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Mathematics Test 139

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Mathematics Test 139
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Weekly Quiz Competition
  • Question 1
    4 / -1

    Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r is equal to

    Solution

    From figure it is clear that ΔPRQ and ΔRSP are similar

     

  • Question 2
    4 / -1

    If a3 + b6 = 2, then the maximum value of the term independent of x in the expansion of (ax1/3 + bx-1/6)9 (a > 0, b > 0) is

    Solution

     

  • Question 3
    4 / -1

    The co-ordinates of the point which divides the line segment joining the points (5,4, 2) and (−1,−2, 4) in the ratio 2 : 3 externally is

    Solution

    Let A (5 , 4 , 2) and B (- 1, - 2, 4) be the given points.
    Let P (x , y, z) be the point, which divides the line segment [AB] in the ratio - 2 : 3

     

  • Question 4
    4 / -1

    The area of an expanding rectangle is increasing at the rate of 48 cm2/sec. The length of the rectangle is always equal to the square of the breadth. At the instant when the breadth is 4.5 cm, The length is increasing at the rate of

    Solution

     

  • Question 5
    4 / -1

    If nP= 1680 and nC= 70, then 69 n + r! is equal to

    Solution

     

  • Question 6
    4 / -1

    Consider the straight line ax + by = c, where a, b, c ∈ Rthis line meets the coordinate axes at A and B respectively. If the area of the ΔOAB, O being origin, is always a constant equal to half, then

    Solution

    Given line is ax + by = c


    ⇒ This line cut the coordinate axes as points 

    Equating this area to half, we get,
    c2 = ab
    Therefore, a, c, b are in GP

     

  • Question 7
    4 / -1

    If for a ΔABC, cot A⋅cot B⋅cot C>0 then the triangle is

    Solution

    cot A⋅cot B⋅cot C>0⇒cot A>0, cot B>0, cot C > 0

    Because two or more of cot A, cot B, cot C cannot be negative at the same time in a triangle, as no two angle could be more than 90 degree.

     

  • Question 8
    4 / -1

    The equation of the smallest circle passing through the intersection of the line x + y = 1 and the circle x+ y2 = 9 is

    Solution

    Any circle passing through the points of intersection of the given line and circle has the equation x2 + y2 - 9 + λ (x + y - 1) = 0. Its centre 


    The circle is the smallest if center 



    Putting this value for λ, the equation of the smallest circle is x2 + y2 - 9 - (x + y - 1) = 0
    ⇒ x2 + y2 - x - y - 8 = 0

     

  • Question 9
    4 / -1

    Solution

     

  • Question 10
    4 / -1

     be two perpendicular unit vectors such that 

    Solution

     

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