Question

Let $f(x)=\left\\{\begin{array}{cll}x^{2} & \text { if } & x \leq x_{0} \\ a x+b & \text { if } & x>x_{0}\end{array}\right.$ The values of the coefficients a and \(\mathrm{b}\) for which the function is continuous and has a derivative at \(\mathrm{x}_{0}\), are (A) \(a=x_{0}, b=-x_{0}\) (B) \(a=2 x_{0}, b=-x_{0}^{2}\) (C) $\mathrm{a}=\mathrm{a}=\mathrm{x}_{\mathrm{e}}^{2}, \mathrm{~b}=-\mathrm{x}_{0}$ (D) \(a=x_{0}, b=-x_{0}^{2}\)

Short answer

The correct answer is Option (B), in which the coefficients are \(a = 2x_0\) and \(b = -x_0^2\). These are the coefficients that make the given piecewise function continuous and differentiable at \(x_0\).

Step-by-step solution

Find the value of the function on both sides of \(x_0\)

Since the function is defined piecewise, we'll have to compute the values on both sides of \(x_0\) using each definition of the function. For \(x \leq x_0\), the function is defined as \(f(x) = x^2\). Thus, at \(x_0\), we have: \(f(x_0^-) = x_0^2\) For \(x > x_0\), the function is defined as \(f(x) = ax + b\). Thus, at \(x_0\), we have: \(f(x_0^+) = ax_0 + b\)

Apply the condition for continuity

For the function to be continuous at \(x_0\), we must have \(f(x_0^-) = f(x_0^+)\). From Step 1: \(x_0^2 = ax_0 + b\)

Find the derivative of the function on both sides of \(x_0\)

Now, let's find the derivative of the function on both sides of \(x_0\). For \(x \leq x_0\), taking the derivative of \(x^2\) with respect to \(x\) gives: \(f'(x) = 2x\) At \(x_0\), we have: \(f'(x_0^-) = 2x_0\) For \(x > x_0\), taking the derivative of \(ax + b\) with respect to \(x\) gives: \(f'(x) = a\) At \(x_0\), we have: \(f'(x_0^+) = a\)

Apply the condition for differentiability

For the function to be differentiable at \(x_0\), we must have \(f'(x_0^-) = f'(x_0^+)\). From Step 3: \(2x_0 = a\)

Solve the system of equations for \(a\) and \(b\)

Now we have two equations for \(a\) and \(b\): (1) \(x_0^2 = ax_0 + b\) (2) \(2x_0 = a\) Substituting (2) into (1) gives: \(x_0^2 = (2x_0)x_0 + b \Rightarrow b = -x_0^2\) So, the coefficients are \(a = 2x_0\) and \(b = -x_0^2\). Comparing with the given options, we find that Option (B) matches the solution we found: \[ a = 2x_0, \quad b = -x_0^2 \]