Question

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x)=\begin{cases} x+a, & x<0 \\ |x-1|, & x \ge 0 \end{cases}$ and $g(x)=\begin{cases} x+1, & x<0 \\ (x-1)^2+b, & x \ge 0 \end{cases}$, where $a, b$ are non-negative real numbers. If $gof(x)$ is continuous for all $x \in R$, then $a+b$ is equal to ____.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

Answer 2

The functions are defined as: $f(x)=\begin{cases} x+a, & x<0 \\ |x-1|, & x \ge 0 \end{cases}$ $g(x)=\begin{cases} x+1, & x<0 \\ (x-1)^2+b, & x \ge 0 \end{cases}$ where $a, b \ge 0$.

First, let's analyze $f(x)$ for $x \ge 0$: $f(x) = |x-1| = \begin{cases} 1-x, & 0 \le x < 1 \\ x-1, & x \ge 1 \end{cases}$ So, $f(x)$ can be written as: $f(x)=\begin{cases} x+a, & x<0 \\ 1-x, & 0 \le x < 1 \\ x-1, & x \ge 1 \end{cases}$

The composite function is $gof(x) = g(f(x))$. For $gof(x)$ to be continuous, we need to check the continuity at points where $f(x)$ might be discontinuous ($x=0, x=1$) and at points where $f(x)$ makes the argument of $g$ equal to the point of discontinuity of $g$ ($f(x)=0$).

Let's find $x$ such that $f(x)=0$:

1. If $x<0$: $x+a=0 \implies x=-a$. This is valid if $-a<0$, i.e., $a>0$.
2. If $0 \le x < 1$: $1-x=0 \implies x=1$. This is not in the interval $[0, 1)$.
3. If $x \ge 1$: $x-1=0 \implies x=1$. This is valid.

So, $f(x)=0$ at $x=1$ and possibly at $x=-a$ (if $a>0$). The critical points for continuity of $gof(x)$ are $x=0, x=1$, and $x=-a$ (if $a>0$).

Case 1: $a > 0$ The critical points are $x=0, x=1, x=-a$.

• Continuity at $x=0$:
  • Left-hand limit: $\lim_{x \to 0^-} gof(x)$. For $x<0$, $f(x)=x+a$. As $x \to 0^-$, $f(x) \to a$. Since $a>0$, $f(x)>0$. $gof(x) = g(f(x)) = g(x+a) = ((x+a)-1)^2+b$. $\lim_{x \to 0^-} gof(x) = (a-1)^2+b$.
  • Right-hand limit: $\lim_{x \to 0^+} gof(x)$. For $x \to 0^+$, $f(x) = |x-1| = 1-x$. As $x \to 0^+$, $f(x) \to 1$. Since $f(x)>0$, $gof(x) = g(f(x)) = g(1-x) = ((1-x)-1)^2+b = (-x)^2+b = x^2+b$. $\lim_{x \to 0^+} gof(x) = 0^2+b = b$.
  • Function value: $f(0)=|0-1|=1$. $gof(0)=g(f(0))=g(1)=(1-1)^2+b=b$. For continuity at $x=0$: $(a-1)^2+b = b \implies (a-1)^2 = 0 \implies a=1$. This is consistent with $a>0$. So, $a=1$.

Now $a=1$. The critical points are $x=0, x=1, x=-1$. $f(x)=\begin{cases} x+1, & x<0 \\ 1-x, & 0 \le x < 1 \\ x-1, & x \ge 1 \end{cases}$

• Continuity at $x=-1$:
  • Left-hand limit: $\lim_{x \to -1^-} gof(x)$. For $x<-1$, $f(x)=x+1$. So $f(x)<0$. $gof(x) = g(f(x)) = g(x+1) = (x+1)+1 = x+2$. $\lim_{x \to -1^-} gof(x) = -1+2 = 1$.
  • Right-hand limit: $\lim_{x \to -1^+} gof(x)$. For $-1 \le x < 0$, $f(x)=x+1$. So $0 \le f(x) < 1$, which means $f(x) \ge 0$. $gof(x) = g(f(x)) = g(x+1) = ((x+1)-1)^2+b = x^2+b$. $\lim_{x \to -1^+} gof(x) = (-1)^2+b = 1+b$.
  • Function value: $f(-1)=-1+1=0$. $gof(-1)=g(f(-1))=g(0)=0+1=1$. For continuity at $x=-1$: $1 = 1+b \implies b=0$. This is consistent with $b \ge 0$. So, $b=0$.

With $a=1$ and $b=0$, we have $a+b=1$. Let's verify continuity at $x=1$. $f(x)=\begin{cases} x+1, & x<0 \\ 1-x, & 0 \le x < 1 \\ x-1, & x \ge 1 \end{cases}$, $g(x)=\begin{cases} x+1, & x<0 \\ (x-1)^2, & x \ge 0 \end{cases}$

• Left-hand limit at $x=1$: $\lim_{x \to 1^-} gof(x)$. For $0 \le x < 1$, $f(x)=1-x$. As $x \to 1^-$, $f(x) \to 0^+$, so $f(x) \ge 0$. $gof(x) = g(f(x)) = g(1-x) = ((1-x)-1)^2 = (-x)^2 = x^2$. $\lim_{x \to 1^-} gof(x) = 1^2 = 1$.
• Right-hand limit at $x=1$: $\lim_{x \to 1^+} gof(x)$. For $x \ge 1$, $f(x)=x-1$. As $x \to 1^+$, $f(x) \to 0^+$, so $f(x) \ge 0$. $gof(x) = g(f(x)) = g(x-1) = ((x-1)-1)^2 = (x-2)^2$. $\lim_{x \to 1^+} gof(x) = (1-2)^2 = 1$.
• Function value: $f(1)=1-1=0$. $gof(1)=g(f(1))=g(0)=0+1=1$. Continuity at $x=1$ is satisfied. Thus, $a=1, b=0$ is a valid solution, giving $a+b=1$.

Case 2: $a = 0$ The critical points are $x=0, x=1$. $f(x)=\begin{cases} x, & x<0 \\ 1-x, & 0 \le x < 1 \\ x-1, & x \ge 1 \end{cases}$

• Continuity at $x=0$:
  • Left-hand limit: $\lim_{x \to 0^-} gof(x)$. For $x<0$, $f(x)=x$. So $f(x)<0$. $gof(x) = g(f(x)) = g(x) = x+1$. $\lim_{x \to 0^-} gof(x) = 0+1 = 1$.
  • Right-hand limit: $\lim_{x \to 0^+} gof(x)$. For $x \to 0^+$, $f(x)=1-x \to 1$. So $f(x)>0$. $gof(x) = g(f(x)) = g(1-x) = ((1-x)-1)^2+b = x^2+b$. $\lim_{x \to 0^+} gof(x) = 0^2+b = b$.
  • Function value: $f(0)=|0-1|=1$. $gof(0)=g(f(0))=g(1)=(1-1)^2+b=b$. For continuity at $x=0$: $1=b$. So, $a=0, b=1$.

Now let's check continuity at $x=1$ with $a=0, b=1$. $f(x)=\begin{cases} x, & x<0 \\ 1-x, & 0 \le x < 1 \\ x-1, & x \ge 1 \end{cases}$, $g(x)=\begin{cases} x+1, & x<0 \\ (x-1)^2+1, & x \ge 0 \end{cases}$

• Left-hand limit at $x=1$: $\lim_{x \to 1^-} gof(x)$. For $0 \le x < 1$, $f(x)=1-x$. As $x \to 1^-$, $f(x) \to 0^+$, so $f(x) \ge 0$. $gof(x) = g(f(x)) = g(1-x) = ((1-x)-1)^2+1 = (-x)^2+1 = x^2+1$. $\lim_{x \to 1^-} gof(x) = 1^2+1 = 2$.
• Right-hand limit at $x=1$: $\lim_{x \to 1^+} gof(x)$. For $x \ge 1$, $f(x)=x-1$. As $x \to 1^+$, $f(x) \to 0^+$, so $f(x) \ge 0$. $gof(x) = g(f(x)) = g(x-1) = ((x-1)-1)^2+1 = (x-2)^2+1$. $\lim_{x \to 1^+} gof(x) = (1-2)^2+1 = 1+1 = 2$.
• Function value: $f(1)=1-1=0$. $gof(1)=g(f(1))=g(0)=0+1=1$. For continuity at $x=1$, we need $2=2=1$, which is false. So, the case $a=0$ does not yield a continuous function $gof(x)$.

The only valid solution is $a=1, b=0$. Therefore, $a+b = 1+0 = 1$.