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Theorems for Continuity

The value of ...

Question

The value of $K$ which the Function
$f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{4}{5})}^{\frac{tan 4 x}{tan 5 x}}, & 0 < x < \frac{π}{2} k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}$
continuous at $x = \frac{π}{2},$ is

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Solution

$\begin{matrix} f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{4}{5})}^{\frac{tan 4 x}{tan 5 x}} 0 < x < \frac{π}{2} k + \frac{2}{5} x = \frac{π}{2} \end{matrix} a t x = \frac{π}{2} L . H . L = {lim}_{x \to \frac{π}{2}} {(\frac{4}{5})}^{\frac{tan 4 x}{tan 5 x}} = k + \frac{2}{5} {lim}_{h \to 0} (\frac{4}{5}) \frac{tan 4 (\frac{π}{2} - h)}{tan 5 (\frac{π}{2} - h)} = k + \frac{2}{5} {lim}_{h \to 0} {(\frac{4}{5})}^{\frac{tan 4 (\frac{π}{2} - h)}{tan 5 (\frac{π}{2} - h)}} = k + \frac{2}{5} 1 = k + \frac{2}{5} k = 1 - \frac{2}{5} k = \frac{3}{5} w h i c h i s t h e r e q u i r e d a n s w e r . \end{matrix}$

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f (x) = \{\begin{matrix} \frac{1 - \sin x}{π - 2 x}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{matrix}

is continuous at

x = \frac{π}{2},

then k = _______________.

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, & - π \leq x \leq \frac{- π}{2} a cos (x) + b, & \frac{- π}{2} < x < 0 - 2 + cos x, & 0 \leq x \leq \frac{π}{2} c sin x, & \frac{π}{2} < x \leq π \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

is continuous in

[- π, π],

then number of points at which

f (x)

is not differentiable is/are

α

β

, so that the function

f (x)

f (x) = - 2 sin x,

- π \leq x \leq - \frac{π}{2}

= α sin x + β,

- \frac{π}{2} < x < \frac{π}{2}

= cos x,

\frac{π}{2} \leq x \leq π

is continuous on

[- π, π]

f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] .

If f(x) is continuous in

[0, \frac{π}{2}]

f (\frac{π}{4}) =

f (x) = \frac{1 - sin x}{(π - 2 x)^{2}},

x \neq \frac{π}{2}

is continuous at

x = \frac{π}{2},

f (\frac{π}{2})

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