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STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
The minimum value of the function $y = 2{x^3} - 21{x^2} + 36x - 20$ is
  • ✓
    $-128$
  • B
    $-126$
  • C
    $-120$
  • D
    None of these
✓

Answer

Correct option: A.
$-128$
a
(a) Given, $f(x) = 2{x^3} - 21{x^2} + 36x - 20$

$f'(x) = 6{x^2} - 42x + 36$

Put $f'(x) = 0$==> $6{x^2} - 42x + 36 = 0$==> ${x^2} - 7x + 6 = 0$

==> ${x^2} - 6x - x + 6 = 0$==> $(x - 1)(x - 6) = 0$==> $x = 1,\,6$

Now, $f''\,(x) = 12x - 42$

$f''\,(1) = - 30 = - ve$ and $f''\,(6) = 30 = + ve$

Hence $x = 6$ is the point of minima

$\therefore$ Minimum value = $f(6) = 2{(6)^3} - 21{(6)^2} + 36 \times 6 - 20$

$\therefore$ $f(6) = - 128$.

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