Question

Which of the following option(s) is(are) false

• A. Rolle's theorem holds for $f\left(x\right)=|x-3|$ in $[2,4]$
• B. Rolle's theorem holds for $f\left(x\right)=x^3-3x$ in $[0,\sqrt{3}]$
• C. Rolle's theorem does not hold for $f\left(x\right)=\cos |x|$ in $[-\frac{\pi }{2},\frac{\pi }{2}]$
• D. Rolle's theorem does not hold for $f\left(x\right)=1-\sqrt[3]{3x^4}$ in $[-1,1]$

Answer 1

Answer: (A), (C), (D)

Rolle's theorem does not hold for $f\left(x\right)=1-\sqrt[3]{3x^4}$ in $[-1,1]$

option $\left(a\right)\to$ False,because $f$ is not differentiable at $x=3$

option $\left(b\right)\to$True
$f\left(0\right)=f\left(\sqrt{3}\right)=0,f$is continuos in $[0,\sqrt{3}]$ and differentiable in $(0,\sqrt{3})$
$f^{\prime }\left(x\right)=0\Rightarrow 3x^2-3=0$
$\Rightarrow x=\pm 1$ and $x=1\in (0,\sqrt{3})$

option $\left(c\right)\to$ False
$f(-\frac{\pi }{2})=f\left(\frac{\pi }{2}\right)=0$
$f\left(x\right)=\cos |x|=\cos x,\forall x\in [-\frac{\pi }{2},\frac{\pi }{2}]$
So, $f$ is continuous in $[-\frac{\pi }{2},\frac{\pi }{2}]$ and differentiable in $(-\frac{\pi }{2},\frac{\pi }{2})$
$f^{\prime }\left(x\right)=0$ $\Rightarrow -\sin x=0$
$\Rightarrow x=0\in (-\frac{\pi }{2},\frac{\pi }{2})$

option $\left(d\right)\to$False
$f(-1)=f\left(1\right)=0$
$f$ is continuous in $[-1,1]$ and differentiable in $(-1,1)$
$f^{\prime }\left(x\right)=0\Rightarrow -\frac{4}{3}\cdot x^{\frac{1}{3}}=0$
$\Rightarrow x=0\in (-\frac{\pi }{2},\frac{\pi }{2})$