IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kishore.subramanian.b

The function (x²-1)|x²-3x+2| +cos(|x|) is not differentiable at

1. -1 2. 0 3. 1 4. 2

srujana

Let
f(x)=(x²-1)|x²-3x+2|, g(x)=cos|x|

f'(x)=((x²-1)|x²-3x+2|)'

=(x²-1) |x²-3x+2|' + |x²-3x+2| (x²-1)'

|x²-3x+2|' exists only when x²-3x+2 is nt eq to 0 (a fn y=|x| is not diff when x=0)

hence f'(x) exists at all values of x except when x²-3x+2=0 or when x=2 or x=1

g(x) =cos|x| or cos x(edited)

which is diff at all x belongs to R

hence the function is not diff at x=1,2 but diff at 0(edited)

srujana

Edited-wrong graph

sboosy

srujana , i think cos|x| is differentiable at x = 0

cos|x| diff

= -sin|x| *|x| /x

both LHD and RHD come out to be -sinx and substituing x=0 , we get LHD

RHD = 0

so i think it is differentiable there

srujana

But it is discontinuous right.....

sboosy

no it is continuous
cos|x|
LHL
cos(-x) = cosx
RHL
cos(x) = cosx
which is nothing but cos 0 = 1
so it is continuous

srujana

well.....im a bit confused......

let us wait for the experts

akhil_o

isnt cos|x| graph same as graph of cos x?
i agree with sboosy that at 0 its differentiable

srujana

@sboosy

can u say the function is continuous at x=0 judging from the 1st graph of mine?

sboosy

dont draw the graph ...which i think is rather misleading u srujana
now what is the condition for continuity?
LHL(at a pt) = RHL(at a pt) = f(the point)
as i have shown above it is satisfied ..which means the function is continuous ...there need be no confusion abt it

srujana

@sboosy

i have got a new doubt now, is the graph of cos x same as graph of cos|x|?????

sboosy

cos|x| = cos(-x) x<0
= cos(x)
cos|x| = cos(x) x> 0
so for both positive and negative cos|x| assumes cos x
so i think it is same

elastiboysai

Nice effort der

bt

graph of cos|x| is same as cos x

so its diff at x=0.

so ans is 1,2

as already pointed out .

srujana

ohk !!!

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