Exercise \( 2-(8 \) points) Consider the function \( g \) defined over IR by: \( g(x)=m x^{2}-(2 m-1) x+m \) where \( m \) is a non-zero real parameter and designate by \( \left(C_{m}\right) \) its representative curve in an orthonormal system \( (O ; \vec{l}, \vec{\jmath}) \). 1) Show that for \( m \neq 0,\left(C_{m}\right) \) passes through a fixed point \( A \) whose coordinates are to be determined. 2) Let \( B \) be the symmetric of \( A \) with respect to the axis of symmetry of the curve \( \left(C_{m}\right) \). Calculate the coordinates of \( B \). 3) 'a) Find the slopes of the tangents at \( A \) and \( B \). b) Find the value of \( m \) if the tangents are perpendicular.

To solve the problem step by step, we will address each part of the exercise systematically. ### Part 1: Show that for \( m \neq 0, (C_{m}) \) passes through a fixed point \( A \) The function is given by: \[ g(x) = m x^{2} - (2m - 1)x + m \] To find a fixed point \( A \) that the curve passes through for all \( m \neq 0 \), we can evaluate \( g(x) \) at a specific value of \( x \). A common choice is \( x = 1 \): \[ g(1) = m(1)^{2} - (2m - 1)(1) + m \] Calculating this gives: \[ g(1) = m - (2m - 1) + m = m - 2m + 1 + m = 1 \] Thus, the point \( A(1, 1) \) is a fixed point through which the curve \( (C_{m}) \) passes for all \( m \neq 0 \). ### Part 2: Calculate the coordinates of \( B \) The axis of symmetry of a quadratic function \( ax^2 + bx + c \) is given by the formula: \[ x = -\frac{b}{2a} \] For our function \( g(x) \): - \( a = m \) - \( b = -(2m - 1) \) Calculating the axis of symmetry: \[ x = -\frac{-(2m - 1)}{2m} = \frac{2m - 1}{2m} \] Now, we need to find the coordinates of point \( B \), which is the symmetric point of \( A(1, 1) \) with respect to the axis of symmetry. The x-coordinate of \( B \) can be calculated as follows: Let \( x_s = \frac{2m - 1}{2m} \) be the x-coordinate of the axis of symmetry. The distance from \( A \) to the axis of symmetry is: \[ d = 1 - x_s = 1 - \frac{2m - 1}{2m} = \frac{2m - (2m - 1)}{2m} = \frac{1}{2m} \] Thus, the x-coordinate of \( B \) is: \[ x_B = x_s - d = \frac{2m - 1}{2m} - \frac{1}{2m} = \frac{2m - 1 - 1}{2m} = \frac{2m - 2}{2m} = \frac{m - 1}{m} \] The y-coordinate of \( B \) is the same as that of \( A \) since the reflection across the axis of symmetry does not change the y-coordinate: \[ y_B = 1 \] Thus, the coordinates of point \( B \) are: \[ B\left(\frac{m - 1}{m}, 1\right) \] ### Part 3: Find the slopes of the tangents at \( A \) and \( B \) To find the slopes of the tangents at points \( A \) and \( B \), we need to calculate the derivative \( g'(x) \): \[ g'(x) = \frac{d}{dx}(m x^{2} - (2m - 1)x + m) = 2mx - (2m - 1) \] Now, we evaluate the derivative at point \( A(1, 1) \): \[ g'(1) = 2m(1) - (2m - 1) = 2m - 2m + 1 = 1 \] So, the slope of the tangent at \( A \) is \( 1 \). Next, we evaluate the derivative at point \( B\left(\frac{m - 1}{m}, 1\right) \): \[ g'\left(\frac{m - 1}{m}\right) = 2m\left(\frac{m - 1}{m}\right) - (2m - 1) = \frac{2m(m - 1)}{m} - (2m - 1) = 2(m - 1) - (2m - 1) = 2m - 2 - 2m + 1 = -1 \] So, the slope of the tangent at \( B \) is \( -1 \). ### Part 3b: Find the value of \( m \) if the tangents are perpendicular Two lines are perpendicular if the product of their slopes is \( -1 \). The slopes we found are \( 1 \) (at \( A \)) and \( -1 \) (at \( B \)). Calculating the product: \[ 1 \cdot (-1) = -1 \] Since the product is already \( -1 \), the tangents at points \( A \) and \( B \) are perpendicular for any non-zero value of \( m \). ### Summary of Results 1. The fixed point \( A \) is \( (1, 1) \). 2. The coordinates of \( B \) are \( \left(\frac{m - 1}{m}, 1\right) \). 3. The slopes of the tangents at \( A \) and \( B \) are \( 1 \) and \( -1 \), respectively, and they are perpendicular for any non-zero \( m \).