Question

please ans differentiability class 12th​

Answer 1

$clearly \: we \: can \: see \: that \: \\ f(x) = \binom{x + sinx \: \: (0 \leqslant x < \frac{\pi}{2}) }{ - x - sinx \: (0 > x < \frac{ - \pi}{2} )}$

Note:-this is the case when we r ignoring the mod.....

we can clearly see,as mod is applied in the question,,,,so,

negative sign will become positive and we can say

|-x|+|-sinx|=x+sinx

so

LHL =RHL

Hence limit will be continuous at

$( \frac{ - \pi}{2} \: \: 0)U(0 \: \: \frac{\pi}{2} )$

now if we differentiate the limit at x =0

then we get

LHL=-1-cos0=-1-1=-2

RHL=1+cos0=2

Clearly we can see that

LHL is not eqaul to RHL

so the limit will not differentiable at x=0

{hope it helps}

Answer 2

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also don't check LHL and RHL of dy/dx of

to make sure that function is differentiable at every point.That isn't always true

ALWAYS CHECK BY FIRST PRINCIPLE