If [t] denotes the greatest integer leq t,the | vedclass

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(B) The function is given by $f(x) = 4|2x + 3| + 9[x + \frac{1}{2}] - 12[x + 20]$.$1$. The term $4|2x + 3|$ is non-differentiable at $2x + 3 = 0$,whic

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$79$

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(B) The function is given by $f(x) = 4|2x + 3| + 9[x + \frac{1}{2}] - 12[x + 20]$.$1$. The term $4|2x + 3|$ is non-differentiable at $2x + 3 = 0$,which gives $x = -\frac{3}{2}$. This is $1$ point.$2$. The term $9[x + \frac{1}{2}]$ is non-differentiable when $x + \frac{1}{2} = k$ for any integer $k$. In the interval $(-20, 20)$,$x + \frac{1}{2} \in (-19.5, 20.5)$. The integers $k$ are $\{-19, -18, \dots, 20\}$. There are $20 - (-19) + 1 = 40$ points. However,we must check if any of these coincide with $x = -\frac{3}{2}$. For $x = -\frac{3}{2}$,$x + \frac{1}{2} = -1$,which is an integer. So,one point is common.$3$. The term $-12[x + 20]$ is non-differentiable when $x + 20 = k$ for any integer $k$. In the interval $(-20, 20)$,$x + 20 \in (0, 40)$. The integers $k$ are $\{1, 2, \dots, 39\}$. There are $39$ points.$4$. Total points of non-differentiability = (Points from $|2x+3|$) + (Points from $[x+1/2]$) + (Points from $[x+20]$) - (Common points).Points from $|2x+3|$: $x = -1.5$ ($1$ point).Points from $[x+1/2]$: $x \in \{-19.5, -18.5, \dots, 19.5\}$ ($39$ points).Points from $[x+20]$: $x \in \{-19, -18, \dots, 19\}$ ($39$ points).Since the sets of points are disjoint,the total number of points is $1 + 39 + 39 = 79$.

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

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