If the function f(x)=cases-x, & x

Question

If the function $f(x)=\begin{cases}-x, & x<1\a+\cos^{-1}(x+b), & 1\le x\le2\end{cases}$ is differentiable at $x=1$, then $\dfrac ab$ is equal to:

Accepted Answer

Continuity: $-1=a+\cos^{-1}(1+b)$ ... (a)

Differentiability: $-1=\dfrac{-1}{\sqrt{1-(1+b)^2}}\Rightarrow\sqrt{1-(1+b)^2}=1\Rightarrow b=-1$.

From (a): $\cos^{-1}(0)=-1-a\Rightarrow\dfrac\pi2=-1-a\Rightarrow a=\dfrac{-\pi-2}{2}$.

$$\frac ab=\frac{\pi+2}{2}$$

Solutions · 1