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Question: The left hand derivative of \(f(x)=[x]\sin \pi x\) at x=k, k is an integer, is A.\({{\left( -1 \ri...

The left hand derivative of f(x)=[x]sin⁡πxf(x)=[x]\sin \pi x at x=k, k is an integer, 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: Use the formula 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} . You might also have to use the identity that \sin \left( k\pi -x \right)=\left\\{ \begin{aligned} & \sin x\text{ k is odd} \\\ & \text{-sinx k is even} \\\ \end{aligned} \right. .

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 with the signs and calculations as in such questions, the possibility of making a mistake is either of the sign or a calculation error. 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.