Question

If [t] denotes the greatest integer ≤ t, then the number of points, at which the function
\(f(x) = 4|2x + 3| + 9\lfloor x + \frac{1}{2} \rfloor - 12\lfloor x + 20 \rfloor\)
is not differentiable in the open interval (–20, 20), is ____ .

Answer 1

Correct Answer: 79

 To determine the number of points where the function \(f(x) = 4|2x + 3| + 9\lfloor x + \frac{1}{2} \rfloor - 12\lfloor x + 20 \rfloor\) is not differentiable in the interval (–20, 20), we need to analyze where each component causes non-differentiability.
1. **Absolute Value Function**: The term \(|2x + 3|\) is not differentiable when its argument is zero. Solving \( 2x + 3 = 0 \) gives \( x = -\frac{3}{2} \).
2. **Floor Functions**: The terms \( \lfloor x + \frac{1}{2} \rfloor \) and \( \lfloor x + 20 \rfloor \) are not differentiable at integer values of their arguments.
For \( \lfloor x + \frac{1}{2} \rfloor \): Non-differentiability occurs at \( x + \frac{1}{2} = n \), where \( n \) is an integer. Thus, \( x = n - \frac{1}{2} \). In the range \( -20 < x < 20 \), \( n \) ranges from \(-19\) to \(19\), giving points \( -\frac{39}{2}, -\frac{37}{2}, \ldots, \frac{37}{2}, \frac{39}{2}\).
For \( \lfloor x + 20 \rfloor \): Non-differentiability occurs at \( x + 20 = m \), \( m \) integer. So \( x = m - 20 \). In the interval for \( x \), \( m \) ranges from \(1\) to \(39\), giving points \(-19, -18, \ldots, 18, 19\).
3. **Combining Points**: Align distinct points of non-differentiability: \( \{-19, -\frac{39}{2}, -\frac{37}{2}, \dots, -\frac{3}{2}, \dots, \frac{37}{2}, \frac{39}{2}, 19\} \). The combinations correspond to distinct values of \( x\).
4. **Count**: Count unique points, totaling 79 non-differentiable points within the specified range.

The number of points in the interval (–20, 20) where \( f(x) \) is not differentiable is \( \boxed{79} \), verified by checking the distinct values of such \( x\).

Answer 2

\(f(x) = 4|2x + 3| + 9\left\lfloor x + \frac{1}{2} \right\rfloor - 12\left\lfloor x + 20 \right\rfloor\)
\(=4|2x+3|+9[x+\frac{1}{2}]−12[x]−240\)
f(x) is non differentiable at x \(= \frac{-3}{2}\)
and f(x) is discontinuous at {–19, –18, ….., 18, 19} as well as 
\(\left\{ -\frac{39}{2}, -\frac{37}{2}, \ldots, -\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, \ldots, \frac{39}{2} \right\}\)
At same point, they are also non differentiable
∴ Total number of points of non differentiability
= 39 + 40
= 79
So, the correct answer is 79.