Let be defined by and be defined by . If both the functions are onto and , then n(S) is equal to: [2025]

Question

Let $f : [0, 3] \to A$ be defined by $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 7$ and $g : [0, \infty) \to B$ be defined by $g (x) = \frac{x^{2025}}{x^{2025} + 1}$ . If both the functions are onto and $S = {x \in Z : x \in A or x \in B}$ , then n(S) is equal to: [2025]

Smart Achievers · Accepted Answer

(3)

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

Now, $f' (x) = 6 x^{2} - 30 x + 36 = 6 (x - 2) (x - 3)$

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

Hence, range $\in$ [7, 35] = A

Also, for range of g(x), g(x) = $1 - \frac{1}{x^{2025} + 1} \in [0, 1) = B$

$∴$ S = {0, 7, 8, ..., 35}

Hence, n(S) = 30

Solution

Q.
Let $f : [0, 3] \to A$ be defined by $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 7$ and $g : [0, \infty) \to B$ be defined by $g (x) = \frac{x^{2025}}{x^{2025} + 1}$ . If both the functions are onto and $S = {x \in Z : x \in A or x \in B}$ , then n(S) is equal to: [2025]

1 36

2 29

3 30

4 31

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