Question

For which interval, the function $f\left(x\right)=\frac{x^2-3x}{x-1}$ satisfies all the conditions of Rolle's theorem.

• A. $[0,3]$
• B. $[-3,0]$
• C. $[1,3]$
• D. For no interval

Answer 1

The correct option is D For no interval

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

Roll's theorem $\to f\left(x\right)=\text{cont}[a,b]\text{diff}(a,b)f\left(a\right)=f\left(b\right)$

$\Rightarrow \exists c\leftarrow (a,b)$

Such that, $f^{{}^{\prime }}\left(c\right)=0$

$f\left(x\right)=\frac{x(x-3)}{x-1}$

$\to \text{In interval}[0,3]\Rightarrow f\left(0\right)=0|f\left(3\right)=0$

$\to \text{In interval}[-3,0]\Rightarrow f(-3)=\frac{-3\times -6}{-3-1}=\frac{-9}{2}|f\left(0\right)=0$

$\to \text{In interval}[1.5,3]\Rightarrow f\left(3/2\right)=\frac{\frac{3}{2}\times \frac{-3}{2}}{\frac{1}{2}}=\frac{-9}{2}|f\left(3\right)=0$

So , in interval [ 0, 3] $\Rightarrow f\left(a\right)=f\left(b\right)$

Now, $f\left(x\right)=\frac{x(x-3)}{x-1}$

$\therefore$ f(x) is not continuous at x = 1

So , correct answer is D for no interval.