If the function fx- -2-cos x-1-x2 x- k x- is continuous at x- then k equals -

The correct option is D 14 lim _ x→π #160;√( 2+cos x #160;) 1 / (π x)^ 2 #160; =lim _ x→π #160; 1+cos x #160;/ (π x)^ 2 (√( 2+cos x ...

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Theorems for Continuity

If the functi...

Question

If the function $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{\sqrt{2 + cos x} - 1}{(π - x)^{2}} & x \neq π k & x = π \end{matrix}$ is continuous at $x = π$ , then $k$ equals :

0

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\frac{1}{2}

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2

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\frac{1}{4}

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k

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

x = \frac{π}{2}

f (x) = \{\begin{cases} \frac{\sin (\cos x) - \cos x}{{(π - 2 x)}^{2}}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{cases}

\frac{1}{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 + 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} k & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

k

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - tan x}{4 x - π}, & x \neq \frac{π}{4} λ, & x = \frac{π}{4} \end{matrix}

x ϵ [0, \frac{π}{2})

f (x)

[0, \frac{π}{2})

λ

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