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The function $f:[0,3] \rightarrow[1,29]$, defined by $f(x)=2 x^{3}-15 x^{2}+36 x+1$, is
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onto but not one-one
Given : $f(x)=2 x^{3}-15 x^{2}+36 x+1$
$\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$
$=6\left(x^{2}-5 x+6\right)=6(x-2)(x-3)$
$\because f^{\prime}(x)>0 \forall x \in[0,2)$ and $f^{\prime}(x) < 0 \forall x \in(2,3)$
$\therefore f(x)$ is increasing on $[0,2)$ and decreasing
on $(2,3)$
$\therefore \quad f(x)$ is many one on $[0,3]$
Also $f(0)=1, f(2)=29, f(3)=28$
$\therefore \quad$ Absolute min $=1$ and Absolute max $=29$
$\therefore \quad$ Range of $f=[1,29]=$ codomain
Hence $f$ is onto.
$\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$
$=6\left(x^{2}-5 x+6\right)=6(x-2)(x-3)$
$\because f^{\prime}(x)>0 \forall x \in[0,2)$ and $f^{\prime}(x) < 0 \forall x \in(2,3)$
$\therefore f(x)$ is increasing on $[0,2)$ and decreasing
on $(2,3)$
$\therefore \quad f(x)$ is many one on $[0,3]$
Also $f(0)=1, f(2)=29, f(3)=28$
$\therefore \quad$ Absolute min $=1$ and Absolute max $=29$
$\therefore \quad$ Range of $f=[1,29]=$ codomain
Hence $f$ is onto.
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