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)
option: e -> -2
ReplyDeleteI too go with e)-2 (2^0.5 * -2^0.5)
ReplyDeleteboth of u are right!
ReplyDeletei also go wid e
ReplyDeletePls explain the solution in simple words
ReplyDeleteSharon, we can assume x=cos A and y=Sin A
ReplyDeletethen 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.....
ReplyDeletetherefore 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)
x and y are 1/root 2...
ReplyDeletemax 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