8條10分 MATHS

(13) From the graph y attains its maximum when x = 1 and therefore, a = 1
So the function is now y = -4(x - 1)2 + b
Sub x = 0:
5 = -4(0 - 1)2 + b
b = 9
(14) First of all, C is x = 1 and y = 4 for vertice.
Also Solving y = 0, we have:
x - 1 = 2 or -2
x = 3 or -1
So A and B are (-1, 0) and (3, 0) respectively.
Therefore, base and height of ABC are both 4 units and hence its area is 8 sq. units.
(44) The x-coordinate of M is in fact half the sum of roots of y = 0 and hence OM = -b/2
(45) Using completing square:
y = x2 - 6x + 10
= (x2 - 6x + 9) + 1
= (x - 3)2 + 1
So y attains its minimum value of 1 when x = 3.
So P is (3, 1)
(46) Let ∠CAB = ∠CDB = ∠CBD = x for the reason of same arc, same angle at circumference.
Then, ∠CDB + ∠CBD + 108 = 180
2x = 72
x = 36
∠AKB = 180 - 20 - x = 124
(47) Let ∠CAD = ∠CAB = x for the reason of same arc, same angle at circumference.
∠BCA = 90 (Angle in semicircle)
∠CAB = 180 - 90 - 68 = 22
x = 22
∠ADC = 180 - 68 = 112 (Opp. angles of cyclic quad.)
∠ACD = 180 - 112 - x = 46
So ∠BCD = 90 + 46 = 136
(48) Let the other terminal of the tangent be S, then ∠BAS = 28 for the reason of angle in alt. segment.
So ∠BAD = 180 - 34 - 28 = 118.
(49) ∠CAB = 38 (Angle in alt. segment)
∠AOB = 76 (Angle at centre = Twice the angle at circumference)
∠OBA = ∠OAB = (180 - 76)/2 = 52
So ∠ABC = 52 - 10 = 42

Q.47 Angle ACB =90 (angle in semi-circle).
Therefore, angle CAB = 180 -90 - 68 = 22 (angle sum of triangle).
Therefore, angle DAC = angle CAB = 22 (equal angle equal arc).
Therefore, angle DAB = angleDAC + angle CAB = 22 + 22 = 44.
Therefore, angle BCD = 180 - angle DAB = 180 - 44 = 136 (opposite angle of cyclic quad).
Q.49 Angle ACB = angle BAT = 38 (angle in alternate segment).
Angle AOB = 2 x angle ACB = 2 x 38 = 76 (angle at centre twice angle at cicumference).
Triangle AOB is an isos. triangle (OA=OB=radius).
Therefore, angle OBA = (180-76)/2 = 52 (base angle of isos. triangle)
Therefore, angle CBA = 52 - angle OBC = 52 - 10 = 42.

2008-06-07 22:12:35 補充：
Q.48 Angle ACB = angle TAB = 34 (angle in alt. segment). Angle DCB = angle DCA + angle ACB = 34 + 28 = 62. Therefore, angle BAD = 180 - 62 = 118 (opposite angle of cyclic quad.)