Question

Let f(x) = [2x² + 1] and \(g(x)=\left\{\begin{matrix}  2x-3,\,x<0&\\2x+3,  x≥0 &\end{matrix}\right.\)where [t] is the greatest integer ≤ t. Then, in the open interval (–1, 1), the number of points where fog is discontinuous is equal to _______.

Answer 1

Correct Answer: 62

\(g(x)=\left\{\begin{matrix}  2x-3,\,x<0&\\2x+3,  x≥0 &\end{matrix}\right.\)

The possible points where fog(x) may be discontinuous are

2(2x – 3)² ∈ I & x ∈ (–1, 0)

2(2x + 3)² ∈ I & x ∈ [0, 1)

x ∈ (–1, 0)	x ∈ [0, 1)
2x – 3 ∈ (–5, –3)	2x + 3 ∈ [3, 5)
2(2x – 3)2 ∈ (18, 50)	2(2x + 3)2 ∈ [18, 50)
So, no. of points = 31	It is discontinuous at all points except x = 0 of no. points = 31

So, the correct answer is: 62.