The range of fx-1-sin x-sin3x-sin5x- x-2-2 is

The correct option is C ( ∞,∞) f(x)=1+sin x/cos^2x f'(x)=cos^2x(cos x)+sin x(2cos xsin x)/cos^4x =cos x(cos^2x+2sin^2x)/cos^4x=1+sin^2xcos^ ...

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Definition of Functions

The range of ...

Question

The range of $f (x) = 1 + sin x + {sin}^{3} x + {sin}^{5} x + \dots . . \infty, \forall$ $x \in (- \frac{π}{2} . \frac{π}{2})$ is

(0, 1)

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(- \infty, \infty)

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(- 2, 2)

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(- 1, 0)

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{3} x}{3 {cos}^{2} x}, if x < \frac{π}{2} a, if x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}}, if x > \frac{π}{2} \end{matrix}

f (x)

x = \frac{π}{2}

p

q

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{3} x}{3 {cos}^{2} x}, & i f x < \frac{π}{2} p & i f x = \frac{π}{2} \frac{q (1 - sin x)}{(π - 2 x)^{2}} & i f x > \frac{π}{2} \end{matrix}

\frac{π}{2}

sin 5 x + sin 3 x + sin x = 0

0 \leq x \leq π

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - {sin}^{3} x)}{3 {cos}^{2} x}, & x < \frac{π}{2} a, & x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}}, & x > \frac{π}{2} \end{matrix}

x = \frac{π}{2}

{(\frac{b}{a})}^{5 / 3}

sin x + sin 3 x + sin 5 x = 0

[\frac{π}{2}, 3 \frac{π}{2}]

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