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(C) Given $f(x)$ is continuous at $x=2$,we have $\lim_{x 	o 2^-} f(x) = f(2)$.$\lim_{x 	o 2^-} (ae^x + [x-2]) = [e^{-2}] - C$$ae^2 + [0^-] = 0 - (1

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$x=0$ only

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(C) Given $f(x)$ is continuous at $x=2$,we have $\lim_{x 	o 2^-} f(x) = f(2)$.$\lim_{x 	o 2^-} (ae^x + [x-2]) = [e^{-2}] - C$$ae^2 + [0^-] = 0 - (1 - 2e^2)$$ae^2 - 1 = -1 + 2e^2 \Rightarrow a = 2$.Now,check continuity at $x=0$:$LHL = \lim_{x 	o 0^-} [e^x] = [e^0] = [1] = 1$.$RHL = \lim_{x 	o 0^+} (2e^x + [x-2]) = 2(1) + [-2] = 2 - 2 = 0$.Since $LHL 
eq RHL$,$f(x)$ is discontinuous at $x=0$.Check continuity at $x=1$:$f(1) = 2e^1 + [1-2] = 2e - 1$.$\lim_{x 	o 1^-} f(x) = 2e - 1$.$\lim_{x 	o 1^+} f(x) = 2e + [1-2] = 2e - 1$.Since $LHL = RHL = f(1)$,$f(x)$ is continuous at $x=1$.Thus,$f(x)$ is discontinuous only at $x=0$.

Let $[t]$ represent the greatest integer not exceeding $t$ and $C=1-2e^2$. If the function $f(x)=\begin{cases} [e^x], & x < 0 \\ ae^x+[x-2], & 0 \leq x < 2 \\ [e^{-x}]-C, & x \geq 2 \end{cases}$ is continuous at $x=2$,then $f(x)$ is discontinuous at

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