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    Mathematics: Xy M M M 2, 3

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    Tomislav Car
    The document contains solutions to two math problems: 1) It finds the two possible values of m that satisfy the equations of two intersecting curves: xy = 8, y = x^2, and y = 1. The solutions are m = 2, 3. 2) It calculates the x-coordinate of the midpoint R of the points of intersection P and Q of a circle and line. The x-coordinate is found to be -3/5.

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    Download as PDF, TXT or read online from Scribd

    Mathematics: Xy M M M 2, 3

    Uploaded by

    Tomislav Car
    42 views1 page
    The document contains solutions to two math problems: 1) It finds the two possible values of m that satisfy the equations of two intersecting curves: xy = 8, y = x^2, and y = 1. The solutions are m = 2, 3. 2) It calculates the x-coordinate of the midpoint R of the points of intersection P and Q of a circle and line. The x-coordinate is found to be -3/5.

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    031

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    The document contains solutions to two math problems: 1) It finds the two possible values of m that satisfy the equations of two intersecting curves: xy = 8, y = x^2, and y = 1. The solutions are m = 2, 3. 2) It calculates the x-coordinate of the midpoint R of the points of intersection P and Q of a circle and line. The x-coordinate is found to be -3/5.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    42 views1 page

    Mathematics: Xy M M M 2, 3

    Uploaded by

    Tomislav Car
    The document contains solutions to two math problems: 1) It finds the two possible values of m that satisfy the equations of two intersecting curves: xy = 8, y = x^2, and y = 1. The solutions are m = 2, 3. 2) It calculates the x-coordinate of the midpoint R of the points of intersection P and Q of a circle and line. The x-coordinate is found to be -3/5.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 1

    EBD_7801

    2019 - 26

    Mathematics (-3m + 15) m


    Þ = -1 Þ m 2 - 5m + 6 = 0
    -18
    37. (b) xy £ 8, 1 £ y £ x2
    Intersection points of xy = 8 and y = 1 is (8, 1); xy = 8 Þ m = 2, 3
    and y = x2 is (2, 4) and y = x2 and y = 1 is (1, 1)
    Y Alternately :- The x-coordinates of intersection
    y = x2 points P and Q of circle and line can be given by
    solving their equations for x.
    (2, 4) Putting y = mx + 1 in (x –3)2 + (y + 2)2 = 25, we
    xy get (x – 3) 2 + (mx + 3) 2 = 25
    = 8 (8, 1) Þ (m2 + 1) x 2 + 6(m - 1) x - 16 = 0
    (1, 1) y=1
    -6( m - 1)
    X' O X Þ x1 + x2 = where x1 and x2 are x-
    m2 + 1
    coordinates of P and Q respectively. As per
    -3
    question, x-coordinate of mid point of PQ is
    Y' 5
    2 8 8 x1 + x2 3 -3( m - 1) -3
    2 8 \ =- Þ =
    Required area = ò x dx + ò dx - ò1dx 2 5 m2 + 1 5
    x
    1 2 1
    2 Þ m2 – 5m + 6 = 0 Þ m = 2, 3
    æ x3 ö 39. (d) S : |z – 2 + i| ³ 5 represents boundary and outer
    = ç ÷ + (8lnx)82 - (x)18
    è 3 ø1 region of circle with centre (2, –1) and radius 5 .
    8 1
    = - + 8 ln 8 - 8ln 2 - (8 - 1) 1
    3 3 z0 Î S, such that is the maximum.
    7 14 | z0 - 1 |
    = + 24 ln 2 - 8 ln 2 - 7 = 16 ln 2 -
    3 3
    \ |z0 –1| is minimum
    \ correct option is (b)
    z0 Î S with |z0 – 1| as minimum will be a point on
    38. (a) Circle (x – 3)2 + (y + 2)2 = 25, with centre C(3, –2) and
    boundary of circle of region S which lies on radius of
    radius 5 is intersected by a line y = mx + 1 at p & Q
    this circle, and passes through (1, 0).
    such that mid point R of PQ has its
    \ z0, 1, 2 – i are collinear, or (x0, y0), (1, 0),
    3
    x-coordinate as - . (2, –1) are collinear.
    5
    æ 3 -3m ö \ Using slopes of paralled lines,
    Let R ç - , + 1÷
    è 5 5 ø y0 -1
    = Þ y0 = 1 – x0
    x0 - 1 2 - 1
    Y

    y = mx + 1 P
    R Q z0
    O (1, 0)
    X' X
    O (2, –1)

    C (3, –2)

    4 - z0 - z0 4 - ( z0 + z0 )
    Now, =
    z0 - z 0 + 2i ( z0 - z0 ) + 2i
    Y' 4 - 2 x0 4 - 2 x0 2(2 - x0 ) 1
    -3m = = = = -i
    +1 + 2 2iy0 + 2i 2i - 2 x0i + 2i
    =
    5 2(2 - x0 )i i
    Then CR ^ PQ Þ ´ m = -1
    3
    - -3 æ 4 - z0 - z0 ö -p
    5 \ Arg ç
    z - z - 2i ÷ = Arg (–i ) = 2
    è 0 0 ø

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