Let f : [ 0 , 3 ] → A be defined by f ( x ) = 2 x 3 − 15 x 2 + 36 x + 7 and g : [ 0 , ∞ ) → B be defined by g ( x ) = x 2025 + 1 x 2025 . If both the functions are onto and S = { x ∈ Z : x ∈ A or x

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Let f : [ 0 , 3 ] → A be defined by f ( x ) = 2 x 3 − 15 x 2 + 36 x + 7 and g : [ 0 , ∞ ) → B be defined by g ( x ) = x 2025 + 1 x 2025 ​ . If both the functions are onto and S = { x ∈ Z : x ∈ A or x ∈ B } , then n ( S ) is equal to :

TestHub · Accepted Answer

as f ( x ) is onto hence A is range of f ( x ) ​ now f ′ ( x ) = 6 x 2 − 30 x + 36 = 6 ( x − 2 ) ( x − 3 ) f ( 2 ) = 16 − 60 + 72 + 7 = 35 f ( 3 ) = 54 − 135 + 108 + 7 = 34 f ( 0 ) = 7 ​ ​ hence range ∈ [ 7 , 35 ] = A also for range of g ( x ) ​ g ( x ) = 1 − x 2025 + 1 1 ​ ∈ [ 0 , 1 ) = B s = { 0 , 7 , 8 , … .35 } hence n ( s ) = 30 ​

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