Given f(x) = b ([x]^2 + [x]) + 1 for x geq -1 | vedclass

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(A) For the function to be continuous at $x = -1$,the left-hand limit,right-hand limit,and the value of the function at $x = -1$ must be equal.First,f

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$a = 2n + (3/2) ; b \in R ; n \in I$

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(A) For the function to be continuous at $x = -1$,the left-hand limit,right-hand limit,and the value of the function at $x = -1$ must be equal.First,find the value of the function at $x = -1$:$f(-1) = b([-1]^2 + [-1]) + 1 = b(1 - 1) + 1 = 1$.Next,find the right-hand limit $(RHL)$ as $x 	o -1^+$:$\lim_{x 	o -1^+} f(x) = \lim_{h 	o 0} b([-1+h]^2 + [-1+h]) + 1 = b((-1)^2 + (-1)) + 1 = b(1-1) + 1 = 1$.Now,find the left-hand limit $(LHL)$ as $x 	o -1^-$:$\lim_{x 	o -1^-} f(x) = \lim_{h 	o 0} \sin(\pi(-1-h+a)) = \sin(\pi(a-1))$.For continuity,$LHL$ = $RHL$ = $f(-1)$,so $\sin(\pi(a-1)) = 1$.This implies $\pi(a-1) = 2n\pi + \frac{\pi}{2}$ for any integer $n \in I$.Dividing by $\pi$,we get $a - 1 = 2n + \frac{1}{2}$,which simplifies to $a = 2n + \frac{3}{2}$.Since $b$ does not affect the limit at $x = -1$ (as the $RHL$ is independent of $b$),$b$ can be any real number $(b \in R)$.

Given $f(x) = b ([x]^2 + [x]) + 1$ for $x \geq -1$ and $f(x) = \sin(\pi(x+a))$ for $x < -1$,where $[x]$ denotes the greatest integer function,for what values of $a$ and $b$ is the function continuous at $x = -1$?

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