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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:


Answer Verified Verified
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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