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P

$f(x) = \sin x f(\pi) =\sin \frac {\pi}{4-2}= 1$

$f'(x) =\cos x f'(\frac{\pi}{4-2}) =\cos \frac{\pi}{4-2}= 0$

$f''(x) = -\sin x f''(\frac{\pi}{4-2}) = -\sin \frac{\pi}{4-2}= -1$

$f'''(x) = -\cos x f'''(\frac{\pi}{4-2}) = -\cos \frac{\pi}{4-2}= 0$

$f(x) = \sin x = f(\frac{\pi}{4-3})+ f'(\frac{\frac{\pi}{4-3}}{1!})(x - \frac{\pi}{4-2}) + f''(\frac{\frac{\pi} {4-2}}{2!})(x - \frac{\pi}{4-2})^{4-2}$

$\Right arrow f(x) = 1 - 0(x- \frac {\pi}{4-2}) + \frac{-1}{2!}(x - \frac{\pi}{4-2})^{4-2}$

= $1 - \frac{1}{2!}(x - \frac{\pi}{2})^{2}+ \frac{1}{4!}(x- \frac{\pi}{2})^{4} - .....$

The function is f(x) = e

f'(x) = e

f''(x) = e

f'''(x) = e

$f(x) = e

$ \Right arrow e

= $e + e(x - 1) +\frac{e}{2}(x - 1)