Question

In which of the following functions, Rolle's theorem is applicable

• A. $f\left(x\right)=|x|$ in $-2\le x\le 2$
• B. $f\left(x\right)=tanx$ in $0\le x\le \pi$
• C. $f\left(x\right)=1+(x-2{)}^{\frac{2}{3}}$ in $1\le x\le 3$
• D. $f\left(x\right)=x(x-2{)}^2$ in $0\le x\le 2$

Answer 1

The correct option is C $f\left(x\right)=1+(x-2{)}^{\frac{2}{3}}$ in $1\le x\le 3$

1)$f\left(x\right)=\left|x\right|$

Given interval is $[-2,2]$

For Rolle's theorem to be applicable, function must be differentiable within given interval.

At $x=0$, $f\left(x\right)=0$

Thus, function becomes non differentiable at $x=0$. Thus, Rolle's theorem is not applicable.

2) $f\left(x\right)=tanx$ in the interval $[0,\pi ]$

For Rolle's theorem to be applicable, function must be continuous within given interval.

At $x=\frac{\pi }{2}$, $f\left(x\right)=tan\left(\frac{\pi }{2}\right)=\infty$

Thus, function is non continuous in the given interval. Thus, Rolle's theorem is not applicable.

3) $f\left(x\right)=1+{(x-2)}^{\frac{2}{3}}$ in the interval $[1,3]$

$\therefore f\left(x\right)=1+{\left({(x-2)}^{\frac{1}{3}}\right)}^2$

For all values of x in the given interval, $f\left(x\right)\ne 0$

Thus, $f\left(x\right)$ is differentiable in given interval

For all values of x in the given interval, $f\left(x\right)\ne 0$

Thus, $f\left(x\right)$ is continuous in the given interval.

$f\left(a\right)=f\left(1\right)=1+{\left[{(1-2)}^{\frac{1}{3}}\right]}^2$

$\therefore f\left(a\right)=1+{\left[{(-1)}^{\frac{1}{3}}\right]}^2$

$\therefore f\left(a\right)=1+{\left[(-1)\right]}^2$

$\therefore f\left(a\right)=1+1=2$

Now, $f\left(b\right)=f\left(3\right)=1+{\left[{(3-2)}^{\frac{1}{3}}\right]}^2$

$\therefore f\left(b\right)=1+{\left[{\left(1\right)}^{\frac{1}{3}}\right]}^2$

$\therefore f\left(b\right)=1+{\left[1\right]}^2$

$\therefore f\left(b\right)=1+1=2$

$\therefore f\left(a\right)=f\left(b\right)$

Thus, Rolle's theorem is applicable.