Question

If the function \(f(x)=\begin{cases}-x, & x\lt 1\\ a+\cos^{-1}(x+b), & 1\le x\le 2\end{cases}\) is differentiable at \(x=1\), then \(\frac ab\) is equal to :

• A. \(\frac{\pi-2}{2}\)
• B. \(\frac{-\pi-2}{2}\)
• C. \(\frac{\pi+2}{2}\)
• D. \(-1-\cos^{-1}(2)\)

Hint:

Differentiability at 1 needs continuity and equal left and right derivatives; the derivative condition fixes \(b\), continuity then fixes \(a\).

Answer 1

The Correct Option is C

Left derivative at 1: \(f'(1^-)=-1\).
Right derivative: \(f'(x)=-\frac{1}{\sqrt{1-(x+b)^2}}\), so \(f'(1^+)=-\frac{1}{\sqrt{1-(1+b)^2}}\).
Equate: \(\frac{1}{\sqrt{1-(1+b)^2}}=1\Rightarrow(1+b)^2=0\Rightarrow b=-1\).
Continuity at \(x=1\): \(-1=a+\cos^{-1}(1+b)=a+\cos^{-1}(0)=a+\frac\pi2\Rightarrow a=-1-\frac\pi2\).
\(\frac ab=\frac{-1-\frac\pi2}{-1}=1+\frac\pi2=\frac{\pi+2}{2}\).
Answer: (3) \(\frac{\pi+2}{2}\)