Simple one ......... !!!


ƒ : R → R is a twice diffentiable fn. such that
Æ’'(x) = Æ’(1-x) for all x \epsilon R
Given Æ’(0) = 1 , Find Æ’(1)

(A) e - 2 (B) ln 2 - 1 (C) sec 1 + tan 1 (D) cos 1 + sin 1

12 Answers

11
rkrish ·

o..thanx bhaiyya...BUT i've done this...and in the same way......this was for others to try !!!

21
tapanmast Vora ·

ek dbt tha : wud this f''(x)=-f'(1-x)=-f(x) not be f''(x)=-f'(1-x)=-f(-x)

Oh sorry did a mist in the signs ....... its perfect ...........

11
rkrish ·

@manish bhaiyya......

a sincere request : pls delete ur post so that others may try.

1357
Manish Shankar ·

ok rkrish

11
rkrish ·

thanx bhaiyya

11
Mani Pal Singh ·

rkish
HINT USE THE BASIC DEFINITION OF DIFFERENTIATION

f'(x)=\lim_{h\rightarrow 0}f(x+h)-f(x)/h

and also use the fact that

f"(x)=-f'(1-x)=-f(x)

I HOPE U COULD DO THIS 1 NOW !!!!!!!!!!!!!

11
Mani Pal Singh ·

did u got it now rkrish

11
rkrish ·

arrey bhai.....see post #2 !!!!

btw...thanx.
Pls proceed and get the ans.

11
virang1 Jhaveri ·

Æ’'(x) = Æ’(1-x)

f(0) = 1
Æ’'(0) = 0
Therefore f(1) = 0

Therefore B) is the answer

11
virang1 Jhaveri ·

Final answer pls

11
rkrish ·

@virang....

Did not understand :

i) Æ’'(0) = 0 [7][7][7]

ii) Therefore B) is the answer [7][7][7]

11
virang1 Jhaveri ·

Its given f(0)=1
Differentiating the function
f`(0) = 0

I think this is a method tell me if my method is wrong

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