The function fx-1-sin x-2x2 x -2 k x-2 is continuous at x-2 then k is equal to

The correct option is B 18 Substitute #160; x=π2 t , lim _t→ 01 sin(π2 t)(2t)^2 Applying L Hospital's rule twice ,we get #160; =1 cos t4 ...

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Existence of Limit

The function ...

Question

The function $f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - sin x}{(π - 2 x)^{2}} & x \neq \frac{π}{2} k & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ then $k$ is equal to

\frac{1}{8}

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

f (x)

x = \frac{π}{2}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - sin x}{(π - 2 x)^{2}} \cdot \frac{log sin x}{log (1 + π^{2} - 4 π x + 4 x^{2})}, & x \neq \frac{π}{2} k, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - sin x)}{(π - 2 x)^{2}}, & x \neq \frac{π}{2} λ, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

λ

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}

x = \frac{π}{2}

a

b

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{2} 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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