Let a Z and [t] be the greatest integer t.

Question

Let $a\in Z$ and $[t]$ be the greatest integer $\le t$. Then the number of points, where the function $f(x)=[a+13\sin x],x\in (0,\pi )$ is not differentiable, is \_\_\_\_\_\_\_\_\_\_.

Answer: 25

Accepted Answer

Given that $f(x)=[a+13\sin (x)]$, where $[t]$ is the greatest integer function and $x\in (0,\pi )$.

The function $[t]$ is not differentiable wherever $t$ is an integer because at these points, the function has a jump discontinuity.

For $f(x)$ to have a point of non-differentiability, the value inside the greatest integer function, i.e., $a+13\sin (x)$, should be an integer.

So, we need to find the values of $x$ in the interval $(0,\pi )$ for which $a+13\sin (x)$ is an integer.

Now, $\sin (x)$ varies from 0 to 1 in the interval $(0,\pi )$, so the maximum value of $13\sin (x)$ in this interval is 13.

For each integer value of $13\sin (x)$ between 0 and 13, we'll have a corresponding value of $x$. There will be two such values of $x$ for each value of $13\sin (x)$ (because of the periodic and symmetric nature of sine function over the interval $(0,\pi )$), except for the maximum value, 13, which will have only one corresponding value of $x$ (namely $x=\frac{\pi }{2}$).

Thus, the total number of integer values of $13\sin (x)$ between 0 and 13 is 13. Excluding the maximum value, we have 12 integer values, each giving rise to two values of $x$. Including the maximum value, which gives one value of $x$, we have :

$12	imes 2+1=25$

Thus, $f(x)$ is not differentiable at 25 points in the interval $(0,\pi )$.

Solutions · 1