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(D) Given that $f^{\prime}(x)$ is a constant for all $x \in R$. Let $f^{\prime}(x) = a$,where $a$ is a constant. Integrating both sides with respect t

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$f(x)$ is continuous on $R$

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(D) Given that $f^{\prime}(x)$ is a constant for all $x \in R$. Let $f^{\prime}(x) = a$,where $a$ is a constant. Integrating both sides with respect to $x$,we get $f(x) = ax + b$,where $b$ is an arbitrary constant. Given $f(0) = 2$,substituting $x = 0$ in $f(x) = ax + b$,we get $a(0) + b = 2$,which implies $b = 2$. Given $f^{\prime}(0) = 1$,since $f^{\prime}(x) = a$ for all $x$,we have $a = 1$. Thus,the function is $f(x) = x + 2$. Since $f(x) = x + 2$ is a polynomial function,it is continuous for all real numbers $x \in R$.

Let $f(x)$ be a real-valued function. If $f^{\prime}(x)$ is a constant for all $x \in R$,$f(0)=2$,and $f^{\prime}(0)=1$,then

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