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6. Application of derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
MCQ
The maximum and minimum values of ${x^3} - 18{x^2} + 96x$ in interval  $(0, 9)$  are
  • A
    $160, 0$
  • B
    $60, 0$
  • ✓
    $160, 128$
  • D
    $120, 28$
✓

Answer

Correct option: C.
$160, 128$
c
(c) Let $y = {x^3} - 18{x^2} + 96x$==> $\frac{{dy}}{{dx}} = 3{x^2} - 36x + 96 = 0$

$\therefore$ ${x^2} - 12x + 32 = 0 \Rightarrow (x - 4)\,(x - 8) = 0$, $x = 4,\,8$

Now, $\frac{{{d^2}y}}{{d{x^2}}} = 6x - 36$

At $x = 4\,\,\,\frac{{{d^2}y}}{{d{x^2}}} = 24 - 36 = - 12 < 0$

At $x = 4$ function will be maximum

and ${[f(4)]_{{\rm{max}}{\rm{.}}}} = 64 - 288 + 384 = 160$

At $x = 8$, $\frac{{{d^2}y}}{{d{x^2}}} = 48 - 36 = 12 > 0$

At $x = 8$, function will be minimum and ${[f(8)]_{{\rm{min}}{\rm{.}}}} = 128.$

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