Functions Question 310
Question: The values of A and B such that the function $ f(x)= \begin{cases} -2\sin x, & x\le -\frac{\pi }{2} \\ A\sin x+B, & -\frac{\pi }{2}<x<\frac{\pi }{2} \\ \cos x, & x\ge \frac{\pi }{2} \\ \end{cases} . $ , is continuous everywhere are
[Pb. CET 2000]
Options:
A) $ A=0,,B=1 $
B) $ A=1,,B=1 $
C) $ A=-1,,B=1 $
D) $ A=-1,,B=0 $
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Answer:
Correct Answer: C
Solution:
For continuity at all $ x\in R, $ we must have $ f( -\frac{\pi }{2} )=\underset{x\to {{(-\pi /2)}^{-}}}{\mathop{lim}},(-2\sin x) $ $ =\underset{x\to {{(-\pi /2)}^{+}}}{\mathop{lim}},(A\sin x+B) $
Þ $ 2=-A+B $ …..(i) and $ f( \frac{\pi }{2} )=\underset{x\to {{(\pi /2)}^{-}}}{\mathop{\lim }},(A\sin x+B) $ $ =\underset{x\to {{(\pi /2)}^{+}}}{\mathop{lim}},(\cos x) $
Þ $ 0=A+B $ …..(ii) From (i) and (ii), $ A=-1 $ and $ B=1 $ .
BETA
