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

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
MCQ
The function $f(x) = 2{x^3} - 15{x^2} + 36x + 4$  is maximum at $x=$ ......
  • ✓
    $2$
  • B
    $4$
  • C
    $0$
  • D
    $3$
✓

Answer

Correct option: A.
$2$
a
(a) $f(x) = 2{x^3} - 15{x^2} + 36x + 4$

==> $f'(x) = 6{x^2} - 30x + 36$.…..$(i)$

We know that for its maximum value $f'(x) = 0.$

$6{x^2} - 30x + 36 = 0$ ==> $(x - 2)(x - 3) = 0$ ==> $x = 2,\,3.$

Again differentiating equation $(i),$ we get

$f''(x) = 12x - 30$ ==> $f''(2) = 24 - 30 = - 6 < 0$.

Therefore $f(x)$ is maximum at $x = 2.$

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