Question

f(x) = $\{\begin{array}{cc}\frac{4-x}{2-\sqrt{x}},& for0<x<4\\ 4,& forx=4\\ 16-3x& for4<x<6\end{array}$
Which of the following properties does $f$ have on the interval $(0,6)$ ?
I. ln $f\left(x\right)$ exists
II. f is continuous
III. f is monotonic

• A. I only
• B. II only
• C. III only
• D. None

Answer 1

Answer: (B)

II only

${lim}_{x\to 4^{-}}f\left(x\right)$

$={lim}_{x\to 4^{-}}\frac{4-x}{2-\sqrt{x}}$

$={lim}_{x\to 4^{-}}(2+\sqrt{x})$

$=2+2=4$.

${lim}_{x\to 4^{+}}f\left(x\right)$

$={lim}_{x\to 4^{+}}16-3x$

$=16-12=4$

Hence $LHL=RHL=$ value of function at $4$.

Hence $f\left(x\right)$ is continuous.

Consider differentiability

${lim}_{x\to 4^{-}}{f}^{\prime }\left(x\right)$

$={lim}_{x\to 4^{-}}\frac{d(2+\sqrt{x})}{dx}$

$=\frac{1}{4}$.

${lim}_{x\to 4^{+}}{f}^{\prime }\left(x\right)$

$={lim}_{x\to 4^{+}}\frac{d(16-3x)}{dx}$

$=-3$

Hence

$LHD\ne RHD$

Therefore the function is not differentiable.

Hence $f\left(x\right)$ is only continuous over the interval of $(0,6)$ and is not differentiable at $x=4.$