The function $f: [0, 3] → [1, 29]$, defined by $f(x)=2x^3-15x^2 + 36x + 1$, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → onto but not one-one
We have,
$f(x)=2x^3-15x^2 + 36x+1$
$∴f'(x) = 6x^2 - 30x + 36 = 6(x − 3) (x − 2)$
Clearly, $f'(x) > 0$ for $x ∈ [0, 2)$ and $f'(x) <0$ for $x ∈ (2, 3)$. So, f(x) is increasing on (0, 2) and decreasing (2, 3).
Also, $f(0)=1, f(2) = 29$ and $f(3) = 28$.
As f is a continuous function. So, it attains every value between its minimum value 1 and maximum value $f(2) = 29$. So, f is onto and it is not one-one.