Question

If the function, $f(x)=x^3-6x^2+ax+b$ satisfies Rolles theorem in the interval [1, 3] and $f\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right)=0,$ then

• A. $a=-11$
• B. $a=-6$
• C. $a=6$
• D. $a=11$

Answer 1

Answer: (D)

We have $f(x)=\{\begin{array}{cc}\frac{|x-4|}{x-4},& x\ne 4\\ 0,& x=4\end{array}$ Now, RHL $\underset{x\to 4^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(4+h)$ $=\underset{h\to 0}{\lim }\frac{|4+h-4|}{4+h-4}=\underset{h\to 0}{\lim }\frac{|h|}{h}$ $=\underset{h\to 0}{\lim }1=1$ and LHL $\underset{x\to 4^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }f(4-h)$ $=\underset{h\to 0}{\lim }\frac{|4-h+4|}{4-h+4}$ $=\underset{h\to 0}{\lim }\frac{|-h|}{-h}=\underset{h\to 0}{\lim }(-1)=-1$ Since, $LHL\ne RHL$ ie, $\underset{x\to 4^{-1}}{\lim }f(x)\ne \underset{x\to 4^{+}}{\lim }f(x)$ Thus, limit does not exist.