Answer
The number for which the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\cos x}{x-\frac{\pi }{2}}\text{if }x\ne \frac{\pi }{2} \\
& \text{1 if }x=\frac{\pi }{2}
\end{align} \right.$ is discontinuous is $ a=\frac{\pi }{2}$.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\cos x}{x-\frac{\pi }{2}}\text{if }x\ne \frac{\pi }{2} \\
& \text{1 if }x=\frac{\pi }{2}
\end{align} \right.$,
Check the discontinuity of the function at $ a=\frac{\pi }{2}$ .
Find the value of $ f\left( x \right)$ at $ a=\frac{\pi }{2}$,
From the definition of the function,
$ f\left( \frac{\pi }{2} \right)=1$
The function is defined at the point $ a=\frac{\pi }{2}$.
Now find the value of $\,\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)$,
Take some points near to the left of $\frac{\pi }{2}$.
Suppose the points are $\frac{\pi }{2}-0.1,\frac{\pi }{2}-0.01\text{ and }\frac{\pi }{2}-0.001$
Take some points near to the right of $\frac{\pi }{2}$.
Suppose the points are $\frac{\pi }{2}+0.001,\frac{\pi }{2}+0.01\text{ and }\frac{\pi }{2}+0.1$
Now evaluate the values of the function at the above chosen points.
As x nears $\frac{\pi }{2}$ from the left or right the value of the function nears $-1$.
Thus $\,\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)=-1$.
From the above steps, $\,\,\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)\ne f\left( \frac{\pi }{2} \right)$.
Thus, the function does not satisfy the third property of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& \frac{\cos x}{x-\frac{\pi }{2}}\text{if }x\ne \frac{\pi }{2} \\
& \text{1 if }x=\frac{\pi }{2}
\end{align} \right.$ is discontinuous at $\frac{\pi }{2}$.
