Add Math (4044/02) (+) x (+) 2. Find the coordinates of the points of intersection of the curve xy 2 the line 2y 1 x 0. [5]
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1 Add Math (444/) Requirement : Answer a questions Tota mars : 7 Duration : hour 45 minutes. Sove the inequaity 5 and represent the soution set on the number ine. [4] 5 4 From the setch on number ine, we get 4 (+) (+) 4 ( ). Find the coordinates of the points of intersection of the curve y y and the ine y. [5] From y, we get y. y y y y When When y, y y y y or y y, the coordinates of the points of intersection of the curve are (, ), (, ). Prepared by Mr Ang, Nov 6
2 Add Math (444/). (i) Differentiate with respect to. [] Hence find and use your resut to evauate, eaving your answer in eact form. [4] (i) Let y, 4 y c c c 5 5 Prepared by Mr Ang, Nov 6
3 Add Math (444/) 4. Find the vaues of for which the ine y 5 4 is tangent to the curve y. [6] Soving simutaneous equations of y 5 4 and y or Prepared by Mr Ang, Nov 6
4 Add Math (444/) 5. The equation of a curve is y. (i) Epress d y d a in the form, where a is an integer to be determined. [4] What can be deduced about the tangent to the curve at? [] (iii) State the range of vaues of for which y is a decreasing function. [] Given that y, (i) y y 6 when, d y. The tangent is a horizonta ine. (iii) for a decreasing function, d y. 6 Since, we get 6, hence or. Prepared by Mr Ang, Nov 6 4
5 6. Without using a cacuator, (i) epress 5 Add Math (444/) in the form a 5 b, where a and b are integers, [4] factorise p 5 q and hence epress in the form c d, where c and d 5 are integers [4] (i) Let f p p q By ong division, we get Therefore,. When p q f p p q p q pq p q p q p q pq, f f p. q. Therefore p q is a factor of Let p 5, q, 5 5 p q p q p q p q pq p q p q p q p q pq Therefore, Prepared by Mr Ang, Nov 6 5
6 Add Math (444/) 7. Soutions to this question by accurate drawing wi not be accepted. y A(, 4) X P O B(7, ) D(.5, 7) C(6, 6) The diagram shows a quadriatera ABCD with vertices A(, 4), B(7, ), C(6, 6) and D(.5, 7). The point X ies on the ine AB such that DX is perpendicuar to AB. The ines DX and AC intersect at the point P. (i) Find the equation of AC. [] Find the coordinates of P. [5] (iii) Find the coordinates of the midpoint of AC. [] (iv) Find the area of triange ADC and hence state the area of triange PDC. [] (i) A(, 4), C(6, 6) Gradient of AC, m AC y4 y A(, 4), B(7, ) and D(.5, 7). Find the coordinates of P. 4 Gradient of AC, m AB 7 Gradient of DX, mdx y 7.5 y 4.5 Prepared by Mr Ang, Nov 6 6
7 Add Math (444/) Point P ies on the intersection of DX and AC, y 4.5, y y P(.5, ) (iii) A(, 4), C(6, 6) midpoint of AC 6 4 6,.5, (iv) A(, 4), C(6, 6) and D(.5, 7) the area of triange ADC unit the area of triange PDC = the area of triange ADC.5 unit Prepared by Mr Ang, Nov 6 7
8 Add Math (444/) 8. (a) Prove that cosec tan sec cosec sec. [] (b) (i) Write cos sin in the form Rcos, where is acute and R. [] By using appropriate identities and the resut from part (i), show that y 6cos 6cos sin may be written in the form y cos 45. [] (iii) Hence find the smaest positive vaue of for which y =. [] (a) L.H.S. = cosec tan sec seccosec = tansec sec cosec = tansec sec cos = tansec sin cos sin = sec sincos = sec sincos cosecsec = R.H.S. (b) (i) cos sin cos sin cos cos sin sin cos cos 45, where tan 45 From (b)(i), y 6cos 6cos sin y cos cos sin y cos sin y cos sin y cos 45 Prepared by Mr Ang, Nov 6 8
9 (iii) Let y, cos 45 cos cos Principa ange, 45 5 Add Math (444/) 45 5 or or 9 the smaest positive vaue of is 45 Prepared by Mr Ang, Nov 6 9
10 9. O Add Math (444/) P(, y) y horizonta ine Q Load In the diagram above, the curve OPQ represents a beam, one end of which is fied horizontay at the origin O, with a oad suspended from the end Q. The beam, of ength and negigibe mass, bends in such a way that, at a distance from O, the beam is defected a distance y downwards. Assume that the ength of the beam is unchanged when the beam is defected. The reationship between y and can be epressed as d y d, for where the coefficient is fied for the given beam and oad. When =, y = and d y. (i) Find an epression for d y in terms of, and. [] The greatest defection occurs at =. Show that the greatest defection is 6 5 times the defection at the midpoint of the beam. [5] The tangent to the curve at the midpoint of the beam maes an ange of.79 with the horizonta. (iii) Find the vaue of. [] (i) d y d y d y d Prepared by Mr Ang, Nov 6
11 Add Math (444/) c When =, d y. c y C 6 When =, y =. C When When y 6,, y 6 5 y When tan.79 8, tan.79 tan.79 4 tan.79 4 tan ( s.f.) Prepared by Mr Ang, Nov 6
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