Fx-1-sin3x3cos2x- if x -2- is continuous at x-2-

Applying L's Hospital on RHL ,we get RHL = qcos x2(π 2x)( 2) Again L's Hospital, we get RHL = qsin x8 Putting x = π2 RHL = q8 Again L's ...

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Continuity in an Interval

fx=1-sin3x3co...

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

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

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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)

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p

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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}

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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} i s c o n t i n o u s a t x = \frac{π}{2} \frac{q (1 - sin x)}{{(π - 2 x)}^{2}}, i f x > \frac{π}{2} \end{matrix}

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

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a

b

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