For x∈(-π)/2,(π)/2, the range of values of fx=2+sin⁡x+sin^3⁡x+sin^5⁡x…..∞ is

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MARKS Content Team · Accepted Answer

Correct answer: (B) -∞,∞. We have, fx=2+(sin⁡x)/(cos^2⁡x) ⇒f^(′) x =(cos^2 x cosx +sinx 2cosxsinx)/(cos^4 x) =(cos⁡xcos^2⁡x+2sin^2⁡x)/(cos^4⁡x)=(1+sin^2⁡x)/(cos^3⁡x) ⇒f^(′) x >0 ∴fx is an increasing function limx→(-π)/2⁡2+(sin⁡x)/(cos^2⁡x)=-∞ and limx→(π)/2⁡2+(sin⁡x)/(cos^2⁡x)=∞ ∴ Range =-∞, ∞

For $x \in (\frac{- π}{2}, \frac{π}{2})$ , the range of values of $f (x) = 2 + \sin x + \sin^{3} x + \sin^{5} x \dots . . \infty$ is

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For x∈-π2,π2, the range of values of fx=2+sin⁡x+sin3⁡x+sin5⁡x…..∞ is

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For $x \in (\frac{- π}{2}, \frac{π}{2})$ , the range of values of $f (x) = 2 + \sin x + \sin^{3} x + \sin^{5} x \dots . . \infty$ is