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Mathematics Test 220

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Mathematics Test 220
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Weekly Quiz Competition
  • Question 1
    4 / -1

    If x and y are two positive numbers such that x + y = a, then the minimum value of

    Solution

     

  • Question 2
    4 / -1

    The negative of q∨ ~ (p ∧ r) is

    Solution

    As we know from De Morgan's law, that ~(p∨q) ≡ ~p∧~q

    Therefore, ~(q∨ ~ (p ∧ r)) ≡ ~q ∧ (~(~(p ∧ r)))

    Hence, ~(q ∨ ~(p ∧ r)) ≡ ~q ∧ (p ∧ r)

     

  • Question 3
    4 / -1

    For hyperbola  which of the following remains constant with change in α?

    Solution

     

  • Question 4
    4 / -1

    The centre of a circle passing through the points (0,0),(1,0) and touching the circle x+ y2 = 9 is

    Solution

    We have,

    x2 + y2 = 9

    Radius, OP = 3 units.

    Hence, radius of required circle = 3/2 units.

    Equation of required circle is

    (x − h)+ (y − k)= 9/4 ...(i)

    This is passing through (0,0) and (1,0), so

    h+ k2 = 9/4 ...(ii)

    (h − 1)+ k2 = 9/4 ...(iii)

    Subtracting (ii) & (iii)ii & iii, we get

    (h − 1)= h2

    ⇒ 2h = 1

    ⇒ h = 12

     

  • Question 5
    4 / -1

    Solution

     

  • Question 6
    4 / -1

    Given  then for non-zero values of x,f(x) equals

    Solution

     

  • Question 7
    4 / -1

    A is a set containing n elements, subset P of A is chosen. The set A is reconstructed by replacing the elements of P, subset Q of A is chosen. The number of ways of selecting P and Q so that P and Q are non intersecting, is

    Solution

    Let A = {a1, a2, a3......an}P, Q are subsets of A elements in P, Q can be chosen in following ways

    Case I: ai ∈ P ai ∈ Q i = 1,2,...n

    Case II: ai ∈ P ai ∉ Q i = 1,2,....n

    Case III: ai ∉ P ai ∈ Q i = 1,2,....n

    Case IV: ai ∉ P ai ∉ Q i = 1,2,...n

    ∵ P, Q are non intersecting sets every element can be chosen as in case (II), case (III), case (IV) i.e. every elements can be chosen in 3 ways. Total number of ways

    = 3 × 3.....n = times = 3n ways

    But one case needs to be excluded when P = ϕ, Q = ϕ

    ∴ Number of ways = 3n − 1

     

  • Question 8
    4 / -1

    If 3a + 2b + 6c = 0, the family of straight lines ax + by + c = 0 passes through a fixed point whose coordinates are given by

    Solution

     

  • Question 9
    4 / -1

    The equation of the common tangent to the equal parabolas y= 4ax and x= 4ay is

    Solution

     

  • Question 10
    4 / -1

    A dice is rolled three times, then the probability of getting a larger number than the previous number each time is

    Solution

    Exhaustive number of cases is 6= 216.

    Now if a larger number appears than the previous number each time, all the three numbers are distinct.

    Now three numbers can be selected from six numbers in 6C3 ways and there is only one way in which three selected numbers can appear.

    Hence, probability is 20/216 = 5/54.

     

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