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) $f(x) = 2{x^3} - 15{x^2} + 36x + 4$
(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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