Q30 / 60 · JEE Advanced 2012 (Paper 1) · Mathematics

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

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

Solution

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.

Hence, the correct answer is option: 'B'.

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