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This is one of our many quizzes on Continuity. Click on this to access more such competitive quizzes on Calculus.

Continuity Quiz 2

mathematics

 Arrange them in the following order: Jump Discontinuity, Removable Discontinuity and then Infinite discontinuity <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5f1178af-b408-40bf-8e7e-4da5b3f035ca4427425638199821954.png"/>

 From the definition of jump continuity $\text{(sudden change in graph)}$, removable continuity $\text{(function not defined at particular point)}$ and infinite continuity $\text{(asymptotes are present)}$ find the graph that is best suitable.

  $\text{The jump discontinuity can be represented as: }\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\text{ f}(x)\ne \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\text{ f}(x) \\ \text{The removable discontinuity can be given as: }\underset{x\to a}{\mathop{\lim }}\,\text{ f}(x)\ne \text{f}(a) \\ \text{In infinite discontinuity, the function }f\left( a \right)\text{ is not defined.} \\ \text{Since the value of the function at }x = a\text{ tends to infinity.} \\ \text{In first fig, }\underset{x\to a}{\mathop{\lim }}\,\text{ f}(x)\ne \text{f}(a),\text{  so it is a removable discontinuity.} \\ \text{In the second fig, } \text{asymptotes are present.} \text{ We know that the function tends to infinity} \text{ when asymptotes are present.} \text{Thus, it has an infinite type of discontinuity.} \\ \text{In third fig, }\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\text{ f}(x)\ne \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\text{ f}(x),\text{ so it is a jump discontinuity.}$

 Find the values of $a$ and $b$ respectively if the curve is continuous: $f\left( x \right)\text{= }\left\{ \begin{align}  & {{\text{x}}^{2}}+a\text{        }; x\le 2 \\  & \text{5               ; 2x}\le \text{3} \\  & \text{bx-7           ; 3x } \\ \end{align} \right.$

 For a curve to be continuous, it follows $\underset{x\to a}{\mathop{\lim }}\,\text{ f}\left( x \right)=f\left( a \right)$. There are two points if break i.e., at $x=2$ and $x=3$. Use above relation to create two linear equation and by solving we can get values of $a$ and $b$ 

  $\text{For a continuous curve, }\underset{x\to a}{\mathop{\lim }}\,\text{ f}\left( x \right)=f\left( a \right) \\ \text{So, }\underset{x\to 2}{\mathop{\lim }}\,\text{ f}\left( x \right)=f\left( 2 \right) \\ \Rightarrow \underset{x\to 2}{\mathop{\lim }}\, 5={{2}^{2}}+a \\ \Rightarrow 5=4+a \\ \Rightarrow a=1 \\ \text{Next,  }  \underset{x\to 3}{\mathop{\lim }}\,\text{ f}\left( x \right)=f\left( 3 \right) \\ \Rightarrow \underset{x\to 3}{\mathop{\lim }}\,\text{ bx-7}=5 \\ \Rightarrow 3b-7=5 \\ \Rightarrow b=4$

 Find the points of discontinuity in the function $f\left( x \right)$ in the range $\left[ -2,5 \right]$. $f\left( x \right)=\left[ \dfrac{|x+1|}{x+1} \right]$ where, $|.|$ represents mod value and $\left[ . \right]$ represents gif.

 Remove the modulus symbol from the given function by taking the piece wise function. Then draw the graph of piecewise function to get the point of discontinuity.

  $f\left( x \right)=\left[ \dfrac{|x+1|}{x+1} \right] \\ \text{Point of discontinuity }=-1 \\ f\left( x \right)\text{= }\left\{ \begin{align}  & \dfrac{\text{x+1}}{\text{x+1}}\text{=1                  };x\ge -1 \\  & \dfrac{-\left( \text{x+1} \right)}{\text{x+1}}\text{=-1 ;x-1} \\ \end{align} \right. \\ \text{ It has only 1 point of discontinuity} \\ $ <br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4fc0f696-22c3-44ee-94a7-9e2beeb6f0171556422847830488431.png"/>

 Find the value of $“a” \text{(“a” has finite value)}$ in the following functions so that the functions become continuous. $f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{\sin \text{ x}}$   

 Solve the given function using standard formulas of limits. If the limit is constant then it is independent of $'a'$ or else substitute some value for $'a'$ to get the limit value.

  $f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{\sin \text{ x}}\text{ is continuous so it should have a finite value at }x=0 \\ \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{\sin \text{ x}}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{{{x}^{2}} \dfrac{\sin \text{ x}}{{{x}^{2}}}} \\ \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{\sin \text{ x}}=\underset{x\to 0}{\mathop{\lim }}\, ax \dfrac{\dfrac{\sin  \left( {{x}^{2}} \right)}{{{x}^{2}}}}{ \dfrac{\sin \text{ x}}{x}}  \\ \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{\sin \text{ x}}=\underset{x\to 0}{\mathop{\lim }}\, ax \times \underset{x\to 0}{\mathop{\lim }}\, \dfrac{\dfrac{\sin  \left( {{x}^{2}} \right)}{{{x}^{2}}}}{ \dfrac{\sin \text{ x}}{x}} \\ \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{\sin \text{ x}}=\underset{x\to 0}{\mathop{\lim }}\, ax \times 1  \\ \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{a \sin  \left( {{x}^{2}} \right)}{\sin \text{ x}}=\underset{x\to 0}{\mathop{\lim }}\, ax=0 \\ \text{Therefore, }f\left( x \right)\text{ is independent of the value of }a$.

 There is a continuous curve. Following operations are performed: The curve was rotated around the y axis. The amplitude of the curve was scaled down by 2. The frequency of the curve was done twice to the current value. The curve was multiplied with ${{e}^{2x}}$. And the curve was rotated again around y axis. This is the resulting curve. Find the nature of the new curve for the domain $ x \ge 0$ and also notify the points of discontinuity if there. 

 Assume any continuous curve. Then apply the given condition that is amplitude called down implies function is divided and then amplitude increased implies that $'x'$ is increased and rotation about $y-\text{axis}$ implies that sign of $'x'$ reversed to find the new curve to get continuity.

  $\text{Take a continuous curve for convenience.} \\ \text{Let the curve be }f\left( x \right)=2 \sin \text{ x} \\ \text{Curve rotated around the y axis means }{f}'\left( x \right)=2\sin  \left( -x \right) \\ \text{Amplitude scaled down by 2 means }{f}''= \dfrac{\text{2 sin}\left( -x \right)}{2}=\text{ sin}\left( -x \right) \\ \text{Frequency is doubled means }{f}'''= \sin \left( -2x \right) \\ \text{Curve was multiplied by }{{e}^{2x}}\text{ results a new curve }={{e}^{2x}}\text{ sin}\left( -2x \right) \\ \text{Finally it rotated again to get the required curve }= {{e}^{-2x}}\text{ sin}\left( 2x \right) \\ \text{For checking the continuity,} \text{ both the curves need to be analysed. } \\ \sin \left( 2x \right)\text{ is continuous and }{{e}^{-2x}}\text{ is also continuous for the given domain} \\ \text{Thus it is known that the product of two } \text{continuous functions results for a continuous function.}$