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Let

$$f(x) = e^{-{1\over x^2}}+\int_0^{\pi x\over2}(1+\sin t)^{1\over2}dt$$

for $x\in(0,\infty)$

Then which of the following are true?

(A) $f′$ exists and is continuous.

(B) $f′′$ exists for all x.

(C) $f′$ is bounded.

(D) there exists $α > 0$ such that $|f(x)| > |f′(x)|$ for every $x$ in $(α,1)$

My approach: I found out $f'$ and $f''$. Since $x>0$, $f''\ne 0$ so (C) is not true.

(A) & (B) seems true. But I can't understand (D).

Mhenni Benghorbal
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Rudstar
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  • According to Mathematica the result is false. Because there are a 'hood of $1$ say $V$ in which the funtion define by $g(x)=|f(x)|-|f'(x)|<0$, $\forall x\in V$. – Valent May 14 '14 at 06:17

1 Answers1

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Use Lagrange mean value theorem.you will get f(c)=1+c-1f'(c) c=alpha. Find f(c)and f'(c). c>0. Put these values in the above and compare.

Anurag
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