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  • Between any two real roots of the equation , the equation <img alt="e^x \cos x+1=0" src="h

Between any two real roots of the equation e^x \sin x-1=0, the equation e^x \cos x+1=0 has

 

Option: 1

at least one root


Option: 2

at most one root
 


Option: 3

exactly one root


Option: 4

no root


Answers (1)

Let f(x)=e^{-x}-\sin x and let \alpha \text { and } \beta be two roots of the equation e^x \sin x-1=0 such that \alpha < \beta . Then, 

\begin{aligned} & \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: e^\alpha \sin \alpha=1 \text { and } e^\beta \sin \beta=1 \\ \\& \Rightarrow \quad e^{-\alpha}-\sin \alpha=0 \text { and } e^{-\beta}-\sin \beta=0\: \: \: \: \: \: \: \: \: \: \: \: \: ...(i) \\ & \end{aligned}

Clearly, f(x) is continuous on [\alpha, \beta] and differentiable on (\alpha, \beta).

Also                              f(\alpha)=f(\beta)=0                      [using Eq.(i)]

Therefore, by Rolle's theorem there exists c \in(\alpha, \beta) such that

\begin{aligned} f^{\prime}(c)=0 & \Rightarrow-e^{-c}-\cos c=0 \\ \\\Rightarrow \quad e^{-c}+\cos c=0 & \Rightarrow \quad e^c \cos c+1=0 \end{aligned}

where c \in(\alpha, \beta)

Posted by

Sumit Saini

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