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DHASHNI
 3 years ago
Best ResponseYou've already chosen the best response.2if f(x)is continuous at x=0...............find a,b

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6Ok so this means we want the function to have the following properties: \[f(\frac{\pi}{2}) \text{ exist}\] \[\lim_{x \rightarrow \frac{\pi}{2}}f(x) \text{ exist}\] Finally, \[\lim_{x \rightarrow \frac{\pi}{2}}f(x)=f(\frac{\pi}{2})\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6so this means also we need \[\lim_{x \rightarrow \frac{\pi}{2}^+}f(x)=\lim_{x \rightarrow \frac{\pi}{2}^}f(x)\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6so for x=pi/2 f(x)=a we will use this later!

DHASHNI
 3 years ago
Best ResponseYou've already chosen the best response.2myini: the question says that f(x) us continuous at x=0

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6\[f(\frac{\pi}{2})=a\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6oh that's interesting we wouldn't care about a then

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6because that part says x>pi/2

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6i think the question had a typeo

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.0Just take a couple of limits using l'hopitals rule. For the first one since \[1\sin^3(\pi/2)=3\cos^2(\pi/2)=0\] the constraints of l'hopital's rule applies for the left limit. Thus, \[\lim_{x\to\pi/2}\frac{1\sin^3x}{3\cos^2x}=\lim \frac{3\sin^2x\cos x}{6\cos x \sin x}= \lim_{x \to \pi/2} 0.5\sin x=0.5=a\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6because there is nothing to do if they aren't talking about pi/2

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.0Similarly, solve for b.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6and see imperialist ignored that continuous part at 0 lol

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6but yeah i think they meant at pi/2

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, me too, I didn't even notice the zero part.

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.0For b, you just do the same thing, except you take the limit as x approaches pi/2 from the right regarding the second half of the piecewise function. Set that equal to 1/2, since you know the limit must be that, and you are done.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6i was trying to make the function continuous everywhere

DHASHNI
 3 years ago
Best ResponseYou've already chosen the best response.2@imper:but the question says that the function is continuous at x=0

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.0It's a typo, it's impossible to make it continuous at x=0.

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.0You already know how x is defined at zero, that can't be changed, no matter what you make a or b.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6its already continuous at 0

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.0True dat, my bad, I didn't look back at the question and assumed sin was on the bottom.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.6thats why the question makes no sense the work is already done i believe it meant what me and imperialist were trying to do

DHASHNI
 3 years ago
Best ResponseYou've already chosen the best response.2thanks guys !!!!!!!!!!!!!!!!
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