Question

The left-hand derivative of $f\left(x\right)=\left[x\right]\sin \pi x$ at $x=k$,where k is an integer and [k]=greatest integer not greater than x,is

• A. $(-1{)}^k(k-1)\pi$
• B. $(-1{)}^{k-1}\cdot (k-1)\pi$
• C. $(-1{)}^{k}k\pi$
• D. $(-1{)}^{k-1}k\pi$

Answer 1

The correct option is A $(-1{)}^k(k-1)\pi$

Given $f\left(x\right)=\left[x\right]\sin \pi x$

$L{f}^{\prime }\left(k\right)=\underset{h\to 0}{lim}\frac{f(k-h)-f\left(k\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{[k-h]\sin \pi (k-h)-k\sin \pi k}{-h}$

$=\underset{h\to 0}{lim}\frac{(k-1)\sin \pi (k-h)-k\left(0\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{(k-1)(sin\pi k-\pi h)}{-h}$

$=(-1{)}^k(k-1)\underset{h\to 0}{lim}\frac{\sin \pi h}{h}$

$=(-1{)}^k(k-1){lim}_{h\to 0}\frac{\sin \pi h}{\pi h}\pi$

$L{f}^{\prime }\left(k\right)=(-1{)}^k(k-1)\pi$