Question

please and differentiability class 12th​

Answer 1

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in second part I have mistakenly written h tends to 0 minus instead of plus

answer is b and d

Answer 2

here,,we can clearly see

Let us check it first between π/2 to -π/2

clearly we can see

from (0 to π/2) ,

f(x)=2+(1/sinx) =RHL

this is considered as RHL

from (-π/2 to 0)

we can see

f(x)=2+(-1/(sinx)=LHL

as mod is applied on it so we can write

2+(-1/sinx)=2+1/sinx ,which is equal to RHL

so,.we can say

RHL=LHL,,,

Hence it will be conntiuous at(-π/2,0)U(0,π/2)

Now,let us take the limit between,(0,π)

now if the limit will be between,0 to π/2

f(x)=2+(1/sinx),,which is considered as LHL

if we take from π/2 to π

the sign of sin will not change,, because in second quadrant it is positive,,so,

f(x)=2+1/sinx

here also,

LHL=RHL

from.this ananlysis we can say

option 2) is right

now,let us check option d

simply put ,x=0 in the RHL,and differentiate it

putting we get

RHL=2+1/sinx

dy/dx=-cosx/sin^2x

putting x=0

we get

RHL=-infinity

similarly doing the same in LHL we will get

LHL=-infinity

so,

option d is incorrect

Hence option b is right