Question

Let f,g : R → R be functions defined by*
$f(x) = \begin{cases} [x], & x < 0 \\\ |1 - x|, & x \geq 0 \end{cases}$
and $g(x) = \begin{cases} e^x - x, & x < 0 \\\ {(x - 1)^2 - 1}, & x \geq 0 \end{cases}$
Where [x] denotes the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly:

• A. one point
• B. two points
• C. three points
• D. four points

Answer 1

Answer: (B)

two points

Answer 2

The correct answer is (B) : two points
$f(x) = \begin{cases} [x], & x < 0 \\\ |1 - x|, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} e^x - x, & x < 0 \\\ {(x - 1)^2 - 1}, & x \geq 0 \end{cases}$
$(fog)(x) = \begin{cases} [g(x)], & g(x) < 0 \\\ 1 - g(x), & g(x) \geq 0 \end{cases}$

Fig. Graph

$h(x) = \begin{cases} |1 + x - e^x|, & x < 0 \\\ 1, & x = 0 \\\ (x - 1)^2 - 1, & 0 < x < 2 \\\ |2 - (x - 1)^2|, & x \geq 2 \end{cases}$
So, x = 0, 2 are the two points where fog is discontinuous.