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STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 6. Application of derivatives4 Marks
MCQ
The maximum value of function ${x^3} - 12{x^2} + 36x +  17$ in the interval $ [1, 10] $ is
  • A
    $17$
  • ✓
    $177$
  • C
    $77$
  • D
    None of these
✓

Answer

Correct option: B.
$177$
b
(b) Let $f(x) = {x^3} - 12{x^2} + 36x + 17$

$\therefore f'(x) = 3{x^2} - 24x + 36 = 0$ at $x = 2,\,6$

Again $f''(x) = 6x - 24$ is $ - ve$ at $x = 2$

So that $f(6) = 17,\;\;f(2) = 49$

At the end points $ = f(1) = 42,\,\,f(10) = 177$

So that $f(x)$ has its maximum value as $ 177.$

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