KVPY SA Exam 2011 Question Paper

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Question : 2 of 80

Marks: +1, -0

Let a, b, x, y be real numbers such that

a^2+b^2=81,

x^2+y^2=121

and

ax+by=99 .

Then the set of all possible values of

ay-bx

is

$(0,9/11]$
$(0,9/11)$
{0}
$[ 9/11,∞)$

Solution:

a^2 + b^2 = 81

x^2 + y^2 = 121

ax + by = 99

(a^2 + b^2) (x^2 + y^2) = (81) (121)

....(1)
and

(ax + by)^2 = (99)^2

....(2)
(1) – (2)

(ay - bx)^2 = 0

ay - bx = 0

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