सिद्ध कीजिए कि महत्तम पूर्णांक फलन $f(x) = [x], 0 < x < 3, x = 1$ तथा $x = 2$ पर अवकलित नहीं है।

Q: सिद्ध कीजिए कि महत्तम पूर्णांक फलन $f(x) = [x], 0 < x < 3, x = 1$ तथा $x = 2$ पर अवकलित नहीं है।

दिया है, $f(x) = [x]$ 1. $f(x)$ का $x = 1$ पर $\text{RHD,}$ $x = 1 + h$ रखने पर $x \rightarrow 1^+ \Rightarrow h \rightarrow 0$ $R f^\ {\prime}(1) = \lim \limits _{h \rightarrow 0} \frac{[1+h]-[1]}{h} = \lim \limits _{h \rightarrow 0} \frac{1-1}{h} = 0 [\because R f^\ {\prime}(x) = \frac{f(x+h)-f(x)}{h}]$ तथा $f(x)$ का $x = 1$ पर $\text{LHD,} x = 1 - h$ रखने पर $x \rightarrow 1^{-} \Rightarrow h \rightarrow 0$ $L f^\ {\prime}(1) = \lim \limits _{h \rightarrow 0} \frac{[1-h]-[1]}{-h} = \lim \limits _{h \rightarrow 0} \frac{0-1}{-h} = \lim \limits _{h \rightarrow 0} \frac{1}{h} = \infty [\because L f^\ {\prime}(x) = \frac{f(x-h)-f(x)}{-h}]$ $\therefore \text{LHD} \neq \text{RHD,}$ अतः $f(x), x = 1$ पर अवकलनीय नहीं है। 2. $x = 2$ पर, $\text{RHD} = R f^\ {\prime}(2) = \lim \limits _{h \rightarrow 0}$ $\frac{f(2+h)-f(2)}{h} [\because R f^\ {\prime}(x) = \frac{f(x+h)-f(x)}{h}]$ $= \lim \limits _{h \rightarrow 0} \frac{(2+h)-2}{h} = \lim \limits _{h \rightarrow 0} \frac{2-2}{h} = 0$ तथा $\text{LHD} = L f^\ {\prime}(2) = \lim \limits _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h} [\because L f^\ {\prime}(x) = \frac{f(x-h)-f(x)}{-h}]$ $= \lim \limits _{h \rightarrow 0} \frac{[2-h]-[2]}{-h}$ $= \lim \limits _{h \rightarrow 0} \frac{1-2}{-h} = \infty$ $\therefore \text{RHD} \neq \text{LHD}$ अतः $f(x), x = 2$ पर अवकलनीय नहीं है।

Question

Exercise-5.2-10

Download our app for free and get started

Solution

दिया है, $f(x) = [x]$

$f(x)$ का $x = 1$ पर $\text{RHD,}$
$x = 1 + h$ रखने पर $x \rightarrow 1^+ \Rightarrow h \rightarrow 0$
$R f^\ {\prime}(1) = \lim \limits _{h \rightarrow 0} \frac{[1+h]-[1]}{h} = \lim \limits _{h \rightarrow 0} \frac{1-1}{h} = 0 [\because R f^\ {\prime}(x) = \frac{f(x+h)-f(x)}{h}]$
तथा $f(x)$ का $x = 1$ पर $\text{LHD,} x = 1 - h$ रखने पर $x \rightarrow 1^{-} \Rightarrow h \rightarrow 0$
$L f^\ {\prime}(1) = \lim \limits _{h \rightarrow 0} \frac{[1-h]-[1]}{-h} = \lim \limits _{h \rightarrow 0} \frac{0-1}{-h} = \lim \limits _{h \rightarrow 0} \frac{1}{h} = \infty [\because L f^\ {\prime}(x) = \frac{f(x-h)-f(x)}{-h}]$
$\therefore \text{LHD} \neq \text{RHD,}$ अतः $f(x), x = 1$ पर अवकलनीय नहीं है।
$x = 2$ पर,
$\text{RHD} = R f^\ {\prime}(2) = \lim \limits _{h \rightarrow 0}$
$\frac{f(2+h)-f(2)}{h} [\because R f^\ {\prime}(x) = \frac{f(x+h)-f(x)}{h}]$
$= \lim \limits _{h \rightarrow 0} \frac{(2+h)-2}{h} = \lim \limits _{h \rightarrow 0} \frac{2-2}{h} = 0$
तथा $\text{LHD} = L f^\ {\prime}(2) = \lim \limits _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h} [\because L f^\ {\prime}(x) = \frac{f(x-h)-f(x)}{-h}]$
$= \lim \limits _{h \rightarrow 0} \frac{[2-h]-[2]}{-h}$
$= \lim \limits _{h \rightarrow 0} \frac{1-2}{-h} = \infty$
$\therefore \text{RHD} \neq \text{LHD}$
अतः $f(x), x = 2$ पर अवकलनीय नहीं है।

Download our app
and get started for free

Similar Questions

Download our appand get started for free

Similar Questions

Download our app
and get started for free