Discussion Forum

1)

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

         $x^{2}+36x+1$, is 

Answer:

Option B

Explanation:

Concept Involved To check nature of the function

(i) One-One To check one-one, we must check whether

 $f'(x) >0$ or $f'(x) <0$ in given domain

(ii) Onto To check onto, we must check
Range = Co-domain

Description of Situation To find range in given domain [a,b] put f'(x)=0 and find $x= \alpha_{1},\alpha_{2},......\alpha_{n}\epsilon[a,b]$

 Now find, 

  {$f(a),f(\alpha_{1}),f(\alpha_{2}),.....f(\alpha_{n}),f(b)}$}

its the greatest and least gives you range.

 $f:[0,3] \rightarrow [1,29]$

     $f(x)=2x^{3}-15x^{2}+36x+1$

 $\therefore$  $f'(x)=6x^{2}-30x+36$

 =$6(x^{2}-5x+6)=6(x-2)(x-3)$

 

 

For given domain [0,3] ,f(x) is increasing  as well as decreasing

$\Rightarrow$ many-one

 Now, put $f'(x)=0 \Rightarrow x=2,3$

 Thus, for range f(0)=1

  f(2)=29,f(3)=28

 $\Rightarrow$   Range  $\epsilon[1,29]$ $ \therefore$  onto