NIELIT 2016 MAR Scientist B - Section B: 10

Question

Maxima and minimum of the function 

$f(x)=2x^3-15x^2+36x+10$ occur; respectively at 

• A. $x=3$ and $x=2$
• B. $x=1$ and $x=3$
• C. $x=2$ and $x=3$
• D. $x=3$ and $x=4$

Answer 1

The points of maxima and minima can be found by differentiating the function w.r.t  $x$ and equating it to zero.

$\therefore$  $f'(x)=6x^{2}-30x+36$

$\Rightarrow$ $6x^{2}-30x+36=0$  $\Rightarrow$  $x^{2}-5x+6=0$  $\Rightarrow$  $(x-2)\times(x-3)=0$

$\Rightarrow$ $x=2$  and $x=3$

Now, we have to find the value of $f''(x)$ at above points.

$\therefore$   $f''(x)=12x-30$

$\therefore$   $f''(2)=12(2)-30=-6$   $\Rightarrow$  at  $x=2$ , function has maxima as  $f''(2)<0$

Also,  $f''(3)=12(3)-30=6$    $\Rightarrow$   at   $x=3$,  function has minima as  $f''(3)>0$

Option C is correct.