# dse 2011 maths Classic List Threaded 10 messages Open this post in threaded view
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## dse 2011 maths

 f(x,y)=(x-y,x+y) what is the range of function... i got the correct answer by elimination... pls tell me why its range is entire plane??? and also question 39
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## Re: dse 2011 maths

 consider cases for X,Y>0 X
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## Re: dse 2011 maths

 In reply to this post by ritu Ritu, I solved the question by actually plotting the function in the three-dimensional plane. You'll see that every point in the plane in a pre-image, so the entire plane is the range. It's rather a silly method, but it works Open this post in threaded view
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## Re: dse 2011 maths

 chinni can u pls explain how u plotted the diagram....
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## Re: dse 2011 maths

 like if i take f(4,8)=(-4,12) so on x axis i take 4,on y axis i take 8....what do u take on z axis????
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## Re: dse 2011 maths

 Its not exactly in terms of z, ritu. I drew a graph like this one below, where the upper red area is the range and the lower area is the domain. The dots in the doamin have corresponding coloured dots in the range. You an see that the pattern is such that for evry value in the domain, the range becomes the entire red area (though I havent shown the other 4 quadrants). Open this post in threaded view
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## Re: dse 2011 maths

 Administrator In reply to this post by ritu We will say that the range is entire plane if we can get as output anything in the plane by using some input combination in the domain. Lets say I want to generate a vector (a, b) in plane then what must be the value of (x, y) such that f(x, y) = (a, b). Well its easy, solving the system x - y = a x + y = b gives us: x = (a+b)/2 y = (b-a)/2 Thus, if we want to generate the output (a, b) in the co-domain there is an input combination in the domain ((a+b)/2, (b-a)/2) and that will do the job for us. Hence, any point in the plane is mapped by some point in the domain. And therefore the range is the entire plane.
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## Re: dse 2011 maths

 Omg, thank you sir Your method is WAY better than picturing the dimensions in my head like that. Thank you