If the curve `y=ax^(3) +bx^(2) +c x ` is inclined at `45^(@)` to x-axis at (0, 0) but touches x-axis at (1, 0) , then

Options

A. a= 1, b = -2, c = 1
B. a = 1, b = 1, c = -2
C. a = -2, b = 1, c = 1
D. a = -1, b = 2, c = 1

Answer

To solve the problem step by step, we will analyze the conditions given for the curve \( y = ax^3 + bx^2 + cx \).

### Step 1: Understand the conditions
The curve is inclined at \( 45^\circ \) to the x-axis at the point \( (0, 0) \). This means that the slope of the curve at this point is \( 1 \) (since \( \tan(45^\circ) = 1 \)).

### Step 2: Find the derivative of the curve
To find the slope of the curve, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3ax^2 + 2bx + c \]

### Step 3: Evaluate the slope at \( (0, 0) \)
Substituting \( x = 0 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(0,0)} = 3a(0)^2 + 2b(0) + c = c \]
Since the slope at \( (0, 0) \) is \( 1 \), we have:
\[ c = 1 \]

### Step 4: Use the second condition (touching the x-axis at \( (1, 0) \))
The curve touches the x-axis at \( (1, 0) \). This means:
1. The point \( (1, 0) \) lies on the curve.
2. The slope of the curve at this point is \( 0 \).

### Step 5: Substitute \( (1, 0) \) into the curve equation
Substituting \( x = 1 \) and \( y = 0 \) into the curve equation:
\[ 0 = a(1)^3 + b(1)^2 + c(1) \]
Substituting \( c = 1 \):
\[ 0 = a + b + 1 \]
This gives us the first equation:
\[ a + b + 1 = 0 \quad \text{(Equation 1)} \]

### Step 6: Find the slope at \( (1, 0) \)
Now, we need to find the slope at \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{(1,0)} = 3a(1)^2 + 2b(1) + c \]
Substituting \( c = 1 \):
\[ \frac{dy}{dx} \bigg|_{(1,0)} = 3a + 2b + 1 \]
Since the slope at this point is \( 0 \):
\[ 3a + 2b + 1 = 0 \quad \text{(Equation 2)} \]

### Step 7: Solve the system of equations
Now we have two equations:
1. \( a + b + 1 = 0 \)
2. \( 3a + 2b + 1 = 0 \)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = -1 - a \]

Substituting \( b \) into Equation 2:
\[ 3a + 2(-1 - a) + 1 = 0 \]
Simplifying:
\[ 3a - 2 - 2a + 1 = 0 \]
\[ a - 1 = 0 \implies a = 1 \]

### Step 8: Find \( b \)
Now substituting \( a = 1 \) back into Equation 1:
\[ 1 + b + 1 = 0 \implies b = -2 \]

### Step 9: Summary of values
We have found:
- \( a = 1 \)
- \( b = -2 \)
- \( c = 1 \)

### Final Answer
The values of \( a \), \( b \), and \( c \) are:
\[ a = 1, \quad b = -2, \quad c = 1 \]