Question

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

• A. $f\left(x\right)=\{\begin{array}{cc}x,& 0\le x<1\\ 0,& x=1\end{array}$ on $[0,1]$
• B. $f\left(x\right)=\{\begin{array}{cc}\frac{\sin x}{x},& -\pi \le x<0\\ 0,& x=0\end{array}$ on $[-\pi ,0]$
• C. $f\left(x\right)=\frac{x^2-x-6}{x-1}$ on $[-2,3]$
• D. $f\left(x\right)=\{\begin{array}{cc}\frac{x^3-2x^2-5x+6}{x-1},& ifx\ne 1\\ -6,& ifx=1\end{array}$ on $[-2,3]$

Answer 1

The correct option is D $f\left(x\right)=\{\begin{array}{cc}\frac{x^3-2x^2-5x+6}{x-1},& ifx\ne 1\\ -6,& ifx=1\end{array}$ on $[-2,3]$

Option A, $f\left(x\right)=\{\begin{array}{cc}x,& 0\le x<1\\ 0,& x=1\end{array}$ on $[0,1]$

We will check continuity at $x=1$

$LHL={lim}_{x\to 1^{-}}f\left(x\right)={lim}_{h\to 0}f(1-h)=1$

Also, $f\left(1\right)=0$

Since, $LHL\ne f\left(1\right)$

f(x) is discontinuous at $x=1$

So, Rolle's theorem does not hold for option A.

Option B,$f\left(x\right)=\{\begin{array}{cc}\frac{\sin x}{x},& -\pi \le x<0\\ 0,& x=0\end{array}$ on $[-\pi ,0]$

We know that ${lim}_{x\to 0}\frac{\sin x}{x}=1$

Also, $f\left(0\right)=0$

f(x) is discontinuous at $x=0$

So, Rolle's theorem does not hold for option B.

Option C, $f\left(x\right)=\frac{x^2-x-6}{x-1}$ on $[-2,3]$

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

f(x) is not defined at $x=1$

So, Rolle's theorem does not hold for option C.

$f\left(x\right)=\{\begin{array}{cc}\frac{x^3-2x^2-5x+6}{x-1},& ifx\ne 1\\ -6,& ifx=1\end{array}$ on $[-2,3]$

$\Rightarrow f\left(x\right)=(x-3)(x+2)$ at $x\ne 1$

Polynomial functions are continuous and differentiable everywhere.

Also, $f(-2)=f\left(3\right)=0$

So, Rolle's theorem holds for option D.