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

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

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)

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option: e -> -2

Nikhil JangiJuly 26, 2009 at 11:36 am

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

abhitsianJuly 27, 2009 at 5:35 am

both of u are right!

RahulJuly 27, 2009 at 6:35 am

i also go wid e

kuldeepJuly 29, 2009 at 3:11 pm

Pls explain the solution in simple words

sharonAugust 4, 2009 at 7:21 am

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

RahulAugust 4, 2009 at 11:06 am

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)

jatinAugust 28, 2009 at 10:05 am

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

gauravJune 21, 2011 at 8:08 am