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Range of values of 

f(x)=1+\sin x+\sin ^3 x+\sin ^5 x+\ldots ; x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) is 

Option: 1

\left ( 0,1 \right )


Option: 2

\left ( 0,2 \right )


Option: 3

\left ( -2,2 \right )


Option: 4

\left ( - \infty, \infty \right )


Answers (1)

best_answer

\begin{aligned} & \because x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \Rightarrow \sin x \in(-1,1) \\ \\& \therefore \quad f(x)=1+\frac{\sin x}{1-\sin ^2 x} \\ \\& \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: =1+\frac{\sin x}{\cos ^2 x} \\ & \end{aligned}

       \begin{aligned} \therefore \quad f^{\prime}(x) & =0+\frac{\cos ^2 x \cdot \cos x-\sin x \cdot 2 \cos x \cdot(-\sin x)}{\cos ^4 x} \\ \\& =\frac{\cos ^2 x+2 \sin ^2 x}{\cos ^3 x}>0 \: \: \: \: \quad \because x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right. \end{aligned}

\Rightarrow\: f(x) is increasing  function 

                       \begin{aligned} \therefore \text { Range } & =\left(f\left(-\frac{\pi}{2}\right), f\left(\frac{\pi}{2}\right)\right) \\ & =\left(\lim _{x \rightarrow-\frac{\pi}{2}} f(x), \lim _{x \rightarrow \frac{\pi}{2}} f(x)\right) \\ & =\left(1-\frac{1}{0}, 1+\frac{1}{0}\right) \\ & =(-\infty, \infty) \end{aligned}

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Ajit Kumar Dubey

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