Question

The points of discontinuity of the function 

\[f\left( x \right) = \begin{cases}2\sqrt{x} , & 0 \leq x \leq 1 \\ 4 - 2x , & 1 < x < \frac{5}{2} \\ 2x - 7 , & \frac{5}{2} \leq x \leq 4\end{cases}\text{ is } \left( \text{ are }\right)\] 

Options

• x = 1 \[x = \frac{5}{2}\] 

• \[x = \frac{5}{2}\] 

• \[x = 1, \frac{5}{2}, 4\]

•  x = 0, 4

Answer 1

\[ x = \frac{5}{2}\]

 If  \[0 \leq x \leq 1\], then   \[f\left( x \right) = 2\sqrt{x}\] .

Since 

\[f\left( x \right) = 2\sqrt{x}\]  is a polynomial function, it is continuous. 

Thus,  

\[f\left( x \right)\]  is continuous for every  \[0 \leq x \leq 1\] .

If  \[1 < x < \frac{5}{2}\]  , then  

\[f\left( x \right) = 4 - 2x\] . Since 

\[2x\] is a polynomial function and 4 is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus,

\[f\left( x \right)\]  is continuous for every 

\[1 < x < \frac{5}{2}\] .

If  \[\frac{5}{2} \leq x \leq 4\] , then  

\[f\left( x \right) = 2x - 7\] Since  

\[2x\] is a polynomial function and 7 is continuous function, their difference will also be continuous.
Thus,

\[f\left( x \right)\] is continuous for every 

\[\frac{5}{2} \leq x \leq 4\] .

Now,
Consider the point  

\[x = 1\] Here,

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left( 2\left( \sqrt{1 - h} \right) \right) = 2\]

\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = \lim_{h \to 0} \left( 4 - 2\left( 1 + h \right) \right) = 2\] 

Also,  

\[f\left( 1 \right) = 2\sqrt{1} = 2\] 

\[\Rightarrow \lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]

Thus

\[f\left( x \right) \text{is continuous at x} = 1\], 

Now,
Consider the point  \[x = \frac{5}{2}\] Here ,

\[\lim_{x \to \frac{5}{2}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{5}{2} - h \right) = \lim_{h \to 0} \left( 4 - 2\left( \frac{5}{2} - h \right) \right) = - 1\]

\[\lim_{x \to \frac{5}{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{5}{2} + h \right) = \lim_{h \to 0} \left( 2\left( \frac{5}{2} - h \right) - 7 \right) = - 2\]

\[\Rightarrow \lim_{x \to \frac{5}{2}^+} f\left( x \right) \neq \lim_{x \to \frac{5}{2}^-} f\left( x \right)\]

Thus, 

\[f\left( x \right) \text{is discontinuous at x} = \frac{5}{2}\]