Question

If $f\left(x\right)=\sqrt{x^2-10x+25}$, then the derivative of f (x) in the interval [0, 7] is
(a) 1
(b) −1
(c) 0
(d) none of these

Answer 1

(d) none of these

$$\begin{gathered} \mathrm{We}\mathrm{have},f\left(x\right)=\sqrt{x^2-10x+25} \\ =\sqrt{{\left(x-5\right)}^2} \\ =\left|x-5\right| \\ =\left\{\begin{array}{l}x-5\mathrm{for}x>5 \\ \\ -\left(x-5\right)\mathrm{for}x<5\end{array}\right. \\ \mathrm{LHD}=\underset{x\to 5^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a} \\ =\underset{x\to 5^{-}}{\lim }\frac{\sqrt{x^2-10x+25}-\sqrt{5^2-10\left(5\right)+25}}{x-5} \\ =\underset{x\to 5^{-}}{\lim }\frac{\left|x-5\right|}{x-5} \\ =\underset{x\to 5^{-}}{\lim }\frac{-\left(x-5\right)}{x-5} \\ =-1 \\ \mathrm{RHD}=\underset{x\to 5^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a} \\ =\underset{x\to 5^{+}}{\lim }\frac{\sqrt{x^2-10x+25}-\sqrt{5^2-10\left(5\right)+25}}{x-5} \\ =\underset{x\to 5^{+}}{\lim }\frac{\left|x-5\right|}{x-5} \\ =\underset{x\to 5^{+}}{\lim }\frac{x-5}{x-5} \\ =1 \\ \mathrm{Here},\mathrm{LHD}\ne \mathrm{RHD} \\ \mathrm{Thus},\mathrm{the}\mathrm{functionis}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=5 \end{gathered}$$