D”PZi MwYZ Q be©vPwb cix¶vi †miv cÖkœ: m„RbkxjD”PZi MwYZ Q mKj †ev‡Wi© kxl¯©...

D”PZi MwYZ mKj †ev‡W©i kxl© ’vbxq ¸i“Z¡c �Y© ¯‹z‡ji wbe©vPwb cix¶vi cÖkœcÎ 1

1. AvBwWqvj ¯‹zj GÛ K‡jR, gwZwSj, XvKv welq †KvW : 1 2 6

mgq ⎯ 3 NÈv D”PZi MwYZ (ZË¡xq) c �Y©gvb ⎯ 75

1. †fbwP‡Îi mvnv‡h¨ †`LvI †h, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 4 A_ev, †h †Kvb mvš� †mU P I Q Gi Rb¨ cÖgvY Ki †h, n(P∪Q) = n(P) + n(Q) − n(P ∩ Q) 2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3×2=6 (K) g‡b Ki, p(x) = ax5 + bx4 + cx3 + cx2 + bx + a, †hLv‡b a, b, c aª“eK Ges a ≠ 0. †`LvI †h, (x − r) hw` p(x) Gi GKwU Drcv`K nq, Z‡e

(rx − 1) I p(x) Gi GKwU Drcv`K| (L) Drcv`‡K we‡k−lY Ki: a6 − 18a3 + 125

(M) AvswkK fMœvs‡k cÖKvk Ki : x4

x2 − a2

3. hw` m, n, k ∈ ô Ges m < n nq, Z‡e †`LvI †h, m + k < n + k n‡e| 4 A_ev, MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h, mKj n ∈ ô Gi Rb¨, 12 + 22 + 32 + ... ... ... + n2 =

n(n + 1) (2n + 1)6

4. hw` a2 + 2 = 3

23 + 3

– 23 Ges a ≥ 0 nq, Z‡e cÖgvY Ki †h, 3a2 + 9a = 8. 5

A_ev, †`LvI †h, alogkb − logkc × blogkc − logka × clogka − logkb = 1

5. F(x) = 1

x − 2 dvsk‡bi †Wv‡gb wbY©q Ki Ges dvskbwU GK-GK wKbv Zv wba©viY Ki| 4

A_ev, S = {(x, y) : x2 + y2 = 9 Ges y ≥ 0} Aš^‡qi †jLwPÎ AsKb Ki Ges Aš^qwU dvskb wKbv Zv †jLwPÎ †_‡K wbY©q Ki|

6. mgvavb Ki : ax − (a3 + a) ax−22 + a2 = 0 (a > 0, a ≠ 1) 4

A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI : (2x + 5) (x − 1) ≤ 0 7. mgvavb Ki : 2x2 + 3xy + y2 = 20 4 5x2 + 4y2 = 41 A_ev, x − y + 3 > 0 Ges 2x − y − 6 ≥ 0 AmgZvhyM‡ji mgvavb †m‡Ui †jLwPÎ AsKb Ki|

8. x Gi Dci wK kZ© Av‡ivc Ki‡j, 1(1 − x) +

1(1 − x)2 +

1(1 − x)3 + ... ... ... Abš� avivi (AmxgZK) mgwó _vK‡e Ges †mB mgwó wbY©q Ki| 4

9. cÖgvY Ki †h, `yBwU wÎfzR m`„k‡KvYx n‡j Zv‡ì Abyiƒc evû¸‡jv mgvbycvwZK n‡e| 6 A_ev, U‡jwgi Dccv`¨wU wjL Ges cÖgvY Ki| 10. ∆ABC Gi ga¨gvÎq G we›`y‡Z wgwjZ n‡j cÖgvY Ki †h, AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2) 4 A_ev, †Kvb e„‡Ëi ewnt¯’ †Kvb we›`y P †_‡K hw` H e„‡Ë GKwU ¯•k©K PT I GKwU †Q`K PBA A¼b Kiv nq, Z‡e cÖgvY Ki †h, PT2 = PA.PB. 11. wÎfz‡Ri D”PZv, f‚wgi Dci ga¨gv Ges f‚wg msjMœ GKwU †KvY †`qv Av‡Q| wÎfzRwU AsKb Ki‡Z n‡e| [AsK‡bi wPý I weeiY Avek¨K] 5 A_ev, wfbœ wfbœ e¨vmva© wewkó Giƒc wZbwU e„Ë AsKb Ki †hb Zviv ci¯•i‡K ewnt¯•k© K‡i| [AsK‡bi wPý I weeiY Avek¨K] 12. †f±i c×wZ‡Z cÖgvY Ki †h, †Kvb PZyfz©‡Ri mwbœwnZ evû¸‡jvi ga¨we›`yi ms‡hvRK †iLvmg�n GKwU mvgvš�wiK Drcbœ K‡i| 4 A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, UªvwcwRqv‡gi Amgvš�ivj evûØ‡qi ga¨we›`yi ms‡hvRK †iLvsk mgvš�ivj evûØ‡qi mgvš�ivj I Zv‡ì

†hvMd‡ji A‡a©K| 13. Dfq cÖvš� eÜ GKwU mge„Ëf‚wgK wmwjÛv‡ii eµZj 2200 eM© †m.wg.| Gi D”PZv 25 †m.wg. n‡j mgMÖZj wbY©q Ki| 4 A_ev, 13 †m.wg. e¨vmva© wewkó GKwU †Mvj‡Ki †K›`ª n‡Z 12 †m.wg. ` �ieZ©x †Kvb we›`yi ga¨ w`‡q e¨v‡mi Dci j¤^ mgZj †MvjKwU‡K †Q`

K‡i| Drcbœ ZjwUi †¶Îdj wbY©q Ki| 14. †h †Kvb wZbwU cÖ‡kœi DËi `vI : 4×3=12 (K) c„w_exi e¨vmva© 6440 wK‡jvwgUvi n‡j c„w_exi Dc‡ii †h `yBwU ¯’vb †K‡›`ª 1′ †KvY Drcbœ K‡i Zv‡ì ` �iZ¦ KZ? (L) GKwU wÎfz‡Ri †KvY¸‡jvi AbycvZ 2 : 5 : 3 Gi e„nËg †Kv‡Yi e„Ëxqgvb wbY©q Ki|

(M) †`LvI †h, secA − secBtanA + tanB =

tanA − tanBsecA + secB

(N) gvb wbY©q Ki : 1 + 2sin60° cos60°sin60° + cos60° +

1 − 2sin60° cos60°sin60° − cos60°

(O) mgvavb Ki : cot2θ + cosec2θ = 3; hLb 0° ≤ θ ≤ 360° 15. wb‡æi MYmsL¨v wb‡elY mviYx †_‡K msw¶ß c×wZ‡Z MvwYwZK Mo wbY©q Ki : 5

xi 1 2 3 4 5 6 7 8 9 10 ƒi 5 20 30 40 50 35 21 12 10 8

A_ev, wb‡æ †Kv‡bv GKwU †kÖYxi wk¶v_©x‡ì cÖvß b¤^‡ii mviwY †Ìqv nj| mviYx †_‡K Mo eëavb wbY©q Ki: cÖvß b¤î 25-34 35-44 45-54 55-64 65-74 75-84 85-94 wk¶v_©x msL¨v 5 10 15 20 30 16 4

welq †KvW : 1 2 6

wbe©vPwb cix¶vi †miv cÖkœ: m„Rbkxj

2 D”PZi MwYZ mKj †ev‡W©i kxl© ’vbxq ¸i“Z¡c �Y© ¯‹z‡ji wbe©vPwb cix¶vi cÖkœcÎ

2. wfKvi“bwbmv b �b ¯‹zj GÛ K‡jR, XvKv mgq ⎯ 3 NÈv D”PZi MwYZ (ZË¡xq) c �Y©gvb ⎯ 75

1. †h †Kvb †mU A, B, C Gi Rb¨ cÖgvY Ki †h, A × (B ∪ C) = (A × B) ∪ (A × C) 4 A_ev, wKQy msL¨K †jv‡Ki g‡a 50 Rb evsjv, 20 Rb Bs‡iwR Ges 10 Rb evsjv I Bs‡iwR ej‡Z cv‡i| yB fvlvi Aš�Z GKwU fvlv KZRb ej‡Z cv‡i? 2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3×2=6 (K) hw` p(x) = x3 + 5x2 + 6x + 8 nq Ges p(x) †K (x – a) Ges (x – b) Øviv fvM Ki‡j GKB fvM‡kl _v‡K, †hLv‡b a ≠ b, Z‡e †`LvI †h, a2 + b2 + ab + 5a + 5b + 6 = 0 (L) hw` (a + b + c) (ab + bc + ca) = abc nq, Z‡e †`LvI †h, (a + b + c)3 = a3 + b3 + c3

(M) x3 + 2x2 + 1x2 + 2x – 3 †K AvswkK fMœvs‡k cÖKvk Ki|

3. hw S = {n : n ∈ ô Ges 22n – 1 + 1 ivwkwU 3 Øviv wefvR¨} nq, Z‡e † LvI †h, S = ô 4 A_ev, MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h, mKj n ∈ ô Gi Rb¨, 12 + 22 + 32 + ........... + n2 =

n(n + 1) (2n + 1)6

4. hw` a2 + 2 = 323 + 3

– 23 Ges a ≥ 0 nq, Z‡e †`LvI †h, 3a3 + 9a = 8 5

A_ev, hw` xy a – 1 = p, xy b – 1 = q, xy c – 1 = r nq, Z‡e †`LvI †h, (b – c) logkp + (c – a) logkq + (a – b) logkr = 0

5. F(x) = 1

x – 3 dvsk‡bi †Wv‡gb wbY©q Ki Ges dvskbwU GK-GK wKbv Zv wba©viY Ki| 4

A_ev, S = {(x, y) : y2 = 9x} Aš^‡qi †jLwPÎ A¼b Ki Ges Aš^qwU dvskb wKbv Zv †jLwPÎ †_‡K wbY©q Ki|

6. mgvavb Ki : ( )x – 13x + 2 + 2 ( )3x + 2

x – 1 = 3 4

A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI : x – 3x – 4 >

x – 2x – 1

7. mgvavb Ki : 2x2 + 3xy + y2 = 20, 5x2 + 4y2 = 41 4 A_ev, y ≤ 2x AmgZvi †jLwPÎ AsKb Ki|

8. x Gi Dci Kx kZ© Av‡ivc Ki‡j 1x + 1 +

1(x + 1)2 +

1(x + 1)3 + ... ... ... Abš� avivi (AmxgZK) mgwó _vK‡e Ges †mB mgwó wbY©q Ki| 4

9. cÖgvY Ki †h, yBwU wÎfzR m „k‡KvYx n‡j Zv‡ì Abyiƒc evû¸‡jv mgvbycvwZK n‡e| 6 A_ev, cÖgvY Ki †h, †Kvb e„‡Ëi `yBwU R¨v hw` e„‡Ëi Af¨š�i¯’ †Kvb we›`y‡Z †Q` K‡i, Z‡e GKwUi AskØ‡qi Aš�M©Z AvqZ‡¶‡Îi †¶Îdj

AciwUi AskØ‡qi Aš�M©Z AvqZ‡¶‡Îi †¶Îd‡ji mgvb| 10. ABCD AvqZ‡¶‡Îi KY©Øq O we›`y‡Z †Q` K‡i‡Q| AvqZ‡¶‡Îi Af¨š�‡i †h †Kv‡bv we›`y P n‡j, cÖgvY Ki †h, PA2 + PB2 + PC2 + PD2 = AC2 + 4PO2 4 A_ev, AB e¨v‡mi Ici Aw¼Z Aa©e„‡Ëi `yBwU R¨v AC I BD ci¯•i P we›`y‡Z †Q` K‡i| cÖgvY Ki †h, AB2 = AC.AP + BD.BP 11. Ggb GKwU e„Ë A¼b Ki‡Z n‡e hv `yBwU wbw`©ó we› y w`‡q hvq Ges hvi †K›`ª GKwU wbw`©ó mij‡iLvq Aew¯’Z| 5 A_ev, wÎfz‡Ri f‚wg, wkit‡KvY I Aci †KvYØ‡qi Aš�i †Ìqv Av‡Q, wÎfzRwU AuvK|

12. A, B, C we› y wZbwUi Ae ’vb †f±i h_vµ‡g a, b, c Ges C we› ywU hw` AB †iLvsk‡K m : n Abycv‡Z Aš�we©f³ K‡i Z‡e cÖgvY Ki †h, c = mb + nam + n 4

A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, UªvwcwRqv‡gi KY©Ø‡qi ga¨we›`yi ms‡hvRK mij‡iLv mgvš�ivj evûØ‡qi mgvš�ivj Ges Zv‡ì we‡qvMd‡ji A‡a©K|

13. Dfq cÖvš� eÜ GKwU mge„Ëf‚wgK wmwjÛv‡ii eµZj 2200 eM© †m.wg.| Gi D”PZv 25 †m.wg. n‡j, mgMÖZj wbY©q Ki| 4

A_ev, GKwU †MvjK AvK…wZi ej GKwU wmwjÛvi AvK…wZi ev‡· wVKfv‡e Gu‰U hvq| ev·wUi AbwaK…Z As‡ki AvqZb 8958 Nb †m.wg. n‡j, ejwUi cwiwa

KZ? 14. †h †Kvb wZbwU cÖ‡kœi DËi `vI : 4×3=12

(K) GKwU e¨w³ e„ËvKvi c‡_ NÈvq 5 wK.wg. †e‡M †`Š‡o 36 †m‡K‡Û Ggb GKwU e„ËPvc AwZµg K‡i, hv e„‡Ë †K‡›`ª 56° †KvY Drcbœ K‡i| e„‡Ëi e¨vm wbY©q Ki|

(L) acosθ – bsinθ = c n‡j †`LvI †h, asinθ + bcosθ = ± a2 + b2 – c2

(M) 7sin2θ + 3cos2θ = 4 n‡j †`LvI †h, tanθ = ± 1

3

(N) gvb wbY©q Ki : sin2 17π18 + sin2

5π8 + cos2

37π18 + cos2 3π

8

(O) mgvavb Ki : cosθ + sinθ = 2; hLb 0° ≤ θ ≤ 360° 15. wb‡Pi MYmsL¨v wb‡ekb mviwY †_‡K msw¶ß c×wZ‡Z Mo wbY©q Ki : 5

xi 1 2 3 4 5 6 7 8 9 10 ƒi 5 20 30 40 50 35 21 12 10 8

A_ev, wb‡Pi MYmsL¨v wb‡ekb mviwb †_‡K Mo eëavb wbY©q Ki| cÖvß b¤î 51-60 61-70 71-80 81-90 91-100 MYmsL¨v 10 15 20 10 5


3. ivRDK DËiv g‡Wj K‡jR, XvKv welq †KvW : 1 2 6


1. †`LvI †h, A × (B ∪ C) = (A × B) ∪ (A × C) 4 A_ev, †h †Kvb mvš� †mU A I B Gi Rb¨ cÖgvY Ki †h, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3 × 2 = 6 (K) x4 – 5x3 + 7x2 – a eûc`xi GKwU Drcv`K x – 2 n‡j a Gi gvb wbY©q Ki| (L) Drcv`‡K we‡k−lY Ki: a(b – c)3 + b(c – a)3 + c(a – b)3

(M) AvswkK fMœvs‡k cÖKvk Ki: x2

x2 – 9

3. hw m, n, k ∈ ô Ges m < n nq, Z‡e † LvI †h, m + k < n + k n‡e| 4 A_ev, †`LvI †h, S = ô †hLv‡b S = {n : n ∈ ô Ges 22n – 1 ivwkwU 3 Øviv wefvR¨}|

4. hw` a2 + 2 = 3 23 +3 −

23 Ges a ≥ 0 nq, Z‡e cÖgvY Ki †h, 3a3 + 9a = 8 5

A_ev, hw` logk(1 + x)logkx = 2 nq, Z‡e x Gi gvb wbY©q Ki|

5. F(x) = (x − 2)2 dvsk‡bi †Wv‡gb wbY©q Ki Ges dvskbwU GK-GK wKbv Zv wba©viY Ki| 4 A_ev, S = {(x, y) : x2 + y2 = 9 Ges y ≥ 0} Aš^qwUi †jLwPÎ AsKb Ki Ges †jLwPÎ †_‡K Aš^qwU dvskb wKbv Zv wba©viY Ki|

6. mgvavb Ki: (1 + x) 13 + (1 – x)

13 = 2

13 4

A_ev, mgvavb Ki: |x|x + x2 = 3.

7. mgvavb Ki: x2 = 3x + 2y, y2 = 3y + 2x 4 A_ev, 2x – 3y – 1 ≥ 0 Ges 2x + 3y – 7 ≤ 0 AmgZvhyM‡ji mgvavb †m‡Ui †jLwPÎ A¼b Ki|

8. a-Gi Dci wK kZ© Av‡ivc Ki‡j 1a + 1 +

1(a + 1)2 +

1(a + 1)3 + ......... Abš� avivi AmxgZK mgwó _vK‡e Ges †mB mgwó wbY©q Ki| 4

9. cÖgvY Ki †h, wÎfz‡Ri †h †Kvb †Kv‡Yi ewnwØ©LÊK wecixZ evû‡K D³ †KvY msjMœ evûØ‡qi Abycv‡Z ewnwe©f³ K‡i| 6 A_ev, cÖgvY Ki †h, wÎfz‡Ri cwi‡K›`ª, fi‡K›`ª Ges j¤^we›`y mg‡iL| 10. ABCD AvqZ‡¶‡Îi KY©Øq O we›`y‡Z †Q` K‡i‡Q| AvqZ‡¶‡Îi Af¨š�‡i †h †Kvb we›`y P n‡j, cÖgvY Ki †h, PA2 + PB2 + PC2 + PD2 = AC2 + 4PO2 4 A_ev, ABC wÎfz‡Ri AC I AB evûi Dci h_vµ‡g BE I CF j¤^| †`LvI †h, ∆ABC t ∆AEF = AB2 t AE2

11. wÎfz‡Ri D”PZv, f‚wg ,mgwØLÊK ga¨gv Ges f‚wg msjMœ GKwU †KvY †Ìqv Av‡Q, wÎfzRwU AuvK| (A¼‡bi wPý I weeiY Avek¨K) 5 A_ev, GKwU wbw`©ó AvqZ‡¶‡Îi †¶Îd‡ji mgvb †¶Îdjwewkó GKwU eM©‡¶Î A¼b Ki‡Z n‡e| (A¼‡bi wPý I weeiY Avek¨K)

12. A I B we›`yi Ae¯’vb †f±i a, b Ges AB †iLvsk C we›`y‡Z m t n Abycv‡Z Aš�we©f³ n‡j, †`LvI †h, C we›`yi Ae¯’vb †f±i c = mb + nam + n 4

A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, UªvwcwRqv‡gi KY©Ø‡qi ga¨we›`yi ms‡hvRK †iLvsk mgvš�ivj evûØ‡qi mgvš�ivj Ges Zv‡ì we‡qvMd‡ji A‡a©K|

13. GKwU mg‡KvYx wÎfz‡Ri `ywU evûi ˆ`N© 5 †m.wg Ges 3.5 †m.wg.| G‡K e„nËi evûi PZzw`©‡K †Nviv‡j †h Nbe¯‘ Drcbœ nq, Zvi AvqZb wbY©q Ki| 4

A_ev, GKwU †MvjK AvK…wZi ej GKwU wmwjÛvi AvK…wZi ev‡· wVKfv‡e Gu‡U hvq| ev·wUi AbwaK…Z As‡ki AvqZb 89 58 Nb †m.wg n‡j,

ejwUi cwiwa KZ?

14. †h †Kvb wZbwU cÖ‡kœi DËi `vI: 4 × 3 = 12 (K) 540 wK‡jvwgUvi ` �‡i GKwU we›`y‡Z †Kvb cvnvo 7′ †KvY Drcbœ Ki‡j cvnvowUi D”PZv KZ?

(L) hw` sinA = 35 , cosB =

1213 Ges A, B abvÍK m�¶¥‡KvY nq, Z‡e tanA + tanB

1– tanA tanB Gi gvb wbY©q Ki|

(M) 1cotθ + secθ = y n‡j, †`LvI †h, cosecθ =

y2 + 1y2 – 1

(N) θ Gi gvb wbY©q Ki: 1 – 2sinθ – 2cosθ + cotθ = 0, †hLv‡b 0° ≤ θ ≤ 360° (O) cÖgvY Ki †h, sin780° cos390° – sin330° cos(–300°) = 1 15. wb‡æi MYmsL¨v wb‡ekb mviYx †_‡K msw¶ß c×wZ‡Z MvwYwZK Mo wbY©q Ki: 5

xi 1 2 3 4 5 6 7 8 9 10 fi 5 20 30 40 50 35 21 12 10 8

A_ev, wb‡P †Kvb GKwU †kÖwYi wk¶v_©x‡ì MwY‡Z cÖvß b¤^‡ii MYmsL¨v wb‡ekb mviYx †Ìqv nj| cÖvß b¤^‡ii Mo eëavb wbY©q Ki| cÖvß b¤î 25-34 35-44 45-54 55-64 65-74 75-84 85-94 wk¶v_©x msL¨v 5 10 15 20 30 16 4


4. Mymensingh Girls' Cadet College, Mymensingh Subject code : 1 2 6

Time ⎯ 3 hours Higher Mathematics Full marks ⎯ 75

1. For any subsets E, F, G of the universal set U show that E × (F ∪ G) = (E × F) ∪ (E × G) 4 Or, In the Humanities section of class nine in a school of 50 students 29 have taken up Civics, 24 have taken up Geography

and 11 have taken up both civics and geography. How many students have taken up neighter Civics nor Geography? 2. Answer any two of the following questions. 3 × 2 = 6 (a) If (x − 2) is a factor of x4 − 5x3 + 7x2 − k then show that k = 4 (b) If (a + b + c) (ab + bc + ca) = abc, then prove that (a + b + c) = a3 + b3 + c3

(c) Resolve into partial fraction: x3

x2 − 9

3. Show that S = ô where S = {n : n ∈ ô and 22n–1 + 1 is divisible by 3} 4

Or, Use the method of mathematical induction for all n ∈ ô. show that 1

1.3 + 1

3.5 + 1

5.7 + .. ...... + 1

(2n − 1)(2n + 1) = n

2n + 1

4. If p − 1 = 323 + 3

13 then show that p3 = 3p2 + 6p + 4 5

Or, If logkay − z =

logkbz − x =

logkcx − y then prove that axbycz = 1

5. Find the domain of the function F(x) = (x + 1)2 and determine the function one-one or not. 4 Or, Sketch the graph of relation, S = {(x, y) : (y − 1)2 = 4x} and determine from the graph whether the relation is a function

or not.

6. Solve: |x|x + x2 = 2

Or, Solve and show the solution set on the number line: x + 2x + 1 >

x − 3x − 4

7. Solve : x + yx − y +

x − yx + y =

103 , x2 − y2 = 3 4

Or, Solve: x − 2

3 = y + 1

4 = z + 4

5 , 2x + 3y − 4z = 13

8. Find the sum of the given infinite series (up to infinity) if exist: 15 −

252 +

453 −

854 4

9. State and prove the Apollonious's theorem 6 Or, Prove that if two chords of a circle cut at a point within it, the rectangles contained by the segment of one is equal to that

contained by the segment of other. 10. A straight line drawn through the vertex A of the parallelogram ABCD meets the side BC at M and the side DC at N. Prove

that BM.DN is constant. 4 Or, AD is the perpendicular from vertex A to the chord BC of a insectors triangle ABC and R is the circum-radius of the

triangle. Prove that AB2 = 2R.AD. 11. To construct a triangle having given the base, the vertical angle and the sum of other two sides. (Denotation and sign of

construction are necessary) 5 Or, To draw a square equal in area to a given rectangle. (Denotation and sign of construction are necessary) 12. If a , b, c are the position vectors of the points A, B, C respectively and if the point C divides the line AB in the ratio m : n

internally, then prove that c = na + mbm + n 4

Or, If a and b are non zero and non-parallel vectors and if ma + nb = 0 then show that m = n = 0 13. Three spherical balls of glass of radii 6, 8, and r cm are melted and formed into a single solid sphere of radius 9 cm. Find the value of r. 4

Or, The length, breadth and height of a rectangular block of iron are 10, 8 and 512 cm respectively. How many spherical shots

of radius 12 cm can be made by melting the block?

14. Answer any three of the following questions : 4 × 3 = 12 (a) The angles of a triangle are in AP. and the largest angle is double the least. Express the angles in radians. (b) What is the height of a hill which subtends angle 7′ at a point 540 km away from the hill?

(c) Prove that tanθ + secθ − 1tanθ − secθ + 1 =

sinθ + 1cosθ

(d) If sinA + sin2A = 1, prove that cos2A + cos4A = 1 (e) Solve : secθ + tanθ = 3 , 0° < θ < 90° 15. From the following frequency distribution table find the arithmetic mean by using short cut method: 5

xi 1 2 3 4 5 6 7 8 9 10 fi 5 20 30 40 50 35 21 12 10 8

Or, The marks obtained in mathematics by 60 students are given below. Find the mean deviation of the marks obtained: Numbers 51-60 61-70 71-80 81-90 91-100 Students 10 15 20 10 5


5. Mirzapur Cadet College, Tangail Subject code : 1 2 6


1. For, A = {a, b} and B = {1, 2}, C = {2, 3} prove that, A × (B ∩ C) = (A × B) ∩ (A × C) 4 Or, State and prove De Morgan's law for two sets A and B. 2. Answer any two of the following questions: 3 × 2 = 6 (a) Show that 2x + 1 is a factor of 4x4 + 12x3 + 7x2 + 3x − 2 (b) If (a + b + c) (ab + bc + ca) = abc, then show that (a + b + c)3 = a3 + b3 + c3

(c) Simplify: a2 + bc

(a − b) (a − c) + b2 + ca

(b − c) (b − a) + c2 + ab

(c − a) (c − b)

3. Show that, the expression x2n − y2n is divisible by (x − y). 4 Or, Using the method of mathematical induction prove that for all n ∈ ô,

1

1.2 + 1

2.3 + 1

3.4 + ............ + 1

n(n + 1) = n

n + 1.

4. If logkay − z =

logkbz − x =

logkcx − y then prove that, ay+zbz+xcx+y = 1 5

Or, If a = xyp−1, b = xyq−1 and c = xyr−1 then show that, aq−rbr−pcp−q = 1 5. Find the domain of the function F(x) = (x − 3)2 and show that it is not one - one function. 4

Or, Sketch the graph of the relation S = {(x, y) : x2

32 + y2

22 = 1}

6. Solve: (4 + x)12 + (11 + x)

12 = (9 + 8x)

12 4

Or, Solve and show the solution set on the number line : |x|x + x2 = 2

7. Solve: xy = yx, x = 2y (where x ≠ 0, y ≠ 0.) 4 Or, Sketch the graph of the relation x + 2y − 4 > 0 and 2x − y − 3 > 0

8. Solve Express into rational fraction: 0.0.1

.2

.3 4

9. Prove that, if straight lines drawn from an angle of a triangle divides the opposite side internally in the ratio of the other two sides, then the straight line is the internal bisector of that angle. 6

Or, Prove that the circumcentre, the centroid and the orthocenter of any triangle are colinear. 10. Prove that the BC of the triangle ABC is divided into three equal parts by the points P and Q. Prove that AB2 + AC2 = AP2 + AQ2 + 4PQ2. 4 Or, In the triangle ABC, AB = AC and AC is produced to D such that AC = CD. Prove that BD2 = 2BC2 + AC2. 11. To construct a triangle having the altitude, the median bisecting the base and an angle adjacent to the base. (Symbols and

description are necessity) 5 Or, To construct a circle this touches a given straight line at a given point in it and passes through another given point

outside that line. (Symbols and description are necessity) 12. Prove with the help of vector method the straight line segment joining the middle points of the adjacent sides of a

quadrilateral form a parallelogram. 4 Or, Prove by the vector method that the straight line joining the middle points of the nonparallel sides of a trapezium is the

sum of the parallel sides. 13. An iron sphere of diameter 4 cm is flattered into a circular iron sheet of thickness 5 cm. What is the radius of the sheet? 4 Or, Three spherical balls of glass of radii 6, 8 and r cm melted and formed into a single solid sphere of radius 9 cm. Find the

value of r. 14. Answer any three of the following questions: 4 × 3 = 12 (a) The radius of a circle is 7 cm. Find the angle subtended at the centre by an arc of 11 cm long. (b) If cotA + cosecA = 3, what is the value of cosA? (c) If acosA − bsinA = c then prove that asinA + bcosA = ± a2 + b2 − c2

(d) Find the value of tan π12 tan

5π12 tan

7π12 tan

11π12

15. Find the mean deviation of yearly income of 40 industries is given below: 5 Income 1 − 50 51 − 100 101 − 150 151 − 200 201 − 250 251 − 300

No. of industry 8 0 10 13 5 4 Or, Find the standard deviation from the following frequency distribution table:

x 0 1 2 3 4 5 6 7 f 4 10 14 18 20 21 7 6


6. Rajshahi Cadet College, Rajshahi Subject code : 1 2 6


1. If A = {1, 2, 3} and B = {2, 3, 4}, then show that P(A) ∩ P(B) = P(A ∩ B) 4 Or, Show that, A × (B ∩ C) = (A × B) ∪ (A × C). 2. Answer any two of the following questions: 3 × 2 = 6 (i) Let P(x) = ax5 + bx4 + cx3 + cx2 + bx + a, where a, b, c are constants and a ≠ 0, show that, if (x − r) is a factor of P(x),

then (rx − 1) also is a factor of P(x).

(ii) If x2 − yz

a = y2 − zx

b = z2 − xy

c ≠ 0, then show that, (a + b + c) (x + y + z) = ax + by + cz.

(iii) Simplify: 1

x + a + 2x

x2 + a2 + 4x3

x4 + a4 + 8x7

a8 − x8 .

3. Use the Method of Mathematical induction to show that, (x2n − y2n) is divisible by (x + y), where n is any natural number. 4 Or, Use the Method of Mathematical Induction to show that, for all n ∈ ô :

1

1.3 + 1

3.5 + 1

5.7 + ......... + 1

(2n − 1) (2n + 1) = n

2n + 1

4. If a2 = b3 then show that ⎝⎛⎠⎞a

b

32 + ⎝⎛⎠⎞b

a

23 = a

12 + b

−13 5

Or, If a3−x .b5x = a5+x .b3x, then show that x logk ⎝⎛⎠⎞b

a = logka.

5. Find the domain of the function F(x) = (x − 1)2 and determine whether the function is one-one or not. 4 Or, Sketch the graph of the relation S and determine from the graph whether the relation is a function, where S = {(x, y) : x2 + y2 − 2x − 4y − 11 = 0}. 6. Solve : a2x − (a3 + a)ax−1 + a2 = 0 (a > 0, a ≠ 1) 4

Or, Find the solution set: x + 2x + 1 >

x − 3x − 4

7. Solve : x2 = 3x + 2y, y2 = 3y + 2x 4 Or, Solve, x + 2y + z = 0, x − 2y − 2z = 0, 3x + y + z = 7. 8. Impose a condition on x under which the infinites series :

1

x + 1 + 1

(x + 1)2 + 1

(x + 1)3 + ... ... ... (up to infinity) will have a sum and find that sum. 4

9. Prove that, in an acute-angled triangle the perpendicular drawn from the vertices to the opposite side bisect the angle of the pedal triangle. 6

Or, Prove that, the rectangle contained by the diagonal of a quadrilateral inscribed in a circle is equal to the sum of the two rectangles contained by its opposite sides.

10. The side BC of the triangle ABC is divided in to three equal parts by the points P and Q. Prove that AB2 + AC2 = AP2 + AQ2 + 4PQ2. 4 Or, The chords AC and BD of the semi-circle described on AB as diameter intersect in P. Prove that, AB2 = AC.AP + BD.BP. 11. To construct a triangle having given the base, the vertical angle and the difference of the other two sides. 5 Or, To construct a circle which touches a given straight line at a given point in it and passes through another given point

outside that line. 12. Prove by vector method that the straight lines joining the middle points of the adjacent sides of a quadrilateral form a

parallelogram. 4 Or, If a and b are non-zero and non-parallel vectors and of ma + nb = 0 then show that m = n = 0. 13. Three spherical balls of glass of radii 6, 8 and r cm. are melted and formed into a single solid sphere of radius 9cm. Find the value of r. 4 Or, A spherical ball of circumference 44cm. exactly fits into a cubical box. Find volume of the box unoccupied. 14. Answer any three of the following questions : 4 × 3 = 12 (a) What is radian? Prove that radian is a constant angle.

(b) Prove that, tan θ + sec θ − 1tan θ − sec θ + 1 =

1 + sin θcos θ

(c) If sec θ + tan θ = x, show that sin θ = x2 − 1x2 + 1

(d) Solve: 3 sin θ + cos θ = 2 (0° < θ < 90°)

(e) If tanθ = 512 and cosθ is negative, evaluate

sin θ + cos(−θ)sec (−θ) + tan θ

15. The frequency distribution table of the obtained marks in Mathematics of 500 students in first year final Examination of a college is given below. Find the arithmetic mean of the obtained marks of the students using shortcut method: 5

Marks obtained 11 - 20 21 - 30 31 - 40 41 - 50 51 - 60 61 - 70 71 - 80 81 - 90 91 - 100 Number of student 5 20 50 75 145 100 80 20 5

Or, Find the mean deviation from the following frequency distribution table: x 5 10 15 20 25 30 35 40 45 50 f 8 12 15 10 20 18 13 8 3 2


7. e¸ov miKvix evwjKv D”P we`¨vjq, e¸ov welq †KvW : 1 2 6


1. hw` A = {a, b} Ges B = {b, 5} n‡j, †`LvI †h, P(A) ∩ P(B) = P(A ∩ B) 4 A_ev, mvwe©K †mU U Gi †h †Kvb Dc‡mU A I B Gi Rb¨ †`LvI †h, A\B = A ∩ B′ 2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3×2=6 (K) hw` P(x) = ax5 + bx4 + cx3 + cx2 + bx + a, †hLv‡b a, b, c aª“eK Ges a ≠ 0. †`LvI †h, (x − r) hw` P(x) Gi GKwU Drcv`K nq, Z‡e (rx−1)

I P(x) Gi GKwU Drcv`K|

(L) hw` 1a3 +

1b3 +

1c3 =

3abc nq, Z‡e †`LvI †h, bc + ca + ab = 0 A_ev, a = b = c

(M) mij Ki: 11 + x +

21 + x2 +

41 + x4 +

81 + x8 +

16x16 − 1

3. †`LvI †h, S = ô †hLv‡b S ={n : n ∈ ô Ges 22n−1 + 1 ivwkwU 3 Øviv wefvR¨}| 4 A_ev, MvwYwZK Av‡ivn c×wZ‡Z †`LvI †h, mKj n ∈ ô Gi Rb¨ 1

1.3 + 1

3.5 + 1

5.7 + ... ... ... 1

(2n−1)(2n+1) = n

2n+1

4. hw` xyz ≠ 0, ax = by = cz Ges b2 = ac nq, Z‡e †`LvI †h, 1x + 1z =

2y 5

A_ev, †`LvI †h, logk x − x2−1

x + x2 – 1 = 2 logk( )x − x2 − 1

5. F(x) = 1

x+2 dvsk‡bi †Wv‡gb wbY©q Ki Ges dvskbwU GK GK wKbv wba©viY Ki| 4

A_ev, S = {(x,y) : (y−1)2 = 4x} Aš^‡qi †jLwPÎ AsKb Ki Ges Aš^qwU dvskb wKbv †jLwPÎ †_‡K wbY©q Ki| 6. mgvavb Ki: 22x − 3.2x+2 = −32 4

A_ev, (2x−3)(x−2)2

x+1 > 0 mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI|

7. mgvavb Ki : xy − x2 − 1 = 0, y2 − xy − 2 = 0 4 A_ev, y ≥ 2x AmgZvi mgvavb †m‡Ui †jL AsKb Ki|

8. x Gi Dci Kx kZ© Av‡ivc Ki‡j 1

x+1 + 1

(x+1)2 + 1

(x+1)3 + ... ... ... Abš� avivi AmxgZK mgwó _vK‡e Ges †mB kZ©vax‡b mgwó wbY©q Ki| 4

9. cÖgvY Ki †h, wÎfz‡Ri †h †Kv‡bv †Kv‡Yi Aš�wØ©LÊK wecixZ evû‡K D³ †KvY msjMœ evûØ‡qi Abycv‡Z Aš�we©f³ Ki| 6 A_ev, cÖgvY Ki †h, wÎfz‡Ri cwi‡K›`ª, fi‡K›`ª I j¤^we›`y mg‡iL| 10. †Kv‡bv ABC wÎfz‡Ri AD I BE ga¨gvØq ci¯•i G we›`y‡Z †Q` K‡i‡Q| G we›`yi ga¨ w`‡q Aw¼Z DE Gi mgvš�ivj †iLvsk AC †K F we›`y‡Z

†Q` K‡i cÖgvY Ki †h, AC = 6EF. 4 A_ev, AB e¨v‡mi Ici AswKZ Aa©e„‡Ëi `yBwU R¨v AC I BD ci¯•i P we›`y‡Z †Q` K‡i| cÖgvY Ki †h, AB2 = AC.AP + BD.BP. 11. †Kvb wÎfz‡Ri D”PZv, f‚wgi Dci ga¨gv Ges f‚wg msjMœ GKwU †KvY †Ìqv Av‡Q| wÎfzRwU AsKb Ki| 5 A_ev, wfbœ wfbœ e¨vmva©wewkó Giƒc wZbwU e„Ë AuvK †hb Zvi ci¯•i‡K ewnt¯•k© K‡i| 12. †f±i c×wZ‡Z cÖgvY Ki †h, †Kv‡bv PZyfz©‡Ri mwbœwnZ evû¸‡jvi ga¨we›`yi ms‡hvRK †iLvskmg�n GKwU mvgvš�wiK Drcbœ K‡i| 4 A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, UªvwcwRqv‡gi KY©Ø‡qi ga¨we›`yi ms‡hvRK mij‡iLv mgvš�ivj evûØ‡qi mgvš�ivj Ges Zv‡ì

we‡qvMd‡ji A‡a©K|

13. 4 †m.wg. e¨v‡mi GKwU †jŠn †Mvj‡K wcwU‡q 23 †m.wg. cyi“ GKwU e„ËvKvi †jŠncvZ cÖ � yZ Kiv nj| G cv‡Zi e¨vmva© KZ? 4

A_ev, GKwU mge„Ëf‚wgK †KvY‡Ki AvqZb v eµZ‡ji †¶Îdj s f‚wgi e¨vmva© r, D”PZv h Ges Aa© kxl©‡KvY α n‡j †`LvI †h, s = πr2

sinα eM©

GKK Ges v = 13 πr3

tanα Nb GKK|

14. †h †Kvb wZbwU cÖ‡kœi DËi `vI :− 4 × 3 = 12 (K) 540 wK.wg. ` �‡i GKwU we›`y‡Z †Kvb cvnvo 7′ †KvY Drcbœ Ki‡j cvnvowUi D”PZv KZ?

(L) cÖgvY Ki †h, sinθ−cosθ+1sinθ+cosθ−1 =

1+sinθcosθ

(M) mgvavb Ki: 3 sinθ + cosθ = 2; 0° < θ < 90° (N) 2(sinθ cosθ + 3 ) = 3 cosθ + 4sinθ ; 0° ≤ θ ≤ 360° kZ© mv‡c‡¶ mgvavb Ki Ges θ Gi m¤¢ve¨ gvb wbY©q Ki|

(O) gvb wbY©q Ki: cos2 π15 + cos2 13π

30 + cos2 16π15 + cos2 47π

30

15. wb‡æi MYmsL¨v wb‡ekb mviwY †_‡K msw¶ß c×wZ‡Z MvwYwZK Mo wbY©q Ki: 5 xi 1 2 3 4 5 6 7 8 9 10 ƒi 5 20 30 40 50 35 21 12 10 8

A_ev, 10g †kÖwYi 60 Rb QvÎxi cÖvß b¤î †Ìqv n‡jv| cÖvß b¤^‡ii Mo eëavb wbY©q Ki| b¤î 51-60 61-70 71-80 81-90 91-100 QvÎx 10 15 20 10 5


8. Pabna Cadet College, Pabna Subject code : 1 2 6


1. If S = {x : x ∈ Ñ and x2 + 1 = 0}, then Find S′ = Ñ\S. What can you say about S ? 4 Or, Show that the sets A = {1, 2, 3, ........., n} and B = {1, 2, 22, .........., 2n–1} are equivalent. 2. Answer any two of the following queations: 3 × 2 = 6 (a) Resolve into factors: 18x3 + 15x2 – x – 2

(b) If x2 – yz

a = y2 – zx

b = z2 – xy

c ≠ 0, then prove that (a + b + c)(x + y + z) = ax + by + cz

(c) Simplify: 1

1 + x + 2

1 + x2 + 4

1 + x4 + 8

1 + x8 + 16

x16 – 1

3. Show that S = ô, where S = {n : n ∈ ô and 22n–1 + 1 is divisible by 3} 4 Or, Use the Method of Mathematical induction to show that, for n ∈ ô,

11.3 +

13.5 +

15.7 + .........+

1(2n – 1)(2n + 1) =

n2n + 1

4. If xyz ≠ 0, ax = by = cZ and b2 = ac. then show that 1x +

1z =

2y 5

Or, If logk(1 + x)

logkx = 2, then show that x = 1 + 5

2

5. F : Ñ → Ñ, F(x) = 1

x – 2 , find the domain and show that the function is one–one. 4

Or, Sketch the graph of the relation S, Where S = {(x, y) : x2

4 + y2

25 = 1}

6. Solve : 41+x + 41–x = 10 4

Or, Solve x(x + 1)

x – 2 > 0 and show the solution set on the number line.

7. Solve : x2 + y2 = 25, xy = 12 4 Or,Draw the graph of the following inequality : 3x – 2y ≤ 12 8. Impose a condition on x under which the infinites series :

1

x + 1 + 1

(x + 1)2 + 1

(x + 1)3 + ... ... ... (up to infinity) will have a sum and find that sum. 4

9. State and prove Apollonius theorem. 6 Or, Prove that, the circum-centre, the centroid and the orthocentre of acute triangle are collinear. 10. BE and CF are perpendiculars respectively on sides AC and AB of the triangle ABC, then prove that ∆ABC t ∆AEF = AB2 t AE2 4 Or, The bisector of ∠A of the triangle ABC intersects BC at D and the circumcircle of ABC at E, prove that, AD2 = AB.AC – BD.DC. 11. Construct a triangle having given the base, the vertical angle and the sum of other two sides. 5 Or, Draw a circle which touches a given circle at a given point and passes through a point outside the circle. 12. Prove, by vector methods that the diagonals of a parallelogram bisect each other. 4 Or, Prove with the help of vectors that a straight line joining the middle points of the diagonal of a trapezium is a parallel to

half the difference of the parallel sides. 13. A spherical ball of circumference 44 cm, exactly fits into a cubical box. Find the volume of the box unoccupied. 4 Or, The length, breadth and height of a rectangular block of copper are respectively 11 meters, 10 meters and 5 meters. It is

melted and formed into sphere of the diameter 50 cm. Find now many spheres can be formed? 14. Answer any three of the following questions : 4 × 3 = 12 (a) Show that radian is a constant angle.

(b) Prove that, tanA + secA – 1tan A – secA + 1 =

sinA + 1cosA

(c) If acosθ – bcosθ = c, then show that acosθ + bsinθ = ± a2 + b2 – c2

(d) Evaluate : sin2 π7 + sin2

5π14 + sin2

8π7 + sin2

9π14

(e) Solve the equation : secθ + tanθ = 3 for 0° ≤ θ ≤ 90° 15. The marks obtained in mathematics by students of a class are given below: 5

Obtain marks 25–34 35– 44 45–54 55–64 65–74 75–84 85–94 Number of students 5 10 15 20 30 16 4

Find the arithmetic mean from the above table : Or, Find the mean deviation from the following frequency distribution table:

x 51– 60 61–70 71–80 81–90 91–100 f 10 15 20 10 5


9. w`bvRcyi wRjv ¯‹zj, w`bvRcyi welq †KvW : 1 2 6


1. hw` S = {x : x ∈ Ñ Ges x(x − 1) = x2 − x} nq, Z‡e S Ges S′ = Ñ\S wbY©q Ki| 4 A_ev, †h †Kvb mvš� †mU A I B Gi Rb¨ cÖgvY Ki †h, n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3×2=6 (K) x4 − 5x3 + 7x2 − a Gi x − 2 GKwU Drcv`K n‡j, †`LvI †h a = 4. (L) hw` (a + b + c)(ab + bc + ca) = abc nq, Z‡e †`LvI †h, (a + b + c)3 = a3 + b3 + c3.

(M) mij Ki : bc(a + d)(a − b) (a − c) +

ca(b + d)(b − a) (b − c) +

ab(c + d)(c − a) (c − b)

3. hw` S = {n : n ∈ ô Ges 2 Øviv n(n + 1) wefvR¨} nq Z‡e †`LvI †h, S = ô. 4 A_ev, MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h mKj n ∈ ô Gi Rb¨ 13 + 23 + 33 + ... ... ... + n3 =

n2(n + 1)2

4

4. hw` x = (a + b)13 + (a − b)

13 Ges a2 − b2 = c3 nq, Z‡e †`LvI †h, x3 − 3cx − 2a = 0 5

A_ev, hw` logk(1 + x)logkx =2 nq Z‡e †`LvI †h, x =

1 + 52

5. S = {(x, y) : x ∈ A, y ∈ A Ges x + y = 1} Aš^qwU‡K ZvwjKv c×wZ‡Z eY©bv Ki Ges †Wvg S †iÄ S wbY©q Ki: †hLv‡b A = {−2, −1, 0, 1, 2} 4 A_ev, S = {(x, y) : y = − 4x2} Gi †jL A¼b Ki Ges Aš^qwU dvskb wKbv Zv †jL wPÎ †_‡K wbY©q Ki|

6. mgvavb Ki : |x|x + x2 = 2 4

A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI : x(x − 4)x − 5 < 0

7. mgvavb Ki : 8yx − y2x = 16, 2x = y2 4

A_ev, mgvavb Ki : x − 23 =

y + 14 =

z + 45 , 2x + 3y − 4z = 13

8. y Gi Dci Kx kZ© Av‡ivc Ki‡j 1y + 1 +

1(y + 1)2 +

1(y + 1)3 + ... ... ... Abš� avivi (AmxgZK) mgwó _vK‡e Ges †mB mgwó wbY©q Ki| 4

9. cÖgvY Ki †h, `yBwU m`„k wÎfz‡Ri †¶ÎdjØ‡qi AbycvZ Zv‡ì †h †Kvb `yB Abyiƒc evûi Dci AswKZ e‡M©i †¶ÎdjØ‡qi Abycv‡Zi mgvb| 6 A_ev, cÖgvY Ki †h, e„‡Ë Aš�wj©wLZ †Kvb PZyfz©‡Ri KY©Ø‡qi Aš�M©Z AvqZ‡¶Î H PZyfz©‡Ri wecixZ evûØ‡qi Aš�M©Z AvqZ‡¶‡Îi mgwói

mgvb| 10. ABC mgwØevû wÎfz‡Ri f‚wg BC-Gi Dci P †h †Kvb we›`y n‡j †`LvI †h, AB2 − AP2 = BP.PC. 4 A_ev, AB e¨v‡mi Dci AswKZ Aa©e„‡Ëi `yBwU R¨v AC I BD ci¯•i P we›`y‡Z †Q` K‡i| cÖgvY Ki †h, AB2 = AC.AP + BD.BP. 11. wÎfz‡Ri f‚wg, wkit‡KvY I Aci `yB evûi Aš�i †Ìqv Av‡Q| wÎfzRwU AuvK‡Z n‡e| [AsK‡bi wPý I weeiY Avek¨K] 5

A_ev, mg‡KvYx wÎfz‡Ri AwZfzR I Aci `yB evûi mgwó †Ìqv Av‡Q| wÎfzRwU AuvK‡Z n‡e| [AsK‡bi wPý I weeiY Avek¨K] 12. †f±‡ii mvnv‡h¨ cÖgvY Ki †h, wÎfz‡Ri †h †Kvb `yB evûi ga¨we›`yØ‡qi ms‡hvRK †iLvsk H wÎfz‡Ri Z…Zxq evûi mgvš�ivj I ˆ`‡N© Zvi

A‡a©K| 4 A_ev, †f±i c×wZ‡Z cÖgvY Ki †h, †Kvb PZyfz©‡Ri KY©Øq ci¯•i‡K mgwØLwÛZ Ki‡j Zv GKwU mvgvš�wiK nq| 13. 4 †m. wg. e¨vmv‡a©i GKwU wb‡iU †MvjK‡K Mwj‡q 5 †m.wg. ewne¨vmva© wewkó I mgfv‡e cyi“ GKwU duvcv †MvjK cÖ � yZ Kiv nj| wØZxq †MvjKwU

KZ cyi“? 4 A_ev, †KvYK AvKv‡ii GKwU Zuveyi D”PZv 7.5 wgUvi| GB Zuvey Øviv 2000 eM©wgUvi Rwg wNi‡Z PvB‡j wK cwigvY K¨vbfvm jvM‡e?

14. †h †Kvb wZbwU cÖ‡kœi DËi `vI : 4×3=12 (K) GKwU wÎfz‡Ri †KvY¸‡jv mgvš�i †kÖYxfz³ Ges e„nËg †KvYwU ¶z`ªZg †KvYwUi wØ¸Y| †KvY¸‡jv‡K †iwWqv‡b cÖKvk Ki|

(L) hw` cosA − sinA = 2 sinA nq Z‡e †`LvI †h, cosA + sinA = 2 cosA.

(M) tanθ + secθ = p n‡j †`LvI †h, sinθ = p2 − 1p2 + 1

(N) cos2 π8 + cos2

3π8 + cos2

5π8 + cos2

7π8 Gi gvb wbY©q Ki|

(O) mgvavb Ki : 2(cos2θ − sin2θ) = 1, †hLv‡b, 0° < θ < 90° 15. †Kvb GjvKvi AbyaŸ© 50 eQi eq‡mi MYmsL¨v wb‡ekb wbæiƒc| Zv‡ì eq‡mi MvwYwZK Mo wbY©q Ki| 5

†kÖwYe¨vwß 15-19 20-24 25-29 30-34 35-39 40-44 45-49 MYmsL¨v 3 13 21 15 5 4 2

A_ev, 60 Rb Qv‡Îi MwY‡Z cÖvß b¤^‡ii wb‡ekb †Ìqv nj| cÖvß b¤^‡ii Mo eëavb wbY©q Ki| b¤î 51-60 61-70 71-80 81-90 91-100 QvÎ msL¨v 10 15 20 10 5

10 D”PZi MwYZ mKj †ev‡W©i kxl© ’vbxq ¸i“Z¡c �Y© ¯‹z‡ji wbe©vPwb cix¶vi cÖkœcÎ 10. K¨v›Ub‡g›U cvewjK ¯‹zj I K‡jR, iscyi welq †KvW : 1 2 6


1. hw` S = {y : y ∈ Ñ Ges y (y – 1) = y2 – y} nq Z‡e S′ = Ñ\S wbY©q Ki| 4 A_ev, †Kvb †kÖwYi 30 Rb Qv‡Îi 20 Rb dzUej Ges 15 Rb wµ‡KU †Lj‡Z cQ› K‡i| cÖ‡Z¨K QvÎ yBwU †Ljvi Aš�Z GKwU †Ljv cQ› K‡i| KZRb

QvÎ yBwU †LjvB cQ› K‡i? 2. †h †Kvb `yÕwU cÖ‡kœi DËi `vI : 3 × 2 = 6 (K) Drcv`‡K we‡k−lY Ki: a3 – a2 – 10a – 8

(L) mij Ki: 11 + x +

21 + x2 +

41 + x4 +

81 + x8 +

16x16 – 1

(M) hw` 1a3 +

1b3 +

1c3 =

3abc nq, Z‡e †`LvI †h, bc + ca + ab = 0 A_ev a = b = c

3. MvwYwZK Av‡ivn c×wZi mvnv‡h † LvI †h, mKj n ∈ ô Gi Rb¨ 13 + 23 + 33 + .................+ n3 = n2(n + 1)2

4 4

A_ev, hw` S = {n : n ∈ ô Ges 22n–1 + 1 ivwkwU 3 Øviv wefvR¨} nq Z‡e †`LvI †h, S = ô

4. hw` x = (a + b) 13 + (a – b)

13 Ges a2 – b2 = c3 nq Z‡e †`LvI †h, x3 – 3cx – 2a = 0 5

A_ev, hw` xya–1 = p, xyb–1 = q, xyc–1 = r nq, Z‡e †`LvI †h, (b – c)logk p + (c – a) logkq + (a – b) logkr = 0

5. hw` A = {–2, –1, 0, 1, 2} nq Z‡e S = {(x, y) : x ∈ A, y ∈ A Ges y = x2} †K ZvwjKv c×wZ‡Z eY©bv Ki Ges †Wvg S I †iÄ S wbY©q Ki| 4 A_ev, S = {(x, y) : x2 + y2 + 2x + 4y + 1 = 0} Aš^‡qi †jLwPÎ AsKb Ki Ges Aš^qwU dvskb wK-bv Zv †jLwPÎ n‡Z wbY©q Ki| 6. mgvavb Ki: |x| + |x + 1| = 5 4

A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI : x + 2x + 1 >

x – 3x – 4

7. mgvavb Ki: 8yx – y2x = 16

2x = y2 4 A_ev, wb‡Pi AmgZv hyM‡ji mgvavb †m‡Ui †jLwPÎ AsKb Ki: 5x – 3y – 9 > 0 Ges 3x – 2y ≥ 0

8. x Gi Dci wK kZ© Av‡ivc Ki‡j 1x + 1 +

1(x + 1)2 +


9. yBwU m „k wÎfz‡Ri †¶ÎdjØ‡qi AbycvZ Zv‡ì †h †Kvb yB Abyiƒc evûi Dci Aw¼Z eM©‡¶‡Îi †¶ÎdjØ‡qi Abycv‡Zi mgvb, cÖgvY Ki| 6 A_ev, cÖgvY Ki †h, e„‡Ë Aš�wj©wLZ †Kv‡bv PZyf©y‡Ri KY©Ø‡qi Aš�M©Z AvqZ‡¶Î H PZyf©y‡Ri wecixZ evûØ‡qi Aš�M©Z AvqZ‡¶‡Îi mgwói mgvb| 10. †Kv‡bv ABC wÎfz‡Ri ga¨gvÎq G we›`y‡Z wgwjZ n‡j cÖgvY Ki †h, AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2) 4 A_ev, AB e¨v‡mi Dci Aw¼Z Aa©e„‡Ëi `yBwU R¨v AC I BD ci¯•i P we›`y‡Z †Q` K‡i| cÖgvY Ki †h, AB2 = AC.AP + BD.BP

11. GKwU wÎfz‡Ri f‚wg, wkit‡KvY Ges Aci `yB evûi Aš�i †Ìqv Av‡Q| wÎfzRwU AsKb Ki| 5 A_ev, Giƒc GKwU e„Ë AsKb Ki hv GKwU wbw`©ó mij‡iLv‡K GKwU wbw`©ó we›`y‡Z ¯•k© K‡i K‡i Ges H †iLvi ewnt¯’ †Kv‡bv we›`y w`‡q hvq|

(Dfq †¶‡Î AsK‡bi wPý I weeiY Avek¨K) 12. †f±‡ii mvnv‡h¨ cÖgvY Ki †h, wÎfz‡Ri †h †Kv‡bv yB evûi ga we› yi ms‡hvRK †iLvsk H wÎfz‡Ri Z…Zxq evûi mgvš�ivj I ˆ`‡N© Zvi A‡a©K| 4 A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, †Kv‡bv PZzfz©‡Ri KY©Øq ci¯•i‡K mgwØLwÊZ Ki‡j Zv GKwU mvgvš�wiK nq|

13. †Kvb Nb‡Ki c„ôZ‡ji K‡Y©i ˆ`N© 8 2 †m.wg. n‡j Zvi K‡Y©i ˆ`N© I AvqZb wbY©q Ki| 4 A_ev, †KvYK AvKv‡ii GKwU Zuveyi D”PZv 5.5 wgUvi| GB Zuvey Øviv 1500 eM© wgUvi Rwg wNi‡Z PvB‡j wK cwigvY Kvbfvm jvM‡e? 14. †h‡Kvb wZbwU cÖ‡kœi DËi `vI : 4 × 3 = 12 (K) †Kv‡bv wÎfz‡Ri †KvY¸‡jv mgvš�i †kÖwYfz³ Ges ¶z`ªZg †KvYwU e„nËg †Kv‡Yi A‡a©K n‡j †KvY¸‡jv‡K †iwWqv‡b cÖKvk Ki|

(L) a cosθ – b sinθ = c n‡j †`LvI †h, a sinθ + b cosθ = ± a2 + b2 – c2 (M) mgvavb Ki: 3(sec2θ + tan2θ) = 5 †hLv‡b 0° ≤ θ ≤ 360°

(N) tanθ + secθ = x n‡j, †`LvI †h, sinθ = x2 – 1x2 + 1

(O) gvb wbY©q Ki: tan π12 tan

5π12 tan

7π12 tan

11π12

15. wb‡æi MYmsL¨v wb‡ekb mviwYi Mo eëavb wbY©q Ki: 5 cÖvß Mo 51 − 60 61 − 70 71 − 80 81 − 90 91 − 100 wk¶v_©xi msL¨v 10 15 20 10 5

A_ev, wb‡Pi mviwY †_‡K MvwYwZK Mo wbY©q Ki: †kÖwY 15 − 19 20 − 24 25 − 29 30 − 34 35 − 39 40 − 44 45 − 49


MYmsL¨v 3 13 21 15 5 4 2

11. Comilla Cadet College, Comilla Subject code : 1 2 6


1. For any sub-set, P, Q, R for universal set U, Prove that P × (Q ∪ R) = (P × Q) ∪ (P × R) 4 Or, Show that the set S = (1, 4, 9, 25, 36 .........} of square of natural numbers, is an infinite set. 2. Answer any two of the following: 3×2=6 (a) Resolve : x3 − 9x2y + 26xy2 − 24y3 into factors.

(b) If 1a3 +

1b3 +

1c3 =

3abc . Then show that (a + b + c)3 = a3 + b3 + c3

(c) Resolve : x3

x2 − 9 into partial fraction.

3. If {n : n ∈ ô and 22n – 1 + 1 is divided by 3} then show that S = ô 4

Or, Prove by the method of mathematical induction 13 + 23 + 33 + ......... + n3 = n2(n + 1)2

4

4. If a2 + 2 = 32/3 + 3

−2/3 then show that 3a3 + 9a − 8 = 0 5

Or, Show that, 1

loga(abc) + 1

logb(abc) + 1

logc(abc) = 1

5. Find the domain of the function f (x) = 1

x − 3 and show that the function is one-one or not. 4

Or, Draw the graph of the function S = {(x, y) : y = − 4x2} and show that the function S is one-one or not. 6. Solve : a2x − (a3 + a)ax − 1 + a2 = 0 4

Or, Solve and show the solution set on the number line, x − 3x − 4 >

x − 2x − 1

7. Solve : x + yx − y +

x − yx + y =

52 , x2 + y2 = 90 4

Or, Draw the graph of the inequality x + y − 4 ≤ 0 and 2x − y − 3 ≥ 0 8. Find the summation of the infinite series: 2 + 0.2 + 0.02 + ... ... ... 4 9. Prove that the ratio of the areas of two similar triangular is equal to the squares on corresponding sides. 6 Or, Prove that the circumcentre, the centroid and the ortho-center of any triangle are collinear. 10. The diagonals of a rectangle ABCD intersect at Q. If P is any point inside the rectangles then prove that PA2 + PB2 + PC2 + PD2 = AC2 + 4PQ2. 4 Or, BE and CF are respectively perpendicular to AC and AB of a triangle ABC, Prove that ∆ABC t ∆AEF = AB2 t AE2. 11. To divide a given line segment internally and exeternally in given ratio. [Construction & description needed] 5 Or, Draw a circle touching a given circle at a given point in it and also touching another circle. [Construction & description needed] 12. If a, b, c are the position vectors of A, B, C respectively and if the point C divides AB in the ratio m : n internally, then show that c =

mb + nam + n 4

Or, Prove with the help of vectors that the straight line joining the middle points of the diagonals of a trapezium is parallel to half the difference of the parallel sides.

13. If the volume of a right circular cone is V, the area of its curved surface is S, radius of the base is r, its height is h and semi-vertical angle is

α, then show that S = πh2 tanα

cosα = πr2

sinα square unit and V = 13 πh3 tan2α =

13 .

πr3

tanα cubic unit. 4

Or, Find the surface and volume of a sphere of radius 6 cm. 14. Answer any three of the following questions : 4×3=12

(a) Prove that tanθ + secθ − 1tanθ − secθ + 1 =

sinθ + 1cosθ = tanθ + secθ

(b) If a sinθ + b cosθ = a2 + b2 − c2 then prove that acosθ − b sinθ = c

(c) Find the value of tan215° + sin335°cos425° + cot125°

(d) Solve : secθ + tanθ = 3 where 0° < θ < 90°

(e) In triangle ABC, show that sin A + B

2 + tanA + B

2 = cos C2 ( )1 + cosec

C2

15. Frequency distribution table men of age under of 50 are is given below. Find their average age. 5 Class 15 − 19 20 − 24 25 − 29 30 − 34 35 − 39 40 − 44 45 − 49

Frequency 3 13 21 15 5 4 2 Or, Find the standard deviation from the following frequency distribution table:

class 5 − 14 15 − 24 25 − 34 35 − 44 45 − 54 55 − 64 65 − 74 75 − 84 frequency 10 20 30 40 50 60 70 80


12. Feni Girls' Cadet College, Feni Subject code : 1 2 6


1. For any subsets A and B of the universal set U, show that : A \ B = A ∩ B′. 4 Or, Show that the set S = {3n : n = 0 or n ∈ ô} is equivalent to ô.. 2. Answer any two of the following questions. 3 × 2 = 6 (a) If (a + b + c) (ab + bc + ca) = a3b3c3 then show that (a + b + c)3 = a3 + b3 + c3

(b) Resolve into factors : 18a3 + 15a2 − a − 2

(c) Resolve into partial fractions : x + 2

x2 − 7x + 12

3. Use the Method of Mathematical Induction to show that, x2n − y2n is divisible by x + y, when n is any natural number. 4

Or, Use the Method of Mathematical Induction to show that, for n ∈ ô, 13 + 23 + 33 + ......... + n3 = n2(n + 1)2

4

4. If x = aq+r bp, y = ar+p bq, z = ap+q br, then xq−r yr−p zp−q = 1 5

Or, Show that 1

loga(abc) + 1

logb(abc) + 1

logc(abc) = 1

5. State the given relation S by roster method and find dom S and range S where A = {−2, −1, 0, 1, 2}, S = {(x, y) : x ∈ A and y2 = x} 4 Or, Sketch the graph of the relation and determine from the graph whether the relation is a function : S = {(x, y) : y = −4x2} 6. Solve: a2x − (a3 + a)ax−1 + a2 = 0, (a > 0, a ≠ 1) 4

Or, Slove: |x|x + x2 = 2

7. Solve: 8yx − y2x = 16, 2x = y2 4 Or, Draw the graph of the solution sets of the following inequalities: x + y − 4 ≤ 0 and 2x − y − 3 ≥ 0 8. Impose a condition on x under which the infinite series 4

1

x + 1 + 1

(x + 1)2 + 1

(x + 1)3 + ....... (up to infinite) will have a sum and find that sum.

9. Prove that 'The external bisector of any angle of a triangle divides the opposite side externally in the ratio of the other two sides'. 6

Or, Prove that 'The ratio of the areas of two similar triangles is equal to the ratio of the squares on corresponding sides'. 10. If P is any point on the base BC of the isosceles triangle ABC, then prove that AB2 − AP2 = BP.PC 4 Or, E is the middle point of the median AD of the triangle ABC and produced BE meets AC at F. Prove that AC = 3AF and

BE = 4EF. 11. To construct a triangle having given the base, the vertical angle and the difference of the other two sides. [Sign & description

are essential] 5 Or, To construct a circle which touching a given circle at a given point in it and also touching another circle. [Sign &

description are essential] 12. Prove with the help of vectors that 'The diagonals of a parallelogram bisect each other'. 4

Or, If D, E, F are the middle points respectively of the sides BC, CA, AB of the triangle ABC, prove that ⎯AD +

⎯BE +

⎯CF = 0

13. Find the surface and volume of a sphere of radius 6 cm. 4 Or, A right circular cone, a semi-sphere and a cylinder of equal heights stand on equal bases. Show that their volumes are in the

ratio 1 t 2 t 3. 14. Answer any three of the following questions : 4 × 3 = 12 (a) The diameter of the wheel of a carriage is 84 cm. and it revolves 6 times per second. What is the speed of the carriage? (b) If sinθ + cosθ = 1, then show that sinθ − cosθ = ± 1. (c) If cotθ + cosecθ = 3, what is the value of cosθ ?

(d) Evaluate: cos2 π8 + cos2 3π

8 + cos2 5π8 + cos2 7π

8

(e) Solve the equation secA + tanA = 3 for 0 ≤ A ≤ 90°. 15. The frequency distribution table of height (in cm) of 100 people are given below. Draw the histogram and frequency

polygon of this frequency distribution. 5 Height (in cm) 146 − 155 156 − 165 166 − 175 176 − 185 186 − 195

Frequency 5 35 25 15 20 Or, Find the standard deviation from the following frequency distribution table :

Class Interval 5 − 14 15 − 24 25 − 34 35 − 44 45 − 54 55 − 64 65 − 74 75 − 84 Frequency 10 20 30 40 50 60 70 80


13. Faujdarhat Cadet College, Chittagong Subject code : 1 2 6


1. If A = {a, b, c}, B = {a, b}, C = {b, c} and D = {a, c} then show that P(A) = {A, B, C, D, {a}, {b}, {c}, ∅} 4 Or, For any finite sets A and B. Prove that n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 2. Answer any two of the following queations. 3 × 2 = 6 (a) Show that a = 4, If (x – 2) is factor of x4 – 5x3 + 7x2 – a. (b) Resolve into factors : a2(b – c) + b2(c – a) + c2(a – b).

(c) Resolve into partial fractions: x3

(x2 – 9)

3. Use the Method of Mathematical induction to show that, for all n ∈ ô, x2n – y2n is divisible by (x + y) 4 Or, Use the Method of Mathematical induction to show that, for all n ∈ ô.

1

1.3 + 1

3.5 + 1

5.7 + ........+ 1

(2n – 1)(2n + 1) = n

2n + 1

4. Simplify : a

32 + ab

ab – b3 – a

a – b 5

Or, If logk(1 + x)

logkx = 2, then show that x = 1 + 5

2

5. Find the domain of the function ƒ(x) = x – 1 and determine whether the function is one–one or not. 4 Or, Sketch the graph of the relation S = {(x, y): y = 3 – 4x – 2x2} 6. Solve : x2 – 6x + 9 – x2 – 6x + 6 = 1 4

Or, Solve : |x|x + x2 = 2.

7. Solve : 18yx – y2x = 81 3x = y2 4 Or, Draw the graph of the solution sets of the inequalities : 3x – 3y > 5 and x + 3y ≤ 9

8. Impose a condition on x under which the infinite series 1

x + 1 + 1

(x + 1)2 + 1

(x + 1)3 + ... ... ... (upto infinity) will have a sum

and find that sum. 4 9. Prove that, The sum of the squares on any two sides of a triangle is equal to twice the square on half the third side together

with twice the square on the median which bisects the third side. 6 Or, Prove that, In an acute angle triangle the perpendiculars drawn from the vertices to opposite side bisect the angle of the

padal triangle. 10. X is any point on the side BC of the triangle ABC, O is a point on the line AX. Prove that ∆AOB t ∆AOC = BX t XC 4 Or, The chords AC and BD of the semicircle described on AB as diameter intersect in P. Prove that AB2 = AC.AP + BD.BP 11. Construct a triangle having given the altitude, the median bisecting the base and an angle adjacent to the base. (Description

and symbol are essential) 5 Or, Draw a cricle touching a given straight line, at a given point in it and also touching another circle. (Description and

symbol are essential) 12. With the help of vectors, prove that the straight line joining the middle points of the non-parallel sides of a trapezium is to

and half the sum of the parallel sides. 4 Or, If a, b, c are the position vectors of A, B, C respectively and if the point C divides AB in the ratio m t n internally, then show

that c = mb + nam + n .

13. An iron sphere of diameter 4cm. is flattened into a circular iron sheet of thickness 23 cm. What is the radius of the sheet? 4

Or, Find the length of the diagonal and the volume of the cube of which the length of the diagonal of a face is 8 2 cm. 14. Answer any three of the following questions : 4 × 3 = 12 (a) If an arc of a circle of radius 28 cm. subtends an angle 45° at the centre of the circle, then find the length of the arc. (b) If sinθ + cosθ = 1, then show sinθ – cosθ = ±1. (c) If cot A + cosecA = 3, what is the value of cosA?

(d) If tanθ = 34 and sinθ is negative, then find the value of

sinθ + cosθsecθ + tanθ .

(e) Solve: 3sinθ + cosθ = 2 2 when θ < 90° 15. Find the arithmetic mean from the following frequency distribution table: 5

xi 1 2 3 4 5 6 7 8 9 10 ƒi 5 20 30 40 50 35 21 12 10 8

Or, The marks obtained in mathematics by 60 students of class IX are given below. Find the standard deviation of the data: Numbers 51–60 61–70 71–80 81–90 91–100 students 10 15 20 10 5


14. PÆMÖvg K¨v›Ub‡g›U cvewjK K‡jR, PÆMÖvg welq †KvW : 1 2 6


1. †`LvI †h, A × (B ∪ C) = (A × B) ∪ (A × C) 4 A_ev, †h †Kvb mvš� †mU P, Q Gi Rb¨ cÖgvY Ki †h, n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q)

2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3×2=6 (K) p(x) = x3 + 5x2 + 6x + 8 †K (x − a) Ges (x − b) Øviv fvM Ki‡j hw` GKB fvM‡kl (†hLv‡b a ≠ b) _v‡K Z‡e cÖgvY Ki †h, a2 + b2 + ab + 5a + 5b + 6 = 0

(L) Drcv`‡K we‡k−lY Ki : (a − b)3 + (b − c)3 + (c − a)3

(M) AvswkK fMœvs‡k cÖKvk Ki : x2

x2 − 16

3. hw` S = {n : n ∈ ô Ges 22n − 1 ivwkwU 3 Øviv wefvR¨} nq Z‡e †`LvI †h, S = ô. 4 A_ev, MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h mKj n ∈ ô Gi Rb¨, 1

1.2 + 1

2.3 + 1

3.4 + ... ... ... + 1

n(n + 1) = n

n + 1

4. hw` a2 + 2 = 323 + 3

−23 nq, Z‡e †`LvI †h, 3a3 + 9a − 8 = 0 5

A_ev, cÖgvY Ki †h, xlogay = yloga

x

5. F(x) = 1

(x − 2)2 Øviv ewY©Z dvsk‡bi †Wv‡gb wbY©q Ki Ges dvskbwU GK-GK wKbv Zv wba©viY Ki| 4

A_ev, S = {(x, y) : y2 = −16x} Aš^‡qi †jLwPÎ A¼b Ki| 6. mgvavb Ki : 32x−2 − 5.3x–2 − 66 = 0 4

A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI : 3x + 12x − 1 >

2x + 13x − 1

7. mgvavb Ki : x + y + z = a + b + c, ax + by + cz = a2 + b2 + c2, xa +

yb +

zc = 3 4

†hLv‡b, a, b, c ci¯•i Ak�b¨ I Amgvb aª“eK| A_ev, x + y − 4 ≤ 0 Ges 2x − y − 3 ≥ 0 AmgZvhyM‡ji mgvavb †m‡Ui †jLwPÎ AsKb Ki| 8. x Gi Dci Kx kZ© Av‡ivc Ki‡j (2x + 1)−1 + (2x + 1)−2 + (2x + 1)−3 + ... ... ... Abš� ¸‡YvËi avivi AmxgZK mgwó _vK‡e Ges †mB mgwó

wbY©q Ki| 4 9. G¨v‡cv‡jvwbqv‡mi Dccv`¨wU wjL Ges cÖgvY Ki| 6 A_ev, cÖgvY Ki †h, m�¶¥‡KvYx wÎfz‡Ri kxl© †_‡K wecixZ evûi Dci Aw¼Z j¤^ Zvi cv` wÎfz‡Ri †KvY‡K mgwØLwÛZ K‡i| 10. ∆ABC I ∆DEF m`„k wÎfzRØ‡qi D”PZv h_vµ‡g AM I DN n‡j cÖgvY Ki †h, AM : DN = AB : DE. 4 A_ev, ABC wÎfz‡Ri ∠A Gi mgwØLÛK BC †K D we›`y‡Z Ges ABC cwie„Ë‡K E we›`y‡Z †Q` K‡i‡Q| cÖgvY Ki †h, AD2 = AB.AC − BD.DC. 11. wÎfz‡Ri f‚wg, wkit‡KvY I Aci `yB evûi Aš�i †Ìqv Av‡Q| wÎfzRwU A¼b Ki| [A¼‡bi wPý I weeiY Avek¨K] 5 A_ev, Giƒc GKwU e„Ë A¼b Ki hv GKwU wbw`©ó e„Ë‡K GKwU wbw`©ó we›`y‡Z ¯•k© K‡i Ges e„‡Ëi ewnt¯’ †Kvb wbw`©ó we›`y w`‡q hvq|

[A¼‡bi wPý I weeiY Avek¨K] 12. †f±‡ii mvnv‡h¨ cÖgvY Ki †h, UªvwcwRqv‡gi KY©Ø‡qi ga¨we› yi ms‡hvRK †iLvsk mgvš�ivj evûØ‡qi mgvš�ivj Ges Zv‡ i we‡qvMd‡ji A‡a©K| 4 A_ev, †f±i c×wZ‡Z cÖgvY Ki †h, mvgvš�wi‡Ki KY©Øq ci¯•i‡K mgwØLwÛZ K‡i| 13. GKwU duvcv †jvnvi †Mvj‡Ki evB‡ii e¨vm 13 †m.wg. Ges †jvnvi †ea 2 †m.wg.| H †Mvj‡K eëüZ †jvnv w`‡q GKwU wb‡iU †MvjK ˆZwi Kiv

nj| Zvi e¨vm KZ n‡e? 4 A_ev, †KvYK AvKv‡ii GKwU Zuveyi D”PZv 7.50 wgUvi| GB Zuvey Øviv 3000 eM©wgUvi Rwg wNi‡Z PvB‡j wK cwigvY K¨vbfvm jvM‡e?

14. †h †Kvb wZbwU cÖ‡kœi DËi `vI : 4×3=12 (K) †iwWqvb †KvY Kv‡K e‡j? cÖgvY Ki †h, †iwWqvb †KvY GKwU aª“e †KvY|

(L) cÖgvY Ki †h, secA − secBtanB + tanA =

tanA − tanBsecA + secB

(M) sinθ + cosθ = 1 n‡j, †`LvI †h, sinθ − cosθ = ± 1

(N) gvb wbY©q Ki : sin2 π4 + sin2

3π4 + sin2

5π4 + sin2

7π4

(O) mgvavb Ki : 5cosec2θ − 7cotθ.cosecθ − 2 = 0 (0° < θ < 360°) 15. †Kvb KviLvbvi AbyaŸ© 50 eQi eq‡mi kÖwgK‡ì eq‡mi MYmsL¨v wb‡ekY wbæiƒc| Zv‡ì eq‡mi MvwYwZK Mo wbY©q Ki| 5

†kÖwY e¨vwß 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 MYmsL¨v 3 13 21 15 5 4 2

A_ev, wb‡Pi MYmsL¨v wb‡ekb mviYx n‡Z Mo eëavb wbY©q Ki : †kÖwY e¨vwß 51-60 61-70 71-80 81-90 91-100 MYmsL¨v 10 15 20 10 5


15. Sylhet Cadet College, Sylhet Subject code : 1 2 6


1. If for any finite sets A & B, show with 'Venn diagram' that, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). 4 Or, Among 50 persons, 35 can speak English, 25 can speak English & Bangla & every one can speak at least one of both

languages. How many can speak Bangla? How many can speak only Bangla? 2. Answer any two : 3 × 2 = 6 (i) Resolve into factors, 18x3 + 15x2 − x − 2.

(ii) Simplify : 1

(x + a) + 2x

(x2 + a2) + 4x3

(x4 + a4) + 8x7

(a8 − x8) .

(iii) If (ab + bc + ca) = abc

(a + b + c) then prove that, (a + b + c) = 3

a3 + b3 + c3

(iv) If (x − 2) is a factor of x4 − 5x3 + 7x2 − a. Then determine the value of 'a'. 3. By Mathematical Induction Formula, prove that (for all n ∈ ô) 4 (xn − yn) is divisible by (x − y) Or, By Mathematical Induction Formula, prove that (for all n ∈ ô)

a + ar + ar2 + ar3 + ... ... ... + arn−1 = a(arn − 1)

r − 1 , r ≠ 1

4. Simplify, a

32 + ab

ab − b3 − a

a − b 5

Or, If a3−x .b5x = a5+xb3x, then show that x logk⎝⎛⎠⎞b

a = logka.

5. If F : Ñ→ Ñ+, F(x) = x2, (i) Dom F & Range F = ? (ii) Is F one-one? justify. 4 Or, Draw the graph of S = {(x, y) : x2 + y2 − 2x − 4y − 11 = 0} & determine is S function or not?

6. Solve, 62x

x − 1 + 5x − 12x = 13 4

Or, Solve : 32x−2 − 5.3x−2 − 66 = 0 7. Solve : x2 − 2xy + 8y2 = 8, 3xy − 2y2 = 4 4

Or, Solve, x + y + z = a + b + c, ax + by + cz = a2 + b2 + c2 & xa +

yb +

zc = 3

(Where a, b & c are non-zero & mutually unequal constants.) 8. Impose a condition on x in which,

1

x + 1 + 1

(x + 1)2 + 1

(x + 1)3 + ... ... ... have the sum up to infinity. What is the sum? 4

9. State and prove that the P tolemy's theorem. 6 Or, Prove that, the ratios of the areas of two similar triangles is equal to the ratio of the squares on corresponding sides. 10. The medians of ∆ABC meet at G. Prove that AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2). 4 Or, If AD⊥BC from the vertex A of equilateral ∆ABC & its radius of circum-circle is R. Then prove that AB2 = 2R.AD. 11. Construct a triangle having given the base, the vertical angle and the sum of the other two sides. 5 Or, Construct a circle which touches a given straight line at a given point & passes through a point outside of it. [Description

& sign of construction are required] 12. If a , b, c are the position vectors of A, B & C respectively & if the point C divides AB in the ratio m t n internally, then

prove that the position vector of C is c = mb + nam + n . 4

Or, Prove with the help of Vectors that the straight line joining the midpoint of the diagonals of a trapezium is parallel to & half the difference of the parallel sides.

13. A spherical box just fits into a cylindrical box, the volume of the portion of the box unoccupied is 89 58 cm3, find the circumference

of the ball. 4 Or, The height of a right circular cone is 24cm & its volume is 1232m3. What is the whole surface area? 14. Answer any three: 4 × 3 = 12 (i) Define Radian. Prove that Radian is a constant angle. (ii) The angles of a triangle are in A.P and the largest angle is double the least. Express the angles in radins & degree.

(iii) If tanθ = 512 & cosθ negative, then

sinθ + cos(−θ)sec(−θ) + tanθ = ?

(iv) If acosA − bsinA = c; then prove that, asinA + bcosA = ± a2 + b2 − c2 . (v) Solve, 4(cos2θ + sin θ) = 5; where 0° ≤ θ ≤ 360°. 15. Determine the Mean in short-cut method. 5

Daily income 91 - 100 81 - 90 71 - 80 61 - 70 51 - 60 41 - 50 31 - 40 No. of people 10 15 08 12 20 13 07

Or, Determine the standard deviation. Class interval 51 - 55 56 – 60 61 - 65 66 - 70 71 - 75 Frequency 05 10 20 15 10


16. Rvjvjvev` K¨v›Ub‡g›U cvewjK ¯‹zj GÛ K‡jR welq †KvW : 1 2 6


1. hw` S = {x : x ∈ Ñ Ges x2 + 2 = 0} nq, Z‡e S Ges S′ = Ñ\S wbY©q Ki| 4 A_ev, †h †Kvb mvš� †mU A I B Gi Rb¨ cÖgvY Ki †h, n(A ∪ B) = n(A) + n(B) − n(A ∩ B) 2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3×2=6 (K) Drcv`‡K we‡k−lY Ki : 18x3 + 15x2 − x − 2

(L) hw` 1a3 +

1b3 +

1c3 =

3abc nq, Z‡e †`LvI †h, bc + ca + ab = 0 A_ev, a = b = c

(M) mij Ki : 1(1 + x) +

2(1 + x2) +

4(1 + x4) +

8(1 + x8) +

16(x16 − 1)

(N) AvswkK fMœvs‡k cÖKvk Ki : x3 + 2x2 + 1 x2 + 2x − 3

3. MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h mKj n ∈ ô Gi Rb¨ 13 + 23 + 33 + ... ... ... + n3 = n2(n + 1)2

4 4 A_ev, †`LvI †h, S = ô †hLv‡b S = {n : n ∈ ô Ges 5n − 2n ivwkwU 3 Øviv wefvR¨}

4. mij Ki : a32 + ab

ab − b3 − a

a − b 5

A_ev, hw` a3−x.b5x = a5+x.b3x nq, Z‡e †`LvI †h, xlogk(b/a) = logk a. 5. S = {(x, y) : x ∈ A, y ∈ A Ges y = x2} AšqwU‡K ZvwjKv c×wZ‡Z eY©bv Ki Ges †Wvg S I †iÄ S wbY©q Ki: †hLv‡b A = {−2, −1, 0, 1, 2} 4 A_ev, S = {(x, y) : x2 + y2 = 9 Ges y ≥ 0} Aš^‡qi †jLwPÎ A¼b Ki Ges Aš^qwU dvskb wKbv Zv †jLwPÎ †_‡K wbY©q Ki|

6. mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI : x(x − 4)x − 5 < 0 4

A_ev, mgvavb Ki : (1 + x)13 + (1 − x)

13 = 2

13

7. mgvavb Ki : x + yx − y +

x − yx + y =

103 , x2 − y2 = 3 4

A_ev, mgvavb Ki: x + y + z = a + b + c ax + by + cz = a2 + b2 + c2

xa +

yb +

zc = 3 [a, b, c, ci¯•i Amgvb Ak�b¨ aª“eK]


1(x + 1)2 +


9. G¨v‡cv‡jvwbqv‡mi Dccv`¨wU wee„wZ I cÖgvY Ki| 6 A_ev, cÖgvY Ki †h, e„‡Ë Aš�wj©wLZ †Kv‡bv PZyfz©‡Ri KY©Ø‡qi Aš�M©Z AvqZ‡¶Î H PZyfz©‡Ri wecixZ evûØ‡qi Aš�M©Z AvqZ‡¶‡Îi mgwói

mgvb| 10. ∆ABC Gi ∠C = GK mg‡KvY| C †_‡K AwZfz‡Ri Dci AswKZ j¤^ CD. cÖgvY Ki †h, CD2 = AD.BD. 4 A_ev, ∆ABC Gi AC I AB evûi Dci h_vµ‡g BE I CF j¤^| †`LvI †h, ∆ABC : ∆AEF = AB2 : AE2. 11. wÎfz‡Ri f‚wg, wkit‡KvY I Aci `yB evûi Aš�i †Ìqv Av‡Q| wÎfzRwU A¼b Ki| [AsK‡bi wPý I weeiY Avek¨K] 5 A_ev, Ggb GKwU e„Ë A¼b Ki hv GKwU wbw`©ó e„Ë‡K GKwU wbw`©ó we›`y‡Z ¯•k© K‡i Ges e„‡Ëi ewnt¯’ †Kvb wbw`©ó we›`y w`‡q hvq|

[A¼‡bi wPý I weeiY Avek¨K]

12. hw A, B, C we› yi Ae ’vb †f±i h_vµ‡g a, b, c nq Ges C we› y hw AB †iLvsk m t n Abycv‡Z Aš�we©f³ K‡i Z‡e † LvI †h, c = na + mbm + n | 4

A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, UªvwcwRqv‡gi KY©Ø‡qi ga¨we›`yi ms‡hvRK mij‡iLv mgvš�ivj evûØ‡qi mgvš�ivj Ges Zv‡ì we‡qvMd‡ji A‡a©K|

13. †Kv‡bv Nb‡Ki c„ôZ‡ji K‡Y©i ˆ`N© 8 2 †m.wg. n‡j, Gi K‡Y©i ˆ`N© I AvqZb wbY©q Ki| 4 A_ev, GKwU mge„Ëf‚wgK wmwjÛv‡ii eµZ‡ji †¶Îdj 160 eM© †m.wg. Ges AvqZb 240 Nb †m.wg.| Gi D”PZv KZ? 14. †h †Kvb wZbwU cÖ‡kœi DËi `vI :− 4 × 3 = 12 (K) c„w_exi e¨vmva© 6440 wK‡jvwgUvi n‡j, c„w_exi Dc‡ii †h `yBwU ¯’vb †K‡›`ª 32″ †KvY Drcbœ K‡i Zv‡ì ` �iZ¦ KZ?

(L) cÖgvY Ki †h, tanθ + secθ − 1tanθ − secθ + 1 =

sinθ + 1cosθ

(M) tanθ = 512 Ges cosθ FYvÍK n‡j †`LvI †h, sinθ + cos(−θ)

sec(−θ) + tanθ = 5126

(N) acosθ − bsinθ = c n‡j, †`LvI †h, asinθ + bcosθ = ± a2 + b2 − c2 (O) mgvavb Ki : 3 sinθ + cosθ = 2, hLb 0° < θ < 90° 15. wb‡Pi mviwY †_‡K Mo wbY©q Ki : 5

†kÖwYe¨vwß 11-20 21-30 31-40 41-50 51-60 61-70 71-80 MYmsL¨v 5 15 8 20 12 8 2

A_ev, wb‡Pi mvwiY †_‡K cwiwgZ eëavb wbY©q Ki : †kÖwYe¨wß 5-14 15-24 25-34 35-44 45-54 MYmsL¨v 10 20 30 40 50


17. nweMÄ miKvwi D”P we`¨vjq, nweMÄ welq †KvW : 1 2 6


1. hw` S = {x : x ∈ Ñ Ges x2 + 3 = 0} nq Z‡e, S Ges S′ wbY©q Ki| 4 A_ev, †h †Kvb mvš� †mU A I B Gi Rb cÖgvY Ki n (A ∪ B) = n(A) + n(B) − n(A ∩ B) 2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3×2=6 (K) hw` p(x) = x3 + 5x2 + 6x + 8 nq Ges p(x) †K x − a Ges x − b Øviv fvM Ki‡j GKB fvM‡kl _v‡K †hLv‡b a ≠ b Z‡e †`LvI †h, a2 + b2 + ab + 5a + 5b + 6 = 0 (L) Drcv`K we‡k−lY Ki : (a − b)3 + (b − c)3 + (c − a)3

(M) AvswkK fMœvs‡k cÖKvk Ki: x3 + 2x2 + 1

x2 + 2x − 3

3. hw m, n, k ∈ ô Ges m < n nq Z‡e †`LvI †h, m + k < n + k n‡e| 4 A_ev, MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h, mKj n ∈ ô Gi Rb¨ 1

1.3 + 1

3.5 + 1

5.7 + ...... + 1

(2n − 1) (2n + 1) = n

2n + 1

4. hw` x = (a + b) 13 + (a − b)

13 Ges a2 − c3 = b2 nq Z‡e †`LvI †h, x = (3cx + 2a)

13 5

A_ev, hw` p = loga(bc), q = logb(ca), r = logc(ab) nq Z‡e †`LvI †h, 1p + 1 +

1q + 1 +

1r + 1 = 1

5. hw` A = {− 2, − 1, 0, 1, 2} nq Z‡e S = {(x, y) : x ∈ A, y ∈ A Ges y2 = x} †K ZvwjKv c×wZ‡Z eY©bv Ki Ges S Gi †Wv‡gb I †iÄ wbY©q Ki| 4

A_ev, S Aš^‡qi †jL A¼b Ki: S = { }(x , y) : x2

4 + y2

25 = 1

6. mgvavb Ki: (1 + x) 13 + (1 − x)

13 = 2

13 4

A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI: 3x + 12x − 1 >

2x + 13x − 1

7. mgvavb Ki: x + 4y = 1, y +

4x = 25 4

A_ev, x + y + z = a + b + c [a, b, c ci¯•i Ak�b¨ aª“eK] ax + by + cz = a2 + b2 + c2

xa +

yb +

zc = 3

8. x Gi Ici Kx kZ© Av‡ivc Ki‡j, 1x + 1 +

1(x + 1)2 +

1(x + 1)3 + ... ... ... Abš� avivi (AmxgZK) mgwó hw` _v‡K Z‡e Zv D‡j−L Ki Ges †mB

kZ©vax‡b mgwó wbY©q Ki| 4 9. wÎfz‡Ri †h †Kvb evûi mgvš�ivj mij‡iLv H wÎfz‡Ri Aci evûØq‡K ev Zv‡ì ewa©ZvskØq‡K mgvb Abycv‡Z wef³ K‡i| 6 A_ev, U‡jwgi Dccv`¨wU wee„Z Ki I cÖgvY Ki| 10. †Kv‡bv ABC wÎfz‡Ri ga¨gvÎq G we›`y‡Z wgwjZ n‡j cÖgvY Ki †h, AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2) 5 A_ev, ABC wÎfz‡Ri AC I AB evûi Dci h_vµ‡g BE I CF j¤^| †`LvI †h, ∆ABC : ∆AEF = AB2 : AE2 11. †Kv‡bv wbw`©ó †iLvsk‡K Giƒcfv‡e Aš�twe©f³ Ki‡Z n‡e †hb mgMÖ †iLvs‡ki I GKwU As‡ki Aš�M©Z AvqZ‡¶‡Îi †¶Îdj Ab¨ AskwUi

Ici eM©‡¶‡Îi †¶Îd‡ji mgvb| 4 A_ev, Ggb GKwU e„Ë A¼b Ki hv GKwU wbw`©ó e„Ë‡K Gi †Kvb wbw`©ó we›`y‡Z Ges Aci GKwU e„Ë‡K †Kv‡bv we›`y‡Z ¯•k© K‡i| 12. †f±i c×wZ‡Z cÖgvY Ki †h, †Kv‡bv PZzfz©‡Ri mwbœwnZ evû¸‡jvi ga¨we›`yi ms‡hvRK †iLvmg�n GKwU mvgvš�wiK Drcbœ K‡i| 4 A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, wÎfz‡Ri GK evûi ga¨we›`y †_‡K Aw¼Z Aci evûi mgvš�ivj †iLv Z…Zxq evûi ga¨we›`yMvgx| 13. †Kv‡bv AvqZ‡¶‡Îi ˆ`N© 10 †m.wg. I cÖ ’ 3 †m.wg.| G‡K e„nËi evûi PZzw`©‡K †Nviv‡j †h Nbe¯‘ Drcbœ n‡e Zvi c„ôZ‡ji †¶Îdj I

AvqZb wbY©q Ki| 4 A_ev, 44 †m.wg. cwiwa wewkó GKwU †MvjK AvK…wZi ej GKwU NbK AvK…wZi ev‡· wVKfv‡e Gu‡U hvq| ev·wUi AbwaK…Z As‡ki AvqZb wbY©q Ki| 14. †h †Kvb wZbwU cÖ‡kœi DËi `vI : 4×3=12

(K) cÖgvY Ki †h, sin θ − cos θ + 1sin θ + cos θ − 1 =

1 + sinθcosθ

(L) tanθ = 512 Ges cos θ FYvÍK n‡j †`LvI †h, sin θ + cos (− θ)

sec (− θ) + tan θ = 5126

(M) gvb wbY©q Ki : sin2 π7 + sin2

5π14 + sin2

8π7 + sin2

9π14

(N) ∆ABC Gi †¶‡Î †`LvI †h, sin A + B

2 + tan A + B

2 = cos C2 ( )1 + cosec

C2

(O) 540 wK.wg. ` �‡i GKwU we›`y‡Z †Kvb cvnvo 7′ †KvY Drcbœ Ki‡j cvnvowUi D”PZv KZ? 15. wb‡æv³ Dcv‡Ëi Rb¨ AvqZ‡jL AuvK: 5

cÖvß b¤î 1 − 10 11 − 20 21 − 30 31 − 40 41 − 50 wk¶v_©xi msL¨v 5 10 25 35 15

A_ev, wb‡æi MYmsL¨v wb‡ekb mviwY †_‡K cwiwgZ eëavb (msw¶ß c×wZ) wbY©q Ki| †kÖwY 5 − 14 15 − 24 25 − 34 35 − 44 45 − 54 55 − 64 65 − 74 75 − 84 MYmsL¨v 10 20 30 40 50 60 70 80


18. h‡kvi wRjv ¯‹zj, h‡kvi welq †KvW : 1 2 6


1. hw` A = {1, 2, 3} Ges B = {3, 4} nq, Z‡e †`LvI †h, P(A) ∩ P(B) = P(A ∩ B) 4 A_ev, P I Q mvš� †mU n‡j, †`LvI †h, n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q)

2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3 × 2 = 6 (K) x4 − 5x3 + 7x2 − 2a eûc`xi GKwU Drcv`K x − 2 n‡j †`LvI †h a = 2. (L) Drcv`‡K we‡k−lY Ki : a6 + 18a3 + 125

(M) mij Ki : a2 + bc(a − b) (a − c) +

b2 + ac(b − c) (b − a) +

c2 + ab(c − a) (c − b)

3. †`LvI †h, S = ô †hLv‡b S = {n : n ∈ ô Ges n(n + 1) ivwkwU 2 Øviv wefvR¨}| 4 A_ev, MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h mKj n ∈ ô Gi Rb¨ 13 + 23 + 33 + .......... + n3 =

n2(n + 1)2

4

4. hw` b = 2 + 223 + 2

13 nq, Z‡e †`LvI †h, b3 + 6b2 + 6b − 2 = 0 5

A_ev, hw` a3−x.b5x = a5+x.b2x nq, Z‡e †`LvI †h, xlogp( )ab = logpa.

5. F(x) = (x − 2)2 Øviv ewY©Z dvsk‡bi †Wv‡gb wbY©q Ki Ges dvskbwU GK-GK wKbv Zv wba©viY Ki| 4 A_ev, S = {(x, y) : y = − 4x2} Aš^‡qi †jLwPÎ Aw¼Z Ki Ges Aš^qwU dvskb wKbv Zv †jL †_‡K wbY©q Ki| 6. mgvavb i : 32x−2 − 5.3x−2 = 66 4

A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI : x(x − 4)x − 5 < 0

7. mgvavb Ki : x + 4y = 1, y +

4x = 25 4

A_ev, x − 23 =

y + 14 =

z + 45 , 2x + 3y − 4z = 15


1(x + 1)2 +

1(x + 1)3 + ............... Abš� avivi (AmxgZK) mgwó _vK‡e Ges †mB k‡Z© avivwUi mgwó

wbY©q Ki| 4 9. cÖgvY Ki †h, `yBwU m`„k wÎfzR‡¶‡Îi †¶ÎdjØ‡qi AbycvZ Zv‡ì †h †Kvb `yB Abyiƒc evûi Dci AswKZ eM©‡¶‡Îi †¶ÎdjØ‡qi Abycv‡Zi

mgvb| 6 A_ev, cÖgvY Ki †h, e„‡Ë Aš�wj©wLZ †Kvb PZyfz©‡Ri KY©Ø‡qi Aš�M©Z AvqZ‡¶Î H PZyfz©‡Ri wecixZ evûØ‡qi Aš�M©Z AvqZ‡¶‡Îi mgwói

mgvb| 10. ABC mgwØevû wÎfz‡Ri f‚wg BC Gi Dci P †h †Kvb we›`y n‡j †`LvI †h, AB2 − AP2 = BP.PC. 4 A_ev, ABC wÎfz‡Ri AC I AB evûi Dci h_vµ‡g BE I CF j¤^| †`LvI †h, ∆ABC : ∆AEF = AB2 : AE2. 11. wÎfz‡Ri D”PZv, f‚wg mgwØLÛK ga¨gv Ges f‚wg msjMœ GKwU †KvY †Ìqv Av‡Q| wÎfzRwU A¼b Ki| [A¼‡bi wPý I weeiY Avek¨K] 5 A_ev, wfbœ wfbœ e¨vmva©wewkó Giƒc wZbwU e„Ë AuvK †hb Zviv ci¯•i‡K ewnt¯•k© K‡i| [AsK‡bi wPý I weeiY Avek¨K] 12. †f±‡ii mvnv‡h¨ cÖgvY Ki †h, wÎfz‡Ri GK evûi ga¨we›`y †_‡K Aw¼Z Aci evûi mgvš�ivj †iLv Z…Zxq evûi ga¨we›`yMvgx| 4 A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, PZyfz©‡Ri ga¨we›`y ‡jv ch©vqµ‡g †hvM Ki‡j Zv GKwU mvgvš�wiK nq| 13. GKwU mge„Ëf‚wgK wmwjÛv‡ii eµZ‡ji †¶Îdj 100 eM© †m.wg. Ges AvqZb 150 Nb †m.wg.| Gi f‚wgi e¨vmva© I D”PZv wbY©q Ki| 4 A_ev, 44 †m.wg. cwiwa wewkó GKwU †MvjK AvK…wZi ej GKwU NbK AvK…wZi ev‡· wVKfv‡e Gu‡U hvq| ev·wUi AbwaK…Z As‡ki

AvqZb wbY©q Ki|

14. †h †Kvb wZbwU cÖ‡kœi DËi `vI :− 4 × 3 = 12 (K) GKwU wÎfz‡Ri †KvY¸‡jv mgvš�i †kÖwYfz³ Ges eüËi †KvYwU ¶z`ªZi †Kv‡Yi wØ¸Y| †KvY¸‡jv‡K †iwWqv‡b cÖKvk Ki| (L) hw` sinθ + cosθ = 1 nq, Z‡e †`LvI †h, sinθ − cosθ = ± 1

(M) tanθ + secθ = x n‡j, †`LvI †h, sinθ = x2 − 1x2 + 1

(N) mgvavb Ki : 3(sec2θ + tan2θ) = 5, hLb 0° ≤ θ ≤ 360°

(O) ABC wÎfz‡Ri †¶‡Î †`LvI †h, sin A + B

2 + tan A + B

2 = cos C2 ( )1 + cosec

C2

15. wb‡æi MYmsL¨v wb‡ekY mviYx †_‡K msw¶ß c×wZ‡Z MvwYwZK Mo wbY©q Ki| 5 cÖvß b¤î 25 − 34 35 − 44 45 −54 55 − 64 65 −74 75 − 84 85 − 94 wk¶v_x© msL¨v 5 10 15 20 30 16 4

A_ev, wb‡æi MYmsL¨v wb‡ekY mviwY †_‡K cwiwgZ eëavb wbY©q Ki| †kÖwYe¨vwß 15 −19 20 − 24 25 − 29 30 − 34 35 − 39 MYmsL¨v 7 13 19 15 6


19. Lyjbv wRjv ¯‹zj, Lyjbv welq †KvW : 1 2 6


1. hw` A = {3, 4}, B = {4, 6} nq, Z‡e †`LvI †h, P(A) ∩ P(B) = P(A ∩ B) 4

A_ev, †h †Kv‡bv mvš� †mU A Ges B Gi Rb cÖgvY Ki †h, n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

2. †h †Kvb `yBwU cÖ‡kœi DËi `vI : 3 × 2 = 6

(K) Drcv`‡K we‡k−lY Ki: y3 − 3y2 + 3y − 9

(L) x2 − yza =

y2 − zxb =

z2 − xyc ∫ 0 nq, Z‡e †`LvI †h, (a + b + c) (x + y + z) = ax + by + cz.

(M) mij Ki : (a + b)2 − ab(b − c) (a − c) +

(b + c)2 − bc(c − a) ( b − a) +

(c + a)2 − ca(a − b) (c − b)

3. † LvI †h, S = ô †hLv‡b, S = {n : n ∈ ô Ges 5n − 3n ivwkwU 2 Øviv wefvR } 4 A_ev, MvwYwZK Av‡ivn c×wZi mvnv‡h¨ †`LvI †h, mKj n ∈ ô Gi Rb¨ 12 + 22 + 32 + ........ + n2 =

n(n + 1) (2n +1)6

4. hw` a2 + 2 = 323 + 3

−23 Ges a ≥ 0 nq, Z‡e †`LvI †h, 3a3 + 9a = 8 5

A_ev, hw` a3 − x b5x = a5 +x b3x nq, Z‡e †`LvI †h, xlogk(ba−1) = logka.

5. F(x) = (x − 2)2 Gi †Wv‡gb wbY©q Ki Ges dvskbwU GK-GK wKbv Zv wba©viY Ki| 4

A_ev, S = {(x, y) : y = 3 − 4x − 2x2} Aš^‡qi †jLwPÎ AsKb Ki| 6. mgvavb Ki : 41 +x + 41−x = 10 4 A_ev, mgvavb Ki Ges mgvavb †mU msL¨v‡iLvq †`LvI: x(x − 1) (x + 2) > 0

7. mgvavb Ki : y + 3 = 4x , x − 4 =

53y 4

A_ev, 3x − 2y ≤ 12 Gi mgvavb †m‡Ui †jLwPÎ AsKb Ki|

8. 8..1

.2 †K g�j`xq fMœvs‡k cÖKvk Ki| 4

9. cÖgvY Ki †h, `yBwU m`„k wÎfzR‡¶‡Îi †¶ÎdjØ‡qi AbycvZ Zv‡ì †h †Kv‡bv `yB Abyiƒc evûi Dci AswKZ eM©‡¶‡Îi †¶ÎdjØ‡qi Abycv‡Zi mgvb| 6

A_ev, e„‡Ë Aš� ©wjwLZ †Kv‡bv PZzfz©‡Ri KY©Ø‡qi Aš�M©Z AvqZ‡¶Î H PZzfz©‡Ri wecixZ evûØ‡qi Aš�M©Z AvqZ‡¶‡Îi mgwói mgvb cÖgvY Ki|

10. ABC wÎfz‡Ri ga¨gvÎvq G we›`y‡Z wgwjZ n‡j, cÖgvY Ki †h, GA2 + GB2 + GC2 = 13 (AB2 + BC2 + CA2) 4

A_ev, ∆ABC Gi ∠C mg‡KvY| C n‡Z AwZf‚‡Ri Dci AswKZ j¤^ CD n‡j, cÖgvY Ki †h, CD2 = AD.BD.

11. wÎfz‡Ri f‚wg, wkit‡KvY I Aci evûØ‡qi mgwó †Ìqv Av‡Q| wÎfzRwU AvuK| (AsK‡bi wPý I weeiY Avek¨K|) 5

A_ev, Giƒc GKwU e„Ë AsKb Ki‡Z n‡e hv GKwU wbw`©ó mij †iLv‡K GKwU wbw`©ó we›`y‡Z ¯•k© K‡i Ges H †iLvi ewnt¯’ †KvY wbw`©ó we›`y w`‡q hvq| (AsK‡bi wPý I weeiY Avek¨K)|

12. †f±‡ii mvnv‡h¨ cÖgvY Ki †h, †Kv‡bv PZzfz©‡Ri KY©Øq ci¯•i‡K mgwØLwÛZ Ki‡j Zv GKwU mvgvš�wiK| 4 A_ev, †f±‡ii mvnv‡h¨ cÖgvY Ki †h, wÎfz‡Ri †h‡Kv‡bv `yB evûi ga¨we›`yØ‡qi ms‡hvRK †iLvsk H wÎfz‡Ri Z…Zxq evûi mgvš�ivj I Zvi A‡a©K| 13. 44 †m.wg. cwiwa wewkó GKwU †MvjK AvK…wZi ej GKwU NbK AvK…wZi ev‡· wVKfv‡e Gu‡U hvq| ev·wUi AbwaK…Z As‡ki AvqZb wbY©q Ki| 4

A_ev, 12, 16, 2r ‡m.wg. e vm wewkó wZbwU KwVb Kv‡Pi ej Mwj‡q 9 †m.wg. e vmva© wewkó GKwU KwVb †Mvj‡K cwiYZ Kiv nj| r Gi gvb wbY©q Ki|

14. †h †Kvb wZbwU cÖ‡kœi DËi `vI :− 4 × 3 = 12

(K) GKwU evjK mvB‡K‡j P‡o e„ËvKvi c‡_ 2 †m‡K‡Û GKwU e„ËPvc AwZµg K‡i| hw` PvcwU †K‡›`ª 28° †KvY Drcbœ K‡i Ges e„‡Ëi e¨vm 180 wgUvi nq, Z‡e evjKwUi MwZ‡eM wbY©q Ki|

(L) 7sin2A + 3cos2A = 4 n‡j, †`LvI †h, cot2A = 3

(M) hw` sinθ − cosθ = 1 nq, Z‡e †`LvI †h, sinθ + cosθ = ± 1

(N) gvb wbY©q Ki : sin2 π7 + sin2 5π

14 +sin2 8π7 + sin2 9π

14

(O) mgvavb Ki : sinx + cosx = 2 (†hLv‡b 0° < x < 90°) 15. ‡Kv‡bv †kÖwYi 50 Rb wk¶v_©xi MwYZ wel‡q cÖvß b¤^‡ii MYmsL¨v wb‡ekb mviwY †Ìqv n‡jv| cÖ Ë Dcv‡Ëi AvqZ‡jL AsKb Ki| 5

cÖvß b¤î 31−40 41−50 51−60 61−70 71−80 81−90 91−100

wk¶v_©xi msL¨v 6 8 10 12 5 7 2

A_ev, 60 Rb wk¶v_©xi MwY‡Z cÖvß b¤^‡ii mviwY †Ìqv n‡jv| cÖvß b¤^‡ii cwiwgZ eëavb msw¶ß c×wZ‡Z wbY©q Ki| b¤î 51−60 61 − 70 71 − 80 81 − 90 91 −100

wk¶v_©xi msL¨v 10 15 20 10 5


20. Barisal Cadet College, Barisal Subject code : 1 2 6


1. Prove that the S = {1, 4, 9, 16, 25, ...........} is an infinite set. 4 Or, For any subsets A, B of the universal set, U. Prove that (A ∪ B)′ = A′ ∩ B′. 2. Answer any two of the following questions: 3 × 2 = 6 (a) Resolve into factors: 18x3 + 15x2 − x − 2 (b) If (a + b + c) (ab + bc + ca) = abc, then prove that (a + b + c)3 = a.3 + b3 + c3.

(c) Simplify: 1

1 + x + 2

1 + x2 + 4

1 + x4 + 8

1 + x8 + 16

x16 − 1

3. Using the method of mathematical induction show that, for all n ∈ ô, 1

1.3 + 1

3.5 + 1

5.7 + ......... + 1

(2n − 1) (2n + 1) = n

2n + 1 4 Or, Using the method of mathematical induction show that for all n ∈ ô, a + ar + ar2 + ar3 + ........... + arn−1 =

a(rn − 1)r − 1

4. If x = (a + b)13 + (a − b)

13 and a2 − b2 = c3 then show that x3 − 3cx − 2a = 0 5

Or, If xya−1 = p, xyb−1 = q, xyc−1 = r, then show that (b − c) logkp + (c − a)logkq + (a − b)logkr = 0 5. Find the domain of the function, F(x) = (x − 1)2 and verify the function is one-one or not? 4 Or, Sketch the graph of the relation S = {(x, y) : y2 = 9x} and determine from the graph whether the graph is a function or not. 6. Solve: 3(9x − 4.3x−1) + 1 = 0 4

Or, Solve: 2x + 4x − 3 ≤

x + 3x − 1

7. Solve: x + yx − y +

x − yx + y =

103 , x2 − y2 = 3 4

Or, Solve: x + y + z = a + b + c, ax + by + cz = a2 + b2 + c2, xa +

yb +

cz = 3

8. Find the sum (if exists) of the infinite series : 12 + 4 + 43 +

49 + ................ 4

9. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on corresponding sides. 6 Or, Prove that in an acute angled triangle the perpendiculars drawn from the vertices to opposite side bisect the angle of the

pedal triangle. 10. The medians of a ∆ABC meet at G. Prove that AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2) 4 Or, ABC is an isosceles triangle is which AB = AC.AD is perpendicular on the side BC and R is the circum-radius of the

∆ABC. Prove that 2R.AD = AB2. 11. Divide a given line segment internally and externally in a given ratio. (Description and denotation of construction are

necessary) 5 Or, Draw a circle which touches a given straight line at a given point in it and passes through another given point outside

that line. (Denotation and description of construction are necessary) 12. If a, b, c are the position vectors of A, B, C respectively and if the point C divides AB in the ratio m t n internally, then

prove that c = mb + nam + n 4

Or, Prove with the help of vectors that if the diagonals of a quadrilateral bisect each other, then it is a parallelogram. 13. Find the length of the diagonal and the volume of the cube of which the length of the diagonal of the cube of which the

length of the diagonal of a face in 8 2 cm. 4

Or, A spherical box is tightly adjusted with a cylindrical box. If the volume of the unoccupied place C be 89 58 cc. then

determine the circumference of the spherical ball. 14. Answer any three of the following questions : 4 × 3 = 12 (a) The angle of a triangle are in A.P. and the largest angle is double the least. Express the angles in radians.

(b) Prove that : tanθ + secθ − 1tanθ − secθ + 1 =

sinθ + 1cosθ

(c) If sinθ + cosθ = 1, then show that sinθ − cosθ = ± 1 (d) If sinθ + tanθ = m, tanθ − sinθ = n, then prove that, m2 − n2 = 4 mn (e) Solve: cosθ + sinθ = 2 , (0° < θ < 90°) 15. Find the arithmetic mean from the following frequency

distribution table by shortcut method: 5 Class 1 2 3 4 5 6 7 8 9 10 Frequency 5 20 30 40 50 35 21 12 10 8

Or, The marks obtained in Mathematics by 60 students are given below. Find the mean deviation of the marks obtained. Marks obtained 51-60 61-70 71-80 81-90 91-100 Number of students 10 15 20 10 5

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