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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
At the point $x = 1$, the given function $f(x) = \left\{ \begin{array}{l}{x^3} - 1;\,\,1 < x < \infty \\x - 1;\,\, - \infty < x \le 1\end{array} \right.$ is
  • A
    Continuous and differentiable
  • B
    Continuous and not differentiable
  • C
    Discontinuous and differentiable
  • D
    Discontinuous and not differentiable
✓

Answer

We have $Rf'(1) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 + h) - f(1)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left\{ {{{(1 + h)}^3} - 1} \right\} - 0}}{h} = 3$
$Lf'(1) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 - h) - f(1)}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left\{ {(1 - h) - 1} \right\} - 0}}{{ - h}} = 1$
$\therefore Rf'(1) \ne Lf'(1)$
$ \Rightarrow f(x)$ is not differentiable at $x = 1.$
Now, $f(1 + 0) = \mathop {\lim }\limits_{h \to 0} f(1 + h) = 0$
and $f(1 - 0) = \mathop {\lim }\limits_{h \to 0} f(1 - h) = 0$
$\therefore f(1 + 0) = f(1 - 0) = f(0)$
$ \Rightarrow f(x)$ is continuous at $x = 1.$
Hence at $x = 1,f(x)$ is continuous and not differentiable.

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