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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan 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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  4. x + y + z = 5
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  3. $\frac{\text{x}}{3}+\frac{\text{y}}{6}+\frac{\text{z}}{6}=1$
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Let $T$ be the temperature of the body at any time $t$ and initial time is taken to be $8 \ p.m.$

Based on the above information, answer the following questions.
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  1. $50^\circ F$
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  4. $98.6^\circ F$
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  4. $98.6^\circ F$
  1. The value of $T$ at any time $t$ is:
  1. $50+20\Big(\frac{1}{2}\Big)^\text{t}$
  2. $50+20\Big(\frac{1}{2}\Big)^\text{t-1}$
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  2. $5:30 \ p.m.$
  3. $6:00 \ p.m.$
  4. $5:00 \ p.m.$
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A function f(x) is said to be continuous in an open interval (a, b), if it is continuous at every point in this interval.
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If function $\text{f}(\text{x})=\begin{cases}\frac{\sin(\text{a}+1)\text{x}+\sin\text{x}}{\text{x}}&,\text{x}<0\\\text{c}&,\text{x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^{\frac{3}{2}}}&,\text{x}>0\end{cases}$ is continuous at x = 0, then answer the following questions.
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  2. $0$
  3. $\frac{1}{2}$
  4. $-\frac{1}{2}$
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  4. Any real number.
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  2. $\frac{1}{2}$
  3. $-1$
  4. $-\frac{1}{2}$
  1. The value of a + c is:
  1. 1
  2. 0
  3. -1
  4. -2
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  2. $ < 1, -2, 2 > $
  3. $ < 2, 1, 2 > $
  4. $ < 2, -1, 2 > $
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  4. None of these
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  1. $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{-2}=\frac{\text{z}-1}{2}$
  2. $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
  3. $\frac{\text{x}+3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
  4. None of these
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  1. $x - 2y + 2z = 0$
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  4. Both $(b)$ and $(c)$
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