If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,\,\, - {x^2},\,{\rm{when\,\, }}x \le 0\\\,\,\,\,\,5x - 4,\,{\rm{when\,\,}}0 < x \le 1\\4{x^2} - 3x,\,{\rm{when\,\, }}1 < x < 2\\\,\,\,\,\,3x + 4,{\rm{when \,\,}}x \ge 2\end{array} \right.$, then
- A
$f(x)$ is continuous $x = 0$
- B
$f(x)$ is continuous $x = 2$
- C
$f(x)$is discontinuous at$x = 1$
- D
✓
Answer
$\mathop {\lim }\limits_{x \to 0 - } f(x) = 0$
$f(0) = 0,\mathop {\lim }\limits_{x \to 0 + } f(x) = - 4$
$f(x)$ discontinuous at $x = 0.$
and $\mathop {\lim }\limits_{x \to 1 - } f(x) = 1$ and $\mathop {\lim }\limits_{x \to 1 + } f(x) = 1,f(1) = 1$
Hence $f(x)$ is continuous at $x = 1$.
Also $\mathop {\lim }\limits_{x \to 2 - } f(x) = 4{(2)^2} - 3\,.\,2 = 10$
$f(2) = 10$ and $\mathop {\lim }\limits_{x \to 2 + } f(x) = 3(2) + 4 = 10$
Hence $f(x)$ is continuous at $x = 2.$
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