If the function f defined on -6- -3 by fx-2cos x-1 x-1- x-4 k- x-4- is continuous- then k is equal to-

The correct option is A 12 ∴ function should be continuous at x=π4 ∴lim_x→x4f(x)=f(π4) ⇒lim_x→x4√(2)cos x 1 x 1=k ⇒lim_x→π4 √(2) ...

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Theorems for Continuity

If the functi...

Question

If the function f defined on $(\frac{π}{6}, \frac{π}{3})$ by $f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{2} cos x - 1}{cot x - 1}, & x \neq \frac{π}{4} k, & x = \frac{π}{4} \end{matrix}$ is continuous, then k is equal to?

\frac{1}{2}

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1

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\frac{1}{\sqrt{2}}

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2

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f

(\frac{π}{6}, \frac{π}{3})

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{\sqrt{2} cos x - 1}{cot x - 1}, & x \neq \frac{π}{4} k, & x = \frac{π}{4} \end{matrix}

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 = \frac{π}{4}

f

f (x) = \frac{1 - \sqrt{2} sin x}{π - 4 x}

x \neq \frac{π}{4}

= \frac{1}{4}

x \neq \frac{π}{4}

f (x) = \frac{\sqrt{2} cos x - 1}{cot x - 1}, x \neq \frac{π}{4}

f (\frac{π}{4})

x = \frac{π}{4}

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

f (x)

[- π, π]

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