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Question: The left hand derivative of \(f(x)=[x]\sin \pi x\) at x=k, k is an integer and [x] is the greatest i...

The left hand derivative of f(x)=[x]sin⁡πxf(x)=[x]\sin \pi x at x=k, k is an integer and [x] is the greatest integer function, is
(a) (−1)k(k−1)π{{\left( -1 \right)}^{k}}\left( k-1 \right)\pi
(b) (−1)k−1(k−1)π{{\left( -1 \right)}^{k-1}}\left( k-1 \right)\pi
(c) (−1)kkπ{{\left( -1 \right)}^{k}}k\pi
(d) (−1)k−1kπ{{\left( -1 \right)}^{k-1}}k\pi

Explanation

Solution

Hint: We know that the left hand derivative of a general function f(x) is given by lim⁡h→0 f(k−h)−f(k)−h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(k-h)-f(k)}{-h} . You should also define sinx for different values of k as \sin \left( k\pi -x \right)=\left\\{ \begin{aligned} & \sin x\text{ k is odd} \\\ & \text{-sinx k is even} \\\ \end{aligned} \right. in order to reach an answer matching with the given options.

Complete step-by-step answer:
We know that the left hand derivative at x=k of a general function f(x) is given by lim⁡h→0 f(k−h)−f(k)−h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(k-h)-f(k)}{-h} .
So, if we put f(x)=[x]sin⁡πxf(x)=[x]\sin \pi x in the above formula, we get
lim⁡h→0 f(k−h)−f(k)−h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(k-h)-f(k)}{-h}
=lim⁡h→0 [k−h]sin⁡(k−h)π−[k]sink⁡π−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{[k-h]\sin \left( k-h \right)\pi -[k]\operatorname{sink}\pi }{-h}
Now we know that the value of sine for a multiple of π\pi is always zero.
=lim⁡h→0 [k−h]sin⁡(k−h)π−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{[k-h]\sin \left( k-h \right)\pi }{-h}
=lim⁡h→0 [k−h]sin⁡(kπ−hπ)−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{[k-h]\sin \left( k\pi -h\pi \right)}{-h}
Now as h is just greater than zero and k is an integer, so [k-h] is the largest integer less than k, which is equal to k-1.
=lim⁡h→0 (k−1)sin⁡(kπ−hπ)−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( k-1 \right)\sin \left( k\pi -h\pi \right)}{-h}
Now we know that \sin \left( k\pi -x \right)=\left\\{ \begin{aligned} & \sin x\text{ k is odd} \\\ & \text{-sinx k is even} \\\ \end{aligned} \right. , and it is equivalent to sin⁡(kπ−x)=(−1)k+1sin⁡x\sin \left( k\pi -x \right)={{\left( -1 \right)}^{k+1}}\sin x . So, using this in our equation, we get
lim⁡h→0 (−1)k+1(k−1)sinh⁡π−h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\left( -1 \right)}^{k+1}}\left( k-1 \right)\sinh \pi }{-h}
=lim⁡h→0 (−1)k(k−1)sinh⁡πh=\underset{h\to 0}{\mathop{\lim }}\,{{\left( -1 \right)}^{k}}\left( k-1 \right)\dfrac{\sinh \pi }{h}
Now we know that the value of lim⁡x→0 sin⁡nxx=n\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin nx}{x}=n .
∴lim⁡h→0 (−1)k(k−1)sinh⁡πh=lim⁡h→0 (−1)k(k−1)π\therefore \underset{h\to 0}{\mathop{\lim }}\,{{\left( -1 \right)}^{k}}\left( k-1 \right)\dfrac{\sinh \pi }{h}=\underset{h\to 0}{\mathop{\lim }}\,{{\left( -1 \right)}^{k}}\left( k-1 \right)\pi
Therefore, the answer to the above question is option (a).

Note: Be careful while defining [x], i.e., greatest integer function, as students generally get confused and make a mistake while defining [x]. You also need to remember the definition of left hand derivative and right hand derivative in terms of limits. Also, try to learn all the identities related to trigonometric ratios and their complements, as they are used very often.