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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
The function f(x) will be discontinuous at x = a if f(x) has
  • Discontinuity of first kind : $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}-\text{h})$ and $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})$ both exist but are not equal. If is also known as irremovable discontinuity.
  • Discontinuity of second kind : If none of the limits $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}-\text{h})$ and $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})$ exist.
  • Removable discontinuity : $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}-\text{h})$ and $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})$ both exist and equal but not equal to f(a).
Based on the above information, answer the following questions.
  1. If $\text{f}(\text{x})=\begin{cases}\frac{\text{x}^2-9}{\text{x}-3},&\text{for x}\neq3\\4,&\text{for x}=3\end{cases},$ then at x = 3
  1. f has removable discontinuity.
  2. f is continuous.
  3. f has irremovable discontinuity.
  4. None of these.
  1. Let $\text{f}(\text{x})=\begin{cases}\text{x}+2,&\text{if x}\leq4\\\text{x}+4,&\text{if x}\geq4\end{cases}$ then at x = 4
  1. f is continuous.
  2. f has removable discontinuit.
  3. f has irremovable discontinuit.
  4. None of thesee.
  1. Consider the function f(x) defined as $\text{f}(\text{x})=\begin{cases}\frac{\text{x}^2-4}{\text{x}-2},&\text{for x}\neq2\\5,&\text{for x}=2\end{cases},$ then at x = 2
  1. f has removable discontinuity.
  2. f has irremovable discontinuity.
  3. f is continuous.
  4. f is continuous if f(2) = 3
  1. If $\text{f}(\text{x})=\begin{cases}\frac{\text{x}-|\text{x}|}{\text{x}},&\text{x}\neq0\\2,&\text{x}=0\end{cases},$ then at x = 0
  1. f is continuous.
  2. f has removable discontinuity.
  3. f has irremovable discontinuity.
  4. None of these.
  1. If $\text{f}(\text{x})=\begin{cases}\frac{\text{e}^\text{x}-1}{\log(1+2\text{x})},&\text{if x}\neq0\\7,&\text{if x}=0\end{cases},$ then at x = 0
  1. fis continuous if f(0) = 2
  2. f is continuous
  3. f has irremovable discontinuity.
  4. f has removable discontinuity.
✓

Answer

  1. (a) f has removable discontinuity.

Solution:

f(3) = 4

$\lim\limits_{\text{x}\rightarrow3}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-9}{\text{x}-3}=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}+3)(\text{x}-3)}{(\text{x}-3)}$

$=\lim\limits_{\text{x}\rightarrow3}(\text{x}+3)=6\because\lim\limits_{\text{x}\rightarrow3}\text{f}(\text{x})\neq\text{f}(3)$

$\therefore$ f(x) has removable discontinuity at x = 3.

  1. (c) f has irremovable discontinuit.

Solution:

$\lim\limits_{\text{x}\rightarrow4^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow4}(\text{x}+2)=4+2=6$

$\lim\limits_{\text{x}\rightarrow4^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow4}(\text{x}+4)=4+4=8$

$\therefore\lim\limits_{\text{x}\rightarrow4^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow4^+}\text{f}(\text{x})$

$\therefore$ f(x) has an irremovable discontinuity at x = 4.

  1. (a) f has removable discontinuity.

Solution:

$\lim\limits_{\text{x}\rightarrow2}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}^2-4)}{(\text{x}-2)}=\lim\limits_{\text{x}\rightarrow2}(\text{x}+2)=4$

and f(2) = 5 (given) $\therefore\lim\limits_{\text{x}\rightarrow2}\text{f}(\text{x})\neq\text{f}(2)$

$\therefore$ f(x) has removable discontinuity at x = 2.

  1. (c) f has irremovable discontinuity.

Solution:

f(0) = 2

$\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}+\text{x}}{\text{x}}=2$

$\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}-\text{x}}{\text{x}}=0$

$\because\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})$

$\therefore$ f(x) has an irremovable discontinuity at x = 0.

  1. (d) f has removable discontinuity.

Solution:

f(0) = 7

$\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\log(1+2\text{x})}=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{\text{e}^\text{x}-1}{\text{x}}\Big)}{\frac{\log(1+2\text{x})}{2\text{x}}\cdot2}=\frac{1}{2}$

$\because\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})\neq\text{f}(0)$

$\therefore$ f(x) has removable discontinuity at x = 0.

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