91Ó°ÊÓ

To differentiate the given function, we'll use the chain rule. Let F(x) be a function such that \[F(x) = f(x+y)\] Now, we differentiate F(x) to get F'(x): \[F^{\prime}(x) = \frac{d}{dx}F(x) = \frac{d}{dx}(f(x+y))\] Using the chain rule, we have, \[F^{\prime}(x) = \frac{df(u)}{du}\frac{du}{dx}\] Where, \(u=x+y\) As we are differentiating with respect to x, \[\frac{du}{dx}=1\] And hence, \[F^{\prime}(x) = f^{\prime} (u) \cdot 1 = f^{\prime}(x+y)\]

We have \(f^{\prime}(x+y)=F^{\prime}(x)\). Now, let's put the value of x = 0 in this equation: \[f^{\prime}(y) = F^{\prime}(0)\] Now we can see that the derivative of the function f(x) is equal to a constant for all values of y, which can be written as: \[f^{\prime}(y) = C\] Where C is a constant.

We have now found the derivative of the function f(x) as \(f^{\prime}(y)=C\). Now, let's integrate the derivative with respect to y: \[f(x) = \int C dy\] \[f(x) = Cx + k\] Here, C and k are constants. Now let's compare it to the given statements: (a): We have already shown that \(f^{\prime}(y) = C\), which means \(f^{\prime}(x) = f^{\prime}(0) \forall x\) (True) (b): Comparing our derived f(x) with the given statement, we have: \[f(x) = Cx\] As k can be 0, this statement is also True. (c): Comparing our derived f(x) with the given statement, we have: \[f(x) = xf(1)\] As \(f(1) = C \cdot 1 + k = C\), the statement is True when k=0. (d): If we put the value of 2x in our derived f(x) function, we get: \[f(2x) = C(2x) + k\] Comparing this to the given statement: \[f(x) = \frac{x}{2}f(2)\] When k=0 and with the value \(f(2) = 2C\), we find that this statement is also True. So, all the given statements (a), (b), (c), and (d) are True.