If the tangent to the curve, `y=x^(3)+ax-b` at the point `(1, -5)` is perpendicular to the line, `-x+y+4=0`, then which one of the following points lies on the curve ?

Options

A. (-2, 1)
B. (-2, 2)
C. (2, -1)
D. (2, -2)

Explanation

To solve the problem step by step, we will follow the given information and derive the necessary equations.

### Step 1: Verify the point lies on the curve
We are given the curve \( y = x^3 + ax - b \) and the point \( (1, -5) \). Since this point lies on the curve, we can substitute \( x = 1 \) and \( y = -5 \) into the equation to find a relationship between \( a \) and \( b \).

\[ -5 = 1^3 + a(1) - b \]
\[ -5 = 1 + a - b \]
\[ a - b = -6 \quad \text{(Equation 1)} \]

### Step 2: Find the slope of the tangent
Next, we need to find the slope of the tangent to the curve at the point \( (1, -5) \). We first differentiate the curve:

\[ \frac{dy}{dx} = 3x^2 + a \]

Now, substituting \( x = 1 \) into the derivative to find the slope at that point:

\[ \frac{dy}{dx} \bigg|_{x=1} = 3(1^2) + a = 3 + a \quad \text{(Slope of tangent)} \]

### Step 3: Find the slope of the line
The line given is \( -x + y + 4 = 0 \). We can rearrange this to find its slope:

\[ y = x - 4 \]

The slope of this line is \( 1 \).

### Step 4: Use the perpendicularity condition
Since the tangent to the curve is perpendicular to the line, the product of their slopes must equal \(-1\):

\[ (3 + a) \cdot 1 = -1 \]

Solving for \( a \):

\[ 3 + a = -1 \]
\[ a = -4 \quad \text{(Equation 2)} \]

### Step 5: Substitute \( a \) back into Equation 1
Now we substitute \( a = -4 \) into Equation 1 to find \( b \):

\[ -4 - b = -6 \]
\[ -b = -6 + 4 \]
\[ -b = -2 \]
\[ b = 2 \]

### Step 6: Write the equation of the curve
Now that we have \( a \) and \( b \), we can write the equation of the curve:

\[ y = x^3 - 4x - 2 \]

### Step 7: Check which points lie on the curve
We need to check which of the given points lies on the curve. We will evaluate the curve equation for each point.

1. **Point (-2, 1)**:
\[ y = (-2)^3 - 4(-2) - 2 = -8 + 8 - 2 = -2 \quad (\text{not } 1) \]

2. **Point (-2, 2)**:
\[ y = (-2)^3 - 4(-2) - 2 = -8 + 8 - 2 = -2 \quad (\text{not } 2) \]

3. **Point (2, -1)**:
\[ y = (2)^3 - 4(2) - 2 = 8 - 8 - 2 = -2 \quad (\text{not } -1) \]

4. **Point (2, -2)**:
\[ y = (2)^3 - 4(2) - 2 = 8 - 8 - 2 = -2 \quad (\text{is } -2) \]

Thus, the point \( (2, -2) \) lies on the curve.

### Final Answer
The point that lies on the curve is \( (2, -2) \).