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Partial variation: x = 0?

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2004 CE Maths Paper 1 Q10It is known that y is the sum of two parts, one part varies as x and the other part varies as the square of x. When x = 3, y = 3 and when x = 4, y = 12.(a) Expressy in terms of x. (4marks)(b) Ifx is an integer and y < 42, find all possible value(s) ofx. (4marks)Ans: y = 2x2– 5x;... 顯示更多 2004 CE Maths Paper 1 Q10It is known that y is the sum of two parts, one part varies as x and the other part varies as the square of x. When x = 3, y = 3 and when x = 4, y = 12.(a) Expressy in terms of x. (4marks)(b) Ifx is an integer and y < 42, find all possible value(s) ofx. (4marks)Ans: y = 2x2– 5x; –3, –2, –1, 0, 1, 2, 3, 4, 5 I know how to do Part (a).For part (b), I got -3.5 < x < 6 so x= –3, –2, –1, 0,1, 2, 3, 4, 5 Myquestion is: is x = 0 rejected? Because,in direct variation, we say that if y varies as x, then x and y must be bothnon-zero (therefore, y = kx where k is a NON-ZERO constant). So,why in this question, x = 0 does not need to be rejected?Thankyou for answering me. 更新: Re MAN: your explanation still cannot solve my problem... sorry...

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You are right that for (a), the solution is y = 2x^2 - 5x. In (b), we then set 2x^2 - 5x < 42. ( 2x + 7 )( x - 6 ) < 0 -3.5 < x < 6 x = -3, -2, -1, 0, 1, 2, 3, 4, 5. Clearly, x = 0 was finally accepted as the solution. It was not rejected because the above definition of Direct Variation is not correct. Its definition is The equation of variation is y = kx, where k is the non-zero variation constant. But, I found that the graph to illustrate Direct Variation can pass through the Origin ( 0, 0 ) with the line situated in Quadrant I only. It means it is possible for ( x, y ) = ( 0, 0 ). We may only say it is relatively "meaningless" to discuss the case when one quantity is zero in value, since the other one is also zero as well. But it is acceptable. In such circumstance, actually the input and output values can be zero in quantity. It accounts why the answer x = 0 does not need rejection. The concern is arisen from the understanding of the basic definition only. A good question to raise here. Hope I can help you.

其他解答:

if y varies as x, then x and y must be bothnon-zero. this is correct. but in this case, you do not need to reject x=0 because 0 is an integer. x=0 is only one of the possible values. when x=0, y=0., it is correct. if y varies as x, then x and y must be bothnon-zero. means y=kx+kx2, the x here can not be zero because x is a variation, when you put different values of x, you will get different values of y.
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