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Problem Of the Day 26.07.09

with 8 comments


If x² + y²= 1 and x, y are real numbers. Let p, q be the largest and smallest possible
value of x + y respectively. Then compute pq
a) 0                    b) 1/2         c) −1/2                          d) 2                           e) −2

8 Responses

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  1. option: e -> -2

    Nikhil Jangi

    July 26, 2009 at 11:36 am

  2. I too go with e)-2 (2^0.5 * -2^0.5)

    abhitsian

    July 27, 2009 at 5:35 am

  3. both of u are right!

    Rahul

    July 27, 2009 at 6:35 am

  4. i also go wid e

    kuldeep

    July 29, 2009 at 3:11 pm

  5. Pls explain the solution in simple words

    sharon

    August 4, 2009 at 7:21 am

  6. Sharon, we can assume x=cos A and y=Sin A
    then x+y=Cos A +sin A=sqrt(2) sin(45+A)
    max value of sin is 1 and min is -1
    hence max of x+y=sqrt(2)
    and min is -sqrt(2)
    and hence product is -2

    Rahul

    August 4, 2009 at 11:06 am

  7. As nothing is said about z i will take z as 2 to be on safer side…..

    therefore the equation represents a circle of radius 1

    and x+y is maximum only when x = y which gives x= +-1/root(2) same with y
    p = max(x+y) = 1/root(2) + 1/root(2) = root(2)
    q = min(x+y) = -1/root(2) -1/root(2) = -root(2)
    pq = -root(2)

    jatin

    August 28, 2009 at 10:05 am

  8. x and y are 1/root 2…
    max value of x+y is 2/root 2
    min value -2/root 2 ,wen x&y are -1/root 2 each
    pq= 2/root 2 * -2/root 2 = -4/2= -2

    gaurav

    June 21, 2011 at 8:08 am


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