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The equation of the line passing through pole and open parentheses 2 comma fraction numerator pi over denominator 3 end fraction close parentheses is

Maths-General

  1. r to the power of 2 end exponent equals 2 r plus 3    
  2. theta equals pi divided by 3    
  3. r S i n space theta equals 4    
  4. r=2    

    Answer:The correct answer is: theta equals pi divided by 3

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    The polar equation of the straight line passing through open parentheses 2 comma fraction numerator pi over denominator 6 end fraction close parentheses and parallel to the initial line is

    The polar equation of the straight line passing through open parentheses 2 comma fraction numerator pi over denominator 6 end fraction close parentheses and parallel to the initial line is

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    The polar equation of y to the power of 2 end exponent equals 4 x is

    The polar equation of y to the power of 2 end exponent equals 4 x is

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    The polar equation of the straight line passing through open parentheses 6 comma fraction numerator pi over denominator 3 end fraction close parentheses and perpendicular to the initial line is

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    The cartesian equation of r equals a s i n space 2 theta is

    The cartesian equation of r equals a s i n space 2 theta is

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    The angle between the lines r left square bracket 2 C o s space theta plus 5 S i n space theta right square bracket equals 3 and r left square bracket 2 s i n space theta minus 5 C o s space theta right square bracket plus 4 equals 0 is

    The angle between the lines r left square bracket 2 C o s space theta plus 5 S i n space theta right square bracket equals 3 and r left square bracket 2 s i n space theta minus 5 C o s space theta right square bracket plus 4 equals 0 is

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    The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| less or equal than k, |y| less or equal than k, |x – y| less or equal than k ; is-

    |x| less or equal than k rightwards double arrow –k less or equal thanless or equal than k ….(1)
    & |y| less or equal thanrightwards double arrow –k less or equal thanless or equal than k ….(2)

    & |x – y| less or equal thanrightwards double arrow |y – x| less or equal thanrightwards double arrow –k less or equal than y – x less or equal thanrightwards double arrow x – k less or equal thanless or equal than x + k ….(3)
    therefore Number of points having integral coordinates
    = (2k + 1)2 – 2
    = (3k2 + 3k + 1).

    The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| less or equal than k, |y| less or equal than k, |x – y| less or equal than k ; is-

    maths-General
    |x| less or equal than k rightwards double arrow –k less or equal thanless or equal than k ….(1)
    & |y| less or equal thanrightwards double arrow –k less or equal thanless or equal than k ….(2)

    & |x – y| less or equal thanrightwards double arrow |y – x| less or equal thanrightwards double arrow –k less or equal than y – x less or equal thanrightwards double arrow x – k less or equal thanless or equal than x + k ….(3)
    therefore Number of points having integral coordinates
    = (2k + 1)2 – 2
    = (3k2 + 3k + 1).
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    A is a set containing n elements. A subset P1 is chosen, and A is reconstructed by replacing the elements of P1. The same process is repeated for subsets P1, P2, … , Pm, with m > 1. The Number of ways of choosing P1, P2, …, Pm so that P1 union P2 union … union Pm= A is -

    Let A = {a1, a2,…..an}.
    For each ai (1 less or equal thanless or equal than n), either ai element of Pj or ai not an element of Pj (1 less or equal thanless or equal than m) . Thus, there are 2m choices in which ai (1 less or equal thanless or equal than  n) may belong to the Pj apostrophes.
    Also there is exactly one choice, viz., ai not an element of Pj for j = 1, 2, …, m, for which ai not an element of P1 union P2 union...union Pm.
    Therefore, ai not an element of P1 union P2 union …. union Pm in (2m – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
    P1, P2, ….. , Pm is (2m – 1)n

    A is a set containing n elements. A subset P1 is chosen, and A is reconstructed by replacing the elements of P1. The same process is repeated for subsets P1, P2, … , Pm, with m > 1. The Number of ways of choosing P1, P2, …, Pm so that P1 union P2 union … union Pm= A is -

    maths-General
    Let A = {a1, a2,…..an}.
    For each ai (1 less or equal thanless or equal than n), either ai element of Pj or ai not an element of Pj (1 less or equal thanless or equal than m) . Thus, there are 2m choices in which ai (1 less or equal thanless or equal than  n) may belong to the Pj apostrophes.
    Also there is exactly one choice, viz., ai not an element of Pj for j = 1, 2, …, m, for which ai not an element of P1 union P2 union...union Pm.
    Therefore, ai not an element of P1 union P2 union …. union Pm in (2m – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
    P1, P2, ….. , Pm is (2m – 1)n
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    physics-

    Two tuning forks P and Q are vibrated together. The number of beats produced are represented by the straight line O A in the following graph. After loading Q with wax again these are vibrated together and the beats produced are represented by the line O B. If the frequency of P is 341 H z comma the frequency of Q will be

    n subscript Q end subscript equals 341 plus-or-minus 3 equals 344 H z or 338 H z
    On waxing Q comma the number of beats decreases hence
    n subscript Q end subscript equals 344 H z

    Two tuning forks P and Q are vibrated together. The number of beats produced are represented by the straight line O A in the following graph. After loading Q with wax again these are vibrated together and the beats produced are represented by the line O B. If the frequency of P is 341 H z comma the frequency of Q will be

    physics-General
    n subscript Q end subscript equals 341 plus-or-minus 3 equals 344 H z or 338 H z
    On waxing Q comma the number of beats decreases hence
    n subscript Q end subscript equals 344 H z
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    If a hyperbola passing through the origin has 3 x minus 4 y minus 1 equals 0 and 4 x minus 3 y minus 6 equals 0 as its asymptotes, then the equation of its tranvsverse and conjugate axes are

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    Which of the following curves represents correctly the oscillation given by y equals y subscript 0 end subscript sin invisible function application open parentheses omega t minus ϕ close parentheses comma blank w h e r e blank 0 less than ϕ less than 90

    Given equation y equals y subscript 0 end subscript sin invisible function application left parenthesis omega t minus ϕ right parenthesis
    At t equals 0 comma blank y equals negative y subscript 0 end subscript sin invisible function application ϕ
    This is case with curve marked D

    Which of the following curves represents correctly the oscillation given by y equals y subscript 0 end subscript sin invisible function application open parentheses omega t minus ϕ close parentheses comma blank w h e r e blank 0 less than ϕ less than 90

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    Given equation y equals y subscript 0 end subscript sin invisible function application left parenthesis omega t minus ϕ right parenthesis
    At t equals 0 comma blank y equals negative y subscript 0 end subscript sin invisible function application ϕ
    This is case with curve marked D
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    A small source of sound moves on a circle as shown in the figure and an observer is standing on O. Let n subscript 1 end subscript comma n subscript 2 end subscript and n subscript 3 end subscript be the frequencies heard when the source is at A comma blank B blankand C respectively. Then

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    Hence n subscript 2 end subscript greater than n subscript 3 end subscript greater than n subscript 1 end subscript
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    The polar equation of the circle with pole as centre and radius 3 is

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    The equation of the circle passing through pole and centre at (4,0) is

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