Question

If the function $f\left(x\right)=4x^3-6x^2+60$ is decreasing, then

[1 mark]

• A. $x\in (-\infty ,0]\cup [1,\infty )$
• B. $x\in [0,1]$
• C. $x\in [1,\infty )$
• D. $x\in (-\infty ,0]$

Answer 1

The correct option is B $x\in [0,1]$

Given : $f\left(x\right)=4x^3-6x^2+16$
Differentiating w.r.t. $x,$ we get
$f^{\prime }\left(x\right)=12x^2-12x$
For decreasing, $f^{\prime }\left(x\right)\le 0$
$\Rightarrow 12(x^2-x)\le 0$
$\Rightarrow x(x-1)\le 0$
$\Rightarrow x\in [0,1]$