A curve `f(x)=x^(3)+ax-b` pass through `p(1,-5)` and tangent to `f(x)` at point `p` is perpendicular to `x-y+5=0` then which of the following point will lie on curve ? A(2,-2) B(2,-1) C(2,1) D(-2,2)

Options

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

Answer

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

### Step 1: Substitute point P(1, -5) into the curve equation
The curve is given by the equation:
\[ f(x) = x^3 + ax - b \]
Since the curve passes through the point \( P(1, -5) \), we substitute \( x = 1 \) and \( f(x) = -5 \) into the equation:
\[ -5 = 1^3 + a(1) - b \]
This simplifies to:
\[ -5 = 1 + a - b \]
Rearranging gives us:
\[ a - b = -6 \quad \text{(Equation 1)} \]

### Step 2: Find the slope of the tangent line
The line given is \( x - y + 5 = 0 \). Rearranging this, we get:
\[ y = x + 5 \]
The slope \( m_1 \) of this line is \( 1 \).

### Step 3: Determine the slope of the tangent to the curve at point P
The slope of the tangent to the curve at any point \( x \) is given by the derivative \( f'(x) \):
\[ f'(x) = 3x^2 + a \]
At point \( P(1, -5) \), we find:
\[ f'(1) = 3(1)^2 + a = 3 + a \]
Since the tangent is perpendicular to the line with slope \( m_1 = 1 \), we have:
\[ m_1 \cdot m_2 = -1 \]
Thus,
\[ 1 \cdot (3 + a) = -1 \]
This leads to:
\[ 3 + a = -1 \]
Solving for \( a \):
\[ a = -4 \]

### Step 4: Substitute \( a \) back into Equation 1 to find \( b \)
From Equation 1:
\[ a - b = -6 \]
Substituting \( a = -4 \):
\[ -4 - b = -6 \]
Rearranging gives:
\[ -b = -6 + 4 \implies -b = -2 \implies b = 2 \]

### Step 5: Write the final form of the function
Now we have:
\[ a = -4, \quad b = 2 \]
Thus, the function becomes:
\[ f(x) = x^3 - 4x - 2 \]

### Step 6: Check which point lies on the curve
We need to check which of the given points lies on the curve \( f(x) \).

1. **Point A(2, -2)**:
\[ f(2) = 2^3 - 4(2) - 2 = 8 - 8 - 2 = -2 \]
This point lies on the curve.

2. **Point B(2, -1)**:
\[ f(2) = -2 \quad \text{(not equal to -1)} \]

3. **Point C(2, 1)**:
\[ f(2) = -2 \quad \text{(not equal to 1)} \]

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

Thus, the only point that lies on the curve is **Point A(2, -2)**.

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