Question

\(y=f(x)\) is a quadratic function passing through (–1, 0) and tangent to it at (1, 1) is \(y=x\). Find x intercept by normal at point (𝛂, 𝛂 + 1), (𝛂 > 0)

• A. 7
• B. -7
• C. 5
• D. -5

Answer 1

The Correct Option is A

To solve this problem, we need to determine the quadratic function \( y = f(x) \), given the conditions, and find the x-intercept of the normal at the point \( (\alpha, \alpha + 1) \) where the function and its tangent are defined.

1. The quadratic function \( f(x) \) passes through the point \((-1, 0)\) and is tangent to the line \( y = x \) at \( (1, 1) \). A general quadratic function can be written as: \(f(x) = ax^2 + bx + c\)
2. Since \( f(x) \) passes through \((-1, 0)\), we substitute these coordinates into the quadratic: \(0 = a(-1)^2 + b(-1) + c\), which simplifies to: \(a - b + c = 0\) (Equation 1).
3. The condition that \( f(x) \) is tangent to \( y = x \) at \( (1, 1) \) gives us two conditions:
   • The function value at \( x = 1 \) must be: \(f(1) = 1 \Rightarrow a(1)^2 + b(1) + c = 1\), simplifying to: \(a + b + c = 1\) (Equation 2).
   • The derivative \( f'(x) \) at \( x = 1 \) must equal the slope of the tangent line \( y = x \) (which is 1). The derivative is: \(f'(x) = 2ax + b\). At \( x = 1 \): \(f'(1) = 2a + b = 1\) (Equation 3).
4. We now solve the system of equations to find \( a \), \( b \), and \( c \):
   • From Equation 3: \(2a + b = 1\)
   • Using Equations 1 and 2:
     1. Combine Equation 1 and 2 by subtraction: \(a - b + c - (a + b + c) = 0 - 1\), leading to \(-2b = -1 \Rightarrow b = \frac{1}{2}\).
     2. Substitute \( b = \frac{1}{2} \) in Equation 3: \(2a + \frac{1}{2} = 1 \Rightarrow 2a = \frac{1}{2} \Rightarrow a = \frac{1}{4}\).
     3. Substitute \( a = \frac{1}{4} \) and \( b = \frac{1}{2} \) into Equation 2: \(\frac{1}{4} + \frac{1}{2} + c = 1\), which gives \(c = \frac{1}{4}\).
   • Thus, the quadratic function is: \(f(x) = \frac{1}{4}x^2 + \frac{1}{2}x + \frac{1}{4}\)
5. We need to find the x-intercept of the normal at \( (\alpha, \alpha + 1) \):
   • The slope of the tangent to \( f(x) \) at \(\alpha\) is \(f'(\alpha) = \frac{1}{2}\alpha + \frac{1}{2}\). The slope of the normal is the negative reciprocal: \(-\frac{2}{\alpha + 1}\)
   • The equation of the normal at \( (\alpha, \alpha + 1) \): \(y - (\alpha + 1) = -\frac{2}{\alpha + 1}(x - \alpha)\). Simplifying this, we have: \(y = -\frac{2}{\alpha + 1}x + \left(\frac{2\alpha}{\alpha + 1} + \alpha + 1\right)\).
   • To find the x-intercept, set \( y = 0 \): \(0 = -\frac{2}{\alpha + 1}x + \left(\frac{3\alpha + 1}{\alpha + 1}\right)\), solving for \( x\) gives us: \(x = \frac{3\alpha}{2}\).
   • We know that the normal x-intercept is 7, thus: \(\frac{3\alpha}{2} = 7 \Rightarrow \alpha = \frac{14}{3}\).

The x-intercept by the normal is thus 7. Therefore, the correct option is 7.