A function f(x) is defined as \[{\text{f}}\left( {\text{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{{\text{e}}^x}}&{{\text{x}} < 1} \\ {\ln {\text{x}} + {\text{a}}{{\text{x}}^2} + {\text{bx}},}&{{\text{x}} \geqslant 1} \end{array}} \right.\] where x\[ \in \] R which one of the following statements is TRUE? A. f(x) is NOT differentiable at x = 1 for any values of a and b B. f(x) is differentiable at x = 1 for the unique values of a and b C. f(x) is differentiable at x = 1 for all the values of a and b such that a + b = e D. f(x) is differentiable at x = 1 for all values of a and b

The correct answer is: $\boxed{\text{D}}$.

A function is differentiable at a point if the limit $\lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$ exists. In this case, we have

$$\lim_{h \to 0} \frac{f(1+h) – f(1)}{h} = \lim_{h \to 0} \frac{e^{1+h} – e}{h} = \lim_{h \to 0} \frac{e^h}{h}.$$

The limit $\lim_{h \to 0} \frac{e^h}{h}$ exists and is equal to $e$, so $f(x)$ is differentiable at $x=1$ for all values of $a$ and $b$.

Here is a brief explanation of each option:

• Option A
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  is incorrect because $f(x)$ is differentiable at $x=1$ for some values of $a$ and $b$, such as $a=1$ and $b=0$.
• Option B is incorrect because $f(x)$ is differentiable at $x=1$ for all values of $a$ and $b$.
• Option C is incorrect because $f(x)$ is differentiable at $x=1$ for all values of $a$ and $b$, not just for the unique values of $a$ and $b$ such that $a+b=e$.
• Option D is correct because $f(x)$ is differentiable at $x=1$ for all values of $a$ and $b$.