Question

-does not exist(the last pic didnt came right) ​

Answer 1

Step-by-step explanation:

We have,

$\tt{f(x) = 1 + \left| sin(x)\right| }$

Now,

$\tt{f(x) = \begin{cases} \sf{1 + sin(x), \: \: \: \: \: \: 0 \leqslant x < \pi } \\ \sf{1 - sin(x), \: \: \: \: \: \: \pi \leqslant x < 2\pi }\end{cases}}$

LHD:

$\displaystyle \sf{ \lim_{h \to0} \dfrac{f(\pi - h) - f(\pi)}{h} }$

$\displaystyle \sf{ = \lim_{h \to0} \dfrac{1 + sin(\pi - h) - 1}{h} }$

$\displaystyle \sf{ = \lim_{h \to0} \dfrac{ sin(\pi - h)}{h} }$

$\displaystyle \sf{ = \lim_{h \to0} \dfrac{ sin( h)}{h} }$

$\sf{ = 1 }$

RHD:

$\displaystyle \sf{ \lim_{h \to0} \dfrac{f(\pi + h) - f(\pi)}{h} }$

$\displaystyle \sf{ = \lim_{h \to0} \dfrac{1 - sin(\pi + h) - 1}{h} }$

$\displaystyle \sf{ = \lim_{h \to0} \dfrac{ - sin(\pi + h) }{h} }$

$\displaystyle \sf{ = \lim_{h \to0} \dfrac{ sin( h) }{h} }$

$\sf{ = 1}$

SINCE, LHD=RHD, hence f'(π) exists

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\red{\rm :\longmapsto\:f(x) = 1 + |sinx| }$

Let us first define the function f(x).

$\red{\begin{gathered}\begin{gathered}\bf\: 1 + |sinx| = \begin{cases} &\sf{1 + sinx \: \: \: when \: 0 \leqslant x \leqslant \pi} \\ \\ &\sf{1 - sinx \: \: \: when \: \pi < x \leqslant 2\pi} \end{cases}\end{gathered}\end{gathered}}$

LEFT HAND DERIVATIVE

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi^-} \frac{f(x) - f(\pi)}{x - \pi}$

$= \displaystyle\lim_{x \to \pi^-} \frac{1 + sinx - (1 + sin\pi)}{x - \pi}$

$= \displaystyle\lim_{x \to \pi^-} \frac{1 + sinx - 1 - 0}{x - \pi}$

$= \displaystyle\lim_{x \to \pi^-} \frac{sinx}{x - \pi}$

To evaluate this limit, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = \pi - h, \: \: as \: x \to \: \pi \: \: so \: h \: \to \: 0}$

$= \displaystyle\lim_{h \to 0} \frac{sin(\pi - h)}{\pi - h - \pi}$

$= \displaystyle\lim_{h \to 0} \frac{sinh}{ - h}$

$= - \: \displaystyle\lim_{h \to 0} \frac{sinh}{ h}$

$\rm \:  =  \: - \: 1$

So,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to \pi^-} \frac{f(x) - f(\pi)}{x - \pi} = \: - \: 1 \: }}$

RIGHT HAND DERIVATIVE

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi^ + } \frac{f(x) - f(\pi)}{x - \pi}$

$= \displaystyle\lim_{x \to \pi^ + } \frac{1 - sinx - (1 + sin\pi)}{x - \pi}$

$= \displaystyle\lim_{x \to \pi^ + } \frac{1 - sinx - 1 - 0}{x - \pi}$

$= \displaystyle\lim_{x \to \pi^ + } \frac{- sinx }{x - \pi}$

To evaluate this limit, we use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = \pi + h, \: \: as \: x \to \: \pi \: \: so \: h \: \to \: 0}$

$= - \: \displaystyle\lim_{h \to 0} \frac{sin(\pi + h)}{\pi + h - \pi}$

$= - \: \displaystyle\lim_{h \to 0} \frac{ - sin(h)}{h}$

$= \: \displaystyle\lim_{h \to 0} \frac{sinh}{h}$

$\rm \:  =  \: 1$

So,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to \pi^ + } \frac{f(x) - f(\pi)}{x - \pi} = \:\: 1 \: }}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: LHD \: \ne \: RHD \: }}$

$\red{\rm \implies\:f(x) \: is \: not \: differentiable \: at \: x = \pi}$

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More to Know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$