Maths

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Maths

anon_econ
Let f be a differentiable function satisfying f'(x)=f'(-x) for all x.
Then
(A) f is an odd function.
(B) f(x)+f(-x)=2f(0) for all x.
(C) 0.5f(x) + 0.5f(y) = f(0.5(x+y)) for all x,y
(D) If f(1) = f(2), then f(-1) = f(-2)
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Re: Maths

duck
Hi Vasudha.. :)

Answer is option D.

Take f(x) = cos x to rule out rest of the options.

:)
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Re: Maths

anon_econ
Thanks duck. But if f(x)=cos x, then f'(x)=-sinx and f'(-x)=sinx. So the initial condition is not satisfied according to me. If i take f(x)=sin x, then f(x)=f'(-x) but i can rule out only the 3rd option.
By the way the question had said that there could be 2 correct options as well..i forgot to mention it.
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Re: Maths

duck
Hi Vasudha.. :)

i interpreted the question in a diff way..

i took f(x)= cos(x) => f'(x)= -sin x
We know, f(-x)=cos(x) => f'(-x) = - sin x
{since, cos x is an even funtion}
we can see f'(x)= f'(-x) = -sin x,  for all x.. so the initial condition is satisfied.
In this way, you can rule out rest of the options.

However, I think it depends on the way we interpret the question.


:)
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Re: Maths

duck
In reply to this post by anon_econ
I think there is some problem in the que

Bcoz if we take f(x)=0, initial condition is met.
and all the options holds good.

:) :)








:)
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Re: Maths

anon_econ
I don't think this logic is correct..all the options might be true for a specific function, but only 1 (or 2) options would be correct for ALL functions (which satisfy the initial condition).
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Re: Maths

duck
The question clearly says:

If a function is differentiable and it satisfies f'(-x)=f'(x) then what would be correct.

So, if we could find a function which satisfies the given condition then that should be the answer.
However, in the question , initial condition is met under many functions and almost every function is coming with different answer. So, there must be some problem with the question.


And one more thing as you pointed out "all the options might be true for a specific function, but only 1 (or 2) options would be correct for ALL functions (which satisfy the initial condition)"

ALL Functions includes that specific function as well for which all options are correct!


:)
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Re: Maths

anon_econ
If i make the claim that all functions satisfy a condition, it obviously implies that any specific function satisfies it.
But if i say that one specific function satisfies a condition, it doesn't mean that all functions have to satisfy the same condition.
So by taking examples of functions I cannot say anything about other functions unless 3 of the options get ruled out with that example.
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Re: Maths

duck
Hi Vasudha.. :)

i completely agree with your claim.
Basically, my point was if i could find a function which satisfies the given condition then, i can rule out options using that.
(Sorry, earlier i wrote " that would be the answer" )

Like if f(x)= x-1
then option A is ruled out.

if f(x)= sinx
then option C is ruled out.

Now, Only options left are B and D.
:)
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Re: Maths

duck





:)
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Re: Maths

anon_econ
In reply to this post by duck
Hey thanks duck. But it is possible that there is only 1 correct option. Plz see if u can think of some function that can rule out either of options B and D. If v can't think of any such function we can't say for sure that both r correct..
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Re: Maths

duck
Right now, I can't think of any function which can rule out either of the two options.
But, i will definitely think about it more... :)

:)