Let `(f(x+y)-f(x))/(2)=(f(y)-1)/(2)+xy`, for all `x,yinR,f(x)` is differentiable and `f'(0)=1.` Let `g(x)` be a derivable function at `x=0` and follows the function rule `g((x+y)/(k))=(g(x)+g(y))/(k),kinR,kne0,2andg'(0)-lambdag'(0)ne0.`
If the graphs of `y=f(x) and y=g(x)` intersect in coincident points then `lambda` can take values

Options

A. -3
B. 1
C. -1
D. 4

Answer

To solve the given problem step by step, we will analyze the equation and derive the necessary conditions for the functions \( f(x) \) and \( g(x) \).

### Step 1: Analyze the given equation
We start with the equation:
\[ \frac{f(x+y) - f(x)}{2} = \frac{f(y) - 1}{2} + xy \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Substitute \( x = 0 \) and \( y = 0 \)
Substituting \( x = 0 \) and \( y = 0 \) into the equation gives:
\[ \frac{f(0+0) - f(0)}{2} = \frac{f(0) - 1}{2} + 0 \cdot 0 \]
This simplifies to:
\[ \frac{0}{2} = \frac{f(0) - 1}{2} \]
Thus, we have:
\[ 0 = f(0) - 1 \implies f(0) = 1 \]

### Step 3: Differentiate the original equation
Next, we differentiate both sides of the original equation with respect to \( y \):
\[ \frac{d}{dy}\left(\frac{f(x+y) - f(x)}{2}\right) = \frac{d}{dy}\left(\frac{f(y) - 1}{2} + xy\right) \]
Using the chain rule on the left side:
\[ \frac{1}{2} f'(x+y) = \frac{1}{2} f'(y) + x \]

### Step 4: Substitute \( y = 0 \)
Now, substitute \( y = 0 \):
\[ \frac{1}{2} f'(x+0) = \frac{1}{2} f'(0) + x \]
This simplifies to:
\[ \frac{1}{2} f'(x) = \frac{1}{2} \cdot 1 + x \quad (\text{since } f'(0) = 1) \]
Thus:
\[ f'(x) = 1 + 2x \]

### Step 5: Integrate to find \( f(x) \)
Integrating \( f'(x) \):
\[ f(x) = \int (1 + 2x) \, dx = x + x^2 + C \]
Using the condition \( f(0) = 1 \):
\[ f(0) = 0 + 0 + C = 1 \implies C = 1 \]
Thus, we have:
\[ f(x) = x^2 + x + 1 \]

### Step 6: Analyze the function \( g(x) \)
Given the function rule for \( g(x) \):
\[ g\left(\frac{x+y}{k}\right) = \frac{g(x) + g(y)}{k} \]
This suggests that \( g(x) \) is a linear function. Let \( g(x) = mx + b \).

### Step 7: Differentiate \( g(x) \)
Differentiating \( g(x) \):
\[ g'(x) = m \]
At \( x = 0 \):
\[ g'(0) = m = \lambda \]

### Step 8: Intersection of graphs
For the graphs of \( y = f(x) \) and \( y = g(x) \) to intersect at coincident points, they must be equal for all \( x \):
\[ x^2 + x + 1 = mx + b \]
This leads to the conditions:
1. Coefficients of \( x^2 \): \( 1 = 0 \) (not possible)
2. Coefficients of \( x \): \( 1 = m \)
3. Constant term: \( 1 = b \)

### Step 9: Conclusion on \( \lambda \)
From the conditions, we find:
\[ \lambda = 1 \]
Thus, the only value that \( \lambda \) can take is:
\[ \lambda = 1 \]