Question

The left−handed derivative of $\mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]\sin \left(\mathrm{\pi x}\right)$ at $\mathrm{x}=\mathrm{k},\mathrm{k}$ is an integer, is

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

Answer 1

The correct option is A

${\left(-1\right)}^{\mathrm{k}}\left(\mathrm{k}-1\right)\mathrm{\pi }$

Explanation of the correct option.

Compute the required value.

Given : $\mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]\sin \left(\mathrm{\pi x}\right)$

$\left[x\right]$ is greater integer function.

If $x$ is just less than $\mathrm{k}$, $\left[\mathrm{x}\right]=\mathrm{k}-1$ ……………….$\left(1\right)$ [Basic use of greatest integer function.]

And if $x$ is just greater than or equal to $\mathrm{k}$, $\left[\mathrm{x}\right]=\mathrm{k}$ ……………….$\left(2\right)$

We know that left-hand derivative is given by,

LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{h\to 0}{\lim }\frac{f(x-h)-f\left(x\right)}{(x-h)-x}$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{h\to 0}{\lim }\frac{f(k-h)-f\left(k\right)}{(k-h)-k}\mathbf{}\mathbf{}\mathbf{(}\mathit{x}\mathbf{=}\mathit{k}\mathbf{)}$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{h\to 0}{\lim }\frac{\left[k-h\right]\sin \left(k-h\right)\mathrm{\pi }-\left[\mathrm{k}\right]\sin \left(\mathrm{\pi k}\right)}{-h}$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to \mathrm{k}}{\lim }\frac{\left(\mathrm{k}-1\right)\sin \left(k-h\right)\mathrm{\pi }-\mathrm{k}\sin \left(\mathrm{\pi k}\right)}{-h}$ [From equation $\left(1\right)$ and $\left(2\right)$ ]

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left(\mathrm{k}-1\right)\sin \left(\mathrm{k}-\mathrm{h}\right)\mathrm{\pi }}{-\mathrm{h}},$ $\left[\sin \left(\mathrm{n\pi }\right)=0\right]$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left(\mathrm{k}-1\right)\sin \left(\mathrm{k\pi }-\mathrm{h\pi }\right)}{-\mathrm{h}}$

Since, $\sin \left(k\mathrm{\pi }-\mathrm{\theta }\right)={(-1)}^{k-1}\sin \left(\theta \right)$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left(\mathrm{k}-1\right){\left(-1\right)}^{\mathrm{k}-1}\sin \left(\mathrm{h\pi }\right)}{-\mathrm{h}}$

Multiply and divide by $\mathrm{\pi },$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left(\mathrm{k}-1\right){\left(-1\right)}^{\mathrm{k}-1}\sin \left(\mathrm{h\pi }\right)}{-\mathrm{h\pi }}\times \mathrm{\pi }$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{k}-1\right){\left(-1\right)}^{\mathrm{k}-1}\pi \underset{\mathrm{h}\to 0}{\lim }\frac{\sin \left(h\mathrm{\pi }\right)}{-\mathrm{h\pi }}$ $\left[\underset{h\to 0}{\lim }\frac{\sin \left(\mathrm{h\pi }\right)}{\mathrm{h\pi }}=1\right]$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{k}-1\right){\left(-1\right)}^{\mathrm{k}-1}\times \mathrm{\pi }\times (-1)$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{k}-1\right){\left(-1\right)}^{\mathrm{k}-1+1}\mathrm{\pi }$

$\Rightarrow$LHD of $\mathrm{f}\left(\mathrm{x}\right)={(-1)}^{\mathrm{k}}\left(\mathrm{k}-1\right)\mathrm{\pi }$

Hence option $\left(A\right)$ is the correct option.