Question

The points of discontinuity of the function\[f\left( x \right) = \begin{cases}\frac{1}{5}\left( 2 x^2 + 3 \right) , & x \leq 1 \\ 6 - 5x , & 1 < x < 3 \\ x - 3 , & x \geq 3\end{cases}\text{ is } \left( are \right)\]  

विकल्प

•  x = 1

• x = 3

•  x = 1, 3

• none of these

Answer 1

x = 3 

If \[x \leq 1\] , then

\[f\left( x \right) = \frac{1}{5}\left( 2 x^2 + 3 \right)\] .

Since 

\[2 x^2 + 3\] is a polynomial function and 

\[\frac{1}{5}\] is a constant function, both of them are continuous. So, their product will also be continuous.

 Thus, 

\[f\left( x \right)\]  is continuous at   \[x \leq 1\] 

If \[1 < x < 3\], then

\[f\left( x \right) = 6 - 5x\] .

Since  

\[5x\] is a polynomial function and  \[6\]  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  \[1 < x < 3\] 

If  \[x \geq 3\], then \[f\left( x \right) = x - 3\]

Since  

\[x - 3\]  is a polynomial function, it is continuous. So,

\[f\left( x \right)\]  is continuous for every   \[x \geq 3\]

Now,

Consider the point   \[x = 1\] . Here,

 \[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left( \frac{1}{5}\left[ 2 \left( 1 - h \right)^2 + 3 \right] \right) = 1\]

\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = \lim_{h \to 0} \left( 6 - 5\left( 1 + h \right) \right) = 1\]

Also,

\[f\left( 1 \right) = \frac{1}{5}\left( 2 \left( 1 \right)^2 + 3 \right) = 1\]

Thus,

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

 Hence

\[f\left( x \right)\] is continuous at  \[x = 1\] .

Now,
Consider the point 

\[x = 3\]. Here,

\[\lim_{x \to 3^-} f\left( x \right) = \lim_{h \to 0} f\left( 3 - h \right) = \lim_{h \to 0} \left( 6 - 5\left( 3 - h \right) \right) = - 9\]

\[\lim_{x \to 3^+} f\left( x \right) = \lim_{h \to 0} f\left( 3 + h \right) = \lim_{h \to 0} \left( \left( 3 + h \right) - 3 \right) = 0\]

Also,

\[f\left( 1 \right) = \frac{1}{5}\left( 2 \left( 1 \right)^2 + 3 \right) = 1\]

Thus,

\[\lim_{x \to 3^-} f\left( x \right) \neq \lim_{x \to 3^+} f\left( x \right)\]

Hence,

\[f\left( x \right)\] is discontinuous at  \[x = 3\] .

So, the only point of discontinuity of 

\[f\left( x \right)\] is  \[x = 3\] .