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# Let $m$ be the value of the left derivative at $x=2$ of the function $f\left( x \right)=\left[ x \right]\sin \left( \pi x \right)$ ( $\text{ [ ] }$ is the usual symbol). Then $\left[ m \right]$ is equal to:  Hint: The value of the left derivative of a function $f\left( x \right)$ at $x=a$ is given as ${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}$ .

Complete step-by-step solution -
The given function is $f\left( x \right)=\left[ x \right]\sin \left( \pi x \right)$ . We are asked to find the value of $\left[ m \right]$ , where $m$ is the value of the left derivative of the function $f(x)$ at $x=2$ and $\text{ [ ] }$ is the greatest integer function.
First of all, we will calculate the value of the left derivative of the function $f\left( x \right)=\left[ x \right]\sin \left( \pi x \right)$ .
We know, the left derivative of a function $f\left( x \right)$ at $x=a$ is given as ${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}$ .
Now, we will calculate the value of the left derivative of the function at$x=2$ .
So, the left derivative of the function at $x=2$ is given as $L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 2-h \right)-f\left( 2 \right)}{-h}$ .
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \left[ 2-h \right]\sin \left( 2-h \right)\pi \right)-\left( \left[ 2 \right]\sin 2\pi \right)}{-h}$
We know, $\sin n\pi =0$ , where $n$ is an integer.
$\Rightarrow L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \left[ 2-h \right]\sin \left( 2-h \right)\pi \right)-0}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \left[ 2-h \right]\sin \left( 2-h \right)\pi \right)}{-h}$
Now, we know, $2-h$ is a number which is slightly less than $2$ . On applying the greatest integer function to this value, it becomes equal to $1$ .
$\Rightarrow L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( 2-h \right)\pi }{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( 2\pi -h\pi \right)}{-h}$
Now, we know, $\sin \left( 2\pi -\theta \right)=-\sin \theta$ . So, the value of $\sin \left( 2\pi -h\pi \right)$ is equal to $-\sin \left( h\pi \right)$ .
$L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\sin h\pi }{-h}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin h\pi }{h}$
We can multiply and divide $\dfrac{\sin \pi h}{h}$ by $\pi$ . We get $L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\pi \sin h\pi }{\pi h}$ which can be written as $L'=\pi \underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin h\pi }{\pi h}$ because $\pi$ is a constant and is independent of $h$ . Now, as $h$ approaches $0$ , $\pi h$ also approach $0$ . So, we can write the limit as $L'=\pi \underset{\pi h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin h\pi }{\pi h}$ .
Now, we know, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ . So, $\underset{\pi h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin h\pi }{\pi h}=1$ .
$\Rightarrow L'=\pi \times 1=\pi$
So, the value of the left derivative of the function $f\left( x \right)=\left[ x \right]\sin \left( \pi x \right)$ at $x=2$ is equal to $\pi$ .
Now, in the question, it is given that the value of the left derivative of the function at $x=2$ is equal to $m$ .
So, we can say that the value of $m$ is equal to $\pi$ .
Now, on applying greatest integer function to $m$, we get
$\left[ m \right]=\left[ \pi \right]=\left[ 3.141 \right]$
So, $\left[ m \right]=3$
Hence, the value of $\left[ m \right]$ is equal to $3$ .

Note: $\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 2-h \right]=1$ because when $h$ approaches $0$ from the right side, its value is slightly greater than $0$ . On subtracting a number near to $0$ from $2$ , the value obtained is slightly less than $2$ . On applying the greatest integer function to this number, it is rounded down to the nearest integer, i.e. $1$ . Students generally make a mistake of writing $\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 2-h \right]=2$ . Such mistakes should be avoided as because of such mistakes, students can end up getting a wrong answer.

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