Question

The function $f(x)=\{\begin{array}{ll}5x-4& \text{for }0<x\le 1\\ 4x^2-3x& \text{for }1<x<2,\\ 3x+4& \text{for }x\ge 2,\end{array}$
then which of the following is not true?

• A. Continuous at $x=1$ and $x=2$
• B. Continuous at $x=1$ but not derivable at $x=2$
• C. Continuous at $x=2$ but not derivable at $x=1$
• D. None of these

Answer 1

Answer: (C)

$f(x)=\{\begin{array}{ll}5x-4& \text{for }0<x\le 1\\ 4x^2-3x& \text{for }1<x<2,\\ 3x+4& \text{for }x\ge 2,\end{array}$
and $f^{\prime }(x)=\{\begin{array}{ll}5& \text{for }0<x<1\\ 8x-3& \text{for }1<x<2\\ 3& \text{for }x>2\end{array}$
$f\left(1^{-}\right)=1;f\left(1^{+}\right)=1;f(1)=1$ and $f^{\prime }\left(1^{-}\right)=5;f^{\prime }\left(1^{+}\right)=5$
$f(2)=f\left(2^{+}\right)=10,f\left(2^{-}\right)=10;f^{\prime }\left(2^{+}\right)=3;f\left(2^{-}\right)=13$
Thus $f(x)$ is continuous at $x=1$ and $x=2,$ differential at $x=1$ but non-differentiable at $x=2$