Q70 / 150 · MHT CET 2024 (15 May Shift 2) · Mathematics
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
- A$1,-1$.
- Correct answerB$-1,1$.
- C1,1 .
- D$-1,-1$.
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 $\mathrm{A}=-1, \mathrm{~B}=1$
From (i) and (ii), we get $\mathrm{A}=-1, \mathrm{~B}=1$
Hence, the correct answer is option: 'B'.