Linear Inequalities

  • Question 1
    CBSEENMA11015476

    Let C be the circle with centre at(1,1) and radius 1. If T is the circle centred at (0,y) passing through origin and touching the circle externally, then the radius of T is equal to

    • fraction numerator square root of 3 over denominator square root of 2 end fraction
    • fraction numerator square root of 3 over denominator 2 end fraction
    • 1/2

    • 1/4

    Solution

    D.

    1/4

    Let the coordinate of the centre of T be (0, K)
    Distance between their centre
    straight k plus 1 space equals space square root of 1 plus left parenthesis straight k minus 1 right parenthesis squared end root space left square bracket space because space straight C subscript 1 straight C subscript 2 space equals space straight k plus 1 right square bracket
rightwards double arrow straight k plus 1 space equals space square root of 1 plus straight k squared plus 1 minus 2 straight k end root
rightwards double arrow straight k plus 12 space equals space square root of straight k squared plus 2 minus 2 straight k end root
straight k squared space plus 1 plus 2 straight k space equals space straight k squared space plus 2 minus 2 straight k
rightwards double arrow space straight k space equals space 1 fourth
    So, the radius of circle T is k i.e, 1/4

    Question 2
    CBSEENMA11015478

    The image of line fraction numerator straight x minus 1 over denominator 3 end fraction space equals space fraction numerator straight y minus 3 over denominator 1 end fraction space equals space fraction numerator straight z minus 4 over denominator negative 5 end fraction in the plane 2x-y+z+3 =0 is the line

    • fraction numerator straight x plus 3 over denominator 3 end fraction space equals space fraction numerator straight y minus 5 over denominator 1 end fraction space equals space fraction numerator straight z minus 2 over denominator negative 5 end fraction
    • fraction numerator straight x plus over denominator negative 3 end fraction space equals space fraction numerator straight y minus 5 over denominator negative 1 end fraction space equals space fraction numerator begin display style straight z plus 2 end style over denominator 5 end fraction
    • fraction numerator straight x minus 3 over denominator 3 end fraction space equals space fraction numerator straight y plus 5 over denominator 1 end fraction space equals space fraction numerator straight z minus 2 over denominator negative 5 end fraction
    • fraction numerator straight x minus 3 over denominator negative 3 end fraction space equals space fraction numerator straight y plus 5 over denominator negative 1 end fraction space equals fraction numerator straight z minus 2 over denominator 5 end fraction

    Solution

    A.

    fraction numerator straight x plus 3 over denominator 3 end fraction space equals space fraction numerator straight y minus 5 over denominator 1 end fraction space equals space fraction numerator straight z minus 2 over denominator negative 5 end fraction

    Here, plane and line are parallel to each other. Equation of normal to the plane through the point (1,3,4) is
    fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y minus 3 over denominator negative 1 end fraction space equals space fraction numerator straight z minus 4 over denominator 1 end fraction space equals space straight k
    Any point in this normal is (2jk+2, -k+3,4+k)
    Then, open parentheses fraction numerator 2 straight k plus 1 plus 1 over denominator 2 end fraction comma space fraction numerator 3 minus straight k plus 3 over denominator 2 end fraction comma space fraction numerator 4 plus straight k plus 4 over denominator 2 end fraction close parentheses lies on the plane.
    rightwards double arrow space 2 left parenthesis straight k plus 1 right parenthesis space minus space open parentheses fraction numerator 6 minus straight k over denominator 2 end fraction close parentheses space plus open parentheses fraction numerator 8 plus straight k over denominator 2 end fraction close parentheses space plus 3 space equals 0
rightwards double arrow straight k space equals negative 2
    hence, point through which this image pass is
    (2k+1,3-k,4+k)
    i.e., (2(-2) + 1, 3+2, 4-2) = (-3,5,2)
    Hence equation of image line is
    fraction numerator straight x plus 3 over denominator 3 end fraction space equals space fraction numerator straight y minus 5 over denominator 1 end fraction space equals space fraction numerator straight z minus 2 over denominator negative 5 end fraction

    Question 3
    CBSEENMA11015480

    A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O n the ground is 45o. It flies off horizontally straight away from the point O. After 1s, the elevation of the bird from O is reduced to 30o. Then, the speed (in m/s of the bird is

    • (40 square root of 2 minus end root space 1

    • 40 left parenthesis square root of 3 space space end root space minus space square root of 2 right parenthesis
    • 20 space square root of 2
    • 20 left parenthesis square root of 3 minus 1 end root right parenthesis

    Solution

    D.

    20 left parenthesis square root of 3 minus 1 end root right parenthesis
    Question 4
    CBSEENMA11015486

    A ray of light along straight x space plus square root of 3 straight y end root space equals space square root of 3 get reflected upon reaching X -axis, the equation of the reflected ray is 

    • straight y equals space straight x plus square root of 3
    • square root of 3 straight y end root space equals space straight x minus square root of 3
    • straight y space equals space square root of 3 straight x end root minus square root of 3
    • square root of 3 straight y end root space equals space straight x minus 1

    Solution

    B.

    square root of 3 straight y end root space equals space straight x minus square root of 3

    Given equation of line
    straight x plus square root of 3 straight y space equals square root of 3 space end root space space space space space.... space left parenthesis straight i right parenthesis
straight y equals space 1 minus fraction numerator straight x over denominator square root of 3 end fraction
    Slope of incident ray is space minus fraction numerator 1 over denominator square root of 3 end fraction 
    So, slope of reflected ray must be fraction numerator 1 over denominator square root of 3 end fraction and the point of incident left parenthesis square root of 3 comma 0 right parenthesis
    So equation of reflected ray
    straight y minus 0 space equals space fraction numerator 1 over denominator square root of 3 end fraction space left parenthesis straight x minus square root of 3 right parenthesis
rightwards double arrow space square root of 3 straight y end root space equals space straight x minus square root of 3

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