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In open interval $\left(0, \frac{\pi}{2}\right)$,
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Solution:
1566 Upvotes
Verified Answer
The correct answer is:
$\cos x+x \sin x>1$
Hint:
$f(x)=\cos x+x \sin x-1$
$\Rightarrow f^{\prime}(x)=-\sin x+\sin x+x \cos x>0 ; x \in\left(0, \frac{\pi}{2}\right)$
$\Rightarrow f(x)$ is increasing function
$\begin{array}{l}
\Rightarrow f(x)>f(0) \\
\cos x+x \sin x-1>0 \\
\cos x+x \sin x>1
\end{array}$
$f(x)=\cos x+x \sin x-1$
$\Rightarrow f^{\prime}(x)=-\sin x+\sin x+x \cos x>0 ; x \in\left(0, \frac{\pi}{2}\right)$
$\Rightarrow f(x)$ is increasing function
$\begin{array}{l}
\Rightarrow f(x)>f(0) \\
\cos x+x \sin x-1>0 \\
\cos x+x \sin x>1
\end{array}$
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