Question

Discuss the continuity and differentiability of $f\left(x\right)=\{\begin{array}{cc}|x-3|,& x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$ at $x=1,3.$

• A. continuous at $x=1$ and $x=3$
• B. discontinuous at $x=1$ and $x=3$
• C. differentiable at $x=1$ and $x=3$
• D. differentiable at $x=1$ but not at $x=3$

Answer 1

Answer: (A), (D)

continuous at $x=1$ and $x=3$
differentiable at $x=1$ but not at $x=3$

$f\left(x\right)=\{\begin{array}{cc}x-3,& x\ge 3\\ 3-x,& 1\le x\le 3\\ x^2/4-3x/2+13/4,& x<1.\end{array}$
Since, $f(1-)=f(1+)$ and $f(3-)=f(3+)$
Therefore, f is continuous at $1$ and $3$
now, $f^{\prime }\left(x\right)=\{\begin{array}{cc}1,& x\ge 3\\ -1,& 1\le x\le 3\\ x/2-3/2,& x<1.\end{array}$
Since, $f^{\prime }(1-)=f^{\prime }(1+)$ and $f^{\prime }(3-)\ne f^{\prime }(3+)$
Therefore, f is differentiable at $1$ but not at $3$
Ans: A,D