Let f(X)={1-sinπx÷1+cos2πx. X<1÷2. { P . X=1÷2. { √2x-1÷√4+√2x-1-2. X>1÷2. Determine the value of p , so that the function is continuous at X=1÷2 . { Is over the whole f(X) and √ is on complete √4+√2x-1

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - sin π x}{1 + cos 2 π x}, & x < \frac{1}{2} p, & x = \frac{1}{2} \frac{\sqrt{2 k + \sqrt{2 x - 1}} - 2}{\sqrt{2 x - 1}}, & x > \frac{1}{2} \end{matrix}

f

x = \frac{1}{2},

k + 12 p

k

f (x)

x = 2.

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 2 x + 1; i f x < 2 k; x = 2 3 x - 1; x > 2 \end{matrix}

x = - \frac{1}{2}

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{4 x^{2} - 1}{2 x + 1} - 2, \end{matrix} \begin{matrix} x \neq - \frac{1}{2} x = - \frac{1}{2} \end{matrix}

f (x) = ⎧ ⎨ ⎩ \begin{matrix} a x - 2 & f o r & - 2 < x < - 1 - 1 & f o r & - 1 \leq x \leq 1 a + 2 (x - 1)^{2} & f o r & 1 < x < 2 \end{matrix}

f (x)

x = - 1

x = 1

f (0)

f (x) = \frac{2 x - {sin}^{- 1} x}{2 x + {tan}^{- 1} x}

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