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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 minimum value of function $f(x) = 3{x^4} - 8{x^3} + 12{x^2} - 48x + 25$ on $[0, 3] $ is equal to
  • A
    $25$
  • ✓
    $-39$
  • C
    $-25$
  • D
    $39$
✓

Answer

Correct option: B.
$-39$
b
(b) $f(x) = 3{x^4} - 8{x^3} + 12{x^2} - 48x + 25$

$\therefore$ $f'(x) = 12{x^3} - 24{x^2} + 24x - 48$

$ = 12[{x^3} - 2{x^2} + 2x - 4]$$ = 12[(x - 2)({x^2} + 2)]$

For maximum and minimum value of the function$f'(x) = 0$

==> $x = 2$. Now $f''(x) = 12[3{x^2} - 4x + 2]$

$\therefore$ $f''(2) = 12\,[12 - 8 + 2] = 72 > 0$

Hence the function is minimum at $x = 2$

Minimum value of $f(x)$on $ [0, 3]$

$ = \min \left\{ {f(0),\,f(2),\,f(3)} \right\}$

$ = \min \left\{ {25,\, - 39,\,16} \right\}$$ = - 39$.

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