Let `(f(x+y)-f(x))/(2)=(f(y)-1)/(2)+xy`, for all `x,yinR,f(x)` is differentiable and `f'(0)=1.`
Range of `y=log_(3//4)(f(x))` is

Options

A. `(-oo,1)`
B. `[3//4,oo`
C. `(-oo,oo)`
D. `R`

Answer

To solve the problem, we start with the given equation:

\[ \frac{f(x+y) - f(x)}{2} = \frac{f(y) - 1}{2} + xy \]

### Step 1: Simplifying the Equation
Multiply both sides by 2:

\[ f(x+y) - f(x) = f(y) - 1 + 2xy \]

### Step 2: Finding \( f(0) \)
Substituting \( x = 0 \) and \( y = 0 \):

\[ f(0+0) - f(0) = f(0) - 1 + 2 \cdot 0 \cdot 0 \]

This simplifies to:

\[ 0 = f(0) - 1 \]

Thus, we find:

\[ f(0) = 1 \]

### Step 3: Finding the Derivative \( f'(x) \)
We know \( f'(0) = 1 \). To find \( f'(x) \), we use the limit definition of the derivative:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

Using our earlier equation, we can substitute \( y = h \):

\[ f(x+h) - f(x) = f(h) - 1 + 2xh \]

So we have:

\[ f'(x) = \lim_{h \to 0} \frac{f(h) - 1 + 2xh}{h} \]

This can be split into two parts:

\[ f'(x) = \lim_{h \to 0} \frac{f(h) - 1}{h} + \lim_{h \to 0} 2x \]

The first limit is \( f'(0) \), which we know is 1. Thus,

\[ f'(x) = 2x + 1 \]

### Step 4: Integrating to Find \( f(x) \)
Now, we integrate \( f'(x) \):

\[ f(x) = \int (2x + 1) \, dx = x^2 + x + C \]

### Step 5: Finding the Constant \( C \)
Using \( f(0) = 1 \):

\[ f(0) = 0^2 + 0 + C = 1 \implies C = 1 \]

Thus, we have:

\[ f(x) = x^2 + x + 1 \]

### Step 6: Finding the Range of \( y = \log_{3/4}(f(x)) \)
Now, we need to find the range of \( y = \log_{3/4}(f(x)) \).

Since \( f(x) = x^2 + x + 1 \), we need to find the minimum value of \( f(x) \):

Completing the square:

\[ f(x) = (x + \frac{1}{2})^2 + \frac{3}{4} \]

The minimum value of \( f(x) \) occurs at \( x = -\frac{1}{2} \):

\[ f\left(-\frac{1}{2}\right) = \frac{3}{4} \]

Thus, \( f(x) \geq \frac{3}{4} \).

### Step 7: Evaluating the Logarithm
Since \( f(x) \) is always greater than or equal to \( \frac{3}{4} \):

\[ y = \log_{3/4}(f(x)) \text{ is defined for } f(x) \geq \frac{3}{4} \]

Now, since \( \log_{3/4}(a) \) is decreasing, we find:

\[ \log_{3/4}\left(\frac{3}{4}\right) = 1 \]

As \( f(x) \to \infty \), \( y \to -\infty \).

### Conclusion
The range of \( y = \log_{3/4}(f(x)) \) is:

\[ (-\infty, 1] \]