Problem Of the Day 26.07.09
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
option: e -> -2
Nikhil Jangi
July 26, 2009 at 11:36 am
I too go with e)-2 (2^0.5 * -2^0.5)
abhitsian
July 27, 2009 at 5:35 am
both of u are right!
Rahul
July 27, 2009 at 6:35 am
i also go wid e
kuldeep
July 29, 2009 at 3:11 pm
Pls explain the solution in simple words
sharon
August 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
Rahul
August 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)
jatin
August 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
gaurav
June 21, 2011 at 8:08 am