Inverse Trigonometric Functions Question 103
Question: If $ {{\sin }^{-1}}( x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-… ) $
$ +{{\cos }^{-1}}( x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}-… )=\frac{\pi }{2} $ for $ 0<| x |<\sqrt{2}, $ then x equals
Options:
A) ½
B) 1
C) -1/2
D) -1
Show Answer
Answer:
Correct Answer: B
$ {{\sin }^{-1}}( x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-… )+{{\cos }^{-1}}( x^{2}-\frac{x^{2}}{2}+\frac{x^{6}}{4}-… )=\frac{\pi }{2} $ This is true only when $ x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-…=x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}… $
$ \Rightarrow \frac{x}{1+\frac{x}{2}}=\frac{x^{2}}{1+\frac{x^{2}}{2}} $ (Common rations are $ -\frac{x}{2}\And -\frac{x^{2}}{2}\And $ |common ratios|<1, in the given interval) $ \frac{2x}{2+x}=\frac{2x^{2}}{2+x^{2}}\Rightarrow x=0orx=1\Rightarrow x=1, $ {x cannot be zero as $ 0<| x |<\sqrt{2} $ }.
BETA
