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Question: Let \[m\] be the value of the left derivative at \[x=2\] of the function \[f\left( x \right)=\left[ ...

Let mm be the value of the left derivative at x=2x=2 of the function f(x)=[x]sin⁡(πx)f\left( x \right)=\left[ x \right]\sin \left( \pi x \right) (  [ ] \text{ [ ] } is the usual symbol). Then [m]\left[ m \right] is equal to:

Explanation

Solution

Hint: The value of the left derivative of a function f(x)f\left( x \right) at x=ax=a is given as L′=lim⁡h→0+ f(a−h)−f(a)−h{{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(x)=[x]sin⁡(πx)f\left( x \right)=\left[ x \right]\sin \left( \pi x \right) . We are asked to find the value of [m]\left[ m \right] , where mm is the value of the left derivative of the function f(x)f(x) at x=2x=2 and  [ ] \text{ [ ] } is the greatest integer function.
First of all, we will calculate the value of the left derivative of the function f(x)=[x]sin⁡(πx)f\left( x \right)=\left[ x \right]\sin \left( \pi x \right) .
We know, the left derivative of a function f(x)f\left( x \right) at x=ax=a is given as L′=lim⁡h→0+ f(a−h)−f(a)−h{{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 atx=2x=2 .
So, the left derivative of the function at x=2x=2 is given as L′=lim⁡h→0+ f(2−h)−f(2)−hL'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 2-h \right)-f\left( 2 \right)}{-h} .
=lim⁡h→0+ ([2−h]sin⁡(2−h)π)−([2]sin⁡2π)−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⁡nπ=0\sin n\pi =0 , where nn is an integer.
⇒L′=lim⁡h→0+ ([2−h]sin⁡(2−h)π)−0−h\Rightarrow L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \left[ 2-h \right]\sin \left( 2-h \right)\pi \right)-0}{-h}
=lim⁡h→0+ ([2−h]sin⁡(2−h)π)−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−h2-h is a number which is slightly less than 22 . On applying the greatest integer function to this value, it becomes equal to 11 .
⇒L′=lim⁡h→0+ sin⁡(2−h)π−h\Rightarrow L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( 2-h \right)\pi }{-h}
=lim⁡h→0+ sin⁡(2π−hπ)−h=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( 2\pi -h\pi \right)}{-h}
Now, we know, sin⁡(2π−θ)=−sin⁡θ\sin \left( 2\pi -\theta \right)=-\sin \theta . So, the value of sin⁡(2π−hπ)\sin \left( 2\pi -h\pi \right) is equal to −sin⁡(hπ)-\sin \left( h\pi \right) .
L′=lim⁡h→0+ −sin⁡hπ−h=lim⁡h→0+ sin⁡hπhL'=\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 sin⁡πhh\dfrac{\sin \pi h}{h} by π\pi . We get L′=lim⁡h→0+ πsin⁡hππhL'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\pi \sin h\pi }{\pi h} which can be written as L′=πlim⁡h→0+ sin⁡hππhL'=\pi \underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin h\pi }{\pi h} because π\pi is a constant and is independent of hh . Now, as hh approaches 00 , πh\pi h also approach 00 . So, we can write the limit as L′=πlim⁡πh→0+ sin⁡hππhL'=\pi \underset{\pi h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin h\pi }{\pi h} .
Now, we know, lim⁡x→0+ sin⁡xx=1\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 . So, lim⁡πh→0+ sin⁡hππh=1\underset{\pi h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin h\pi }{\pi h}=1 .
⇒L′=π×1=π\Rightarrow L'=\pi \times 1=\pi
So, the value of the left derivative of the function f(x)=[x]sin⁡(πx)f\left( x \right)=\left[ x \right]\sin \left( \pi x \right) at x=2x=2 is equal to π\pi .
Now, in the question, it is given that the value of the left derivative of the function at x=2x=2 is equal to mm .
So, we can say that the value of mm is equal to π\pi .
Now, on applying greatest integer function to mm, we get
[m]=[π]=[3.141]\left[ m \right]=\left[ \pi \right]=\left[ 3.141 \right]
So, [m]=3\left[ m \right]=3
Hence, the value of [m]\left[ m \right] is equal to 33 .

Note: lim⁡h→0+ [2−h]=1\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 2-h \right]=1 because when hh approaches 00 from the right side, its value is slightly greater than 00 . On subtracting a number near to 00 from 22 , the value obtained is slightly less than 22 . On applying the greatest integer function to this number, it is rounded down to the nearest integer, i.e. 11 . Students generally make a mistake of writing lim⁡h→0+ [2−h]=2\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.