ap calculus bc mcq practice test b\n25. let f be the function defined above, where c and d are constants. if…

Explanation:

Step1: Check continuity at x = 2

For a function to be differentiable at a point, it must be continuous at that point. So, $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{+}}f(x)$. $\lim_{x\rightarrow2^{-}}(cx + d)=2c + d$ and $\lim_{x\rightarrow2^{+}}(x^{2}-cx)=4 - 2c$. Then $2c + d=4 - 2c$.

Step2: Check differentiability at x = 2

The left - hand derivative $f_{-}'(x)=c$ for $x\leq2$ and the right - hand derivative $f_{+}'(x)=2x - c$ for $x > 2$. At $x = 2$, $f_{-}'(2)=c$ and $f_{+}'(2)=4 - c$. Since the function is differentiable at $x = 2$, $c=4 - c$. Solving $c=4 - c$ gives $2c=4$, so $c = 2$.

Step3: Find the value of d

Substitute $c = 2$ into the continuity equation $2c + d=4 - 2c$. We get $2\times2 + d=4-2\times2$. $4 + d=4 - 4$, so $d=-4$.

Step4: Calculate c + d

$c + d=2+( - 4)=-2$.

Answer:

B. -2