Tiêu đề Polynomial Varsity Practice Sheet Solution.pdf ✅

eûc`x I eûc`x mgxKiY  Varsity Practice Sheet Solution 1 04 eûc`x I eûc`x mgxKiY Polynomial and Polynomial Equation weMZ mv‡j DU-G Avmv cÖkœvejx 1. k Gi †Kvb gv‡bi Rb ̈ 2x2 – (k + 1)x + k = 0 Gi GKwU g~j Aci g~‡ji wecix‡Zi wZb ̧‡Yi mgvb n‡e? [DU 23-24] 3 4 5 6 DËi: 6 e ̈vL ̈v: GKwU g~j  n‡j Aci g~j 3   . 3  = k 2  k = 6 2. 1 x + a – bx = 0 mgxKi‡Yi g~jØq mgvb n‡j †KvbwU mwVK? [DU 21-22] a 2 – 4b = 0 b 2 – 4a = 0 b 2 + 4a = 0 a 2 + 4b = 0 DËi: a 2 + 4b = 0 e ̈vL ̈v: 1 x + a – bx = 0  1 + ax – bx2 = 0  bx2 – ax – 1 = 0 kZ©vbymv‡i, D = 0  (– a)2 – 4  b  (– 1) = 0  a 2 + 4b = 0 3. x 2 – 2x + 1 = 0 mgxKiYwUi g~j؇qi wÎNvZ Gi mgwó n‡jv- [DU 20-21; Agri. Guccho 20-21] – 3 3 – 2 2 DËi: 2 e ̈vL ̈v: g~jØq  I  n‡j,  +  = 2 Ges  = 1   3 +  3 = ( + ) 3 – 3( + ) = 23 – 3  1  2 = 2 4. x 2 – 5x – 1 = 0 mgxKi‡Yi g~jØq n‡Z 2 Kg g~jwewkó mgxKiY n‡jvÑ [DU 18-19; RU 22-23] x 2 + 2x + 3 = 0 x 2 – x – 7 = 0 x 2 – 5x + 6 = 0 6x2 – 5x + 1 = 0 DËi: x 2 – x – 7 = 0 e ̈vL ̈v: cÖ`Ë mgxKi‡Y x Gi e`‡j x + 2 ewm‡q, (x + 2)2 – 5(x + 2) – 1 = 0  x 2 + 4x + 4 – 5x – 10 – 1 = 0  x 2 – x – 7 = 0 5. f(x) = 1 + x3 eμ‡iLvwUi mv‡_ x A‡ÿi †Q`we›`yi msL ̈vÑ [DU 18-19] 0 1 2 3 DËi: 1 e ̈vL ̈v: x A‡ÿ y = 0  0 = 1 + x3  x 3 = – 1  x 3 = (– 1)3      x – 1 3 = 1  x – 1 = 3 1  x – 1 = 1, ,  2  x = – 1, – , –  2 GLv‡b, – 1 GKwU ev ̄Íe g~j wKš‘ – , –  2 Aev ̄Íe g~j|  †Q`we›`y msL ̈v 1wU| 6. 3x3 – 1 = 0 Gi g~j ̧wj , ,  n‡j,  3 +  3 +  3 Gi gvbÑ [DU 16-17] – 1 0 1 3 1 DËi: 1 e ̈vL ̈v: 3x3 – 1 = 0  x 3 = 1 3  x =     1 3 1 3 ,     1 3 1 3  ,     1 3 1 3  2   3 +  3 +  3 = 1 3 + 1 3  3 + 1 3  6 = 1 3 + 1 3 + 1 3 = 1
2  Higher Math 2nd Paper Chapter-4 7. 2x2 – 7x + 5 = 0 mgxKi‡Yi g~jØq  Ges ; x2 – 4x + 3 = 0 mgxKi‡Yi g~jØq  Ges  n‡j, ( + ) : ( – ) = ? [DU 15-16; RU 09-10] 6 : 5 5 : 6 11 : 1 1 : 6 DËi: 11 : 1 e ̈vL ̈v:  = 5 2 Ges  = 3   = 5 2  3 = 5 6    = 5 6    = 6 5   +   –  = 6 + 5 6 – 5 [†hvRb-we‡qvRb K‡i]  ( + ) : ( – ) = 11 : 1 8. 3x2 – kx + 4 = 0 mgxKiYwUi GKwU g~j AciwUi 3 ̧Y n‡j, k Gi gvbÑ [DU 14-15] 8 – 8 8  8 DËi:  8 e ̈vL ̈v: awi, mgxKiYwUi g~jØq , 3  .3 = 4 3   2 = 4 9   =  2 3 Ges  + 3 = k 3  4       2 3 = k 3  k = 4       2 3  3  k =  8 9. ev ̄Íe mnMwewkó GKwU wØNvZ mgxKiY MVb Ki‡Z n‡e hvi GKwU g~j – 1 + – 5 n‡eÑ [DU 13-14; RU 13-14; CU 22-23] x 2 + x + 6 = 0 x 2 + 6 = 0 x 2 + 2x + 6 = 0 x 2 – 2x + 6 = 0 DËi: x 2 + 2x + 6 = 0 e ̈vL ̈v: mgxKi‡Yi GKwU g~j, x = – 1 + – 5  x + 1 = – 5  (x + 1)2 = – 5  x 2 + 2x + 1 = – 5  x 2 + 2x + 6 = 0 10. x 2 – 5x + c = 0 mgxKi‡Yi GKwU g~j 4 n‡j, Aci g~jwU KZ? [DU 08-09; JU 14-15; JnU 11-12;] 4 1 – 4 – 1 DËi: 1 e ̈vL ̈v: awi, Aci g~j  g~j؇qi †hvMdj =  + 4 = 5   = 1 weMZ mv‡j GST-G Avmv cÖkœvejx 1. x 2 – 11x + a = 0 I x 2 – 14x + 2a = 0 mgxKiY؇qi GKwU mvaviY g~j _vK‡j, a = ? [GST 23-24] 0, 24 0, – 24 1, – 1 – 2, 1 DËi: 0, 24 e ̈vL ̈v:  2 – 11 + a = 0  2 – 14 + 2a = 0 3 – a = 0 (we‡qvM K‡i)   = a 3  a 2 9 – 11  a 3 + a = 0  a 2 – 33a + 9a = 0  a 2 – 24a = 0  a = 0, 24 Shortcut: a1x 2 + b1x + c1 = 0 Ges a2x 2 + b2x + c2 = 0 mgxKiY؇qi,  GKwU mvaviY g~j _vK‡j, (a1b2 – a2b1)(b1c2 – b2c1) = (c1a2 – c2a1) 2  `ywU g~jB mvaviY n‡j, a1 a2 = b1 b2 = c1 c2 2. hw` x 2 – bx + c = 0 Gi g~jØq μwgK c~Y©msL ̈v nq, Z‡e b 2 – 4c Gi gvb KZ? [GST 23-24] 1 2 3 4 DËi: 1 e ̈vL ̈v:  +  = b,  = c   –  =  1  ( – ) 2 = 1  ( + ) 2 – 4 = 1  b 2 – 4c = 1 3. a Gi †Kvb †Wv‡g‡bi Rb ̈ x 2 + ax + 3 = 0 Gi g~jØq ev ̄Íe I Amgvb n‡e? [GST 22-23] (– 2 3 2 3) (–  – 2 3) (–  – 2 3)  (2 3 ) (2 3 ) DËi: (–  – 2 3)  (2 3 )
eûc`x I eûc`x mgxKiY  Varsity Practice Sheet Solution 3 e ̈vL ̈v: kZ©vbymv‡i, D  0  a 2 – 4  1  3  0  a 2 – 12  0  a 2 – (2 3) 2 > 0  (a + 2 3)(a – 2 3)  0 0 – 2 3 2 3  a  (–  – 2 3)  (2 3 ) 4. †Kvb mgxKi‡Yi GKwU g~j 2 + i 3 ? [GST 21-22] x 2 + 4x – 7 = 0 x 2 – 3x + 2 = 0 x 2 – 4x + 7 = 0 x 2 – 4x – 7 = 0 DËi: x 2 – 4x + 7 = 0 e ̈vL ̈v: mgxKi‡Yi GKwU g~j, x = 2 + i 3  x – 2 = i 3  (x – 2)2 = 3i2  x 2 – 4x + 4 = – 3  x 2 – 4x + 7 = 0 5. hw` [a, b] e ̈ewa‡Z f(x) GKwU Awew”Qbœ dvskb nq, †hLv‡b f(a)f(b)  0, Z‡e D3 e ̈ewa‡Z f(x) = 0 mgxKi‡Yi ev ̄Íe g~j _vK‡e- [GST 21-22] †Rvo msL ̈K gvÎ 2wU gvÎ 1wU we‡Rvo msL ̈K DËi: †Rvo msL ̈K e ̈vL ̈v: f(a)f(b) > 0 n‡e hw` f(a) > 0, f(b) > 0 nq A_ev f(a) < 0, f(b) < 0 nq| f(x) = 0 mgxKi‡Yi ev ̄Íe g~j _vK‡e hw` dvskbwU x Aÿ‡K †Q` K‡i| X X O Y (a, f(a)) (b, f(b)) (a, f(a)) (b, f(b)) wPÎ-I wPÎ-II O GLb y = f(x) dvsk‡bi [a, b] e ̈ewa‡Z f(a), f(b) > 0 Ges ev ̄Íe g~j m¤úbœ nIqvi Rb ̈ dvskb‡K (a, f(a)) we›`y n‡Z x Aÿ‡K †Q` K‡i y A‡ÿi FYvZ¥K w`‡K Avm‡Z n‡e Ges cybivq x Aÿ‡K †Q` K‡i y A‡ÿi abvZ¥K w`‡K (b, f(b)) we›`y‡Z †h‡Z n‡e (wPÎ-I)| Avevi, y = f(x) dvsk‡bi [a, b] e ̈ewa‡Z f(a), f(b) < 0 Ges ev ̄Íe g~j m¤úbœ nIqvi Rb ̈ dvskb‡K (a, f(a)) we›`y n‡Z y A‡ÿi abvZ¥K w`‡K Avm‡Z n‡e Ges cybivq y A‡ÿi FYvZ¥K w`‡K (b, f(b)) we›`y‡Z †h‡Z n‡e (wPÎ-II)| d‡j dvskbwU‡K cÖwZ‡ÿ‡Î †RvomsL ̈K evi x Aÿ‡K †Q` Ki‡Z n‡”Q|  D3 mgxKi‡Yi †Rvo msL ̈K ev ̄Íe g~j _vK‡e| 6. k Gi gvb KZ n‡j, (k + 1)x2 + (k + 1)x + 1 = 0 mgxKi‡Yi g~j ̧wj KvíwbK n‡e? [GST 21-22] – 1  k  3 – 3  k  1 1  k  3 1  k  3 DËi: – 1  k  3 e ̈vL ̈v: kZ©vbymv‡i, D  0  (k + 1)2 – 4(k + 1)  1  0  k 2 + 2k + 1 – 4k – 4  0  k 2 – 2k – 3  0  k 2 – 3k + k – 3  0  k(k – 3) + 1(k – 3)  0  (k – 3)(k + 1)  0 – 1 0 3  – 1  k  3 7. †Kvb k‡Z© x 3 – mx2 + nx + r = 0 mgxKi‡Yi `yBwU g~‡ji mgwó k~b ̈ n‡e? [GST 20-21] mn – r = 0 mn + r = 0 mr + n = 0 mr – n = 0 DËi: mn + r = 0 e ̈vL ̈v: awi, g~jÎq , ,  cÖkœg‡Z,  +  = 0 GLb,  +  +  = m  0 +  = m   = m Ges,  +  +  = n   + ( + ) = n   + .0 = n   = n Avevi,  = – r  n.m = – r  mn + r = 0 weMZ mv‡j Agri.-G Avmv cÖkœvejx 1. x 2 – 2x + 1 = 0 mgxKiYwUi g~j؇qi wÎNvZ Gi mgwó n‡jv- [Agri. Guccho 20-21; DU 20-21] – 3 3 – 2 2 DËi: 2 e ̈vL ̈v: g~jØq  I  n‡j,  +  = 2 Ges  = 1   3 +  3 = ( + ) 3 – 3( + ) = 23 – 3  1  2 = 2 2. hw` x 2 – 5x + k = 0 mgxKiYwUi GKwU g~j 4 nq, Zvn‡j k Gi gvb Ges Ab ̈ g~jwU KZ? [Agri Guccho 19-20; CU 21-22] 0, 0 4, 1 – 4, – 1 4, – 1 DËi: 4, 1
4  Higher Math 2nd Paper Chapter-4 e ̈vL ̈v: awi, Aci g~j  g~j؇qi †hvMdj =  + 4 = 5   = 1 g~j؇qi ̧Ydj =   4 = k  1  4 = k  k = 4 3. †Kvb k‡Z© ax2 + bx + c ivwkwU GKwU c~Y©eM© n‡e? [Agri. Guccho 19-20] 4ac = b2 4ac  b 2 4ac  b 2 ac = b DËi: 4ac = b2 e ̈vL ̈v: kZ©vbymv‡i, D = 0  b 2 – 4ac = 0  4ac = b2 Note: ax2 + bx + c = 0 mgxKi‡Yi c„_vqK/wbðvqK, D = b2 – 4ac  D = 0 n‡j, mgxKi‡Yi g~jØq ev ̄Íe I mgvb n‡e Ges ax2 + bx + c ivwkwU c~Y©eM© n‡e |  D > 0 n‡j, mgxKi‡Yi g~jØq ev ̄Íe I Amgvb n‡e|  D > 0 Ges c~Y©eM© n‡j, mgxKi‡Yi g~jØq g~j` I Amgvb n‡e|  D < 0 n‡j, mgxKi‡Yi g~jØq Aev ̄Íe/RwUj n‡e| weMZ mv‡j JU-G Avmv cÖkœvejx 1. k Gi gvb KZ n‡j, x 2 – 6x – 1 + k(2x + 1) = 0 mgxKi‡Yi g~j `ywU mgvb n‡e? [JU 22-23] 3 A_ev 6 2 A_ev 5 2 A_ev 6 3 A_ev 5 DËi: 2 A_ev 5 e ̈vL ̈v: x 2 – 6x – 1 + 2kx + k = 0  x 2 + (2k – 6)x + (k – 1) = 0 kZ©vbymv‡i, D = 0  (2k – 6)2 – 4  1  (k – 1) = 0  4k2 – 24k + 36 – 4k + 4 = 0 4k2 – 28k + 40 = 0  k 2 – 7k + 10 = 0  k 2 – 5k – 2k + 10 = 0  k(k – 5) – 2(k – 5) = 0  (k – 2)(k – 5) = 0  k = 2 A_ev 5 2. x 2 + x + 1 = 0 mgxKi‡Yi g~j ̧‡jvi cÖK...wZ †KvbwU? [JU 22-23] ev ̄Íe I mgvb ev ̄Íe I Amgvb Aev ̄Íe I Amgvb Aev ̄Íe I mgvb DËi: Aev ̄Íe I Amgvb e ̈vL ̈v: mgxKi‡Yi wbðvqK/c„_vqK, D = b2 – 4ac = 1 2 – 4  1  1 = 1 – 4 = – 3 < 0  g~jØq Aev ̄Íe I Amgvb| 3. 6x2 – 5x + 3 = 0 mgxKi‡Yi g~jØq  I  n‡j, 1  + 1  Gi gvb †KvbwU? [JU 22-23] – 5 3 – 5 12 5 12 5 3 DËi: 5 3 e ̈vL ̈v:  +  = 5 6 Ges  = 3 6 = 1 2  1  + 1  =  +   = 5 6 1 2 = 5 3 4. 2x2 – 7x + k = 0 mgxKiYwUi GKwU g~j 3 n‡j, k Gi gvb KZ? [JU 21-22] 5 4 2 3 DËi: 3 e ̈vL ̈v: x = 3 mgxKi‡Y ewm‡q cvB, 2  3 2 – 7  3 + k = 0  18 – 21 + k = 0  k = 3 5. 4x2 – kx + 5 = 0 mgxKi‡Yi GKwU g~j AciwUi wØ ̧Y n‡j, k Gi gvb KZ? [JU 21-22] 3 10 80 2 10 10 DËi: 3 10 e ̈vL ̈v: awi, g~jØq , 2 GLb, .2 = 5 4   2 = 5 8   = 5 2 2 Avevi,  + 2 = – – k 4  3 = k 4  3  5 2 2 = k 4  k = 3 10