Inverse Trigonometric Functions Question 83
Question: $ {{\sin }^{-1}}( a-\frac{a^{2}}{3}+\frac{a^{3}}{9}+… )+{{\cos }^{-1}}(1+b+b^{2}+…)=\frac{\pi }{2} $ when
Options:
A) $ a=-3 $ and $ b=1 $
B) $ a=1 $ and $ b=-\frac{1}{3} $
C) $ a=\frac{1}{6} $ and $ b=\frac{1}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
The given relation is possible when $ a-\frac{a^{2}}{3}+\frac{a^{3}}{9}+…=1+b+b^{2}+… $ Also $ -1\le a-\frac{a^{2}}{3}+\frac{a^{3}}{9}+…\le 1 $ and $ -1\le 1+b+b^{2}+…\le 1 $
$ \Rightarrow | b |<1\Rightarrow | a |<3 $ and $ \frac{a}{1+\frac{a}{3}}=\frac{1}{1-b} $
$ \Rightarrow \frac{3a}{a+3}=\frac{1}{1-b}. $
There are infinitely many solutions. But in the given options, it is satisfied only when $ a=1 $ and $ b=-\frac{1}{3}. $
BETA
