Continuity and Differentiability — Class 12 Maths Solution

We have, $f(x) = \frac{1}{{x + 2}}$
therefore,$y = f\{ f(x)\}$
$= f\left( {\frac{1}{{x + 2}}} \right) = \frac{1}{{\frac{1}{{x + 2}} + 2}}$

$= \frac{1}{{1 + 2x + 4}} \cdot (x + 2) = \frac{{(x + 2)}}{{(2x + 5)}}$

So, the function $y$ will not be continuous at those points, where it is not defined as it is a rational function.

Therefore, $y = \frac{{x + 2}}{{(2x + 5)}}$ is not defined, when $2x + 5 = 0$

$\Rightarrow x = \frac{{ - 5}}{2}$

Therefore we can say that $y$ is discontinuous at $x = \frac{{ - 5}}{2}$.