Consider the function f(x)=\ array cc -2 x & if x ≤-(π)/(2) A x+B &...

\because f(x)=\left\{$\begin{array}{ccc}-2 \sin x & \text{\text{ if } } & x \leq-\pi / 2 \\ A \sin x+B & \text{\text{ if } } & -\pi / 2 < x < \pi / 2 \\ \cos x & \text{\text{ if } } & x \geq \pi / 2\end{array}$\right.$\begin{aligned} & \because \mathrm{f}(\mathrm{x}) \text{ is continuous every where: } \\ & \therefore \lim _{\mathrm{x} \rightarrow \pi / 2} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \pi / 2} \mathrm{f}(\mathrm{x}) \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow \pi / 2} \cos \mathrm{x}=\lim _{\mathrm{x} \rightarrow \pi / 2} \mathrm{ A} \sin \mathrm{x}+\mathrm{B} \\ & \Rightarrow \mathrm{ A}+\mathrm{B}=0 \end{aligned}$Also; $\lim _{x \rightarrow-\pi / 2^{+}} f(x)=\lim _{x \rightarrow-\pi / 2^{-}} f(x)\begin{aligned} & \Rightarrow \lim _{x \rightarrow-\pi / 2} \mathrm{ A} \sin \mathrm{x}+\mathrm{B}=\lim _{\mathrm{x} \rightarrow-\pi / 2}-2 \sin \mathrm{x} \\ & \Rightarrow-\mathrm{A}+\mathrm{B}=2 \end{aligned}$$$ is from eq(1) \& (2) we get, A=-1