IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - IIT JEE 2007 question on continuity and diffrentiability(match the column)

dark_fantasy

The match the coulmn had the option xIxI. according to me, the graph is continous and diffrentiable in (-1,1), but at x=0, the slope is equal to 0.

Hence it is not strictly increasing( for a strictly increasing function f(x), f'(x)>0 for all points in its domain). please tell me if iam right. because of this option, i am losing 6 valuable marks.

vinod

Do check this solution from FIIT-JEE :

dark_fantasy

The graph is like a S shaped curve, similar to y=x³.

ie.

f(x)= x²x>=0

-x² x<0

f'(x)= 2x x>=0

-2x x<0;

at x=0, f'(0+)=0, and f'(0-)=0. this shows that in (-1,1) the f'(x) becomes zero once. for a function to be strictly increasing f'(x)>0, ie. it is never equal to zero. therefore the fiitjee solution is wrong. someone let me know if i am right.

amangem

see i dont noe wether i am exactly correct but tis is what i can think from the definition.

see what is a strictly increasing function ??? not in terms of derivative but otherwise ??? it is a function which increases as x inxreases or decreases as x decreases . so at x=0 , where the derivative is rightly 0 , luk carefully at the behaviour of the function in the neighbourhood of 0 . for x>0 as we move towards right the function also increases . the reverse is true if the if we consider the case when x<0 . so what do we see ??? derivative at A POINT is zero . and in the neighbourhood of the point the function assumes unique behaviour . that is it is increasing . so it can be rightly said that the funcyion is strictly increasing even though the derivative is zero at one point .

and as far as the derivative test is concerned , see as far as the word strict is concerned ....the actual statement should be that the derivative of the function at whichever point is zero , those point(s) shud not form a part of an interval .

so the answer , as far as i can thinjk according to this explaination is right that is X|X| is strictly increasing .

i hope u get my point .

and dont bother now abt the six marks . u cant do anything abt it . only focus on the concept now which may be useful further .
wishing u a very best of luck
cheers

dark_fantasy

But a strictly function's slope never does become zero. this function is called a non-decreasing function. here the word STRICTLY is being used, therefore the slope has to be always greater than zero. check out any maths book, there this is specified. i checked in arihant maths. if they had not specified strictly, it would had been right.

and as we move towards the right of 0, the value of the function is constant for a short time, similarly as we move left.{(0-h,0+h) h

0}. therefore it does not increase with the increase of the value of x for atleast a short interval, which cannot be overlooked.

brutus_genius

i guess amangem is right ...somehow by definition

amaron

"For a strictly increasing function f(x), f'(x)>0 for all points in its domain".
The above statement is WRONG. the actual statement is "IF f'(x)>0 for all points in its domain THEN f(x) is surely a strictly increasing function".

But if f(x) is known to be a strictly increasing function, it is not necessary that f'(x)>0.

example. f(x)= x³ is a strictly increasing function. but at x=0 f'(x)=0.
similarly x|x|.

That solves the problem!

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