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 $x^3 - 18x^2 + 96x$ in $[0, 9]$ is:

A
$126$
✓
$0$
C
$135$
D
$160$

✓

Answer

Correct option: B.

$0$

We have, $f(x) = x^3 - 18x^2 + 96x$
$\therefore f'(x) = 3x^2 - 36x + 96$
$f'(x) = 0$
$\therefore 3x^2 - 36x + 96 = 0$
$\Rightarrow 3(x^2 - 12x + 32) = 0$
$\Rightarrow (x - 8)(x - 4) = 0$
$\Rightarrow x = 4, 8$
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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