Question:

Step 1: Continuity at $x=0$ For $f(x)$ to be continuous at $x=0$, the left-hand limit, right-hand limit, and the function value at $x=0$ must be equal. The function is defined as: $$f(x)=\begin{cases} \sin^{-1}x+a \cdot \cos(\pi x), & -1<x<0 \\ b \cdot \cos^{-1}x+\sin(\pi x), & 0\leq x<1 \end{cases}$$ Left-Hand Limit (LHL) at $x=0$: $$ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\sin^{-1}x+a\cos(\pi x)) = \sin^{-1}(0)+a\cos(0) = 0+a(1) = a $$ Right-Hand Limit (RHL) at $x=0$: $$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (b\cos^{-1}x+\sin(\pi x)) = b\cos^{-1}(0)+\sin(0) = b(\pi/2)+0 = b\pi/2 $$ For continuity, LHL = RHL: $$ a = \frac{b\pi}{2} \quad \cdots(1) $$ Step 2: Differentiability at $x=0$ For $f(x)$ to be differentiable at $x=0$, the left-hand derivative (LHD) and right-hand derivative (RHD) must be equal. First, find the derivative of $f(x)$ for $x \neq 0$. For $-1 < x < 0$: $$ f'(x) = \frac{d}{dx}(\sin^{-1}x+a\cos(\pi x)) = \frac{1}{\sqrt{1-x^2}} - a\pi\sin(\pi x) $$ Left-Hand Derivative (LHD) at $x=0$: $$ f'(0^-) = \lim_{x \to 0^-} \left(\frac{1}{\sqrt{1-x^2}} - a\pi\sin(\pi x)\right) = \frac{1}{\sqrt{1-0}} - a\pi\sin(0) = 1 - 0 = 1 $$ For $0 < x < 1$: $$ f'(x) = \frac{d}{dx}(b\cos^{-1}x+\sin(\pi x)) = -\frac{b}{\sqrt{1-x^2}} + \pi\cos(\pi x) $$ Right-Hand Derivative (RHD) at $x=0$: $$ f'(0^+) = \lim_{x \to 0^+} \left(-\frac{b}{\sqrt{1-x^2}} + \pi\cos(\pi x)\right) = -\frac{b}{\sqrt{1-0}} + \pi\cos(0) = -b + \pi(1) = \pi - b $$ For differentiability, LHD = RHD: $$ 1 = \pi - b $$ Solving for $b$: $$ b = \pi - 1 $$ Substitute the value of $b$ into equation (1) to find $a$: $$ a = \frac{(\pi-1)\pi}{2} $$ Step 3: Calculate $2a+b$ Substitute the derived values of $a$ and $b$: $$ 2a+b = 2\left(\frac{(\pi-1)\pi}{2}\right) + (\pi-1) $$ $$ 2a+b = \pi(\pi-1) + (\pi-1) $$ $$ 2a+b = \pi^2 - \pi + \pi - 1 $$ $$ 2a+b = \pi^2 - 1 $$