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Consider the following functions

$f(x)=\begin{cases} 1, & \text{if } \mid x \mid \leq 1 \\ 0, & \text{if } \mid x \mid >1 \end{cases}.$   and  $g(x)=\begin{cases} 1, & \text{if } \mid x \mid \leq 2 \\ 2, & \text{ if } \mid x \mid > 2 \end{cases}. $

Define $h_1(x) = f(g(x))$ and $h_2(x) = g(f(x))$. Which of the following statements is correct?

  1. $h_1$ and $h_2$ are continuous everywhere
  2. $h_1$ is continuous everywhere and $h_2$ has discontinuity at $\pm1$
  3. $h_2$ is continuous everywhere and $h_1$ has discontinuity at $\pm2$
  4. $h_1$ has discontinuity at $\pm 2$ and $h_2$ has discontinuity at $\pm1$.
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I can rewrite f(x)f(x) and g(x)g(x) as :-

f(x)={1,0,0≤x≤1∪−1≤x<0x>1∪x<−1f(x)={1,0≤x≤1∪−1≤x<00,x>1∪x<−1

g(x)={1,2,0≤x≤2∪−2≤x<0x>2∪x<−2g(x)={1,0≤x≤2∪−2≤x<02,x>2∪x<−2

Here, ff has jump discontinuity at ±1±1 and gg has jump discontinuity at ±2±2.

Now, For h1=f(g(x)),h1=f(g(x)),

f(−1≤g(x)≤1)=1f(−1≤g(x)≤1)=1 when −2≤x≤2−2≤x≤2

and f(g(x)>1∪g(x)<−1)=0f(g(x)>1∪g(x)<−1)=0 when x>2∪x<−2x>2∪x<−2

So, h1=f(g(x))h1=f(g(x)) has discontinuity at ±2±2

Now, For h2=g(f(x)),h2=g(f(x)),

g(−2≤f(x)≤2)=1g(−2≤f(x)≤2)=1 when −∞<x<∞−∞<x<∞

So, h2=g(f(x))h2=g(f(x)) is continuous everywhere.
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Sorry for the small image, but the graphs will look something like this.

We can cleary see that h2 is continuous everywhere, but h1 is broken @ x = 2 and x  =-2.

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