Equation of a line which is tangent to both the curve `y=x^2+1\ a n d\ y=x^2` is `y=sqrt(2)x+1/2` (b) `y=sqrt(2)x-1/2` `y=-sqrt(2)x+1/2` (d) `y=-sqrt(2)x-1/2`

Options

A. `y=sqrt2x-(1)/(2)`
B. `y=sqrt2x+(1)/(2)`
C. `y=-sqrt2x+(1)/(2)`
D. `y=-sqrt2x-(1)/(2)`

Explanation

To find the equations of the lines that are tangent to both curves \( y = x^2 + 1 \) and \( y = x^2 \), we will follow these steps:

### Step 1: Set up the equations of the curves
The equations of the curves are:
1. \( y = x^2 + 1 \) (upper curve)
2. \( y = x^2 \) (lower curve)

### Step 2: Assume the equation of the tangent line
Let the equation of the tangent line be:
\[ y = ax + b \]
where \( a \) is the slope and \( b \) is the y-intercept.

### Step 3: Find the intersection with the upper curve
Set the tangent line equal to the upper curve:
\[ ax + b = x^2 + 1 \]
Rearranging gives:
\[ x^2 - ax + (1 - b) = 0 \]
For this quadratic equation to have equal roots (indicating tangency), the discriminant must be zero:
\[ (-a)^2 - 4(1 - b) = 0 \]
This simplifies to:
\[ a^2 - 4 + 4b = 0 \quad \text{(1)} \]

### Step 4: Find the intersection with the lower curve
Set the tangent line equal to the lower curve:
\[ ax + b = x^2 \]
Rearranging gives:
\[ x^2 - ax + b = 0 \]
Again, for tangency, the discriminant must be zero:
\[ (-a)^2 - 4b = 0 \]
This simplifies to:
\[ a^2 - 4b = 0 \quad \text{(2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( a^2 + 4b = 4 \) (from (1))
2. \( a^2 - 4b = 0 \) (from (2))

From equation (2), we can express \( b \):
\[ b = \frac{a^2}{4} \]

Substituting this into equation (1):
\[ a^2 + 4\left(\frac{a^2}{4}\right) = 4 \]
This simplifies to:
\[ a^2 + a^2 = 4 \implies 2a^2 = 4 \implies a^2 = 2 \implies a = \pm \sqrt{2} \]

### Step 6: Find the values of \( b \)
Substituting \( a^2 = 2 \) back into \( b = \frac{a^2}{4} \):
\[ b = \frac{2}{4} = \frac{1}{2} \]

### Step 7: Write the equations of the tangent lines
Thus, we have two tangent lines:
1. For \( a = \sqrt{2} \):
\[ y = \sqrt{2}x + \frac{1}{2} \]
2. For \( a = -\sqrt{2} \):
\[ y = -\sqrt{2}x + \frac{1}{2} \]

### Final Result
The equations of the tangent lines that are tangent to both curves are:
- \( y = \sqrt{2}x + \frac{1}{2} \)
- \( y = -\sqrt{2}x + \frac{1}{2} \)