Title Differentiation Practice Sheet Solution HSC FRB 25.pdf ✅

AšÍixKiY  Final Revision Batch '25 1 Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 1 2 2 1 1 1 1 1 2 2022 1 2 1 2 1 1 2 1 2 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 4 4 2 3 2 2 3 2 4 2022 5 4 5 5 4 5 4 4 5 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| `„k ̈Kí-1: g(x) = sinx `„k ̈Kí-2: x = tan 2y [XvKv †evW©- Õ23] (K) x Gi mv‡c‡ÿ logx a Gi AšÍiR wbY©q Ki| (L) `„k ̈Kí-1 Gi Av‡jv‡K 1 + 2g(x) + 3[1 – {g(x)} 2 ] Gi 0  x   2 e ̈ewa‡Z Pig gvb wbY©q Ki| (M) `„k ̈Kí-2 Gi Av‡jv‡K cÖgvY Ki †h, (1 + x2 ) 2 y2 + 2(1 + x2 )xy1 = 1 mgvavb: (K) awi, y = logxa = logxe  logea = 1 logex  lna = ln a  1 lnx = lna (lnx)–1 x Gi mv‡c‡ÿ AšÍixKiY K‡i cvB, dy dx = lna d dx (lnx)–1 = lna(– 1) (lnx)– 1 – 1 d dx (lnx) = – lna(lnx)–2  1 x = – lna x(lnx) 2 (Ans.) (L) †`Iqv Av‡Q, g(x) = sinx g‡b Kwi, y = 1 + 2g(x) + 3[1 – {g(x)}2 ] = 1 + 2 sinx + 3 [1 – sin2 x] = 1 + 2 sinx + 3 cos2 x  dy dx = 0 + 2 cosx + 3  2 cosx(– sinx) = 2 cosx – 6 sinx cosx d 2 y dx2 = 2(– sinx) – 6{sinx (– sinx) + cosx  cosx} = – 2sinx – 6(cos2 x – sin2 x) Piggv‡bi Rb ̈, dy dx = 0  2 cosx – 6sinx cosx = 0  cosx(1 – 3 sinx) = 0 nq, cosx = 0  cosx = cos  2  x =  2 A_ev, 1 – 3 sinx = 0  1 = 3 sinx  sinx = 1 3  x = sin–11 3 GLv‡b, cosx = 0  cos2 x = 0  1 – sin2 x = 0  sin2 x = 1  sinx = 1     ⸪ 0  x   2  ZvB FYvZ¥K eR©b K‡i cosx = 0 Ges sinx = 1 n‡j, d 2 y dx2 = – 2  1 – 6(0 – 1) = 4 > 0  x =  2 Gi Rb ̈ y dvsk‡bi jNygvb we` ̈gvb| AZGe, dvskbwUi jNygvb = 1 + 2 sin 2 + 3cos2 2 = 1 + 2  1 + 3  0 = 3 (Ans.) Avevi, x = sin–11 3  sinx = 1 3 n‡j, d 2 y dx2 = – 2sinx – 6(cos2 x – sin2 x) = – 2 sinx – 6(1 – 2sin2 x)
2  Higher Math 1st Paper Chapter-9     d 2 y dx2 x = sin–1 1 3 = – 2  1 3 – 6     1 – 2  1 9 = – 16 3 < 0  x = sin–11 3 Gi Rb ̈ y dvskbwUi ̧iægvb we` ̈gvb| AZGe, dvskbwUi ̧iægvb = 1 + 2 sin sin–11 3 + 3       1 –     sinsin–11 3 2 = 1 + 2  1 3 + 3       1 –     1 3 2 = 13 3 (Ans.) (M) †`Iqv Av‡Q, x = tan 2y  2y = tan–1 x  2y = (tan–1 x)2  2 dy dx = d dx (tan–1 x)2  2y1 = 2tan–1 x  d dx tan–1 x  y1 = tan–1 x  1 1 + x2  (1 + x2 )y1 = tan–1 x (1 + x2 )y1 = 2y (1 + x2 ) 2 y 2 1 = 2y  (1 + x2 ) 2  2y1y2 + y 2 1  2(1 + x2 ) 2 – 1 (0 + 2x) = 2y1  (1 + x2 ) 2 y2 + 2(1 + x2 )xy1 = 1 (Proved) 2| f(x) = (ax)n ; g(x) = 17 – 15x + 9x2 – x 3 [ivRkvnx †evW©- Õ23] (K) y = sin x n‡j, dy dx wbY©q Ki| (L) wjwgU Gi mvnv‡h ̈ f(x) Gi AšÍiR wbY©q Ki| (M) g(x) dvskbwU †Kvb e ̈ewa‡Z n«vm cvq Ges e„w× cvq Zv wbY©q Ki| mgvavb: (K) †`Iqv Av‡Q, y = sin x  dy dx = d dx ( sin x) = d dx (sin x) 1 2 = 1 2 sin x  d dx (sin x) = 1 2 sin x  (cos x)  d dx ( x) = 1 2 sin x  (cos x)  1 2 x = cos x 4 x sin x = cos x 4 xsin x (Ans.) (L) †`Iqv Av‡Q, f(x) = (ax)n = an x n  f(x + h) = an (x + h)n AšÍi‡Ri g~j wbqgvbyhvqx, d dx f(x) = lim h  0 f(x + h) – f(x) h = lim h  0 a n (x + h) n – a n x n h = an lim h  0 (x + h) n – x n h = an lim h  0       x     1 + h x n – x n h = an x n lim h  0     1 + h x n – 1 h = an x n lim h  0       1+n h x + n(n–1) 2! ( ) h x 2 + n(n–1) (n–2) 3! ( ) h x 3 + ... – 1 h = a n x n lim h  0 h{ } n x + n(n – 1) 2!  h x 2 + n(n – 1) (n – 2) 3!  h 2 x 3 + .... h = an x n lim h  0 { } n x + n(n – 1) 2!  h x 2 + n(n – 1) (n – 2) 3!  h 2 x 3 + .... = a n x n     n x + 0 + 0 + .... = nan x n–1 (Ans.) (M) †`Iqv Av‡Q, g(x) = 17 – 15x + 9x2 – x 3  g(x) = – 3x2 + 18x – 15 †h mKj e ̈ewa‡Z g(x) e„w× cÖvß †mLv‡b g(x) > 0  – 3x2 + 18x – 15 > 0  x 2 – 6x + 5 < 0  x 2 – 5x – x + 5 < 0  (x – 1) (x – 5) < 0 Line test: – – + – + + 1 5  AmgZvwU mZ ̈ n‡e hw` 1 < x < 5 nq| A_©vr, 1 < x < 5 e ̈ewa‡Z g(x) e„w× cÖvß| (Ans.) Avevi, g(x) n«vmcÖvß n‡j, g(x) < 0  – 3x2 + 18x – 15 < 0  x 2 – 6x + 5 > 0  (x – 1) (x – 5) > 0 Line test: – – + – + + 1 5  AmgZvwU mZ ̈ n‡e hw` x < 1 A_ev x > 5 nq| A_©vr, x < 1 A_ev x > 5 e ̈ewa‡Z g(x) n«vm cÖvß| (Ans.) 3| `„k ̈Kí-1: f(x) = 3x2 + 2x + 7 `„k ̈Kí-2: g(x) = 54x – (2x – 7)3 [ivRkvnx †evW©- Õ23] (K) †`LvI †h, lim x  0 1 + 2x – 1 – 3x x = 5 2 (L) `„k ̈Kí-1 G y = f(x) eμ‡iLvi (2, 23) we›`y‡Z ̄úk©K I Awfj‡¤^i mgxKiY wbY©q Ki| (M) `„k ̈Kí-2 n‡Z dvskbwUi ̧iægvb I jNygvb wbY©q Ki|
AšÍixKiY  Final Revision Batch '25 3 mgvavb: (K) L.H.S = lim x  0 1 + 2x – 1 – 3x x = lim x  0 ( 1 + 2x – 1 – 3x)( 1 + 2x + 1 – 3x) x( 1 + 2x + 1 – 3x) = lim x  0 ( 1 + 2x) 2 – ( 1 – 3x) 2 x( 1 + 2x + 1 – 3x) = lim x  0 1 + 2x – 1 + 3x x( 1 + 2x + 1 – 3x) = lim x  0 5x x( 1 + 2x + 1 – 3x) = lim x  0 5 1 + 2x + 1 – 3x = 5 1 + 0 + 1 – 0 = 5 2 = R.H.S (Showed) (L) awi, y = f(x) = 3x2 + 2x + 7  dy dx = 6x + 2 (2, 23) we›`y‡Z dy dx =     dy dx (2,23) = 6  2 + 2 = 14 (2, 23) we›`y‡Z ̄úk©‡Ki mgxKiY, y – y1 =     dy dx (2,23) (x – x1)  y – 23 = 14(x – 2)  14x – y – 5 = 0 (Ans.) (2, 23) we›`y‡Z Awfj‡¤^i mgxKiY, y – y1 = – 1     dy dx (2,23) (x – x1)  y – 23 = – 1 14 (x – 2)  x + 14y – 324 = 0 (Ans.) (M) †`Iqv Av‡Q, g(x) = 54x – (2x – 7)3  g(x) = 54 – 6(2x – 7)2  g(x) = – 24(2x – 7) ̧iægvb I jNygv‡bi Rb ̈, g(x) = 0  54 – 6(2x – 7)2 = 0  9 – (4x2 – 28x + 49) = 0  9 – 4x2 + 28x – 49 = 0  x 2 – 7x + 10 = 0  x 2 – 5x – 2x + 10 = 0  (x – 2) (x – 5) = 0 nq, x – 2 = 0 A_ev, x – 5 = 0  x = 2  x = 5 x = 2 n‡j, g(x) = – 24(2  2 – 7) = – 24(– 3) = 72 > 0  x = 2 Gi Rb ̈ g(x) Gi jNygvb we` ̈gvb|  jNygvb, g(2) = 54  2 – (2  2 – 7)3 = 135 (Ans.) Avevi, x = 5 n‡j, g(5) = – 24(2  5 – 7) = – 24  3 = – 72 <0  x = 5 Gi Rb ̈ g(x) Gi ̧iægvb we` ̈gvb|  ̧iægvb, g(5) = 54  5 – (2  5 – 7)3 = 243 (Ans.) 4| f(x) = a5x, g(x) = x 2a + y 2a – 1 [h‡kvi †evW©- Õ23] (K) lim x  a x 3 2 – a 3 2 x – a Gi gvb wbY©q Ki| (L) wjwg‡Ui mvnv‡h ̈ f(x) Gi AšÍiR wbY©q Ki| (M) cÖgvY Ki †h, g(x) = 0 eμ‡iLvi †h‡Kv‡bv we›`y‡Z ̄ck©K Øviv Aÿ؇qi LwÛZvs‡ki †hvMdj GKwU aaæeK| mgvavb: (K) lim x  a x 3 2 – a 3 2 x – a = lim x  a (x ) 1 2 3 – (a ) 1 2 3 x – a = lim x  a ( x) 3 – ( a) 3 x – a = 3.( a) 3 – 1    ⸪ lim  x  p x n – p n x – p = npn – 1 = 3( a) 2 = 3a (Ans.) (L) †`Iqv Av‡Q, f(x) = a5x  f(x + h) = a5(x + h) AšÍi‡Ri g~j wbqg Abyhvqx, f(x) = d dx f(x) = lim h  0 f(x + h) – f(x) h = lim h  0 a 5(x + h) – a 5x h = lim h  0 a 5x(a 5h – 1) h = a5x lim h  0       1 + 5h lna + (5h lna) 2 2! + (5h lna) 3 3! + ... – 1 h = a 5x lim h  0 h       5 lna + h(5 lna) 2 2! + h 2 (5 lna) 3 3! + .... h = a 5x lim h  0     5 lna + h(5 lna) 2 2! + h 2 (5 lna) 3 3! + .... = a 5x [5 lna + 0 + 0 + .....] = 5a5x lna (Ans.) (M) †`Iqv Av‡Q, g(x) = x 2a + y 2a – 1 Avevi, eμ‡iLvi mgxKiY, g(x) = 0  x 2a + y 2a – 1 = 0  x 2a + y 2a = 1  x + y = 2a .....(i)