Continuity of Composite Function - Practice Questions & MCQ
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MEDIUMLet (where, [.] denotes the greatest integer function) and
For $a, b>0$, let
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}\frac{\tan ((a+1) x)+b \tan x}{x}, x<0 \\ 3^{, \mathrm{x}=0} \\ \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, x>0\end{array}\right.$
be a continuous function at $x=0$. Then $\frac{b}{a}$ is equal to
Let $f$ and $g$ be two functions defined by
$
f(x)=\left\{\begin{array}{ll}
x+1, & x<0 \\
|x-1,| & x \geq 0
\end{array} \text { and } g(x)=\left\{\begin{array}{cl}
x+1, & x<0 \\
1, & x \geq 0
\end{array}\right.\right.
$
Then (gof) $(x)$ is
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Concepts Covered - 1
Continuity of Composite Function
If the function $f(x)$ is continuous at the point $x=a$ and the function $y=g(x)$ is continuous at the point $x=f(a)$, then the composite function $y=(g \circ f)(x)=g(f(x))$ is continuous at the point $x=a$.
Consider the function $f(x)=\frac{1}{1-x}$, which is discontinuous at $x=1$ If $g(x)=f(f(x))$
$g(x)$ will not be defined when $f(x)$ is not defined, so $g(x)$ is discontinuous at $x=1$
Also $g(x)=f(f(x))$ is discontinuous when $f(x)=1$
i.e. $\frac{1}{1-x}=1 \Rightarrow x=0$
We ca check it by finding $g(x), \quad g(x)=\frac{1}{1-f(x)}=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$
It is discontinuous at $\mathrm{x}=0$
So, $g(x)=f(f(x))$ is discontinuous at $x=0$ and $x=1$
Now consider,
$
\begin{aligned}
\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x}))) & =\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right) \\
& =\frac{1}{1-\frac{\mathrm{x}-1}{\mathrm{x}}}=\mathrm{x}
\end{aligned}
$
seems to be continuous, but it is discontinuous at $x=1$ and $x=0$ where $f(x)$ and $f(f(x))$ respectively are not defined.
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