$$\begin{gathered} f\mathrm{(x)}=\{\begin{array}{ll}{C}_1{e}^x-C_2{\log }_e\left(\frac{1}{x}\right)& x>0\\ 3& \text{otherwise}\end{array} \\ \end{gathered}$$

\text{<br>}

\text{Where }{C}_1,C_2\in R.\text{ If }f\text{ is continuous at }x=0\text{, then }{C}_1+C_2\text{ is \_\_\_\_\_.}

The correct answer is 3.

To find the value of C₁ + C₂, we use the definition of continuity at a point. For the function f(x) to be continuous at x = 0, the left-hand limit, the right-hand limit, and the value of the function at x = 0 must all exist and be equal:
$$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$$

First, from the definition of the function, for x ≤ 0 (which falls under the "otherwise" case), we have:
$$f\left(0\right)=3$$

Therefore, the left-hand limit is:
$$\underset{x\to 0^{-}}{lim}f\left(x\right)=3$$

Next, we evaluate the right-hand limit as x approaches 0 from the positive side (x > 0):
$$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}[C_1{e}^x-C_2{log}_e\left(\frac{1}{x}\right)]$$

Let us analyze the behavior of the terms in the limit as x → 0⁺:
1. The exponential term behaves as:
$$\underset{x\to 0^{+}}{lim}{e}^x=e^0=1$$
2. The logarithmic term behaves as:
As x → 0⁺, the term $\frac{1}{x}$ approaches +∞. Consequently, ${log}_e\left(\frac{1}{x}\right)\to +\infty$.

For the right-hand limit to exist and be equal to the finite value 3, the coefficient of the divergent term must be zero. Therefore, we must have:
$$C_2=0$$

Substituting C₂ = 0 back into the limit expression gives:
$$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\left[C_1{e}^x\right]=C_1·1=C_1$$

Now, matching the right-hand limit to the value of the function at x = 0 for continuity:
$$C_1=3$$

Finally, we calculate the sum C₁ + C₂:
$$C_1+C_2=3+0=3$$