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Question: The left-hand derivative of f(x) = [x] sin (πx) at x = k, k an integer, is...

The left-hand derivative of f(x) = [x] sin (πx) at x = k, k an integer, is

A

(−1)k (k − 1)π

B

(−1)k −1(k − 1)π

C

(−1)k kπ

D

(−1)k −1kπ

Answer

(−1)k −1(k − 1)π

Explanation

Solution

LD⁡at x=k=\operatorname { LD } _ { \text {at } \mathrm { x } = \mathrm { k } } = lim⁡h→0f(x)−f(k−h)−h\lim_{h \rightarrow 0}\frac{f(x) - f(k - h)}{- h}

(k = integer)

= lim⁡h→0[k]sin⁡kπ−[k−h]sin⁡(k−h)π−h\lim_{h \rightarrow 0}\frac{\lbrack k\rbrack\sin k\pi - \lbrack k - h\rbrack\sin(k - h)\pi}{- h}

= lim⁡h→0(k−1)sin⁡(k−h)πh\lim_{h \rightarrow 0}\frac{(k - 1)\sin(k - h)\pi}{h}

[∴ sin kπ = 0]

= lim⁡h→0(k−1)sin⁡(kπ−hπ)h[sin⁡(nk−θ)=(−1)n−1sin⁡θ]\lim_{h \rightarrow 0}\frac{(k - 1)\sin(k\pi - h\pi)}{h}\left\lbrack \sin(nk - \theta) = ( - 1)^{n - 1}\sin\theta \right\rbrack

= lim⁡h→0(k−1)(−1)k−1sin⁡hπhπ×π\lim_{h \rightarrow 0}\frac{(k - 1)( - 1)^{k - 1}{\sin h}\pi}{h\pi} \times \pi

= p (k − 1)(−1)k −1

∴ B is the correct alternative.