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A math problem in the simulation problem of senior high school entrance examination
(1) Let X= 1, Y= 1 and m=5/3. Let Y=0 and X=5/2. So d (5/2,0)

(2) make Y=aX? +bX+c, substituting O (0 0,0), A (1, 1) and d (5/2,0), we get

c=0,b=5/3,a=-2/3

So Y=-2/3X? +5/3 times

(3) let p (x, y), 1 < x < 5/2, 0 < y < 1, and the vertical cross p is the OD cross q. ..

S=SOAB+SOAPQ+SPQD

= 1/2× 1× 1+ 1/2×(y+ 1)×(x- 1)+ 1/2×y×(5/2-x)

=0.5×( 1-x? +5/2 times)

=-0.5×(x-√5/2)? +9/8

Therefore, when x=√5/2, the maximum value of S =9/8.