Choose the correct answer from the given four options : The small… - Vidyadip

MCQ

Choose the correct answer from the given four options : The smallest value of the polynomial $x^3-18 x^2+96 x$ in $[0,9]$ is :

A
$126$
✓
$0$
C
$135$
D
$160$

✓

Answer

Correct option: B.

$0$

We have, $f(x)=x^3-18 x^2+96 x$
$\therefore f^{\prime}(x)=3 x^2-36 x+96$
$f^{\prime}(x)=0$
$\therefore 3 x^2-36 x+96=0$
$\Rightarrow 3\left(x^2-12 x+32\right)=0$
$\Rightarrow(x-8)(x-4)=0$
$\Rightarrow x=4,8$
For least value of $f(x)$ in $[0,9]$, we should find the value of function at $x=0,$
$4,8,9$
$f(0)=0$
$f(4)=4^3-18 \times 4^2+96 \times 4=64-288+384=160$
$f(8)=8^3-18 \times 8^2+96 \times 8=128$
$f(9)=9^3-18 \times 9^2+96 \times 9=729-1458+864=195$
Thus, absolute minimum value of $f$ on $[0,9]$ is $0$ occurring at $x=0$.

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