The left-hand derivative of f(x)=[x] (π x) at x=k where k is an integer and [x] is the greatest integer function, is

Correct answer: A. (-1)^ k (k-1) π f(x)=[x] (π x) at x=k Left hand derivative, _ h arrow 0 (f(k)-f(k-h))/(h) (k-integer) = _ h arrow 0 ([k] k π-[k-h] (k-h) π)/(h) aligned & = _ h arrow 0 (-(k-1) (k-h) π)/(h) & k π=0 and (k π-θ)=(-1)^ k-1 θ & = _ h arrow 0 -(k-1)-(1)^ k-1 π h π × π aligned aligned & = _ h arrow 0 -(k-1)(-1)^ k-1 π h π × π & =π(k-1)(-1)^k aligned

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Continuity and Differentiability

NDA 2017 · Mathematics · Continuity and Differentiability

The left-hand derivative of f(x)=[x] (π x) at x=k where k is an...

The left-hand derivative of

f(\mathrm{x})=[\mathrm{x}] \sin (\pi \mathrm{x})

\mathrm{x}=\mathrm{k}

where k is an integer and

[\mathrm{x}]

is the greatest integer function, is

Options

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

Correct answer

A. $(-1)^{\mathrm{k}}(\mathrm{k}-1) \pi$

Step-by-step solution

f(x)=[x] \sin (\pi x) \text{ at } x=k

Left hand derivative,

\lim _{h \rightarrow 0} \frac{f(k)-f(k-h)}{h}

(k-integer)

=\lim _{h \rightarrow 0} \frac{[k] \sin k \pi-[k-h] \sin (k-h) \pi}{h}

\begin{aligned} & =\lim _{h \rightarrow 0} \frac{-(k-1) \sin (k-h) \pi}{h} \\ & \sin k \pi=0 \text{\text{ and } } \sin (k \pi-\theta)=(-1)^{k-1} \sin \theta \\ & =\lim _{h \rightarrow 0} \frac{-(k-1)-(1)^{k-1} \sin \pi}{h \pi} \times \pi\end{aligned}

$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{-(k-1)(-1)^{k-1} \sin \pi}{h \pi} \times \pi \\ & =\pi(k-1)(-1)^k\end{aligned}

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