If the function $f(x)= \begin{cases}-2 \sin x & \text {, if } x \leq \frac{-\pi}{2} \\ A \sin x+B & , \text…

If the function $f(x)= \begin{cases}-2 \sin x & \text {, if } x \leq \frac{-\pi}{2} \\ A \sin x+B & , \text { if } \frac{-\pi}{2} \lt x \lt \frac{\pi}{2} \\ \cos x & , \text { if } x \geq \frac{\pi}{2}\end{cases}$ is continuous everywhere, then the values of A and B are respectively
  1. $1,-1$.
  2. $-1,1$.
  3. 1,1 .
  4. $-1,-1$.

Solution

Since $\mathrm{f}(x)$ is continuous everywhere. $\therefore \quad \mathrm{f}(x)$ is continuous at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$. $\begin{aligned} \therefore \quad & \lim _{x \rightarrow \frac{-\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{-\pi^{+}}{2}} \mathrm{f}(x) \\ & \Rightarrow \lim _{x \rightarrow \frac{-\pi}{2}}(-2 \sin x)=\lim _{x \rightarrow \frac{-\pi}{2}}(\mathrm{~A} \sin x+\mathrm{B}) \\ \Rightarrow & -2(-1)=\mathrm{A}(-1)+\mathrm{B} \\ & \Rightarrow-\mathrm{A}+\mathrm{B}=2...(i) \end{aligned}$ $\begin{aligned} & \text { Also, } \lim _{x \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \mathrm{f}(x) \\ & \Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}(\mathrm{~A} \sin x+\mathrm{B})=\lim _{x \rightarrow \frac{\pi}{2}}(\cos x) \\ & \Rightarrow \mathrm{A}(1)+\mathrm{B}=0 ...(ii)\\ & \Rightarrow \mathrm{~A}+\mathrm{B}=0 \end{aligned}$
From (i) and (ii), we get $\mathrm{A}=-1, \mathrm{~B}=1$

Asked in: MHT CET 2024 (15 May Shift 2)

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