[ ] ( ) + % x 2 1 $ x ( ) ( ) 1 n. ( ) 1 ( sin x) ( ) sin x NSW HSC Mathematics Exam Worked Solutions. ( x + 4) 2. Question 1.

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1 008 NSW HSC Mathematics Eam Worked Solutions Question a cos π sig. fig s b + + c n n + n + n n + d 7 e n n n + n + n n n + n + n n + 7 or 7 0. or f The formula for the sum of n terms of an arithmetic series is given by S n n [ a + n d ] If n, a, and d, then S [ + ] [ ] [ + 80] So the sum of the first terms is 90. Question a i ii d d [ + 9 ] d [ d + ] d d log e [ ] log e d [ d ] + d d log [ e ] + log e log e + log e + iii d " sin d + d d [ sin ] + d d [ + ] sin + cos + cos + sin + sin + pasthsc.com.au Page of

2 008 NSW HSC Mathematics Eam Worked Solutions Question continued b Midpoint of, c i and, 8 +, + 8 ",, Equation of line through, gradient of is given by: ii y y y + + y 0 with d ln + + c + π sec d 0 Question a i gradient of BC: π [ tan] 0 tan π tan0 tan π 0 0 gradient of AD: y 0 y so gradient Since the gradient of BC is the same as the gradient of AD, the lines are parallel. Question continued a ii D is the intersection of the line y and the line y And so D is the point,. iii BC 0 + iv perp. dist v Area BC + AD square units. pasthsc.com.au Page of

3 008 NSW HSC Mathematics Eam Worked Solutions Question continued b i d d log e cos [ ] cos d d cos [ ] cos sin sin cos tan Question a ii π 0 tan d π 0 d tan π [ log e cos ] 0 log e cos π log e cos0 [ ], + log e log e. * - [ ] log e 0 log e log e RTP: ΔXYR is isosceles. Now YRX QRX Since XR bisects PRQ Also, YXR QRX Alternate s on parallel lines YRX YXR both equal to QRX ΔXYR is isosceles two equal angles. b i height 0. 8 mm. ii 0. n > 00. n > 8 log e [. n ] > log e 8 n log e. > log e 8 n > log e 8 log e. n >. So the zoom function must be applied at least times. pasthsc.com.au Page of

4 Question continued c i 0, 008 NSW HSC Mathematics Eam Worked Solutions y ii 8 y iii So the focal length is units. The focus will be units above the verte. the verte has coordinates 0, iv Area 8 y 8 y 8 y 8 y dy dy units. pasthsc.com.au Page of

5 008 NSW HSC Mathematics Eam Worked Solutions Question a dy d y If 0, y 7 So sin sin d y + cos + c cos0 + c 7 + c 7 + c c So the equation of the curve is y + cos + c i A 000 ii e k e k " k ln k k k ln ln ln k 0.98 b i common ratio. For a limiting sum, the common ratio must be between and. ii ie < < < < iii di ds If s then di ds I 000e 0.98 s e 0.98 s e e So the light intensity is decreasing at a rate of approimately 99 lu per metre, metres below the surface of the lake. pasthsc.com.au Page of

6 008 NSW HSC Mathematics Eam Worked Solutions Question a Note that ± π sin for π π sin sin ± sin ± domain π π b i 0 m/s. π, π are both outside the π, ii When t 0 seconds. π iii Acceleration will be equal to zero when the gradient of the curve is zero. ie when t seconds. iv d [ ] [ ] So the particle travels 9 metres. c Question 7 a V π π d d π π[ ] π π π + π 7π d So the volume of the solid is 7π units. log e log e log e log e Let u log e u u u u 0 u u + 0 So u or u. log e or log e e! or e!! so e! or!! pasthsc.com.au Page of

7 008 NSW HSC Mathematics Eam Worked Solutions Question 7 continued b i If the sector were a full circle, then θ π, and πr 0π r 0π π c i For the arc length to remain constant, then as θ decreases, r must increase. Since π is the maimum value for θ, must be the minimum value for r. ie r. ii If r, then First game X G θ 0π θ 0π π θ Area π 0π 0π units. Second game X G X G Third game X G X G c ii P Gabrielle wins iii P three games Question 8 a i For y intercepts, 0. ie y ii 0 So the y intercept is at 0,0. For intercepts, f 0. ie So 0 or 8 0. If 0, then 0. If 8 0 then 8 and ± 8 ± So the intercepts are at 0,0,, 0 and, 0. f a a 8a f a a 8 a Since f a function. a 8a f a f a, f is an even pasthsc.com.au Page 7 of

8 008 NSW HSC Mathematics Eam Worked Solutions Question 8 continued a iii Stationary points occur when f 0 ie , or a iv f 0 0 f 8 f So the stationary points are at 0,0,, and,. To determine their nature: f f 0 0 < 0 So 0,0 is a maimum turning point. f > 0 So, is a minimum turning point. Since f, point. is an even function, is also a minimum turning pasthsc.com.au Page 8 of

9 008 NSW HSC Mathematics Eam Worked Solutions Question 8 continued b i R.T.P: CD BE CD AB opp sides of a parallelogram BE AB sides of a square CD BE both equal to AB b ii R.T.P: BD EH In Δ s BCD and HBE: BH CB sides of a square BE CD from part i EBH ABC angles at a point 80 ABD CBD DCB 80 CDB CBD angle sum of a triangle But CDB ABD alternate angles on parallel lines DCB 80 ABD CBD EBH from above ΔBCD ΔHBE SAS BD EH corresponding sides in congruent triangles Question 9 a i P neither owns a mobile phone ii P owns a mobile and used it in class pasthsc.com.au Page 9 of

10 008 NSW HSC Mathematics Eam Worked Solutions Question 9 continued M b i A ii A 0 A A.00 M [ ].00 M M M.00 M A M.00 M.00 M M A M A n n M n n M.00n n M.00n M M M pasthsc.com.au Page 0 of

11 008 NSW HSC Mathematics Eam Worked Solutions Question 9 continued c i ii d d f k b k b f But f b f b, so k b b b k b k b + c + c * k -, b b b + c/ +. k b b + c k b b c c k b b k b b k b b + k b b 0 c 0 so f k b f + 0 k b k b d k b d k b + c k b + c pasthsc.com.au Page of

12 008 NSW HSC Mathematics Eam Worked Solutions Question 9 c part ii continued But b,0 is a point on f, so therefore f b 0. ie k b b b + c 0 kb + c 0 c kb f k b k b 0 0 f 0 kb kb So the beam is kb Question 0 a Area of rectangle 7 log e Equation of curve in terms of y: 0 kb units below the -ais. y log e e y e y + log e So shaded area 7log e e y + 0 dy 7log e e y log [ + y] e 0 7log e e log e [ + log e e ] 7log e + log e log e So the eact value of the shaded area is log e units. pasthsc.com.au Page of

13 008 NSW HSC Mathematics Eam Worked Solutions Question 0 continued b i Note that ΔKOJ ΔPOM since POM and KOJ are vertically opposite, and OMP and OJK are alternate angles on parallel lines. To calculate the length of MP: MP s MP Area l s l corresponding sides in similar triangles s sinα + l s + l + * s + l s l l s + l sinα sinα s + l l + sinα s + l l sinα s + l l sinα s l + l sinα,. sinα - sinα ii To find the value of that minimises A, we first find da d. da d d * d s l + -, l sinα/ +. s sinα d d l + l s sinα + l d [ d ]* s sinα l pasthsc.com.au Page of

14 008 NSW HSC Mathematics Eam Worked Solutions Question 0 b part ii continued If da 0, then d s sinα l 0 l 0 l l l note that the negative case can be ignored as is a length To show that this gives a minimum value for A, we check d A d : d A d d d * l -, s sinα / +. s sinα d d l * s sinα l sl sinα Given that s, l, and α are all positive, sl sinα is always positive. So when l, da d 0 and d A is positive. d Therefore the value of that makes A as small as possible is l. pasthsc.com.au Page of

15 Question 0 continued b iii MP If 008 NSW HSC Mathematics Eam Worked Solutions l, then MP s l s l l l s l l l s l l l l s pasthsc.com.au Page of

HIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 2/3 UNIT (COMMON) Time allowed Three hours (Plus 5 minutes reading time)

HIGHER SCHOOL CERTIFICATE EXAMINATION 998 MATHEMATICS / UNIT (COMMON) Time allowed Three hours (Plus 5 minutes reading time) DIRECTIONS TO CANDIDATES Attempt ALL questions. ALL questions are of equal value.