HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. B n+ 8 n+ 4 ( ) ( ) n+ n+ 6n+ 6n+ (6n+ ) (6n+ ). D α β x x α x β ( x) α x β β x α x + β x β ( α + β ) x β β x α + β. C 6 4 h h k k ( 4 ) 6( ) h k h + k ( h + k)( h k) 6( h + k) ( h + k)( h k 6)
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 4. A x + 7 x 7 x 7 x + 7 (x + 7)(x 7) (x + 7)(x 7) (x 7) (x + 7) (x + 7)(x 7) x 7 x 7 9x 49 4 9x 49 4 (9x 49) 4 49 9x. A y 6 ( x 6) x + < 0 ( 6) 6 i.e. the grph opens downwrds i.e. B is not true put x 0 y 6 (0 6) 0 6 i.e. C is not true put (0, 0) into the eqution R.H.S. 6 (0 6) 0 L.H.S. 0 i.e. R.H.S. L.H.S. i.e. D is not true put y 0 6 ( x 6) 0 6 ( x 6) x 6 4 or x 6 4 x 0 or x i.e. the grph cuts the x-xis t two points i.e. A is true
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 6. D For L : slope of L > 0 i.e. < 0 b x-intercept < 0 i.e. b < 0 b y-intercept For L : slope of L c > 0 i.e. c < 0 y-intercept d > 0 i.e. d > 0 x-intercept From the grph slope of L > c < c > c c < i.e. I is true d c > slope of L y intercept of L > y intercept of L b > d b < d d > b i.e. II is not true x intercept of L > x intercept of L b d > c bc < d i.e. III is true
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 7. D f x x x + ( ) f m m m + ( ) ( ) ( ) m m + m + + (4 4 ) 4 m m + m + 4 m m + 6 6 8. C g( x ) is divisible by x i.e. g () 0 8 7 () + () + b 0 b...() When g( x ) is divisible by x + Reminder g( ) 8 7 ( ) + ( ) + + b + ( ) b 9. D The required interest % $00000 ( + ) $00000 $678 (correct to the nerest dollr) 4
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 0. B 4b 4 b : b 4 : : b 4 : : c : : b 4 : : c 4 : 0 : b : c 4 : : 0 Let 4k, b k, c 0k + b b + c 4k + ( k) k + (0 k) k k. D k u Let w, where k is non-zero constnt v wv u wv u w v u 4 k k k which is constnt
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. A Put n 4 + 6 4 89 +...() 4 Put n + 4 4 +...() Sub. () into () 89 + + 68 4 4 4 4 4 4 + Put n + 4 4 + Put n + + 8. C x x x x 9 0 x x Or 4x + 9 < 4x < 8 x < The combined result is x 6
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 4. B Let S cm be the ctul re of ABCD And T cm be the ctul re of EFGH Then x S T (4 ) (6 ) S < (4 + ) (6 + ) 9. S < 9...() ( ) ( ) T < ( + ) ( + ). T < 6. 6. < T...() () + () 9. 6. < S T < 9.. < x < 7 7
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. D E B D C A AB DE 7 cm ABC AC 7 8 cm The volume of the prism ( 8) 70 cm 8
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 6. A BE : EC : Therefore, BE : BC : ( + ) :8 i.e. BE : DA :8 FBE ~ FDA i.e. FB BE FE FD DA FA 8 re of FDA FD re of FBA FB re of FDA 8 0 re of FDA 9 cm re of FEB FE re of FAB FA re of FEB 0 8 re of FEB 7 cm re of BCD re of BAD 9 + 0 cm re of qudrilterl CDFE 7 7 cm 9
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 7. B G AD 9 + 9 48 cm OC AD Therefore, EF 4 9 cm OEA 48 AE DE 4 cm OA 8 + 4 0 cm OBG OB OA 0 cm BG EF cm BG sin BOG OB 0 BOG 0 re of sector OBC 0 π (0 ) 60 7 π cm 0
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 8. B AEC nd AFC AC AC (common side) AE AF (given) CAE CAF Therefore, Hence, And (prop. of rhombus) AEC AFC (SAS) 0 CAE CAF 4 ACE ACF BEC + (ext. of ) 76
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 9. D Ech interior ngle of the regulr pentgon ( ) 80 08 80 08 DCE DEC 6 Due to symmetry EDA EAD 6 CDF CFD 6 + 6 7 CDF 80 6 7 7 CFD i.e. CD CF (sides opp. eq. s) i.e. I is true FDE FED 6 Hence FD FE ( sides opp. eq. s ) AD CE AD FD CE FE AF CF BF BF AB CB ABF CBF i.e. II is true (common side) (prop. of regulr pentgon) (SSS) ABF CBF (corr. s, s) CF CD CB 08 4 CFB CBF (bse s, isos. s) 4 AFB CFB (corr. s, s) 4 AFB + EAF 4 + 6 90 i.e. III is true
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 0. B Let l cm be the length of the squre BEF BF 4 cm FBE ~ FCD FB BE FC CD 4 l l l 4l l FE BE FD CD 4 FD DF cm
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. C Given: AE EF FB i.e. AF BE AFD AF cos β DF AF DF cos β...() AD tn β AF AD AF...() tn β BEC BE cosα CE BE CE cos α...() BC tnα BE BC BE...(4) tnα Compring () nd () BE AF CE cosα DF cos β i.e. II is true Compring () nd (4) BE AF BC AD tnα tn β BC tn β AD tnα i.e. III is true If I is true AF sinα BE sin β sinα BE sin β AF sinα sin β α β But α nd β my not be the sme i.e. I my not be true 4
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. B Let x CED BD DE Therefore, DBE DEB x (bse s, isos. ) DAC DBC x ( s in the sme segment) ADB x + x (ext. of ) x ABD 0 + 66 + x + x 80 ( sum of ) x 84 x 8. B 4. A DOC 7 7 90 EOD 07 7 90 i.e. EOC 80 EC 6 + COD CD 6 + 0 EOD ED + Therefore, the perimeter of CDE + 0 + 4
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. D slope of L slope of L i.e. L / / L The locus of P is the stright line prllel to L nd L nd midwy between L nd L i.e. slope of the locus of P y intercept of L 7 y intercept of L i.e. 4 7 + ( ) 4 7 y intercept of locus of P 8 the eqution of the locus of P: 7 y x + 8 4x 8y + 7 0 6. C For L : x intercept 9 y intercept 4 slope For L : 4 slope ( ) 4 y intercept Eqution: y ( x 0) 4 x 4y + 48 0 x intercept 6 The required re [9 ( 6)] 0 6
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 7. C C x y x y : + 0 + 0 + 6 0 6 x + y 6x + y + 0 centre (, ), which lies in fourth qudrnt i.e. B nd D re not true 6 44 rdius () + ( ) Distnce between the origin nd the centre of C ( 0) + ( 0) > 0 44 rdius i.e. the origin lies outside C i.e. A is not true The circumference of C 44 π 8.64 < 0 i.e. C is true 8. A 4 4 4 4 4 6 4 6 4 4 4 7 4 6 6 7 The required probbility 0 4 7
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 9. C The totl number of pges of the 0 mgzines 0 0 The totl number of pges of the first 6 mgzines 6 08 648 Hence, the totl number of pges of the remining 4 mgzines 0 648 67 The men number of pges of the remining 4 mgzines 67 4 68 0. A (0 + ) + (0 + ) Q 0 + Q (60 + b) + (60 + b) 60 + b IQR (60 + b) (0 + ) 0 + b b If b 0, cn be, 6, 7, 8 or 9 If b, cn be 6, 7, 8 or 9 If b, cn be 7, 8 or 9 If b, cn be 8 or 9 If b 4, cn be 9 only If b > 4, there is no solution for Hence 0 b 4, 9 And b 9 i.e. I nd II re true but III is not true 8
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. C y p( x) y q( x) The grph of y q( x) cn be obtined by reflecting the grph of y p( x) long the x-xis nd trnslted to the right by 4 units i.e. q( x) p( x 4) OR The grph of y p( x) cn be obtined by reflecting the grph of y q( x) long the x-xis nd trnslted to the left by 4 units i.e. p( x) q( x + 4) hence, y q( x) is y f ( x) nd the grph of y p( x) is the grph of y f ( x + 4) 9
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. C According to the shpe of the grphs, > nd b > i.e. I is true p q put x into y log x p log Put x into y log b x q log b From the grph, q < p i.e. log < b i.e. < b i.e. II is not true x x c 0
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution put x c into y log x AC log c Put x c into y log b x BC log b c AB AC BC BC BC log c log log c log c log c log b log c log b log b log b log log b i.e. III is true b b c. D y kx log y log ( kx ) 4 4 log y log k + log x 4 4 4 Put log 4 x nd log 4 y log k +...() 4 Put log4 x 9 nd log4 y 6 6 log k + 9...() 4 () () 8 4 log 4 k + log4 k k 4 8
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 4. C A( 4, ) D(, 4) C(, 4) B(4, ) Let C x + 6y + 4 At A( 4, ), C ( 4) + 6() + 4 90 At B(4, ), C (4) + 6( ) + 4 78 At C(, 4), C () + 6( 4) + 4 At D (, 4), C () + 6(4) + 4 4 Therefore, the lest vlue is 78
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. B S( n) 6n n T () S() 6() i.e. II is true T ( n) S( n) S( n ) n n n n 6 [6( ) ( )] n n n n + n + 6 (6 6 ) n n n + n 6 6 7 n 7 Put T ( n ) n 7 n 9 Which is not positive integer i.e. is not term of the sequence i.e. I is not true T ( n + ) T ( n) ( n + ) 7 (n 7) n + 7 n + 7 Which is constnt i.e. T ( n ) is n A.S. with common difference i.e. the sequence is not geometric sequence i.e. III is not true
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 6. A m m n n 4 + + is equivlent to m + m n + n 4 4 Which could be interpreted s: m nd n re the distinct roots of the qudrtic eqution sum of roots product of roots ( m + )( n + ) m + n mn + ( m + n) + 4 7 + ( ) + 4 8 4 mn 7 x + x 4 7. D 4 6 i + i + 4i + i + 6i i () + ( i) + 4( ) + ( i) + 6() i 4 i i ( i) + i i + i ( + i i i ) i ( + i) + i i.e. rel prt is 4
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 8. B x 6cos cos + x 6cos cos 0 x x (cos x )(6 cos x + ) 0 cos x or cos x 6 x 0 x 80 α or x 80 + α i.e. there re roots 6 (where α cos ) 9. B 7 x 4 ACB 4 ( in lt. segment) CD AB (given) Therefore, CAD ACB 4 (eq. chord, eq. s) ACD x ( in lt. segment) ACE 7 + x + 4 + x 80 ( sum of ) x 84 x 4 CDA x + 7 (ext. of ) 4 ABC 80 ADC (opp. s, cyclic qud.) 66
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 40. A Let H be the orthocenter nd R (0, k) Put y 0, x i.e. P (, 0) put x 0, y i.e. Q (0, ) As OP QR, OP is n ltitude of PQR i.e. orthocenter H lies on OP i.e. orthocenter H (0, 0) y Q(0, ) O H (0, 0) P(, 0) x R(0, k) RH PQ slope of RH slope of PQ 0 k 0 0 0 0 k ( ) 0 k 6
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 4. D Y Let Y be the projection of X on the plne ABGF Then θ XBY YAB YA DX 9 cm BY 9 + cm ABD BD + 8 08 cm BXD BX ( 08) + 9 7 cm BXY cosθ 7 4. A The required number C + C 89 4 7
HKDSE08 Mthemtics (Compulsory Prt) Pper Solution 4. C The required probbility 4 6 4 4 4 + + + +... 6 6 6 6 6 6 6 6 4 6 0 44. B Let σ be the stndrd devition 46 68. σ σ 0 The stndrd score of Susn 68 0.6 4. A Let,,, 49 be the terms in the A.S. And let d be the common difference of the A.S. The first 7 terms re,,, 4,, 6, 7 The lst 7 terms re 4, 44, 4, 46, 47, 48, 49 But 4 + 4d, 44 + 4d, 4 + 4d, 46 4 + 4d, 47 + 4d, 48 6 + 4d, 49 7 + 4d i.e. The lst 7 terms could be obtined by dding the constnt 4d to ech term in the first 7 terms i.e. the vrince is unchnged i.e. the vrince of the lst 7 terms is 9 8