Question ID: #489
Let $f:[0,3]\rightarrow A$ be defined by $f(x)=2x^{3}-15x^{2}+36x+7$ and $g:[0,\infty)\rightarrow B$ be defined by $g(x)=\frac{x^{2025}}{x^{2025}+1}$.
If both the functions are onto and $S=\{x\in Z : x\in A \text{ or } x\in B\}$, then $n(S)$ is equal to:
- (1) 30
- (2) 36
- (3) 29
- (4) 31
Solution:
For $f(x)$ to be onto, set $A$ must be its Range.
$f'(x) = 6x^2 – 30x + 36 = 6(x^2 – 5x + 6) = 6(x-2)(x-3)$.
Critical points in $[0,3]$ are $x=2$ and $x=3$.
Evaluate $f(x)$ at endpoints and critical points:
$$f(0) = 7$$
$$f(2) = 2(8) – 15(4) + 36(2) + 7 = 16 – 60 + 72 + 7 = 35$$
$$f(3) = 2(27) – 15(9) + 36(3) + 7 = 54 – 135 + 108 + 7 = 34$$
Minimum value is 7, Maximum value is 35.
Range $A = [7, 35]$.
Integers in $A$: $\{7, 8, \dots, 35\}$. Number of integers = $35 – 7 + 1 = 29$.
For $g(x)$ to be onto, set $B$ must be its Range.
$g(x) = \frac{x^{2025}}{x^{2025}+1} = 1 – \frac{1}{x^{2025}+1}$.
As $x$ goes from $0$ to $\infty$:
At $x=0, g(0) = 0$.
As $x \to \infty, g(x) \to 1$.
Range $B = [0, 1)$.
Integer in $B$: $\{0\}$ (since 1 is not included).
Set $S = \{x \in Z : x \in A \cup B\}$.
Integers in $A \cup B = \{0\} \cup \{7, 8, \dots, 35\}$.
Since 0 is not in $A$, all elements are distinct.
$n(S) = 1 (\text{from } B) + 29 (\text{from } A) = 30$.
Ans. (1)
Was this solution helpful?
YesNo