Tiêu đề Polynomial & Digestion and absorption-Daily-20 (Set-A)-With Solve.pdf ✅

3 15. 1 x – 1 x – a = 1 b mgxKi‡Yi g~jØq  I  n‡jÑ ( + ) = a  = ab 1  + 1  = 1 b All DËi: All e ̈vL ̈v: 1 x – 1 x – a = 1 b  x – a – x x(x – a) = 1 b  – ab = x2 – ax  x 2 – ax + ab = 0   +  = a   = ab  1  + 1  =  +   = a ab = 1 b 16. hw` x abvZ¥K c~Y©msL ̈v nq Ges x 3 – x †K 3 Øviv fvM Ki‡j fvM‡kl k nq, Z‡e k Gi gvbÑ k = 1 k < 1 k > 1 k = 0 DËi: k > 1 e ̈vL ̈v: awi, f(x) = x3 – x G‡K (x – 3) Øviv fvM Ki‡j k = f(3)  k = 33 – 3 = 24 > 1 17. x 2 + px + q = 0 mgxKi‡Y g~j؇qi cv_©K ̈ 1 n‡j, p 2 + 4q2 = ? (q + 1)2 (2q – 1)2 (2q + 3)2 (2q + 1)2 DËi: (2q + 1)2 e ̈vL ̈v: g~jØq ,  + 1   +  + 1 = – p 2 = – p – 1  = – 1 2 (p + 1)  ( + 1) = q   2 +  = q  1 4 (p + 1)2 – 1 2 (p + 1) = q  p 2 + 2q + 1 – 2p – 2 = 4q  p 2 = 4q + 1  p 2 + 4q2 = 4q + 1 + 4q2 = (2q + 1)2 18. y = x 3 – 1 mgxKiYwU x Aÿ‡K KqwU we›`y‡Z †Q` K‡i? 0 1 2 3 DËi: 1 e ̈vL ̈v: x 3 – 1 = 0  (x – 1)(x2 + x + 1) = 0 ⸪ x2 + x + 1 = 0 Gi †Kv‡bv ev ̄Íe mgvavb †bB  x 3 – 1 = 0 Gi ev ̄Íe mgvavb 1wU  x Aÿ‡K 1wU we›`y‡Z †Q` Ki‡e| 19. x 3 + px + q = 0 mgxKi‡Yi GKwU g~j Aci `ywUi AšÍ‡ii wØ ̧Y n‡j, wb‡Pi †Kvb g~jwU mwVK? 13q 3p – 13q 3p 3q 13p None of these DËi: 13q 3p e ̈vL ̈v: x 3 + px + q = 0 g~j , , 2( – )  +  + 2( – ) = 0   = 3 ..... (i)  + 2( – ) + 2( – ) = p   + 2( 2 –  2 ) = p  .3 + 2( 2 – 9 2 ) = p  – 13 2 = p ..... (ii) cybivq, .2( – ) = – q  2.3( – 3) = – q  12 3 = q .... (iii)  (iii)  (ii)  12 13  = – q p   = – 13q 12p   = 3 = – 39q 12p  2( – ) = 2    – 13q 12p + 13q 12p = 13q 3p 20. x 2 + px + 12 = 0 mgxKi‡Yi GKwU g~j 2 n‡j I x 2 + px + q = 0 mgxKi‡Yi g~j ̧‡jv mgvb n‡j, q Gi gvb KZ? – 16 49 4 4 16 DËi: 16 e ̈vL ̈v: 2 2 + 2p + 12 = 0  p = – 8  x 2 – 8x + q = 0 (– 8)2 – 4q = 0  4q = 64  q = 16 21. x 2 – px + 2p = 0 mgxKi‡Yi g~j؇qi AšÍi 3 n‡j, p Gi gvbØq‡K g~j a‡i MwVZ mgxKiY †KvbwU? x 2 – 9x + 9 = 0 x 2 + 9x – 9 = 0 x 2 – 8x – 9 = 0 x 2 + 8x + 9 = 0 DËi: x 2 – 8x – 9 = 0 e ̈vL ̈v: g~jØq ,  + 3  +  + 3 = p  2 = p – 3   = 1 2 (p – 3)