y=x is tangent to the parabola `y=ax^(2)+c`.
If a=2, then the value of c is (a) 1 (b) `-1/2` (c) `1/2` (d) `1/8`

Options

A. 1
B. `-1//2`
C. `1//2`
D. `1//8`

Explanation

To solve the problem, we need to find the value of \( c \) such that the line \( y = x \) is tangent to the parabola \( y = 2x^2 + c \).

### Step 1: Identify the equations
The equation of the parabola is given as:
\[ y = 2x^2 + c \]
The equation of the tangent line is:
\[ y = x \]

### Step 2: Find the derivative of the parabola
To find the slope of the parabola at any point, we differentiate the equation of the parabola with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 + c) = 4x \]

### Step 3: Set the slope of the tangent equal to the slope of the parabola
Since the line \( y = x \) has a slope of 1, we set the derivative equal to 1:
\[ 4x = 1 \]
Solving for \( x \):
\[ x = \frac{1}{4} \]

### Step 4: Find the corresponding \( y \) value on the tangent line
Substituting \( x = \frac{1}{4} \) into the equation of the tangent line:
\[ y = x = \frac{1}{4} \]

### Step 5: Substitute \( x \) and \( y \) into the parabola's equation
Now we substitute \( x = \frac{1}{4} \) and \( y = \frac{1}{4} \) into the parabola's equation to find \( c \):
\[ \frac{1}{4} = 2\left(\frac{1}{4}\right)^2 + c \]
Calculating \( \left(\frac{1}{4}\right)^2 \):
\[ \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
So,
\[ \frac{1}{4} = 2 \cdot \frac{1}{16} + c \]
This simplifies to:
\[ \frac{1}{4} = \frac{2}{16} + c \]
\[ \frac{1}{4} = \frac{1}{8} + c \]

### Step 6: Solve for \( c \)
To isolate \( c \), we subtract \( \frac{1}{8} \) from \( \frac{1}{4} \):
\[ c = \frac{1}{4} - \frac{1}{8} \]
Finding a common denominator (which is 8):
\[ c = \frac{2}{8} - \frac{1}{8} = \frac{1}{8} \]

### Conclusion
Thus, the value of \( c \) is:
\[ \boxed{\frac{1}{8}} \]