonal of parallelepiped.

    Using distance formula we can find PQ as:

    \(P Q =\sqrt{(5-3)^{2}+(8-6)^{2}+(10-8)^{2}}\)

    \( PQ=\sqrt{2^{2}+2^{2}+2^{2}}\)

    \(PQ=\sqrt{12}\)

    \(PQ=2 \sqrt{3} \)

    \(\therefore\) Length of diagonal of the parallelepiped is \(2 \sqrt{3}\) units.

  • Question 5
    3 / -1

    Period of cot 3 x - cos (4x + 3) is

    Solution

    Period of cot 3x is π/3 and period of cos (4x + 3) is π/2

    ⇒ L.C.M. is π, hence the answer.

  • Question 6
    3 / -1

    Three identical dice are rolled. The probability that the same number will appear on each of them is

    Solution

    Exhaustive number of cases = 63 = 216.

    The same number can appear on each of the dice in the following ways: (1, 1, 1), (2, 2, 2).... (6, 6, 6).

    So, favourable number of cases = 6.

    Hence, required probability = 6/216 = 1/36.

  • Question 7
    3 / -1

    Solution

  • Question 8
    3 / -1

    Let f(x) = [x], then f′(1) = ?

    Solution

  • Question 9
    3 / -1

    α and β lie between 0 and π/4, cos(α + β) = 12/13 and sin(α - β) = 3/5.

    sin 2α =

    Solution

  • Question 10
    3 / -1

    If n(U) = 20, n(A) = 12, n(B) = 9, n(AB) = 4, where U is the universal set, A and B are subsets of U, then n | (A U B)o| is equal to

    Solution

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