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The value of $a$ and $b$ such that the function $f(x)=\left\{\begin{array}{ll}-2 \sin x, & -\pi \leq x \leq-\frac{\pi}{2} \\ a \sin x+b, & -\frac{\pi}{2} < x < \frac{\pi}{2} \text { is continuous } \\ \cos x, & \frac{\pi}{2} \leq x \leq \pi\end{array}\right.$
in $[-\pi, \pi]$, are
Options:
in $[-\pi, \pi]$, are
Solution:
1106 Upvotes
Verified Answer
The correct answer is:
$-1,1$
For continuity in $[-\pi, \pi]$, we must have
$$
\begin{array}{l}
\text { At } x=-\frac{\pi}{2}, f\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{-}}(-2 \sin x) \\
=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{+}}(a \sin x+b) \\
\Rightarrow \quad 2=-a+b
\end{array}
$$
At $x=\frac{\pi}{2}$,
$$
f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(a \sin x+b)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} \cos x
$$
$$
\Rightarrow \quad 0=a+b
$$
$\cdots$
On solving Eqs. (i) and (ii) we get, $a=-1, b=1$
$$
\begin{array}{l}
\text { At } x=-\frac{\pi}{2}, f\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{-}}(-2 \sin x) \\
=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{+}}(a \sin x+b) \\
\Rightarrow \quad 2=-a+b
\end{array}
$$
At $x=\frac{\pi}{2}$,
$$
f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(a \sin x+b)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} \cos x
$$
$$
\Rightarrow \quad 0=a+b
$$
$\cdots$
On solving Eqs. (i) and (ii) we get, $a=-1, b=1$
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