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Let $f(x)=e^x \cos x+1$. Which of the following statements is always true?
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Verified Answer
The correct answer is:
Between any two consecutive roots of $f(x)=0$ there is always a roots of $e^x \sin x=0$
Let $f(x)=\cos x+e^{-x}$
and $\alpha$ and $\beta$ be two roots of the question
$e^x \cos x+1=0$ such that $\alpha < \beta$, then $\cos \alpha+e^{-\alpha}=0$
and $\cos \beta+e^{-\beta}=0$
clearly, $f(x)$ is continous on $[\alpha, \beta]$ and
differentiable on $(\alpha, \beta)$ also $f(\alpha)-f(\beta)$
[using Eq. (i)]
By Rolle's theorem there exists $\mathrm{C} \in(\alpha, \beta)$ such that
$f^{\prime}(c)=0$
$\Rightarrow-\sin c-e^{-c} \Rightarrow 0 e^c \sin c+1=0$
$\Rightarrow x=c$ is a root of $e^x \sin x+1=0$
and $\alpha$ and $\beta$ be two roots of the question
$e^x \cos x+1=0$ such that $\alpha < \beta$, then $\cos \alpha+e^{-\alpha}=0$
and $\cos \beta+e^{-\beta}=0$
clearly, $f(x)$ is continous on $[\alpha, \beta]$ and
differentiable on $(\alpha, \beta)$ also $f(\alpha)-f(\beta)$
[using Eq. (i)]
By Rolle's theorem there exists $\mathrm{C} \in(\alpha, \beta)$ such that
$f^{\prime}(c)=0$
$\Rightarrow-\sin c-e^{-c} \Rightarrow 0 e^c \sin c+1=0$
$\Rightarrow x=c$ is a root of $e^x \sin x+1=0$
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