Tiêu đề Differentiation Varsity Practice Sheet Solution.pdf ✅

09 AšÍixKiY Differentiation weMZ mv‡j DU-G Avmv cÖkœvejx 1. lim x  –  x 2 + x x + 1 = ? [DU 23-24] 1 –   – 1 DËi: – 1 e ̈vL ̈v: lim x  –  x 2 + x x + 1 = lim x  –  x 2     1 + 1 x x     1 + 1 x = lim x  –  – x 1 + 1 x x     1 + 1 x = lim x  –  – 1 1 + 1 x 1 + 1 x = – 1 2. y = ln(2 – x) n‡j, dy dx = ? [DU 23-24] 1 2 – x 2 2 – x x 2 – x 1 x – 2 DËi: 1 x – 2 e ̈vL ̈v: y = ln(2 – x)  dy dx = 1 2 – x (– 1) = 1 x – 2 3. y = x – 2 lnx n‡j, d 2 y dx2 Gi gvb KZ? [DU 22-23] x – 4 lnx – 2x– 2 – 3x– 4 6x– 4 lnx – 5x– 4 6x – 4 lnx – 2x– 2 – 3x– 4 x – 4 lnx – 2x– 2 + 3x– 4 DËi: 6x– 4 lnx – 5x– 4 e ̈vL ̈v: y = x –2 lnx  dy dx = x – 2 . 1 x + lnx(– 2)x– 3 = x – 3 – 2x– 3 lnx = x – 3 (1 – 2lnx)  d 2 y dx2 = x – 3     0 – 2 x + (1 – 2lnx)(– 3)x– 4  d 2 y dx2 = 6x– 4 lnx – 5x– 4 4. lim x  1 2x3 – (2k + 1)x 2 + 2x + k x – 1 = – 6 n‡j, k Gi gvb KZ? [DU 22-23] 1 – 1 3 – 1 2 DËi: 3 e ̈vL ̈v: lim x  1 2x3 – (2k + 1)x 2 + 2x + k x – 1 = – 6  lim x  1 6x2 – 4kx – 2x + 2 1 = – 6 [L Hôpital’s Rule]  6 – 4k – 2 + 2 = – 6  4k = 12  k = 3 5. †Kvb e ̈ewa‡Z f(x) = x x 2 + 1 μgea©gvb? [DU 21-22] (– , 0) (– 1, 1) (0, ) (– 1, ) DËi: (– 1, 1) e ̈vL ̈v: f (x) = (x 2 + 1) – x . 2x (x 2 + 1) 2 = 1 – x 2 (x 2 + 1) 2 f (x)  0  1 – x 2 (x 2 + 1) 2  0  x 2 – 1 (x 2 + 1) 2  0  (x + 1) (x – 1) (x 2 + 1) 2  0 ⸪ (x2 + 1)2  0  (x + 1)(x – 1)  0  – 1  x  1  e ̈ewa (– 1, 1) Note: dy dx > 0 n‡j dvskb μgea©gvb n‡e Ges dy dx < 0 n‡j, dvskb μgn«vmgvb n‡e|
2  Higher Math 1st Paper Chapter-9 6. †Kvb dvskbwUi Rb ̈ (1 – x 2 ) d 2 y dx2 – x dy dx = 2 mZ ̈? [DU 21-22] y =cos–1 x y = (cos–1 x)2 y = sin–1 x y = tan–1 x DËi: y = (cos–1 x)2 e ̈vL ̈v: Option Test K‡i, y = (cos–1 x)2 dy dx = 2cos–1 x    –  1 1 – x 2  1 – x 2 y1 = – 2cos–1 x  (1 – x 2 )y1 2 = 4(cos–1 x)2  (1 – x 2 ) y1 2 = 4y  (1 – x 2 )2y1y2 – 2xy1 2 = 4y1  (1 – x 2 ) d 2 y dx2 – x dy dx = 2 7. k Gi †Kvb gv‡bi Rb ̈ x = 1 we›`y‡Z f(x) = x2 + k x Gi jNygvb cvIqv hv‡e? [DU 21-22] 0 – 1 2 1 DËi: 2 e ̈vL ̈v: f(x) = x2 + k x  f (x) = 2x – k x 2  f (1) = 2  1 – k 1 2 = 2 – k jNygvb ev ̧iægv‡bi Rb ̈, f (1) = 0  2 – k = 0  k = 2 8. d dx (cos2 (lnx)) = ? [DU 20-21] – sin(2lnx) 2 – 2cos(lnx) x – sin(2ln x) x – 2xcos(lnx)sin(lnx) DËi: – sin(2lnx) x e ̈vL ̈v: d dx (cos2 (lnx)) = – 2cos(lnx).sin(lnx).1 x = – sin(2ln x) x 9. lim x  0 1 – cosx sin2 2x Gi gvb n‡eÑ [DU 20-21] 1 4 1 8 1 2 1 DËi: 1 8 e ̈vL ̈v: lim x  0 1 – cosx sin2 2x = lim x  0 sinx 2sin2x.cos2x.2 [L Hôpital’s Rule] = lim x  0 sinx 4  2sinx.cosx . lim x  0 1 cos2x = 1 8 . lim x  0 1 cos2x = 1 8 10. hw` y = kx(2x + 3) eμ‡iLvi g~jwe›`y‡Z ̄úk©KwU x A‡ÿi mv‡_ 30 †KvY Drcbœ K‡i, Zvn‡j k Gi gvb n‡eÑ [DU 20-21] 1 3 3 1 3 1 2 DËi: 1 3 e ̈vL ̈v: y = kx(2x + 3)  y = 2kx2 + 3kx  dy dx = 4kx + 3k g~jwe›`y‡Z Xvj = 3k  3k =  tan30  k =  1 3 11. lim x  –  x 2 + 2x – x Gi gvb n‡jvÑ [DU 19-20] 1  –  – 1 DËi: 1 e ̈vL ̈v: lim x  –  x 2 + 2x – x = lim x  –  x 2     1 + 2 x – x = lim x  –  – x 1 + 2 x – x = 1 [GLv‡b, x Gi gvb FYvZ¥K Amx‡gi w`‡K nIqvq x 2 Gi eM©g~j – x n‡e] 12. y = tan–1 1 + x 1 – x n‡j, dy dx = KZ? [DU 19-20; Agri. Guccho 20-21; JU 19-20] 1 1 – x 2 1 1 + x x 1 + x 1 1 + x2 DËi: 1 1 + x2
AšÍixKiY  Varsity Practice Sheet Solution 3 e ̈vL ̈v: y = tan–1 1 + x 1 – x = tan–1 1 + x 1 – 1.x = tan–1 1 + tan–1 x =  4 + tan–1 x  dy dx = 0 + 1 1 + x2 = 1 1 + x2 13. y = 1 + x 1 – x n‡j, dy dx Gi gvbÑ [DU 18-19] – 2 (x – 1) 2 2 1 – x 2 2 (1 – x) 2 2x (1 – x) 2 DËi: 2 (1 – x) 2 e ̈vL ̈v: y = 1 + x 1 – x dy dx = (1 – x) d dx (1 + x) – (1 + x) d dx (1 – x) (1 – x) 2 = (1 – x) + (1 + x) (1 – x) 2 = 2 (1 – x) 2 14. lim x  0 sinx tan–1 (3x) = ? [DU 18-19] 0 1 3 1 3 DËi: 1 3 e ̈vL ̈v: lim x  0 sinx tan–1 (3x) = lim x  0 sinx x . lim x  0 3x tan–1 (3x)  1 3 = 1 3 15. y = x 4 – 4x3 + 4x2 + 5 n‡j, jwNô gvbÑ [DU 17-18] 4 5 6 8 DËi: 5 e ̈vL ̈v: y = x 4 – 4x3 + 4x2 + 5  y1 = 4x3 – 12x2 + 8x = 0  x(4x2 – 12x + 8) = 0 x = 0 A_ev, 4x2 – 12x + 8 = 0  y2 = 12x2 – 24x + 8  y2 (0) = 8  0 jNygvb Av‡Q|  f(0) = 04 – 4  0 3 + 4  0 2 + 5 = 5 16. lim x  –  2x2 + 3x + 5 3x2 + 5x – 6 Gi gvbÑ [DU 17-18] 3 5 – 5 6 2 3 – 2 3 DËi: 2 3 e ̈vL ̈v: lim x  –  2x2 + 3x + 5 3x2 + 5x – 6 = lim x  –  x 2     2 + 3 x + 5 x 2 x 2     3 + 5 x – 6 x 2 = 2 3 Shortcut: cÖ‡kœ x Gi m‡ev©”P NvZ = 2  lim x   2x2 + 3x + 5 3x2 + 5x – 6 = j‡ei x 2 Gi mnM n‡ii x 2 Gi mnM = 2 3 17. x Gi †Kvb gv‡bi Rb ̈ y = x + 1 x eμ‡iLvwUi Xvj k~b ̈ n‡e? [DU 16-17, 13-14; Agri. Guccho 21-22, 20-21; JnU 10-11; CU 02-03]  3 2  2 1  1 DËi:  1 e ̈vL ̈v: y = x + 1 x y1 = 1 – 1 x 2 = 0  x 2 = 1  x =  1 18. hw` y = sin–1 (sin x) nq, dy dx Gi gvb †KvbwU? [DU 16-17; JU 22-23] sinx cosx x 1 DËi: 1 e ̈vL ̈v: y = sin–1 (sinx) = x  dy dx = dx dx = 1 19. lim x  0 e cosx cosx Gi gvbÑ [DU 16-17] e 1 1 e 0 DËi: e e ̈vL ̈v: lim x  0 e cosx cosx = e cos0 cos0 = e
4  Higher Math 1st Paper Chapter-9 20. k Gi †Kvb gv‡bi Rb ̈ y = kx(1 – x) eμ‡iLvi g~jwe›`y‡Z ̄úk©KwU x A‡ÿi mv‡_ 30 †KvY Drcbœ K‡i? [DU 15-16; GST 23-24] 3 1 3 3 2 1 DËi: 1 3 e ̈vL ̈v: y = kx (1 – x)  y = kx – kx2  y1 = k – 2kx g~jwe›`y‡Z Xvj, y1 = k  k =  tan30  k =  1 3 21. lim x  0 sin7x – sinx sin6x = ? [DU 15-16] 7 6 – 7 6 1 – 1 DËi: 1 e ̈vL ̈v: lim x  0 sin7x – sinx sin6x = lim x  0 2cos4x sin3x sin6x = lim x  0 2cos4x . lim x  0 sin3x 3x . 3x . lim x  0 6x sin6x . 1 6x = 1 Shortcut: lim x  0 sinax – sinbx sincx – sindx = a – b c – d  lim x  0 sin7x – sinx sin6x = 7 – 1 6 – 0 = 1 22. e xy+1 = 5 n‡j, dy dx Gi gvb- [DU 14-15; Agri. Guccho 20-21; IU 18-19;] ln5 xy ln5 – x 2 – y x – ln5 y DËi: – y x e ̈vL ̈v: e xy+1 = 5  ln e xy+1 = ln5  xy + 1 = ln5  x dy dx + y = 0 [x Gi mv‡c‡ÿ AšÍixKiY K‡i]  dy dx = – y x 23. x = 0 we›`y‡Z y = x + ex Gi †jLwP‡Î ̄úk©‡Ki mgxKiY n‡eÑ [DU 14-15] y = x y = x + 1 y = 2x + 1 y = 2x DËi: y = 2x + 1 e ̈vL ̈v: y = x + ex  dy dx = 1 + ex x = 0 n‡j, dy dx = 2  y = 0 + e0 = 1  ̄úk©‡Ki mgxKiY: y – 1 = 2(x – 0)  y = 2x + 1 24. lim x   x 2 + 6x 2x2 + 5 = ? [DU 14-15] 0 3 2 1 2 1 DËi: 1 2 e ̈vL ̈v: lim x   x 2 + 6x 2x2 + 5 = lim x   x 2     1 + 6 x x 2     2 + 5 x 2 = 1 2 Shortcut: cÖ‡kœ x Gi m‡ev©”P NvZ = 2  lim x   x 2 + 6x 2x2 + 5 = j‡ei x 2 Gi mnM n‡ii x 2 Gi mnM = 1 2 25. hw` x n + yn = an nq, Zvn‡j dy dx =? [DU 13-14; NSTU 19-20]     x y n    –  x y n –     x y n–1 –     x y n+1 DËi: –     x y n–1 e ̈vL ̈v: x n + yn – a n = 0  dy dx = – fx fy = – nxn–1 nyn–1 = –     x y n–1 26. lim x  0 cosx – 1 x 2 = ? [DU 13-14; JU 14-15] – 7 – 1 2 1 2 1 DËi: – 1 2