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Mathematics Test - 5

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Mathematics Test - 5
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Solution

  • Question 2
    1 / -0

    If P(1, 0), Q(– 1, 0) and R(2, 0) are three given points, then the locus of point S, which satisfies the relation (SQ)2 + (SR)2 = 2(SP)2, is (where SQ, SR and SP are line segments)

    Solution

    Let the point be (h, k)

    So, SQ2 = (h + 1)2 + k2

    SR2 = (h - 2)2 + k2

    SP2 = (h - 1)2 + k2

    Substituting in the given equation and replacing the values of h and k by x and y, we get

    x = - 3/2 (A straight line parallel to y-axis).

  • Question 3
    1 / -0

    The orthocenter of the triangle with vertices

    Solution

  • Question 4
    1 / -0

    The orthocenter of the triangle formed by the lines xy = 0 and x + y = 1 is

    Solution

    The triangle formed by the given lines is a right angle triangle right angled at (0, 0). Hence, it is the orthocenter.

  • Question 5
    1 / -0

    The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x - 2y = 7. Then PQRS must be a

    Solution

    The slope of the first line: - 1/3

    Slope of second line is: 3

    Hence the diagonals are perpendicular, which is the property of a Rhombus.

  • Question 6
    1 / -0

    If the vertices P, Q, R of a triangle PQR are rational points, then which of the following points of the triangle PQR is not always a rational point?

    Solution

    Incentre of is notΔPQR always a rational point.

  • Question 7
    1 / -0

    If f(x) = xa + ax and g(x) = xa – ax, find the value of f(x) + g(x).

    Solution

    f(x) + g(x) = 2xa

  • Question 8
    1 / -0

    Find the period of the given function:

    f(x) = tan(x/3) + sin(2x)

    Solution

    Given, f(x) = tan(x/3) + sin(2x).

    Here, tan(x/3) is periodic with period 3π and sin2x is periodic with period π.
    Hence, f(x) will be periodic with period 3π.

  • Question 9
    1 / -0

    Solution

    We know that amount = P(1+ r/100)n. Substituting r = 12 and n = 3 in it, we get A = P[1.12]3 or A/P = 1.404.

  • Question 10
    1 / -0

    If

    L(X, Y) = Square of the larger number,

    S(X, Y) = Square of the smaller number,

    MODD (X, Y) = Absolute value of the difference (X - Y) and

    X * Y = Product of X and Y,

    then the value of MODD [{L (1, 2) * S(1, - 3)}, {S(0, 2) * MODD (3, - 4)}] *MODD[{S(3, -1) * L(1, - 2)}, { MODD (1, 2) * L(1, 2)}] will be

    Solution

  • Question 11
    1 / -0

    Solution

  • Question 12
    1 / -0

    If f(x) = log (1 + x) - log (1 - x) and log (g(x)) = log (3x + x3) - log (1 + 3x2), then what is the value of f(g(x))?

    Solution

  • Question 13
    1 / -0

    Directions: Refer to the following functions:

    A(x, y, z) = Max {max (x, y), min (y, z), min (x, z)}

    B(x, y, z) = Max {max (x, y), min (y, z), max (x, z)}

    C(x, y, z) = Max {min (x, y), min (y, z), min (x, z)}

    D(x, y, z) = Min {max (x, y), max (y, z), max (x, z)}

    Max (x, y, z) = Maximum of x, y and z.

    Min (x, y, z) = Minimum of x, y and z.

    Assume x, y and z as distinct integers.

    For what condition, will A(x, y, z) be equal to Max (x, y, z)?

    Solution

    Let us assign some values to x, y and z.

    Put x = 1, y = 2, z = 3.

    Now, A(x, y, z) = Max (2, 2, 1) = 1.

    B (x, y, z) = Max (2, 2, 2) = 2

    C (x, y, z) = Max (1, 2, 1) = 1

    D (x, y, z) = Min (2, 3, 1) = 3

    Max (x, y, z) = Min (1, 2, 3) = 1

    Min (x, y, z) = Max (1, 2, 3) = 3

    They will be equal only when z has the minimum value among the three, and x or y has the maximum value.

    1. You might have checked by options and option 1 satisfies, but option 2 also satisfies.

    2. You might have checked by options and option 2 satisfies, but option 1 also satisfies.

    3. We can simply ignore this option as for x = 1, y = 2 , z = 3, it does not satisfy.

  • Question 14
    1 / -0

    Solution


  • Question 15
    1 / -0

    Solution

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