Test the continuity of a function at x = 2 {cases} 5/2-x, & x 2 {cases}

Correct Answer is : Discontinuous at x = 2

Correct Answer is : Discontinuous at x = 2

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UPSC NDA Math Model Paper 3

Question : 118 of 120

Marks: +1, -0

\begin{cases} \frac{5}{2}-x, & x<2 \\ 1, & x=2 \\ x-\frac{3}{2}, & x>2 \end{cases}

Solution:

f(x)=\begin{cases} \frac{5}{2}-x, & x<2 \\ 1, & x=2 \\ x-\frac{3}{2}, & x>2 \end{cases}

\lim\limits_{\alpha\rightarrow 0} f(x-\alpha)=\frac{5}{2}-x

For

x=2

\Rightarrow \lim\limits_{\alpha\rightarrow 0} f(2-\alpha)=\lim\limits_{\alpha\rightarrow 0}\left[\frac{5}{2}-(2-\alpha)\right]

\Rightarrow \lim\limits_{\alpha\rightarrow 0} f(2-\alpha)=\lim\limits_{\alpha\rightarrow 0}\left(\frac{1}{2}+\alpha\right)=\frac{1}{2}

\lim\limits_{\alpha\rightarrow 0} f(x+\alpha)=x-\frac{3}{2}

For

x=2

\Rightarrow \lim\limits_{\alpha\rightarrow 0} f(2+\alpha)=\lim\limits_{\alpha\rightarrow 0}\left[(2+\alpha)-\frac{3}{2}\right]

\Rightarrow \lim\limits_{\alpha\rightarrow 0} f(2+\alpha)=\lim\limits_{\alpha\rightarrow 0}\left(\frac{1}{2}+\alpha\right)=\frac{1}{2}

f(x)=f(2)=1

\lim\limits_{\alpha\rightarrow 0} f(2-\alpha)=\lim\limits_{\alpha\rightarrow 0} f(2+\alpha) \neq f(2)

\therefore

The function is not continuous at

x=2

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