Title উচ্চতর গণিত সমাধান.pdf ✅

Kauser’s Math Academy D”PZi MwYZ m„Rbkxj g ̈vwRK wkU cÖ ̄‘ZKviK : KvDQvi Avn‡g` Lvb (01946-775569) GmGmwm- 24/25 AbjvBb ‡Kvm© wd - 310 UvKv aib bs-1 : (dvsk‡bi †Wv‡gb wbY©q) 1. f(x) = 5x  1 Gi †Wv‡gb wbY©q Ki| [e.‡ev.-20] ‡`Iqv Av‡Q, f(x) = √5x − 1 x Gi †hme ev ̄Íe gv‡bi Rb ̈ f(x) Gi ev ̄Íe gvb cvIqv hvq †m ̧‡jvB f(x)Gi †Wv‡gb| f(x) = √5x − 1 ∈ R n‡e hw` I †Kej hw` 5x − 1 ≥ 0 nq ev, 5x ≥ 1 A_©vr x ≥ ଵ ହ nq| ∴ f(x) Gi †Wv‡gb f = ቄx: x ∈ R এবং x ≥ ଵ ହ ቅ nq| 2. f :  f(x) = 2x  5 2x + 3 Gi †Wv‡gb wbY©q Ki| [h.‡ev.-20] 3. g(x) = 1 1  2x Gi †Wv‡gb wbY©q Ki| [w`.‡ev.-20]
Kauser’s Math Academy D”PZi MwYZ m„Rbkxj g ̈vwRK wkU cÖ ̄‘ZKviK : KvDQvi Avn‡g` Lvb (01946-775569) GmGmwm- 24/25 AbjvBb ‡Kvm© wd - 310 UvKv 4. f(x) = 2x 1  3x Gi †Wv‡gb wbY©q Ki| [X.‡ev.-19] 5. F(x) = 2x  3 3x + 2 n‡j, F Gi †Wv‡gb wbY©q Ki| [h.‡ev.-19] 6. F(x) = 1  x n‡j, F(x) Gi †Wv‡gb wbY©q Ki| [h.‡ev.-19]
Kauser’s Math Academy D”PZi MwYZ m„Rbkxj g ̈vwRK wkU cÖ ̄‘ZKviK : KvDQvi Avn‡g` Lvb (01946-775569) GmGmwm- 24/25 AbjvBb ‡Kvm© wd - 310 UvKv aib bs-2 : (GK-GK dvskb msμvšÍ) 1. f(x) = 3x + 5 4x  3 , n‡j, dvskbwU GK-GK wKbv Zv wba©viY Ki| [Kz.‡ev.-20] 2. f :  f(x) = 2x  5 2x + 3 n‡j, f GKwU GK-GK dvskb wKbv hvPvB Ki| [h.‡ev.-20]
Kauser’s Math Academy D”PZi MwYZ m„Rbkxj g ̈vwRK wkU cÖ ̄‘ZKviK : KvDQvi Avn‡g` Lvb (01946-775569) GmGmwm- 24/25 AbjvBb ‡Kvm© wd - 310 UvKv 3. f(x) = 2x + 3 x  2 GKwU dvskb †hLv‡b x  2, f(x) Øviv ewY©Z dvskbwU GK-GK wKbv Zv wbY©q Ki|[wm.‡ev.-20]
Kauser’s Math Academy D”PZi MwYZ m„Rbkxj g ̈vwRK wkU cÖ ̄‘ZKviK : KvDQvi Avn‡g` Lvb (01946-775569) GmGmwm- 24/25 AbjvBb ‡Kvm© wd - 310 UvKv