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Mathematics Test 15
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Mathematics Test 15

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  • Question 1/8
    4 / -1

    The image of the point (–1, 3, 4) in the plane x – 2y = 0 is

    Solutions

    Required image lies on the line through A (- 1,3,4) and perpendicular to x-2y = 0 that is on the line

     

  • Question 2/8
    4 / -1

    Solutions

     

  • Question 3/8
    4 / -1

    The area of the region described by A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 – x} is

    Solutions

     

  • Question 4/8
    4 / -1

    The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is

    Solutions

     

  • Question 5/8
    4 / -1

    The lines 2x − 3y = 5 and 3x − 4y = 7 are diameters of a circle having area as 154 sq units. Then the equation of the circle is

    Solutions

    Intersection of diameter is the point (1, − 1) 
    πs2 = 154 
    ⇒ s2 = 49 
    (x − 1)2 + (y + 1)2 = 49 

  • Question 6/8
    4 / -1

    If |z + 4| ≤ 3, then the maximum value of |z + 1| is

    Solutions

    From the Argand diagram maximum value of |z + 1| is 6.


    Alternative : 
    |z + 1| = |z + 4 – 3| £ |z + 4| + |–3| = 6. 
    Hence, option A is correct.

  • Question 7/8
    4 / -1

     exceeding x is

    Solutions

     

  • Question 8/8
    4 / -1

    The degree and order of the differential equation of the family of all parabolas whose axis is x−axis, are respectively

    Solutions

    Equation y2 = 4a (x − h) 
    2yy1 = 4a ⇒ yy1 = 2a 
    yy2 = y12 = 0. 

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