91Ó°ÊÓ

We are given the function \(y=f(x)+ax+b\). Let's first differentiate this function with respect to \(x\): $$ \frac{dy}{dx} = \frac{d}{dx}(f(x)+ax+b). $$ Using the sum and constant rule of differentiation, we get $$ \frac{dy}{dx} = \frac{df(x)}{dx} + a. $$

Since we are interested in finding if a relative minimum exists at \(x=0\), let's find the first derivative when x=0: $$ \left.\frac{dy}{dx}\right|_{x=0} = \left.\frac{df(x)}{dx}\right|_{x=0}+a. $$ We are given that function \(f(x)\) has a relative minimum at \(x=0\). It means that the first derivative of \(f(x)\) at \(x=0\) is also 0: $$ \left.\frac{df(x)}{dx}\right|_{x=0}=0. $$ Thus, $$ \left.\frac{dy}{dx}\right|_{x=0} = 0+a = a. $$

Next, let's find the second derivative of the function with respect to \(x\): $$ \frac{d^2y}{dx^2} = \frac{d^2}{dx^2}(f(x)+ax+b). $$ Using the constant rule of differentiation, we get $$ \frac{d^2y}{dx^2} = \frac{d^2f(x)}{dx^2}. $$

To find out if there is a relative minimum at \(x=0\), we can use the second derivative test. If the second derivative is positive at \(x=0\), then there is a relative minimum. Let's find the second derivative at x=0: $$ \left.\frac{d^2y}{dx^2}\right|_{x=0} = \left.\frac{d^2f(x)}{dx^2}\right|_{x=0}. $$ Since we were given that \(f(x)\) has a relative minimum at \(x=0\), the second derivative of \(y\) at \(x=0\) is positive as well: $$ \left.\frac{d^2y}{dx^2}\right|_{x=0}>0. $$ As we found earlier, \(\left.\frac{dy}{dx}\right|_{x=0}=a\). The only condition needed for \(y=f(x)+ax+b\) to have a relative minimum at \(x=0\) is for the first derivative to be zero at \(x=0\). So, \(a\) must be equal to 0. There is no condition on \(b\) as it does not affect the minimum value. Therefore, the correct option is: (B) all \(b\) if \(a=0\).