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

The Correct Option is B

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.