Question

The function $f:[0,3]$ $\to$ $[1,29]$, defined by $f\left(x\right)=2x^3-15x^2+36x+1$, is

• A. one-one and onto.
• B. onto but not one-one.
• C. one-one but not onto.
• D. neither one-one nor onto.

Answer 1

The correct option is B onto but not one-one.

$f\left(x\right)=2x^3-15x^2+36x+1,f:[0,3]\to [1,29]$
1. Check for one-one mapping-
$f\left(x\right)$ is a cubic polymonial. First derivative of $f\left(x\right)$ gives $f^{\prime }\left(x\right)=6x^2-30x+36$.
Solve $f^{\prime }\left(x\right)=0$ to get the critical points.
$x^2-5x+6=0$
$(x-3)(x-2)=0\Rightarrow x\in \{3,2\}$
$f\left(x\right)$ is increasing for $x<2$ or $x>3$ $(\because f^{\prime }(x)>0)$ and $f\left(x\right)$ is decreasing for $2<x<3$ $(\because f^{\prime }(x)<0)$.
Hence, $f\left(x\right)$ is not strictly increasing or strictly decreasing in the entire domain. So $f\left(x\right)$ is not one-one.
2. Check for onto mapping-
$f\left(x\right)$ has maximum at $x=2$
$f\left(2\right)=29$, which is maximum value of $f$.
$(\because$ in the given domain $[0,3]$, f is increasing in $[0,2)$ and decreasing in $(2,3]$.)
$f\left(0\right)=1$ and $f\left(3\right)=28$
Hence, range of $f\left(x\right)=[1,29]$ which is equal to the co-domain. Hence, $f\left(x\right)$ is onto.