Piecewise Functions - 3

First of all, since $|x- c| -d$ is a piecewise function as well, we must explicitly define it.

$|x - c| - d = \begin{cases} -x + c - d& x < c \\ x - (c+d) & x \geq c \end{cases}$

Its graph is $\text{V}$ shaped, and the coordinates of the vertex are $(c, -d)$ .

$\sin x$ is always differentiable in its domain, but $|x- c| -d$ has one point of non-differentiabilty at its vertex, $x =c$ . To make $f$ differentiable, $c$ must lie between $a$ and $b$ .

At $a$ and $b$ , $f$ will be differentiable only if left derivative is equal to right derivative.

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At $a$ :

Left derivative $= \cos a$ . Right derivative $= -1$ .

$\implies \cos a = -1$

$\implies a = (2k + 1) \pi$ , where $k \in \mathbb{Z}$

Similarly, at $b$ :

Left derivative $= -1$ . Right derivative $= \cos b$

$\implies \cos b = 1$

$\implies b = 2n \pi$ , where $n \in \mathbb{Z}$

Observe $\sin a$ and $\sin b$ are both $0$ , so $f(a)$ and $f(b)$ are both $0$ .

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Since $(b, 0)$ satisfies $f(x) = x - (c+d)$

$\implies 0 = b - ( c + d)$

$\implies b = c + d$

Thus, the required expression $\lvert \lfloor a + b + c + d \rfloor\rvert = \lvert \lfloor a + b + b \rfloor \rvert$

$= \lvert \lfloor a + 2b\rfloor \rvert$

$= \lvert \lfloor (2k+1 + 4n) \pi \rfloor \rvert$

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Since $\lvert \lfloor (2k+1 + 4n) \pi \rfloor \rvert$ must be made minimum,

$\implies \lfloor (2k+1 + 4n) \pi \rfloor$ must be closest to $0$ .

Note: We have a condition $a < b$

$\implies (2k+1 ) < 2n \implies k < n$ .

On observation, we see that for $n= 1$ , $k = -2$ , the above expression is closest to $0$

After substituting these values, we get

$\lvert \lfloor (2 \times (-2) +1 + 4 \times 1) \pi \rfloor \rvert = \lvert \lfloor \pi \rfloor \rvert = \lvert 3 \rvert = 3$

Therefore the minimum value of $\lvert \lfloor a + b+ c + d \rfloor \rvert = \boxed{3}$