Lf the function fx-kcos x-2x- x-2 3 at x-2is continuous at x-2 then k-

The correct option is C 6 On applying the limits, since the function is of the form 00 , We apply L Hospital's rule, lim_ x →π2 #160; ksin x 2 ...

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Theorems for Continuity

lf the functi...

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lf the function $f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, & x \neq \frac{π}{2} 3 & a t x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ then $k =$

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f (x) =

\frac{k cos x}{π - 2 x}

x \neq \frac{π}{2}

3

x = \frac{π}{2}

x = \frac{π}{2}

k

k

x = \frac{π}{2}

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, if x \neq \frac{π}{2} 3, if x = \frac{π}{2} \end{matrix}

k

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, i f x \neq \frac{π}{2} 3, i f x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

f (x) = \frac{1 - tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{4}]

f (x)

[0, \frac{π}{2}]

f (\frac{π}{4})

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}

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