Title Functions and Graph Engineering Practice Sheet Solution (HSC 26).pdf ✅

dvskb I dvsk‡bi †jLwPÎ  Engineering Practice Sheet Solution (HSC 26) 3 GLb, awi, †h‡Kvb x1, x2  A Gi Rb ̈ f(x1) = f(x2) n‡e hw` I †Kej hw`, 3x1 + 2 7x1 – 3 = 3x2 + 2 7x2 – 3  (3x1 + 2)(7x2 – 3) = (3x2 + 2)(7x1 – 3)  21x1x2 + 14x2 – 9x1 – 6 = 21x1x2 + 14x1 – 9x2 – 6  23x2 = 23x1  x1 = x2  f(x) dvskbwU GK GK| (Showed) (ii) †_‡K cvB, f –1 (x) = msÁvwqZ n‡e hw`, 7x – 3  0 nq,  x  3 7 dvsk‡bi †iÄ = –      3 7 Avevi, †Kv‡Wv‡gb B = –      3 7  †Kv‡Wv‡gb = †iÄ dvskbwU mvwe©K (Showed) 9| f(x) = 2x3 + 3 Ges g(x) = 3 x – 3 2 n‡j, †`LvI †h, (fog)(x) = (gof)(x) [BUET 03-04] mgvavb: L.H.S = fog(x) = f(g(x)) = f      3 x – 3 2 = 2      3 x – 3 2 3 + 3 = 2          x – 3 2 1 3 3 + 3 = x – 3 + 3 = x R.H.S = (gof)(x) = g(f(x)) = g(2x3 + 3) = 3 2x3 + 3 – 3 2 = (x ) 3 1 3 = 3  L.H.S = R.H.S (Showed) weMZ mv‡j KUET-G Avmv cÖkœvejx 10| hw` A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5} Ges f(x) = x + 1 Øviv f : A  B msÁvwqZ nq, Z‡e f Gi †Wv‡gb Ges †iÄ wbY©q Ki| [KUET 05-06] mgvavb: f(x) = x + 1 ; f(1) = 1 + 1 = 2 ; f(2) = 2 + 1 = 3 f(3) = 3 + 1 = 4 ; f(4) = 4 + 1 = 5 myZivs dvskbwUi †Wv‡gb = {1, 2, 3, 4} Ges †iÄ = {2, 3, 4, 5} (Ans.) 11| (a) GKwU dvskb f: R  R Giƒcfv‡e msÁvwqZ n‡q‡Q †h, f(x) = x2 + 1, f –1 (5) Gi gvb wbY©q Ki| (b) cÖgvY Ki †h, (A – B)  (A – C) = A – (B  C), †hLv‡b, A, B I C wZbwU †mU| [KUET 04-05, RUET 07-08] mgvavb: (a) awi, f –1 (5) = x  f(x) = 5  x 2 + 1 = 5  x 2 = 4  x =  2  f –1 (5) = {2, – 2} (Ans.) (b) †`Iqv Av‡Q, evgcÿ = (A – B)  (A – C) = {x : x  (A – B) Ges x  (A – C)} = {x : x  A Ges x  B Ges x  A Ges x  C} = {x : x  A Ges x  B Ges x  C} = {x : x  A Ges x  (B  C)} = {x : x  A – (B  C)}  (A – B)  (A – C)  A – (B  C) GKB fv‡e, Wvbcÿ = A – (B  C) = {x : x  A Ges x  (B  C)} = {x : x  A Ges x  B Ges x  C} = {x : x  A Ges x  B Ges x  A Ges x  C} = (A – B)  (A – C) = evgcÿ  A – (B  C)  (A – B)  (A – C)  (A – B)  (A – C) = A – (B  C) (Proved) 12| hw` f(x) = ln     1 + x 1 – x nq, Z‡e †`LvI †h, f     2x 1 + x2 = 2f(x) [KUET 04-05] mgvavb: L.H.S. = ln      1 +  2x 1 + x2 1 – 2x 1 + x2 = ln     1 + x2 + 2x 1 + x2 – 2x = ln     1 + x 1 – x 2 = 2 ln     1 + x 1 – x = 2f(x) = RHS. (Showed) weMZ mv‡j RUET-G Avmv cÖkœvejx 13| f(x) = x2 + 3x + 1 Ges g(x) = 2x – 3 n‡j (gof) (2) Ges (fog) (2) wbY©q Ki| [RUET 08-09, 12-13, CUET 11-12] mgvavb: (gof) (x) = g(f(x)) = g(x2 + 3x + 1) = 2x2 + 6x – 1  (gof) (2) = 19 (fog) (x) = f(g(x)) = f(2x – 3) = (2x – 3)2 + 3(2x – 3) + 1  (fog) (2) = 1 + 3 + 1 = 5