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There is a function f(x), such that f(0) = 1 and f ' (0)= -1 and f(x) is positive for all values of x. Then,

a) f"(x) < 0 for all x

b) -1 <  f'' (x) < 0 for all x

c) -2 < f '' (x) < -1 for all x

d) None of the above
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2 Answers

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Take an example, f(x) = x- x + 1 so from this example we get f'''(x) = 2. So D) none of these is answer

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any other way of solving??
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I know this method only. But it was simple bcoz  f(0) =1 means function is f(x)= something + 1. Then it was given f'(0)=-1 which means function can bef(x)= something -x + 1 and to contradict options I took x2. Somebody post other answer if there is some theorem or method for this question.

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since f’(x)=  -1.

so the value of f’’(x) will be equal to 0.

so the given option d is correct.