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Solution


Q.

Let f : [0, 3] →A be defined by f(x)=2x3–15x2+36x+7 and g : [0, ∞)→B be defined by g(x)=x2025x2025+1. If both the functions are onto and S={x∈Z : x∈A or x∈B}, then n(S) is equal to:          [2025]
1 36  
2 29  
3 30  
4 31  

Ans.

(3)

As f(x) is onto, hence A is range of f(x).

Now, f'(x)=6x2–30x+36=6(x–2)(x–3)

for extremum, f(2) = 16 – 60 + 72 + 7 = 35; f(3) = 54 – 135 + 108 + 7 = 34; f(0) = 7; F(1) = 30.

Hence, range ∈ [7, 35] = A

Also, for range of g(x), g(x) = 1-1x2025+1∈[0,1)=B

 ∴  S = {0, 7, 8, ..., 35}

Hence, n(S) = 30