If $f:R ightarrow R$ given by $f(x) = \left\{ \begin{array}{r} 2\cos{x,\ if\ \ x \leq - \frac{\pi}{2}} \ a + \sin{x + b,\ if - \frac{\pi}{2}

Question

If $f:R ightarrow R$ given by $f(x) = \left\{ \begin{array}{r} 2\cos{x,\ if\ \ x \leq - \frac{\pi}{2}} \ a + \sin{x + b,\ if - \frac{\pi}{2} < x < \frac{\pi}{2}} \ 1 + \cos^{2}{x,\ \ if\ x \geq \frac{\pi}{2}} \end{array} ight.\$ is a continuous Function on $R$, then $(a,\ b)$ is equal to

Dr. Abubacker · Accepted Answer

(a)
LHL$= \lim_{h ightarrow 0}f\left( - \frac{\pi}{2} - h ight) = \lim_{h ightarrow 0}2\cos{\left( - \frac{\pi}{2} - h ight) = 0}$
RHL$= \lim_{h ightarrow 0}f\left( - \frac{\pi}{2} + h ight) = \lim_{h ightarrow 0}2\ a\sin{\left( - \frac{\pi}{2} + h ight) + b}$
$= - a + b$
Since, function is continuous.
$	herefore\$ RHL=LHL $\Rightarrow \ \ \ a = b$
From the given options only (a) $ie,\ \ \left( \frac{1}{2},\frac{1}{2} ight)$ satisfies this condition

Q. If $f:R \rightarrow R$ given by $f(x) = \left\{ \begin{array}{r} 2\cos{x,\ if\ \ x \leq - \frac{\pi}{2}} \\ a + \sin{x + b,\ if - \frac{\pi}{2} < x < \frac{\pi}{2}} \\ 1 + \cos^{2}{x,\ \ if\ x \geq \frac{\pi}{2}} \end{array} \right.\$ is a continuous Function on $R$, then $(a,\ b)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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