Let f : R $\to$ R and g : R $\to$ R be defined as
$$f(x) = \left\{ {\matrix{
{x + a,} & {x < 0} \cr
{|x - 1|,} & {x \ge 0} \cr
} } \right.$$ and
$$g(x) = \left\{ {\matrix{
{x + 1,} & {x < 0} \cr
{{{(x - 1)}^2} + b,} & {x \ge 0} \cr
} } \right.$$,
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $\in$ R, then a + b is equal to ____________.
Answer (integer)
1
Solution
$$g[f(x)] = \left[ {\matrix{
{f(x) + 1} & {f(x) < 0} \cr
{{{(f(x) - 1)}^2} + b} & {f(x) \ge 0} \cr
} } \right.$$<br><br>$$g[f(x)] = \left[ {\matrix{
{x + a + 1} & {x + a < 0\& x < 0} \cr
{|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr
{{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr
{{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr
} } \right.$$<br><br>$$g[f(x)] = \left[ {\matrix{
{x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr
{|x - 1| + 1} & {x \in \phi } \cr
{{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr
{{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr
} } \right.$$<br><br>$$g[f(x)] = \left[ {\matrix{
{x + a + 1} & {x \in ( - \infty , - a)} \cr
{{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr
{{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr
} } \right.$$<br><br>g(f(x)) is continuous.
<br><br>At x = $-$a
<br><br>-a + a + 1 = (-a + a - 1)<sup>2</sup> + b
<br><br>$\Rightarrow$ 1 = b + 1
<br><br>$\Rightarrow$ b = 0
<br><br>at x = 0
<br><br>(a $-$1)<sup>2</sup> + b = (|0 - 1| - 1)<sup>2</sup> + b
<br><br>$\Rightarrow$ (a $-$1)<sup>2</sup> + b = b
<br><br>$\Rightarrow$ a = 1<br><br>$\Rightarrow$ a + b = 1
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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